College Physics August 2016

Volume Three Classical Electromagnetism

And Optics

byDavid Michael Judd

College Physics 3rd Edition

Volume Three Classical Electromagnetism and Optics

C David Michael Judd

PREFACE

You are beginning an intensive study of introductory electricity and magnetism. As you may already know from your study of classical mechanics, there is nothing particularly easy about physics. Many students find electricity and magnetism even more challenging than mechanics. The major reason for this is that electromagnetic phenomena arise from electric charges. The complicating problem with electric charges is that they are too small to be seen with the unaided eye.

In classical mechanics, we dealt with the physics of slowly moving, massive, physical things. In electromagnetism, we deal with the physics produced by electrons, protons, photons and other objects the existence of which you must take on faith. By that I simply mean that I can not bring these objects into the classroom and toss them around for you so that you can see how they behave.

Electromagnetism, then, turns out to be much more abstract than mechanics was. We are going to attempt to build a complex model of electric charges and their interactions. Your mastery of this model is, in my humble opinion, the key to your success in this discipline. To solve problems, you are going to have to be able to use the “model.” The model is quantitative, and depends, in large part, on several important vector quantities like the so-called electric and magnetic fields.

So, please be prepared to spend a significant amount of time on task. A good rule of thumb is to plan on spending from two to three hours outside of class for every hour in class. As you can see, that is from nine to twelve hours a week just on physics. If you do not have this much time each week to devote to physics, it will be difficult to do well in this class. So, reevaluate the idea of taking twenty credit hours, working full time, being student government president, having an extensive social life, and....

I make the assumption that you are in this class because in some sense it is important to the profession you have chosen to pursue. You must learn to establish priorities and manage your time effectively if you are to succeed in this class; just as you must if you are to succeed in your chosen profession. Also, remember that it is likely that you will have to be making similar time expenditures in your other classes! Enough said.

Having noted the importance of time doing physics, I must also warn you that it is possible to spend a great deal of time on physics and still not do well. I have had students who have spent large amounts of time on physics and still not done well. It is crucial that the time you spend on physics be focused and productive. What do I mean? I will give you an example. Many students suffer from the mistaken notion that all they have to do to learn physics is read the textbook. After all, reading was usually the most important activity they performed to succeed in prior classes. Do

i

not misunderstand me here, you must read the textbook! You must wrestle with the material in the text. However, the most important task is that you must learn and understand the electric and magnetic interaction model! To master introductory electricity and magnetism and solve the problems involved will almost surely prove impossible without a clear understanding of this abstract model. Having said that, know that the best way to learn to grasp the model is to work on physics problems! You must solve many problems. I will have much more to say about problem-solving skills as we proceed.

Your grade in this class will be determined by how well you solve electricity and magnetism problems on the exams. So, when you do work on solving problems, try and do them under test conditions. Never spend more than twenty minutes on a single problem. Move on if you do not find a solution after twenty minutes. Remember, it is helpful to practice for your exams.

I am going to do the best job I can in trying to clarify the concepts we use in physics. I will attempt to prepare you to have successful problem-solving experiences. I will be happy to meet with you outside of class to work with you on this material if needed. You do not need to get tutors! You need to see me! I am the best tutor in the world for this class. And my price is right. (Tutoring is free with me! I also have a pretty good idea of what the instructor of this class might ask on an exam!) Having said this, however, you must also make a commitment if you are to succeed. I can not learn the physics for you. Good luck, and try to learn to enjoy the remarkable physical world within which you are embedded and about which the science of physics tries to gain some insight.

ii

TABLE OF CONTENTS

Preface i

Table of Contents iii

Part OneElectrostatics

Chapter Thirty-Three: Electric Charge 33-1

Chapter Thirty-Four: The Electric Force 34-1

Chapter Thirty-Five: The Electric Field 35-1

Chapter Thirty-Six: Electric Flux and Gauss’ Law 36-1

Chapter Thirty-Seven: The Work Done in Moving Electric Charges 37-1

Part TwoElectrodynamics

Chapter Thirty-Eight: Electric Current and Conductors 38-1

Chapter Thirty-Nine: Electrical Circuit Elements 39-1

Chapter Forty: Direct Current Circuits 40-1

Part ThreeMagnetism

Chapter Forty-One: The Magnetic Force 41-1

Chapter Forty-Two: The Magnetic Field 42-1

Chapter Forty-Three: Magnetic Flux and Ampere’s Law 43-1iii

Chapter Forty-Four: Changing Magnetic Flux and Induced Electric Fields 44-1

Chapter Forty-Five: Magnetic Materials 45-1

Chapter Forty-Six: Alternating Current Circuits 46-1

Part Four Maxwell’s Equations, Electromagnetic Waves and Optics

Chapter Forty-Seven: Light as an Electromagnetic Wave 47-1Chapter Forty-Eight: The Propagation of Light 48-1

Chapter Forty-Nine: Reflection and Refraction at Plane Surfaces 49-1

Chapter Fifty: Images Formed by a single Reflection or Refraction 50-1

Chapter Fifty-One: Lenses and Optical Instruments 51-1

Chapter Fifty-Two: Illumination and Color 52-1

Chapter Fifty-Three: Interference and Diffraction53-1

Chapter Fifty-Four: Polarization 54-1

AppendicesAppendix A: Mathematical Review

A.1: Powers Of Ten And Scientific NotationA.2: Some Algebra Rules; Includes The Quadratic FormulaA.3: Some Plane And Solid GeometryA.4: Some TrigonometryA.5: Time Independent Equation Of Motion With Constant AccelerationA.6: The Vector Cross ProductA.7: Some Calculus Results

Appendix B: Values Of Physical ConstantsB.1: Universal Physical ConstantsB.2: Some Other Useful Physical ValuesB.3: The Greek Alphabet

iv

PART ONE

Electrostatics

Chapter 33

ELECTRIC CHARGE

We have already discussed at some length the physical properties of slowly moving big things. We have discussed the fact that these physical things are extended and we have made the assumption that we can measure this extension. Remember, the SI standard for extension is the meter. We also noted that these physical things have mass, and, we also assumed that we can measure this mass. The SI standard for mass is the kilogram. And, of course, we noted that physical things have the physical property that I have called temporal duration. Measuring the temporal duration of physical processes is an indispensable activity of experimental physics. The SI standard for temporal duration is the second. In this chapter, you will be introduced to yet another important property of physical things, the property of electric charge. The SI standard for electric charge is the Coulomb.

Maybe the most important working theory in all of the physical sciences is that physical things are made up of large numbers of very small objects called atoms. This theory is called the atomic theory. In turn, atoms are themselves made up of even more "fundamental" objects called elementary particles. Understanding how these elementary particles interact is central to the science of physics.

Even though physicists have identified a bewilderingly large number of so-called elementary particles, in this investigation, we will be interested primarily in four such particles. In Table 33-1 below, you will find listed some of the important properties of these four elementary particles.

Table 33-1Properties of Some Elementary Particles

Particle Symbol Rest Mass Electric Charge

electron

protonneutronphoton

e−

pn

γ

9.10938 ×10− 31 kg1.67265 × 10−27 kg1.67495 × 10−27 kg

0

+ 1.602 × 10−19 C00

−1.602 × 10−19 C

33-2

To better introduce three of these elementary particles, we return to the atom. If we were to take a survey of the atoms of the visible universe, we would find that out of every one hundred atoms, approximately ninety-four of them would be hydrogen. Out of the remaining six atoms, about five of them would be helium. Obviously, then, only about one atom in one hundred atoms is anything other than hydrogen or helium. This means, among other things, that the Earth is a very interesting place chemically, and very atypical. It also means that much of the very “stuff” of which you are made is exceedingly rare.

In Figure 33-2 below, is a pictorial representation of monatomic hydrogen. This atom is made up of two elementary particles, a proton and an electron. These two particles are electrically charged, and, by an extraordinary coincidence, carry exactly the same magnitude of electric charge. By convention, the electron is said to have negative electric charge and the proton has positive electric charge. The measured value of the magnitude of the charge on an electron and proton is

e = 1.60217653× 10−19 C . (33-1)

We can also construct a simple model of monatomic helium as shown below in Figure 33-3.This form of helium is made up of two electrons, two protons, and two neutrons. All of the atoms of the universe are made up of various combinations of these three elementary particles.

This property of electric charge is important because physical things which possess this property interact, that is, they exert forces on each other. Electrons and protons interact electrically with other electrons and protons. This interaction between electrically charged objects is one manifestation of the electromagnetic force. In this chapter, we want to gain a clearer understanding of what we mean by electric charge.

In physics, we use the phrase “electric charge” in two different ways. First, we use the phrase to identify a specific physical property. We say that an electron or a proton is “electrically charged.” Electrons and protons possess this intrinsic property of being electrically charged. A neutron, on the other hand, does not possess this property. A neutron is not electrically charged. Since the neutron is without electric charge, it can not interact electromagnetically; it can not “feel the electromagnetic force.”

There is, however, a second sense in which we use the phrase electric charge. Say we have a small piece of ordinary matter in front of us. If we were to count the number of protons in the piece, we would get an integer, say N p . Likewise, we could count the number of electrons

contained in the piece of matter and get an integer value, say Ne− . Since each proton and each

electron is electrically charged, we could add up the total electric charge on the piece of matter. We would call this total the electric charge on the piece of matter; we signify this charge with Q. Its value would be given by

Q = N pe − Ne− e = N p − Ne−( ) e = Ne , (33-2)

33-3

where N is an integer,  , −3, −2, −1, 0, 1, 2, 3.  . The electric charge on material

objects is quantized, and its value is always some integer multiple of e.

Monatomic Hydrogen

Monatomic Helium

Figure 33 - 2

Figure 33 - 3

e−

p

e−

e−

n

npp

33-4

In general, the net electric charge on an atom is zero as there are as many protons as there are electrons. An atom becomes electrically charged, however, when there are more protons than electrons, or more electrons than protons. Such an atom is said to be ionized.

I hope that it has not escaped your attention that electric charge comes in two “flavors,” positive and negative. This bifurcation of electric charge has some wonderfully interesting consequences. One such consequence is that--unlike the gravitational interaction between masses--electrical interactions can be attractive or repulsive. You have probably been taught earlier in school that like charges repel and unlike charges attract. Mass, on the other hand, comes in only one “flavor,” and, as such, mass interactions are always attractive. Because the electrical interaction admits of repulsion, there is variation in the physical world. The universe would be an overwhelmingly boring, monotonous, homogeneous place if there were no repulsive electromagnetic force.

CONDUCTORS AND INSULATORS

You are, no doubt, familiar with various kinds of metal. It turns out that metals have special properties that make them very important for this investigation. One important property of metals is that they are good conductors of electricity. What do I mean? Metals, like most other materials you are familiar with, are usually not electrically charged. That is, there are as many electrons in the metal as there are protons. However, if we were to place an excess electric charge--say in the form of ′ N electrons--in the center of a solid piece of metal, then, in a very short period of time all ′ N

of the excess electrons would be on the surface the metal. The migration of these ′ N electrons is driven by their repulsive interactions, and, as this migration is virtually unimpeded, the metal is said to be a good conductor of electric charge.

In general then, we can classify materials as either conductors, insulators, or semiconductors. Insulators are materials within which excess charges do not move freely. Some examples include glass, rubber and wood. Semiconductors are materials which normally behave like insulators, but which, under the proper conditions, can behave like a conductor.

Imagine that we have a solid, sphere of copper sitting on an insulating stand, see Figure 33-4 below. This sphere is initially uncharged as the number of protons is identical to the number of electrons. That is,

N p = Ne− . (33-3)

If we placed another ′ N e− electrons in the center of the sphere, then in a very short period of time,

all of these “excess” electrons would be at rest, uniformly distributed on the surface of the sphere. So, only for a very short time are there electric charges in motion. Electrostatics deals with physical systems within which the electric charges are stationary.

33-5

Figure 33-4

QL = ′Q = − ′N e−( ) e

t = tL ′ t = tL + trt = to = 0

Q ′L = ′Q = − ′Ne−( ) eQo = 0

′ Q

Figure 33-5

′ Q 2

′ Q 2

33-6

Next, imagine that we connect the sphere with excess charge ′ Q to an identical, uncharged sphere. The connection is made by means of a conductor, see Figure 33-5 above. What will happen to the excess charge? The conductor provides a path along which the excess charge can move. So, a short time after the connection is made, the final charge distribution will find ′Q / 2 on the surface of each sphere.

Figure 33-6

Figure 33-7

33-7

Frictional processes can cause a material to become charged. For example, if you rub a glass rod with silk, the silk will become negatively charged and the glass positively charged. If, after rubbing, one were to place the glass rod close to an uncharged metallic sphere, an induced charge distribution would appear on the surface of the sphere, see Figure 33-6 above. If one were then to connect the metallic sphere to the Earth via a conductor, then the positive charge would run off to the ground, and leave the metallic sphere negatively charged. See Figure 33-7 above. The ground

symbol is . Similarly, if one were to rub a hard rubber rod with wool, the rod would

become negatively charged. In turn, it could be used to induce charge on a neutral metal sphere. One could also transfer the charge on the rod by making contact with an uncharged sphere.

Electrons and protons are not the only elementary particles that have electric charge. Below in Table 33-2 is an incomplete list of some of these particles along with their electric charge.

Table 33-2Electric Charges of Some Elementary Particles

______________________________________________________________________________Particle Name, Symbol Electric Charge

______________________________________________________________________________

Photon, γ 0Neutrino, ν 0Electron, e− − ePositron, e+ + e

Muon, µ − ePion, π o 0Pion, π + + ePion, π − − eProton, p + eNeutron, n 0Delta, Δ+ + + 2eDelta, Δ+ + eDelta, Δo 0Delta, Δ− − e

33-8

The elementary particles listed above are studied by high energy particle physicists using particle accelerators. As a result of all of the experimental work done by chemists and physicists, we know that electric charge is a conserved quantity. The total electric charge before any chemical reaction or nuclear reaction is identical to the total electric charge after the reaction. As an historical note, that kite-flying rascal from Philadelphia, Benjamin Franklin, first proposed this idea of electric charge conservation. We now have overwhelming experiment confirmation of his insight.

It is going to prove useful to describe electric charges as either point-like or as charge distributions “smeared” over space. When we say an electric charge is point-like, we mean the dimensions of the charged object itself are very small compared to the distances to those points where we may be interested in the electrical influence of the point charges. We did a similar thing with the gravitational force. Recall that we could treat the gravitational interaction between the Earth and the Sun as point-like, because the distance between the Earth and the Sun is about five orders of magnitude greater than the dimensions of either the Earth or the Sun. However, there are times where electric charge is distributed over dimensions large enough where we cannot reasonably treat them as point-like. This will all make more sense later when we have occasion to make these distinctions more concrete.

33-9

Summary of Chapter 33

The physical world is made up of elementary particles. These particles posses many physical properties. One such possible property is electric charge. Elementary particles like the photon and the neutron do not have this property and are, therefore, electrically neutral. Other particles like the electron and proton do have this property. The electric charge on the proton has exactly the same magnitude as that of the electron. The charge on the proton is related to that of the electron by

Qp = −Qe− = e = 1.6022 ×10−19 C . (33-4)

When we say that an ordinary macroscopic object is electrically charged, we mean that the number of protons in the object, N p , and the number of electrons in the object, Ne− , are not the

same. The electric charge Q on such an object would then be given by

Q = N p − Ne−[ ] e . (33-5)

The total amount of electric charge is conserved. Electric charge is neither created or destroyed as the result of any reaction or physical process.

Frictional processes can cause objects to become electrically charged. For example, if one rubs a glass rod with silk, the silk will become negatively charged and the rod positively charged. Electrons are transferred from the rod to the silk. The deficit of electrons generated on the glass leaves it positively charged. (The law of conservation of electric charge is not violated here.)

Electrically charged objects that are placed close to secondary objects can induce electric charge distributions on those secondary objects. This process allows us to control the relative charges on interacting bodies in the laboratory.

Chapter 34

ELECTRIC FORCE

Physical things that are electrically chargedexert electrical forces on each other.

COULOMB’S LAW

Figure 34-1The Electric Force

r12

F12

E

F21

E

q1 q2

Two positive point charges q1 and q2 are fixed in space as represented above in Figure 34-1.

These charges, both positive, will repel each other. The charges will exert forces on each other that lie along the line that joins them. Repulsive forces act so as to increase the distance between the charges. The magnitude of these repulsive electrical forces is given by

F12E =

kq1q2

r122 = F21

E , (34-1)

where k is a constant of proportionality and r12 is the distance between the two charges. Note that

r12 = r21 . (34-2)

The constant of proportionality k has a measured value of

k =1

4πεo

= 8.99 ×109  N ⋅m2 ⋅C−2 , (34-3)

34-2

where εo is a constant called the permittivity of free space. In this context, free space means that

there are not any other electric charges in the space between charges q1 and q2 . The superscript E

signifies that this is an electrical force. The subscript 12 signifies that the force is exerted on charge one by charge two. In a similar fashion, subscript 21 signifies that the force is exerted on charge two by charge one. These forces are equal in magnitude but directed in opposite directions, that is

F12

E = −F21

E . (34-4)

Example One:

Two point charges are related by q1 = 3 / 2( ) q2 = −18.00 × 10− 9 C . These charges have

positions given by

r1 = − 5.0 m i + 4.0  m j and

r2 = 6.0 m i + 2.0 m j , respectively. We

want to find the electric force, magnitude and direction, exerted on charge two. Using the force diagram below, we can write

F21

E = F21E F21

E . (1)

The magnitude is given by

y j( )

x i( )5 m

5 m

− 2 m

− 5 m

q1

q2θ2

F12

E

F21

E

2 m

11m

r21 = 125 m

r1

r2

34-3

F21E =

kq2q1

r212 =

8.99 × 109  N ⋅m2 ⋅C−2( ) −18 × 10− 9 C( ) −12 × 10−9 C( )125 m( )2

= 1.55 ×10− 8 N . (2)There are two ways that we could represent the direction of this force. The most useful way is

in terms of a unit vector, namely

F21E = cosθ2 i − sinθ2 j =

11125

i −2125

j . (3)

In example three, I will demonstrate the usefulness of this form of writing the direction of the force.A second, probably more intuitive way of expressing the direction, is to note that this vector

points into the fourth quadrant. So, θ2 is the angle this vector dips below the positive branch of the

x-axis, and is given by

θ2 = tan−1 2 / 11[ ] = 10.3° . (4)

Example Two:Two electrons are fixed at a separation distance r12 . Because each electron has mass, they

interact gravitationally. Also, since each electron has electric charge, they interact electrically. We want to find the ratio of the magnitude of the electrical force exerted on one electron to the magnitude of the gravitational force exerted on the same electron.

r12

F12

E

F21

E−e

F12

G

F21

G −e

We can write

F12E

F12G =

ke2 / r122

GMe−2 / r12

2 =kG

eM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

34-4

=8.99 × 109  N ⋅m2 ⋅C−2( )

6.67 × 10−11  N ⋅m2 ⋅kg−2( )1.602 × 10−19  C( )9.109 × 10−31  kg( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

= 4.17 ×1042 . (1)

Example Three:So far, we have looked at electrical interactions between two point charges only. In this

problem, I want to look at how we can solve a problem involving three point charges. Assume that

three identical positive charges q1 = q2 = q3 = q = 12.00 × 10−9 C are fixed at the vertices of an

equilateral triangle of side length s = 0.750 m , as represented in the diagram below. We wish to

find the net electrical force, magnitude and direction exerted on charge q3 . First, note that charge

three will have a force exerted on it by charge one and another force exerted on it by charge two. Second, it turns out that when we measure these things in the laboratory, the net force turns out to equal the vectorial sum of the individual forces.

q2q1

q3

s

s

s

F32

E

F31

Ey j( )

x i( )

θ1θ2

F3 , tot

E

34-5

So, for this charge configuration, we have

F3 , tot

E =F31

E +F32

E , (1)

where

F31

E = F31E F31

E =kq3q1

r312 cosθ1 i + sinθ1 j⎡⎣ ⎤⎦

=8.99 ×109  N ⋅m2 ⋅C−2( ) 12 × 10−9 C( )2

0.750m( )2 cos 60° i + sin60° j⎡⎣ ⎤⎦

= 2.301 ×10−6 N( ) cos60° i + sin60° j⎡⎣ ⎤⎦

= 1.151 ×10−6 N i + 1.993× 10−6 N j . (2)

Also, we have

F32

E = F32E F32

E =kq3q2

r322 − cosθ2 i + sinθ2 j⎡⎣ ⎤⎦

=8.99 ×109  N ⋅m2 ⋅C−2( ) 12 × 10−9 C( )2

0.750m( )2 − cos 60° i + sin60° j⎡⎣ ⎤⎦

= 2.301 ×10−6 N( ) −cos 60° i + sin 60° j⎡⎣ ⎤⎦

= −1.151× 10−6 N i + 1.993× 10−6 N j . (3)

Therefore, the total or net force on charge three is given by

F3 , tot

E =F31

E +F32

E

= 1.151−1.151( ) i + 1.993 + 1.993( ) j⎡⎣ ⎤⎦ × 10−6 N = 3.99 ×10− 6 N j . (4)

34-6

Example Four:

Three point charges are related by q1 = 2q2 = − 3q3 = 18 × 10−6 C . The charges are

fixed at positions, respectively,

r1 = 3.0m i + 4.0m j ,

r2 = − 5.0m i + 3.0 m j and

r3 = − 3.0m i − 5.0m j . We wish to find the net force, magnitude and direction, exerted on

charge q3 . Using the diagram below, we can write,

F3 , tot

E =F31

E +F32

E =F31

E F31E +F32

E F32E

=kq3q1

r312 cosθ1  i + sinθ1   j( ) +

kq3q2

r322 − cosθ2  i + sinθ 1  j( )

= kq1

3⎛⎝⎜

⎞⎠⎟

q1

r312

⎛⎝⎜

⎞⎠⎟

6r31

 i +9r31

  j⎛⎝⎜

⎞⎠⎟

+q1

3⎛⎝⎜

⎞⎠⎟

q1

2⎛⎝⎜

⎞⎠⎟

1r32

2

⎛⎝⎜

⎞⎠⎟

−2r32

 i +8

r32

  j⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ . (1)

q2

q1

q3

y j( )

x i( )

F31

E

F32

E

θ1θ2

r 31=

117 m

68m

=r

32

F3 , tot

E

6m

9m

2m

8m

θ tot

34-7

So,

F3 , tot

E = kq12 2

r313  i +

3r31

3   j⎛⎝⎜

⎞⎠⎟

+ −1

3r323  i +

43r32

3   j⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

= kq12 2

r313 −

13r32

3

⎛⎝⎜

⎞⎠⎟ i +

3r31

3 +4

3r323

⎛⎝⎜

⎞⎠⎟  j

⎡

⎣⎢

⎤

⎦⎥

= 8.99 ×109  N ⋅m2 ⋅C−2( ) 18 × 10−6 C( )2

×2

117m( )3/2 −1

3 68m( )3/ 2

⎛

⎝⎜

⎞

⎠⎟   i +

3117m( )3/ 2 +

43 68m( )3/2

⎛

⎝⎜

⎞

⎠⎟   j

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 0.00287 N i + 0.0138 N j . (2)

The magnitude of the net force is given by

F3,tot

E =F3 , tot

E = 0.00287 N( )2 + 0.0138 N( )2 = 0.0141 N . (3)

A unit vector to represent the direction of the net force can be found using

F3 , totE =

F3 , tot

E

F3 , tot

E=

0.00290.0141

 i +0.01380.0141

  j = 0.2057 i + 0.9787   j

= cosθ tot  i + sinθtot   j , (4)

where

θ tot = cos−1 0.2057[ ] = 78° . (5)

One also could have used a less algebraic and more straightforward approach:

F31

E = F31E F31

E =kq3q1

r312 cosθ1 i + sinθ1 j⎡⎣ ⎤⎦

=8.99 ×109  N ⋅m2 ⋅C−2( ) − 6 × 10−6 C( ) 18 × 10−6 C( )

117 m( )2

6117

i +9

117j⎡

⎣⎢⎤⎦⎥

34-8

= 4.603× 10−3 N i + 6.905 ×10−3 N j . (6)

Further

F32

E = F32E F32

E =kq3q2

r322 cosθ2 i + sinθ2 j⎡⎣ ⎤⎦

=8.99 ×109  N ⋅m2 ⋅C−2( ) − 6 × 10−6 C( ) 9 × 10−6 C( )

68 m( )2 −268

i +868

j⎡⎣⎢

⎤⎦⎥

= −1.732 ×10− 3N i + 6.926 × 10− 3N j . (7)

Therefore, the total or net force on charge three is given by

F3 , tot

E =F31

E +F32

E

= 4.603 −1.732( ) i + 6.905 + 6.926( ) j⎡⎣ ⎤⎦ × 10−3 N

= 2.87 × 10− 3 N i + 13.83 ×10− 3 N j . (8)

The magnitude then would be

F3,tot

E =F3 , tot

E = 2.87( )2 + 13.83( )2 ×10−3 N = 1.41×10−2 N , (9)

while a unit vector to represent the direction would by

F3 , totE =

F3 , tot

E

F3 , tot

E=

0.00290.0141

 i +0.01380.0141

  j = 0.2057 i + 0.9787   j . (10)

One must choose one’s “poison” and strive to get adept at solving problems involving vector addition!

Example Five:Consider the following situation. Each of two identical spheres has a mass M = 0.050 kg .

Each sphere is attached to a light string of length = 1.50 m , as represented in the diagram below. The strings are fixed to a frictionless pivot. Initially the masses are electrically neutral and allowed to hang and rest in contact with each other. Later, a positive electric charge Q is placed on

34-9

each sphere causing the spheres to swing away form each other in a vertical plane. They come to equilibrium with each string making an angle θ = 15° to the vertical. We wish to determine the magnitude of the electric charge Q placed on each sphere.

Equilibrium requires that the forces directed left must equal in magnitude the forces directed to the right. This means

T sinθ = kQ2 / 2 sinθ( )2 . (1)

Also, the magnitude of the forces directed upward must equal the magnitude of the forces directed downward. So, we have

T cosθ = Mg . (2)Dividing equation (1) by (2) gives us

T sinθT cosθ

= tanθ =kQ2

Mg 2 sinθ( )2 . (3)

Solving for Q we find

Q =42 Mgsin2θ tanθ

k

=4 1.5m( )2 0.05 kg( ) 9.81 m ⋅s−2( ) sin15°( )2 tan15°( )

8.99 × 109  N ⋅m2 ⋅C−2( )

= 2.97 × 10− 6 C = Q . (4)

y ˆ j ( )

x ˆ i ( )

2 sinθ

θθθ T

F E

Mg

Q Q

T cosθ

T sinθ

34-10

THE BOHR THEORY OF THE HYDROGEN ATOM

At the beginning of the last century, physicists and chemists were able to experimentally determine that the material world is made up of very small objects called atoms. Serious work was undertaken to discover the structure of these objects. The most common element in the visible universe is hydrogen. The structure of monatomic hydrogen can be described as a single electron moving “in an orbit” around a solitary, fixed proton, as represented in Figure 34-2 below.

This state of affairs, however, seems to contradict a very important experimental fact. In the laboratory, whenever electric charges are accelerated, they give off energy in the form of electromagnetic radiation (light). The electron orbiting the proton is moving on a curved path so it is accelerating and should be losing kinetic energy and, after a very short time, it should spiral into the proton. However, atoms keep their structure for long periods of time and, thereby, contradict this empirical finding.

The irony is that at that moment in history when scientists could prove the existence of atoms, they did not have a theory that could explain how it was even possible for an atom to exist for any appreciable time. For the first thirty years of the last century, physicists and chemists worked very hard on a new theory of the atom to remedy this problem, and others.

Niels Bohr, a Danish physicist, proposed that there are stable orbits for electrons and that as long as the electrons are in these “allowed” orbits, they do not radiate electromagnetic energy even though they are accelerating.

Figure 43-2Monatomic Hydrogen in the Bohr Model

e−

p

v

FE

rn

34-11

As an electron orbits a proton in the nth “allowed” orbit, it has some angular momentum given by

Ln = rn Me− vn . (34-5)

Bohr assumed that in order for the orbits to be stable, the angular momentum should be some integer multiple of Plank’s constant. (Plank’s constant is the fundamental constant associated with elementary particles--nature at the smallest scale.) So, he argued

Ln = rn Me− vn = nh

2π⎡⎣⎢

⎤⎦⎥

= n , (34-6)

where n = 1,2,3,4,. Plank’s constant is signified by h , and the other constant , pronounced “h bar”, has a value measured to be

=

h2π

=6.626 × 10−34  J ⋅s( )

2π= 1.0546 × 10− 34  J ⋅ s . (34-7)

The orbit of the electron is assumed, for simplicity, to be circular. The radially directed force responsible for this circular motion is the electrical force. So, we can write

ke2

rn2 =

Me−

vn2

rn

. (34-8)

Combining equations (34-6) and (34-8) gives us

ke2

rn2 =

Me−

rn

nrnM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

. (34-9)

The radius of the nth stable orbit is given by

rn = n2 2

ke2 Me−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (34-10)

The so-called ground state obit is the orbit of the electron when it is closest to the proton and can be found using equation (34-10) by setting n = 1. This radius, also called the Bohr radius, has a value

rBohr =

2

ke2 Me−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

34-12

=1.0546 ×10− 34  J ⋅s( )2

8.99 × 109  N ⋅m2 ⋅C−2( ) 1.602 × 10−19  C( )29.109 × 10−31  kg( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 5.29 × 10−11 m ≡ .529 A

, (34-11)

where, recall, A

is an angstrom with a value A

= 1.0 ×10−10 m .

Superposition Principle

Coulomb’s law can be generalized for N discrete electric charges. If we have N electric point

charges q1, q2 , q3 ,  , qi ,  , q j ,  , qN , then the net electric force exerted on the j th

charge is simply the vectorial sum of all of the charge interactions. We can write

Fj ,tot

E =Fji

E

i=1i≠ j

N

∑ =Fj1

E +Fj2

E +Fj 3

E ++FjN

E . (34-12)

It was, of course, the superposition position that we used to do Example Four.

Example Six:

Monatomic HeliumMonatomic helium is represented in the diagram below. The electrons are assumed to

“orbit“ the fixed nucleus on a circular orbit of r = 3.10 × 10−11 m . The nucleus is made up of

two protons and two neutrons. Its mass is Mnuc = 6.67 × 10−27  kg and may be treated as at

rest at the center of the orbit. The electric charge of the nucleus may also be treated as a point charge. We want to calculate the following:

a) the magnitude of the net electric force on each electron.b) the speed of the each electron in its orbit.c) the orbital period of each electron.

a) The electric force on one electron is due to its electrical interaction with the nucleus toward which it is attracted, the dominant interaction, and its interaction with the second electron away from which it is repelled. We can write

34-13

Fe− ,totE =

ke 2e( )r 2 −

ke e( )2r( )2 =

ke2

r2 2 −14

⎡⎣⎢

⎤⎦⎥

=74

ke2

r2

=74

8.99 ×109  N ⋅m2 ⋅C−2( ) 1.602 × 10−19  C( )2

3.10 × 10−11 m( )2 = 4.20 ×10−7 N . (1)

b) As the orbit is circular, we can write

74

ke2

r 2 =M

e−v2

r . (2)

Therefore,

v =74

ke2

Me−

r

=74

8.99 × 109  N ⋅m2 ⋅C−2( ) 1.602 × 10−19  C( )2

9.109 × 10−31  kg( ) 3.10 ×10−11 m( ) = 3.78 × 106  ms

. (3)

c) As the speed for a circular orbit is constant, the period can be found using

τ =2πr

v= 2π

3.10 × 10−11 m

3.78 × 106  ms

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= 5.15 × 10−17 s . (4)

e−

−v

re−

v

ppn

n

34-14

Summary of Chapter 34Coulomb’s Law:

Two electric point charges q1 and q2 are separated by a distance r12 . These electric charges

exert electrical forces on each other such that

F12

E = F12E F12 = −

F21

E = F21E F21 . (34-13)

The magnitudes of the two electrical forces are equal and have a value given by

F12E =

kq1q2

r122 = F21

E , (34-14)

where k is the electrical constant which has a value of

k =1

4πεo

= 8.99 ×109  N ⋅m2 ⋅C−2 , (34-15)

and εo is another constant called the permittivity of free space. (Here free space means there are

no other electrical charges between q1 and q2 that could alter the force.)

These mutual electrical forces are directed oppositely to one another and long the line that joins them. We have

F12E = − F21

E . (34-16)

In Figure 34-3 below, we show the three possible sign configurations for two interacting charges.

Figure 34-3

r12

q1 q2

q1q2

q1 q2

F12

E

F21

E

F12

E

F21

E

F12

E

F21

E

34-15

Superposition Principle:

Experimental work has verified that for N point charges, the net or total electric force on any single charge is the vector sum of the forces exerted on it by all of the other charges. We can write

Fi ,tot

E =Fij

E

j=1j≠i

N

∑ =Fi1

E +Fi2

E ++FiN

E . (34-17)

This physical state of affairs is represented below, in Figure 34-4.

Figure 34-4

Fi2

E

q1

q2

qi

q4

qN

ri2

ri1ri4

riN

Fi1

E

Fi4

E

FiN

E

Fi ,tot

E

Chapter 35

THE ELECTRIC FIELD

Electric charges give rise to electric fields.

Our overall aim is to understand qualitatively and quantitatively the electromagnetic interaction; the interaction between objects that are electrically charged. This interaction is comprised of electrical forces and magnetic forces. The modern physical view of electrical and magnetic interactions is based on the notions of an electric field and a magnetic field. In this chapter, we wish to focus on the concept of an electric field.

The root problem concerns how it is that objects that are not in contact with each other can influence each other. The electric field is a theoretical construct which attempts to account for the mutual influences of charged objects not in actual physical contact.

A small positive point-like test charge qt is fixed at a point P the position of which is

rP ,

as represented below in Figure 35-1. A second positive point charge q is fixed a position given by

rq . According to Coulomb’s law, the electrical force on the test charge is given by

Fqt ,q

E =kqt qrPq

2   rPq . (35-1)

Figure 35-1Coulomb’s Law Revisited

x i( )

y j( )

z k( )

Fqt ,q

E

rq

q

rPq

rP

qt

35-2

The test charge qt interacts electrically with the second charge q . If charge q did not exist,

then there would be no electrical force on the test charge. With that in mind, I want to rewrite equation (35-1) in a slightly different way. We have

Fqt ,q

E = qtkqrPq

2   rPq

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (35-2)

Let us see if we can divine the meaning of the bracketed term in equation (35-2). The unit vector rPq signifies the direction from charge q to the location of the test charge qt . The distance

between the two charges is given by rPq2 . Of course, the electric constant k is a not so subtle

reminder that this interaction is taking place in a vacuum; a space with out any other charges to modify the interaction. The term in the bracket is what we are going to call the electric field. We

will signify the electric field with E . We can now give a technical definition to the electric field:

E = limit

qt→0

Fqt ,q

E

qt

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (35-3)

The electric field is what “remains” when the test charge is removed.We can now rethink the meaning of an electric force on some electric point charge Q . The

electric force on charge Q is given by

FE = Q

E , (35-4)

where E is the electric field at the position of Q due to all of the charges in the universe except

Q .

This view is supposed to solve the problem of electrical action at a distance. You see, Q is

not interacting with remote charges, but with the electric field that is local to the charge Q ! It is difficult to see how this has solved anything. To this author, it only appears to have “displaced” the problem. The modern idea of the electric field, all physical fields in fact, has undergone a drastic reinterpretation. Originally, the field sought to explain how remote charges could influence each other. Charges were considered the “source” of electric fields. However, the modern view defines the field as more primordial that the charges. It could not have been otherwise. The field, in order to mediate physical interactions, itself had to become physical, with all of the properties once attributed only to physical things: momentum, mass, energy and so on. Regardless of the actual ontic status of a field, it is an exceptionally useful model for framing the electromagnetic force.

35-3

THE ELECTRIC FIELD PRODUCED BY A SOLITARY POINT CHARGE

Equation (35-2) holds the key to calculating the electric field at some point P due to some solitary point charge q . We write

E =

E E = E   E =

kqr 2  r

⎡⎣⎢

⎤⎦⎥

. (35-5)

Equation (35-5) is extremely important for our discussion and it is absolutely imperative that you understand it!

The unit vector r always points from the charge producing the electric field to the point at which we wish to calculate the electric field. If the source charge is positive then the electric field will be directed in the same direction as r . However, if the source charge is negative, the electric

field will be directed opposite direction of r . These states of affairs are represented below in Figure 35-2. So, in general, then,

E ≡r  , if  q is positive

− r, if  q  is negative⎧⎨⎩

. (35-6)

The notation used in equation (35-5) takes care of this nicely.The distance between the source charge q and the point P is r . As with most calculations,

we use a Cartesian coordinate system, and, in that system

r = r  r = r  r = rP −rq = xP  i + yP   j + zP   k⎡⎣ ⎤⎦ − xq  i + yq   j + zq  k⎡⎣ ⎤⎦

= xP − xq( ) i + yP − yq( )  j + zP − zq( ) k . (35-7)

Therefore,

r •r = r 2 = xP − xq( )2

+ yP − yq( )2+ zP − zq( )2

. (35-8)

One more comment, physicists often find it more convenient to write equation (35-5) as

E =

kqr 2  rr

⎡⎣⎢

⎤⎦⎥

=kqr3  r⎡

⎣⎢⎤⎦⎥

= kqrP − rqrP − rq

3

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

. (35-9)

35-4

Figure 35-2The Direction of the Electric Field Produced by a Solitary Point Charge

x i( )y j( )

z k( )P

q

rq

rP r

r = rP −rq

E

x i( )y j( )

z k( )P

q

rq

rP r

E

35-5

We now have a very powerful vector notation that we can use to tackle the most involved problem. However, at the introductory level, it is uncommon to have problems involving more than four point charges. Usually, all one needs to do to calculate an electric field at some point P due to some point charge q is to correctly calculate the magnitude of the electric field and a correct unit vector to describe the direction. Recall,

E =

E E . (35-10)

The magnitude of the electric field of a solitary point charge is given by

E =E =

kqr2 . (35-11)

Also, recall that from our earlier work in vectors, a correct unit vector can always be written as

E = cosθx  i + cosθy   j + cosθz   k . (35-12)

This state of affairs, for a negative source charge, is represented below in Figure 35-3.

Figure 35-3A Unit Vector for the Direction of the Electric Field at a Point P

x i( )

y j( )

z k( )P

q

θx

θz

θy

i

j

k

E

35-6

Example One:

A positive charge of q = 6.00× 10−9 C is fixed at the origin of a Cartesian coordinate

system, as represented in the diagram below. We wish to find the magnitude and the direction of the electric field produced by this charge at a point P the position of which is

rP = 2 m i + 3m j .

First, we carefully draw a diagram representing all of the pertinent information. Please pay close attention to the Cartesian coordinate system that you use. (Too many students try to skip this part of problem solving and move directly to the manipulation of equations as if physics were simply an exercise in applied algebra.) Work at trying to grasp what is going on physically. A carefully drawn diagram can help!

The distance from the charge to the point is given by

rP = 2 m( )2 + 3m( )2 = 13 m ≈ 3.6 m , (1)

The magnitude of the electric field is given by

E =kqrP

2 =8.99 ×109  N ⋅m2 ⋅C−2( ) 6.00 ×10− 9 C( )

13 m( )2 = 4.15 N ⋅C−1 . (2)

Using equation (35-4), the unit vector that represents the direction of the electric field can be written as

E = cosθx i + cosθy j = 2 / 13( ) i + 3 / 13( )  j . (3)

x ˆ i ( )

y ˆ j ( )

q

r =13

m

E

5 m

3m

5 m

− 5 m

θx

θy

2 mP

35-7

Therefore,

θx = cos −1 213

⎡⎣⎢

⎤⎦⎥

= 56.3° , (4)

while

θy = cos− 1 313

⎡⎣⎢

⎤⎦⎥

= sin−1 213

⎡⎣⎢

⎤⎦⎥

= 33.7° . (5)

Example Two:

A negative charge q = − 6.00× 10− 9 C is fixed at the origin of a Cartesian coordinate

system, as represented in the diagram below. We wish to find the magnitude and the direction of

the electric field produced by charge q at a point P where

rp = − 5 m i + 4 m j .

The distance from point charge q to point P is given by

rP = 5 m( )2 + 4 m( )2 = 41 m ≈ 6.4 m , (1)

The magnitude of the electric field, then, is

E =kqrP

2 =8.99 × 109  N ⋅m2 ⋅C−2( ) −6.00 × 10− 9 C( )

41 m( )2 = 1.32 NC

. (2)

x ˆ i ( )

y ˆ j ( )

q

r =41m

E

5 m

4 m

5 m

− 5 m

θx

θy 5 mP

35-8

A unit vector that represents the direction of the electric field is given by

E = cosθx i − sinθx j =541 i −

441  j = 0.7809  i − 0.6247  j

= cosθx i + cosθ y j . (3)

Therefore: θx = cos −1 541

⎡⎣⎢

⎤⎦⎥

= 38.7° , (4)

while , θy = cos− 1 −441

⎡⎣⎢

⎤⎦⎥

= 128.7° . (5)

Example Three:

A negative point charge q = − 4.50× 10−9 C is fixed at a point the position of which is

rq = 3 m i + 4 m j , as represented below in the diagram. We wish to find the electric field at

point P the position of which is

rP = − 2 m i + 2 m j .

The diagram clearly indicates that vector r runs from charge q to point P . It should also be

obvious that r is the distance from charge q to point P , and that E = − r . So our first task is to

find r . To that end, we write

r = r  r = rP −rq = − 2 m i + 2 m j⎡⎣ ⎤⎦ − 3m i + 4 m j⎡⎣ ⎤⎦ =−5 m  i − 2 m   j . (1)

Therefore, the magnitude of r is given by

r = r = 5 m( )2 + 2 m( )2 = 29  m ≈ 5.4 m . (2)

A unit vector that represents the direction of r can also be used to signify the direction of the

direction of the electric field produced by charge q at point P . We have

r =rr

= −529  i −

229  j =− 0.9285  i − 0.3714  j . (3)

As was pointed out earlier,

35-9

E = − r =529 i +

229  j = cosθx  i + sinθx   j , (4)

where

θx = cos −1 529

⎡⎣⎢

⎤⎦⎥

= 21.8° . (5)

The magnitude of the electric field at point P due to charge q is given by

E =E =

kqr2 =

8.99 ×109  N ⋅m2 ⋅C−2( ) − 4.50 ×10− 9 C( )29  m( )2 = 1.40 

NC

. (6)

x ˆ i ( )

y ˆ j ( )

E

5 m− 5 m

q

rq

rP

r

θxP

rq

rP

r =rP

−rq

O

O

4m

35-10

THE ELECTRIC FIELD PRODUCED BY N POINT-LIKE ELECTRIC CHARGES

We have found it convenient to define the electric field at some point P due to some point charge q as

E =

E E = E   E =

kqr2  r , (35-13)

where

r = r  r = rP −rq , (35-14)

and

rP is the position of the point where we wish to calculate the electric field and

rq is the

position of the electric charge that produces the electric field. To define the total electric field at some point P produced by N point charges, we use what

physicists call the superposition principle. To find the net or total electric field at point P , we must add vectorially the electric field contribution of each electric charge. This vectorial sum is the net or total electric field. Mathematically, we can write the superposition principle as

Etot =

Etot Etot =

Ei

i=1

N

∑ =E1 +

E2 ++

EN , (35-15)

where

Ei = kqi / ri

2( )  ri , (35-16)

and

ri = ri   ri = r1 ri = rP − rqi . (35-17)

Example Four:

Three positive point charges of magnitude q1 = q2 = q3 = q = 8.75 ×10− 9 C are fixed at

three corners of a square of side length s = 0.250 m , as represented in the diagram below. We wish to find the net electric field at the fourth corner.

The symmetry of the charge distribution makes this problem a little more straightforward. We can write

E1 = −

kq1

s2  i = −8.99 × 109  N ⋅m2 ⋅C−2( ) 8.75× 10− 9 C( )

0.250 m( )2   i = −1,259 NC  i   . (1)

35-11

Also,

E2 =

kq2

s 2( )2

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥  −

22  i +

22  j

⎡

⎣⎢

⎤

⎦⎥ =

2kq2

4 s2

⎡

⎣⎢

⎤

⎦⎥ −  i +   j⎡⎣ ⎤⎦

=2 8.99 ×109  N ⋅m2 ⋅C−2( ) 8.75 ×10− 9 C( )

4 0.25 m( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−  i +   j⎡⎣ ⎤⎦

= − 445 NC i + 445 

NC  j . (2)

The third charge gives rise to

E3 =

kq3

s2   j =8.99 × 109  N ⋅m2 ⋅C−2( ) 8.75× 10−9 C( )

0.250  m( )2   j = 1,259  NC  j   . (3)

x ˆ i ( )

y ˆ j ( )

E1

q2q3

E2

E3

q1

s = r3

r1 = s

s

sr2 = s 2

Enet

θx

θy

35-12

To find the net electric field, we use the superposition position. We have

Etot =

E1 +

E2 +

E3

= −1,259  NC

− 445NC

⎡⎣⎢

⎤⎦⎥ i + 1,259  

NC

+ 445NC

⎡⎣⎢

⎤⎦⎥  j

= − 1,704 

NC  i + 1,704 

NC

j =Etot . (4)

The magnitude of the net electric field would be

Etot =

Etot = 1,704  N / C( )( )2 + 1,704 N / C( )( )2 = 2,410  N / C( ) . (5)

A unit vector to represent the direction of the net electric field can by found using

Enet =EnetEnet

== −1,7042,410

 i +1,7042,410

j =− 0.7071 i + 0.7071  j

= cosθx   i + cosθy   j , (6)

where θx = cos −1 −0.7071( ) = 135° , (7)

while θy = cos−1 0.7071( ) = 45° . (8)

EXAMPLE FIVE

Fixed at a position given by

rq1= 4.00  m  i + 5.00  m   j is a positive point charge of

magnitude q1 = 90 × 10−9 C . A second charge, q2 = − 70 ×10−9 C , is fixed at a position given

by

rq2= −  2.00 m  i + 3.00  m   j . We want to find the net electric field at point P the

position of which is

rP = −  5.00  m  i − 4.00  m   j .

We begin by finding

r1 . We can write

r1 = rP −rq1

= −  5.00 m  i − 4.00  m   j⎡⎣ ⎤⎦ − 4.00 m  i + 5.00 m   j⎡⎣ ⎤⎦

= − 9  m  i − 9 m   j . (1)

35-13

Equation (1) implies

r1 = 9 m( )2 + 9 m( )2 = 9 2  m , (2)

and

E1 =

kq1

r12 − cos 45°  i − sin45°  j( )

=8.99 × 109  N ⋅m2 ⋅C−2( ) 90 ×10−9 C( )

9 2  m( )2 −cos 45°  i − sin45°  j( )

= −3.532 NC i − 3.532 

NC  j . (3)

q14,5( )

q2−2, 3( )

P −5,−4( )9m

9m

r 1=

29m(

)

3m

7m

E1

E2

Enet

x ˆ i ( )

y ˆ j ( )

r 2=

58 m

35-14

Next, we find

r2 . We write

r2 = rP − rq2= − 5.00 m  i − 4.00 m   j⎡⎣ ⎤⎦ − − 2.00 m  i + 3.00 m   j⎡⎣ ⎤⎦

= − 3 m  i − 7 m   j . (4)

Equation (4) implies

r2 = 3 m( )2 + 7 m( )2 = 58  m , (5)

and

E2 =

kq2

r22 cos 66.8°  i + sin66.8°  j( )

=8.99 × 109  N ⋅m2 ⋅C−2( ) −70 × 10−9 C( )

58  m( )2 cos 66.8°  i + sin66.8°  j( )

= 4.274 NC i + 9.973 

NC  j . (6)

Using the superposition principle, the vector representing the net electric field at point P is

Etot =

E1 +

E2 = −  3.532 

NC

+ 4.274 NC

⎡⎣⎢

⎤⎦⎥ i + − 3.532 

NC

+  9.973 NC

⎡⎣⎢

⎤⎦⎥  j

= 0.742  N / C( ) i + 6.441  N / C( )  j . (7)

The magnitude of the total electric field is

Etot = 0.742( )2 + 6.441( )2  NC

= 6.484 NC

. (8)

The angles that tell us about the direction of the net electric field are given by

θx = cos −1 Ex / E[ ] = cos −1 0.742 / 6.484[ ] = 83.4° , (9)

and

θy = cos−1 Ey / E⎡⎣ ⎤⎦ = cos −1 6.441 / 6.484[ ] = 6.6° . (10)

35-15

THE ELECTRIC DIPOLE

Any time we have two point charges of equal charge magnitude and opposite sign, we have what is called an electric dipole. The charges are separated by some distance we signify with . It is useful to set up an axis that runs through both charges called an alignment axis. A second axis, called the transverse axis, is oriented as a perpendicular bisector of the alignment axis. We place the origin of a Cartesian coordinate system at the intersection of the axes. One such possible arrangement is represented below in Figure 35-3.

Point P1 is an arbitrary point on the alignment axis, in this case the x-axis, such that x > a ;

we want to look at points “outside” the dipole. Point P2 is an arbitrary point on one transverse

axis, in this case the y-axis, such that y > 0 . (The z-axis is also transverse.) We want to derive

an expression for the electric field produced by the electric dipole at points P1 and P2 . (It would

be an instructive exercise for the student to derive an expression for the electric field at any point!)

Figure 35-3An Electric Dipole

x i( )

y j( )

q − q

P2

P1

a a x − a

x

y2 + a2 = r

y

E+

E−

E+

E−θ θ

Enet

Enet

θθ

35-16

At point P1 , we have

Etot =

E+ +

E− =

kqx + a( )2  i

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ − kq

x − a( )2   i⎡

⎣⎢⎢

⎤

⎦⎥⎥

= kq1

x + a( )2 −  1

x − a( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥   i = kq

x − a( )2 − x + a( )2

x + a( )2 x − a( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥   i

= kqx2 − 2ax + a2 − x2 − 2ax − a2

x + a( )2 x − a( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥   i

= −

2kq 2a( ) xx4 − 4a2 x2 − a4  i = −

2kqxx4 − 4a2 x2 − a4  i . (35-18)

At point P2 , we have, by symmetry,

E+ = E− = kq / r2 , (35-19)

and

Enet = 2E+ , x  i = 2 E+  cosθ[ ]  i . (35-20)

So, we can write

Enet = 2

kqy2 + a2  

ay2 + a2

⎡

⎣⎢⎢

⎤

⎦⎥⎥ i =

kq 2a( )y2 + a2( )3/ 2  i =

kq

y2 + a2( )3/ 2   i . (35-21)

Figure 35-4The Electric Dipole Moment

p = p p = −  q  iq − q

= 2a

We now want to define a new vector quantity called the electric dipole moment. We have

p = pp = q  p . (35-22)

35-17

The magnitude of the dipole moment is the product of the distance between the two charges, ,and the magnitude of each charge of the electric dipole, q ; that is,

p = q . (35-23)The direction of the electric dipole moment is the direction from the negative to the positive charge of the dipole. For the electric dipole given above

p = pp = − q i = −2qa i . (35-24)

We can now generalize our results. For a point on the alignment axis, the electric field generated by an electric dipole looks like

Edipole  alignment =

2kxpx4 − 4a2x2 − a4 . (35-25)

However, for a point on the transverse axis, the electric field looks like

Edipole   transverse = −

k p

y2 + a2( )3/ 2 . (35-26)

It is of considerable interest to find out what happens to equations (35-25) if x >> a , and to equation (35-26) if y >> a . We begin by rewriting equation (35-25) getting

Edipole  alignment =

2kxpx4 1− 2a / x( )2 − a / x( )4⎡⎣ ⎤⎦

=2k p

x3 1− 2a / x( )2 − a / x( )4⎡⎣ ⎤⎦. (35-27)

Now, if x >> a , then equation (35-27) approximates

Edipole  alignment ≈

2k px3 . (35-28)

We can do the same kind of analysis with equation (35-26) and find

Edipole   transverse = −

kp

y2 1+ a2 / y2( )( )⎡⎣

⎤⎦

3/2  = −kp

y3 1 + a2 / y2( )( )3/ 2   . (35-29)

Of course, when y >> a , then equation (35-21) approximates

Edipole   transverse ≈ −

k py3 . (35-30)

We have spent some time looking at the electric dipole and you may be wondering: “Why?”An electric dipole is electrically neutral; just as a normal atom is. So at large distances the atom and

35-18

the electric dipole have no electrical influence; they fall off as the distance cubed. However, on a small enough scale, the electric dipole, just like an atom, can have profound electrical influences. Virtually every process that is going on in any living organism involving energy transfer is a result of the electrical interactions between objects that are electrically neutral on the macroscopic scale. On the molecular scale, however, there are strong electric fields and bewilderingly complex electromagnetic interactions going on. In trying to model these interactions, physicists began with a simplification like the electric dipole. One can modify this model to account for more complexity by introducing quadrupoles, octupoles, and so on.

THE ELECTRIC FIELD PRODUCED BY CONTINUOUS CHARGE DISTRIBUTIONS

Figure 35-5Volume, Surface and Linear Charge Distributions

P

P

P

volume

surface

linedq

dq

dq

r

rr

r

r

r

dV

dA

d

35-19

Finding the net electric field produced by N electric point charges can be quite tedious work. However, if one pays close attention to all of the details and avails oneself of our powerful vector notation, the procedure is at least straightforward. To master the process does, however, require some practice.

Before I leave this subject of the electric field, I feel I must remind one that there is an entire class of problems that we have not addressed. So far we have looked only at the electric field produced by point-like electric charges. In the physical world, however, there are very many important situations where electric charges are “smeared out” over space and can not be treated reasonably as point-like. Solving such problems, however, requires the use of calculus; and calculus is beyond the mathematical level of this text.

The basic idea involved in solving for the electric field produced by a continuous charge distribution, is, essentially, to breakup the continuous distribution into an infinite number of infinitely small charges each of which can be treated as point-like. So, in keeping with the basic theme of this chapter, we say that an infinitesimal electric charge dq gives rise to an infinitesimal

electric field dE at some point P that is given by

dE =

kdqr 2  r . (35-31)

There are three basic ways electric charge can be “smeared.” It can be distributed over a volume, or over a surface, or along a line. This state of affairs is represented above in Figure 35-5. The value we assign to the infinitesimal charge dq will depend on how the charge is smeared. We use:

dq =ρ dV  , volume distribution

σ  dA , surface distribution

λ  d  , linear distribution

⎧

⎨⎪

⎩⎪ , (35-32)

where ρ signifies the volume charge density--the electric charge per unit of volume; σ signifies

the surface charge density--the electric charge per unit area; and λ signifies the linear charge

density--the electric charge per unit length. (ρ , σ , and λ are the lowercase Greek letters rho,

sigma and lambda, respectively.) Of course, dV signifies the infinitesimal volume, while dA

signifies the infinitesimal area, and d signifies the infinitesimal length.It is useful for you to know the electric field produced by some very common continuous

charge distributions. I am going to use calculus to solve for the electric field of these distributions. Even though you may not know how to do calculations yourself, I think it can be instructive to follow the reasoning involved in solving these problems.

35-20

Example Six: A Uniform Charge Distribution on an Infinitely Long ConductorA very thin conductor of length L lies along the x-axis of a Cartesian coordinate system, as

represented in the diagram below. A positive charge Q is “smeared” uniformly over the

conductor. We want to calculate the electric field at some arbitrary point P on the y-axis, where they-axis is assumed to be a perpendicular bisector of the x-axis. Since the charge is uniformly distributed along the line, the linear charge density λ is given by

λ = Q / L . (1)

The infinitesimal electric charge dq would give rise to an infinitesimal electric field dE at

some point P , where

dE =

k dqr2 r , (2)

where dq = λ  dx . (3)

The distance from the infinitesimal charge dq to point P is given by

r = x2 + y2 . (4)

As the charge distribution is positive, then the direction of the electric field at point P is

r = − cosθ  i + sinθ   j = −x

x2 + y2 i +

ax2 + y2

  j . (5)

x ˆ i ( )

y ˆ j ( )

x dx

dq

P

dE

r = x2 + y2

θ

θ

−dE cosθ  i

dE sinθ   j

y

L

r

35-21

Substitution of equations (3) through (5) into equation (2) gives us

dE =

k λ dxx2 + y2 −

xx2 + y2

 i +a

x2 + y2  j

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (6)

Of course to get the net electric field at point P we must add up all of the infinitesimal contributions. When we do, letting the length of the conductor go to infinity, we get

E = d

E∫ = −kλ

x dx

x2 + y2( )3/ 2−∞

∞

∫  i + kλadx

x2 + y2( )3/ 2−∞

∞

∫   j . (7)

The first integral on the right-hand side of equation (7) vanishes by symmetry. For every dQ right of the y-axis generating a negative x-component of electric field, there is an identical one

left of the y-axis generating a positive x-component canceling all x-directed components. The second integral, however gives us

E = kλ a

dx

x2 + y2( )3/ 2−∞

∞

∫   j =2kλ

y  j =

2kλy

j . (8)

Figure 35-6The Electric Field Due to a Uniform Line Charge Distribution

P

E

r⊥

L

+Q

r⊥

35-22

The important point here is that for a uniform charge distribution on a very, very long straight wire, the electric field can be described by equation (8). I wish, however, to write our result in a slightly different form. I will use the following notation, see Figure 35-6 above.

E = E E =

2kλr⊥

r⊥ . (9)

In equation (9), r⊥ represents the perpendicular distance, the shortest distance, from the line charge distribution to the point of interest at which we wish to calculate the electric field. The unit vector ˆ r ⊥ points away from the line charge distribution. If the line charge is positive, then the

direction of the electric field will be the same as ˆ r ⊥ . However, if the line charge is

negative, then the electric field will be directed opposite to ˆ r ⊥ :

ˆ E =ˆ r ⊥ , for a positive λ

- ˆ r ⊥ , for a negative λ

⎧ ⎨ ⎩

. (10)

Example Seven:A Uniform Charge Distribution on a Circular Conductor

Another common continuous electric charge distribution is that of a positive electric charge q uniformly distributed over a circular ring of radius a , as represented in the diagram below. We wish to find the electric field produced at an arbitrary point P on an axis which passes through the center of the circle and is perpendicular to the plane of the circle; in this instance, the z-axis.

The infinitesimal charge is distributed over a circular arc of length

d = a dϕ . (1)Therefore,

dq = λ  d = q / 2π a[ ] a dϕ[ ] = q / 2π( ) dϕ . (2)

The infinitesimal electric field produced at point P due to dq is given by

dE =

k dqr 2 r =

kq2π

dϕz2 + a2( ) − sinθ  r⊥ + cosθ  k⎡⎣ ⎤⎦

=kq2π

dϕz2 + a2( ) −

az2 + a2

  r⊥ +z

z2 + a2 k

⎡

⎣⎢

⎤

⎦⎥ . (3)

35-23

Adding all of the infinitesimal contributions, we have

E = d

E∫ = −

kqa

2π z2 + a2( )3/2 dϕ  0

2π

∫ r⊥ +kqz

2π z2 + a2( )3/ 2 dϕ  0

2π

∫ k . (4)

Again, by symmetry, the first integral on the right-hand side of equation (4) vanishes. So, we are left with

E =

kqz

2π z2 + a2( )3/2 dϕ  0

2π

∫ k =kqz

2π z 2 + a2( )3/ 2 2π[ ]k =kqz

z2 + a2( )3 /2   k . (5)

This is equivalent to:

E =

kqz

z 2 + a2( )3/2  k ≡2πakλz

z2 + a2( )3 /2   k . (6)

(Of course if the “smeared” charge were negative, then E = −k . )

P

dϕ

dq

a

z

x i( )

y j( )

z k( )

r = z2 + a2θ

ϕ

θ dE

r

ϕ

r⊥

d ′q

d′E

−dE sinθ r⊥

dE cosθ  k

35-24

Summary of Chapter 35

Point Charges

Electric charges can be point-like charges or “smeared” out over space. Represented belowin Figure 35-7 is a positive point-like electric charge Q fixed at the origin of a Cartesian coordinate

system. This charge will produce an electric field at a point P given by

E =

kQr 2 r , (35-33)

where r is the distance from the charge to point P , and r is a unit vector pointing from the charge

to point P . The electric constant of proportionality is k is the value of which is

k =1

4πεo

= 8.99 ×109  N ⋅m2 ⋅C−2 , (35-34)

and εo is the so-called permittivity of free space.

Figure 35-7The Electric Field Produced by a Positive Point-Like

Electric Charge Fixed at the Origin of a Cartesian coordinate System

z ˆ k ( )

y ˆ j ( )

x ˆ i ( )

r

P

Q

r

E

35-25

If electric charge Q is negative, as represented below in Figure 35-8, then we can write, using equation (35-33),

E =

k − Q( )r2 r =

k Qr2 − r( ) . (35-35)

So, the direction of the electric field at point P is opposite of that produced by the positive charge above. So in general, for a point charge, the magnitude of the electric field is given by

E =k Qr2 , (35-36)

while the direction of the electric field is given by

E =r , for a positive Q

- r , for a negative Q⎧⎨⎩

. (35-37)

Figure 35-8The Electric Field Produced by a Negative Point-Like

Electric Charge Fixed at the Origin of a Cartesian coordinate System

z ˆ k ( )

y ˆ j ( )

x ˆ i ( )

r

P

Q

r

E

35-26

For a point-like electric charge q that is not located at the origin but located at a point the

position of which is

rq , then the electric field at some point P the position of which is

rP is given

by

E = E E =

kqr2 r , (35-38)

where, as illustrated below in Figure 35-9,

r = r  r = rP −rq . (35-39)

Whether the charge q is located at the origin or not, the physical meaning of the vector r is that

it runs from the source charge to the point at which we wish to calculate the electric field. The magnitude r is the distance from the source charge to point P. The unit vector r signifies the direction pointing from the source charge to point P.

Figure 35-9The Electric Field Produced by a Positive

Point-Like Electric Charge Not Located at the Origin

rP

z ˆ k ( )

y ˆ j ( )x ˆ i ( )

rq

r

E

P

q

35-27

If there are N point-like electric charges, then the total electric field at some point P in space

the position of which is

rP is given by the vectorial sum of each of the individual electric fields.

This is called the superposition principle. We have

Etot =

Ei

i =1

N

∑ =E1 +

E2 ++

EN , (35-40)

where

Ei =

kqi

ri2  ri , (35-41)

andri = rP − rqi

. (35-42)

Continuous Charge Distributions

Very Long, Straight, Uniform Distribution of Electric Charge

If electric charge Q is uniformly distributed over a very long straight wire of length , then the electric field produced can be expressed by

ELSC =

2kλr⊥

r⊥ , (35-43)

where λ is the linear charge density given by

λ =

Q

, (35-44)

and r⊥ is the shortest distance from point P to the wire, and is called the perpendicular distance.

(See Figure 35-10 below.) The direction of the electric field is away from the wire is λ is positive,

or toward the wire, if λ is negative. We can codify these results on the direction:

ELSC =r⊥ , for a positive λ

- r⊥ , for a negative λ

⎧⎨⎩

. (35-45)

35-28

Figure 35-10The Electric Field of a Uniform

Charge Distribution along a Very Long Line

P

E

r⊥

L

+Q

r⊥

Figure 35 - 11Electric Field Produced by Uniform Charge Distribution

On a Circle of Radius a

P

E

x ˆ i ( )

y ˆ j ( )

z ˆ k ( )

+  λ

a

z

k

If λ  were negative, then E  would be in the  - k  direction.

i j

35-29

Uniform Charge Distribution over a Circle of Radius aIf , as represented above in Figure 35-11, we have a circular ring of radius a with a uniform

linear charge density λ on the ring, then the electric field on the axis of symmetry at an arbitrary point P a distance z from the center of the ring is given by

E =

2π kλaz

z 2 + a2( )3/2 k . (35-46)

If, however,

λ =q

2πa , (35-47)

then

E =

kqz

z 2 + a2( )3/2 k . (35-48)

Chapter 36

ELECTRIC FLUX AND GAUSS’ LAW

Electric charges give rise to electric fields. Our ability to understand electrical interactions quantitatively is determined, in large part, by our ability to calculate the electric field. In the last chapter, we did a complicated calculation to determine the electric field that is produced by a uniform line charge distribution of λ Coulombs per meter. That calculation would have been impossible without using calculus. Even with calculus, the derivation was not trivial.

In this chapter, we want to use another technique to find the electric field. This alternative method is called Gauss’ Law. This “law” was discovered by one of the greatest mathematical physicists ever; Karl Friedrich Gauss, 1777-1855. To use this method effectively, one must recognize symmetries in charge distributions, and understand the geometry associated with a specific kind of symmetry.

AREA AS A VECTOR

In physics, it turns out that how an area is oriented in space can be very important. To help us describe the orientation of an area, we treat it as a vector. We can write an area as a vector using

A =

A A = A A , (36-1)

where A , the magnitude of the area , is equal to what we normally think of as the area. The unit

vector A represents the direction of area vector and allows us to describe its orientation in space. Below, in Figure 36-1, I have represented four area unit vectors. The first area unit vector is

associated with a circle. The circle is itself a boundary that encloses the area of the circle. This area also happens to be planar. There are two senses in which one could go around the boundary of the circle. The small gray arrowheads are intended to signify the sense in which we “go around” this

circle. The unit vector Acircle is perpendicular to the plane of the circle, in the right hand sense.

(The right hand sense is the sense in which the fingers of the right hand naturally curl around the extended right thumb.) For the circle indicated, using the Cartesian coordinate system shown, we have

Acircle = Acircle Acircle = πR2 k , (36-2)

36-2

In a similar fashion, the area vector representing the square is given by

Asquare = A Asquare = − s2 k . (36-3)

Note how in going in the opposite sense around the square than we did around the circle, we have changed the direction of the area unit vector for the square.

Figure 36-1Examples of Area Unit Vectors

R

Acircle

s

s

Asquare

x i( )y j( )

z k( )

∧dAsphere

r r⊥ ∧dAcylinder

36-3

The other two area area vectors are different. First, they deal with what we call closed surfaces; in this case we have a sphere and a cylinder. (Closed surfaces enclose a volume. The square and the circle do not enclose a volume.) The first closed surface is a sphere. Here we have represented an infinitesimal spherical surface area. The unit vector for a closed surface is positive if it points outward and negative if it points inward. For the sphere, the unit vector representing the surface area is always aligned with the radial unit vector and always points away from the center of the sphere. So, we have for the sphere

dAsphere = dAsphere dAsphere = dAsphere r , (36-4)

where we are using the unit vector r to signify away from the center of the sphere.The area unit vector for the cylinder is associated with an infinitesimal area on the lateral

surface of the cylinder. (We are now ignoring the top and bottom surfaces.) We can write this area vector as

dAcylinder = dAcylinder dAcylinder = dAcylinder r⊥ , (36-5)

where we are using the unit vector r⊥ to signify away from the axis of symmetry of the cylinder.

(A unit vector − r⊥ is used to signify the direction toward the axis of symmetry of the cylinder,)

ELECTRIC FLUX

In physics, we have may kinds of quantities that are called fluxes. A flux exists wherever a vector quantity is “popping through” an area. For example, water flowing out of a faucet. We can represent this as a flux. The velocity vector of the water would literally pop through the area of the opening of the faucet.

Consider a rectangular area that is oriented vertically in space and facing west like that represented below in Figure 36-2. Also, assume that this rectangle is located in a region of space where there is a uniform electric field directed east. The electric flux is given by

ΦE =

E •A . (36-6)

Using the rules for the dot product, we find

ΦE =

E •A = E E⎡⎣ ⎤⎦ • A A⎡⎣ ⎤⎦ = EA E • A⎡⎣ ⎤⎦ = EA cos∠bet . (36-7)

For this specific flux, we would have

ΦE = −E w( ) , (36-8)

as the angle between the direction of the area and the direction of the electric field is one hundred eighty degrees, and the cosine of one hundred eighty degrees is a negative one. The electric flux

36-4

can be positive or negative. Positive values indicate that the angle between the direction of the area and the direction of the electric field is less than ninety degrees. Negative values indicate that the angle between the direction of the area and the direction of the electric field is greater than ninety degrees. Finally, note that if the electric field were directed north rather than east, then the electric field would not “pop” through the rectangle at all and there would be no electric flux. That is, the electric flux is zero when the angle between the direction of the area and the direction of the electric field is ninety degrees.

What is one to do if the electric field is not uniform but varies from place to place? Then one must use calculus to solve the problem. We would breakup the area into essentially an infinite number of infinitesimal areas, dA , and calculate the electric flux for each such infinitesimal area. We would then have an infinite number of infinitesimal electric flux values. Finally, we would add

up all of the electric flux values to get the total electric flux. (The integral sign, ∫ , is how the mathematician signifies that he is adding up an infinite number of infinitesimal values. The rules for adding such values are not quite the same as the rules for adding finite values. One learns these rules in integral calculus.)

ΦE = dΦE∫ =

E • d

A∫ . (36-9)

Figure 36-2

East ˆ i ( )North ˆ j ( )

Up ˆ k ( )

E

A

w

dA

36-5

SYMMETRIES

So far, we have only looked at planar areas. One could also have electric flux through the surface area of a closed volume. For example, consider a sphere of radius R. The sphere divides the world into three regions: inside the sphere, on the surface of the sphere and outside of the sphere all together.

The sphere has what we call point symmetry. Every point on the surface of the sphere is exactly the same distance from the center of the sphere. As you walk around on the surface of the sphere, everything “looks” the same.

Another kind of symmetry deals with a right-circular cylinder. All points on the lateral surface of the cylinder are exactly the same perpendicular distance from the major axis of symmetry of the cylinder. So, as you walk around on the lateral surface of the cylinder, everything “looks” the same. The cylinder has what we call line symmetry.

Point symmetry and line symmetry are important for our analysis of the electric field because we are often dealing with point-like electric charges, and with charges distributed over straight line conducting wires. We can exploit these symmetries in our analyses of the electric field.

R

r⊥

Cylinders Have Line Symmetry

Spheres Have Point Symmetry

Figure 36 - 3

36-6

GAUSS’ LAW

We found out in the last chapter that a single, stationary, point charge Q produces an electric

field at some point P given by

E =

kQr 2 r , (36-10)

where the magnitude of the field is given by

E =kQr 2 , (36-11)

and the direction is given by

E =r , if q is positive− r , if q is negative

⎧⎨⎩

, (36-12)

where, recall, r is a unit vector that points away from the charge Q toward point P , the point where we wish to calculate the electric field.

Assume that a positive point-like electric charge Q is fixed at the origin of a Cartesian

coordinate system, as represented below in Figure 36-4. If we construct a sphere of radius R centered on the origin of this Cartesian coordinate system, then at each point of the surface of our constructed sphere, the magnitude of the electric field produced by this charge would be given by equation (36-11). Note also that if we wanted to divide this surface up into an infinite number of

infinitesimal areas, dA , then we could represent the direction of each area with a unit vector parallel

to r .So, at every point on the surface of the sphere, the magnitude of the electric field is constant.

At every point on the surface of the sphere, the electric field is directed away from the center of the

sphere, in the r direction. For any infinitesimal area dA on the surface of the sphere, we can see

that it is also directed away from the center of the sphere, that is∧

dA = r . (36-13)

The amount of electric flux through each dA would be given by

dΦE =

E • d

A = E   r[ ] • dA  r[ ] = E  dA r • r

= E dA cos 0°( ) = E dA . (36-14)

36-7

To find the total electric flux through the surface of the sphere, we need to add up all of the infinitesimal fluxes. We have

ΦE = dΦE∫ = E dA =∫ E dA∫ = E Atotal surface . (36-15)

I hope you do not feel like I have just slipped something past you when you were not watching.

Equation (36-15) is saying that there is an electric flux dΦE through each dA the value of which

is E dA, and these are the values we wish to add up. However, since the value of E is the same for each one we wish to add. The total sum, then, is just as if we multiplied the total area by E, that is

E dA∫ . For example, if we had the following finite number of terms to add

EA1 + EA2 + EA3 + EA4 + EA5 ++ EAN , (35-16)

then we could write the sum as

E A1 + A2 + A3 + A4 + A5 + ...+ AN[ ] = E Aii =1

N

∑ = E Atotal . (36-17)

Figure 36-4

x ˆ i ( )

z ˆ k ( )

y ˆ j ( )R

rdA

dA

Q

36-8

This is the kind of thing we have done in equation (36-15) only we are adding infinitesimal quantities and must use the rules of integral calculus.

So, just what is it that we have found? Let us see. We know the E value on the surface, and we know the total surface area of a sphere. So we can write

ΦE = E Atotal =kQR2

⎡⎣⎢

⎤⎦⎥

4π R2⎡⎣ ⎤⎦ = 4π kQ =Qεo

, (36-18)

where, Q is the electric charge inside the sphere, and εo is the permittivity of free space. Gauss’ law tells us that anytime we have a closed surface, the total electric flux through that surface is simply the total electric charge inside divided by εo . Using calculus notation, we write this as

ΦE =E • dA∫ =

Qinside

εo

. (36-19)

The fancy symbol ∫ is telling us to add up the electric flux over a closed surface.

This result really makes calculating the electric flux through a closed surface very simple indeed. However, this is not its primary benefit for the science of physics. Its primary benefit for us is that it gives us yet another way to find the electric field. When can we use Gauss’ law to find the electric field? Whenever we have symmetrical electric charge distributions! Let us do an example to illustrate.

Example One:A Straight Line, Uniform Charge Distribution

Assume that we have a very long, straight conductor of length L with an electric charge Q uniformly distributed along the wire, as represented in the diagram below. We want to use Gauss’ law to determine the electric field at some arbitrary point P off of the wire.

Since we have a line charge distribution, we know that a cylinder has the same symmetry. So, we construct a “virtual” cylinder with its axis of symmetry coaxial with the conductor. We want the radius of the cylinder to be r⊥ so that the point of interest, lies on the lateral surface of the“virtual” cylinder. (I use the word virtual simply to let you know that the cylinder is not real, but merely a mental device to help us calculate. We call these virtual closed surfaces Gaussian surfaces.)

Our Gaussian cylinder is made up of three areas, the top, the bottom and the lateral surface. As the direction of the electric field must everywhere point away from the wire, in what I

call the ˆ r ⊥ direction, then there will be no flux through the top or the bottom as E and ˆ A top are

36-9

perpendicular, as are E and ˆ A bot . The only electric flux will be through the lateral surface.

So, from Gauss’ law, equation (36-19), we can write

E Alat = Qinside

εo

. (1)

Solving equation (36-19) for the magnitude of the electric field, we find

E =Qinside

εo Alat

= Q / L( )εo 2π r⊥( ) =

12πεo

λr⊥

=2kλr⊥

, (2)

where we have defined the linear charge density as

λ = Q / L( ) . (3)

Using Gauss’ Law with a Uniform Line Charge Distribution

ik

j

r⊥

Top

Lateral SurfaceBottom

r⊥

ˆ E = ˆ r ⊥ = ˆ A lat

ˆ A top

ˆ A bot P

Q

36-10

The direction of the electric field is, of course, ˆ r ⊥ ; that is, the electric field at every point in space points away from the line charge. This result is what we found in the last chapter but we had to use the calculus to get it. This time, we have used Gauss’ law.

Example Two:A Uniform Spherical Charge Distribution

Assume that we have an electric charge Q uniformly distributed over the entire volume of a sphere of radius R, as represented in the diagram below. We want to use Gauss’ law to calculate

the electric field at points P and ′P where:

a) P is located in the regime where r > R .

b) ′P is located n the regime where r < R .

a) For point P , where r > R , we use a concentric sphere of radius r so that the point of interest lies on the Gaussian surface. Using equation (36-19), we can write

E Asurface = Qinside

εo

, (1)

and, therefore,

E =Qinside

εoAsurface

=Q

εo 4π r 2( ) =1

4πεo

Qr2 =

kQr 2 , (2)

for any r such that r > R , in other words, for any point outside of the spherical charge distribution. This result is identical to what we have already found to be the case for an electric field due to a point-like electric charge. Outside of a uniform spherical charge distribution, the electric field acts as if all of the charge were at the center of the sphere. The sphere has point symmetry.

b) For a point ′P where r < R , we use a concentric sphere of radius r so that the point of interest lies on the Gaussian surface. Using equation (36-19), we can write

E Asurface = Qinside

εo

, (3)

and, therefore, E =Qinside

εoAsurface

=′Q

εo 4π r 2( ) =1

4πεo

r / R( )3 Qr 2 =

kQR3

⎡⎣⎢

⎤⎦⎥

r , (4)

where we have used the fact that the charge ′ Q is some fractional part of the total charge Q , and

36-11

that the fraction is the ratio of the respective volumes. So, we have

′Q =Vinside

Vtotal

Q =4 / 3( )πr3

4 / 3( )πR3 Q = r / R( )3 Q . (5)

Using Gauss’ Law with a Uniform Spherical Charge Distribution

R

r

rQ

′Q

′P

P

36-12

In Example Two we found that for a uniform spherical charge distribution the magnitude of the electric field inside the sphere is directly proportion to the radial distance of the point of interest. Outside the sphere, the magnitude of the electric field is inversely proportional to the square of the radial distance. If we graph the magnitude of this electric field, we get a graph like the one shown below in Figure 36-5.

Figure 36-5The Magnitude of the Electric Field ProducedBy a Uniform Spherical Charge Distribution

r

E r( )kQR2

R 2R 3R

inside outside

36-13

Example Three:The Interior of a Conductor in Equilibrium

A solid metal sphere of radius R carries a positive electric charge Q that is at rest and uniformly distributed on its surface. We want to use Gauss’ law to determine the electric field at an arbitrary point P inside the sphere. We construct a concentric sphere of radius r , where r < R ,

as our Gaussian surface, as represented in the diagram below. Note that the arbitrary point P lies on the Gaussian surface.

Using equation (36-19), we can write

E Asurface = Qinside

εo

, (1)

E = Qinside

εo Asurface

= 0ε o 4π r 2( ) = 0 . (2)

This is a very important result. The electric field interior to a conductor in equilibrium--charges are not moving in the conductor--is zero!

Using Gauss’ Law Inside a Conductor in Equilibrium

r

R

Q

P

36-14

Example Four:The Electric Field Close to the Surface of a Conductor in Equilibrium

Assume we have a positive charge Q in equilibrium and uniformly distributed over the surface of a conductor, as represented in the diagram below. We create a closed Gaussian surface in the shape of cylinder, with half of the cylinder embedded inside the conductor. We found in example three that the electric field inside the conductor is zero so there can be no electric flux through the “submerged” part of the cylinder. Above the surface of the conductor, the electric field will be directed perpendicular to the top surface of the conductor, and, therefore, will not “pop through” the lateral surface of the Gaussian cylinder. So, the only area associated with flux is the top of the cylinder. Using equation (36-19), we can write

E Atop =Qinside

εo

=σAtop

εo

. (1)

Therefore,

E =σεo

, (2)

where σ is the surface charge density of the conductor. So, very close to the surface of a conductor in equilibrium, regardless of its shape, the magnitude of the electric field is given by equation (2).

Using Gauss’ Law Close to the Surface of a Conductor in Equilibrium

P

E

Q

′ Q

36-15

Summary of Chapter 36

An area can be represented as a vector. It has the form

A = A A , (36-20)

where

A =A , (36-21)

and is simply the measure of the actual area. The unit vector ˆ A represents the direction of the area

vector. The unit vector A is perpendicular to the plane of the area and its sense is right-handed in terms of the sense that we move around the boundary of the circle. For example, we can represent the area of the circle represented in Figure 36-5 below as

A = π R2 A . (36-22)

(The arrowheads are used to signify the sense in which we go around the circle. It is this sense--the right-hand rule--that determines which of the two line segments that are perpendicular to the plane of the area we will use. Later, we will see the importance of this distinction.)

Figure 36-5Representing the Area of a Circle as a Vector

A

R

A

R

36-16

If the electric field is constant over an area, then the electric flux through that area is defined by

ΦE =

E •A = E( ) A( ) cos∠bet( ) . (36-23)

If the electric field over an area is not constant, then we must use integral calculus to solve for the electric flux, that is

ΦE =

E • d

A∫ . (36-24)

Spheres have point symmetry, while cylinders have line symmetry. Gauss’ law gives us a method to determine the magnitude of the electric field by exploiting such symmetries. To use Gauss’ law, one must first determine if there is any symmetry in the charge distribution. If so, then one designs a closed Gaussian surface made up of one or more surfaces over which the magnitude of the electric field is constant--or zero. The electric flux through the entire closed surface is given by

ΦE = EA = Qinside

ε o

, (36-25)

where A represents only the area of the Gaussian surface through which there is flux. We can now solve for the magnitude of the electric field getting

E = Qinside

εo A . (36-26)

We found, using Gauss’ law, the following results:Sphere:For a sphere of radius R throughout which an electric charge Q is uniformly distributed, then at any point a distance r from the center of the sphere, the magnitude of the electric field is given by:

E =kQr2 , for r > R , (36-27)

and

E =kQR3

⎡⎣⎢

⎤⎦⎥

r , for r < R . (36-28)

Line Charge:For a very long, straight conductor of length over which an electric charge Q is uniformly distributed, the magnitude of the electric field at any point a perpendicular

36-17

distance r⊥ from the line charge is given by

E =2kλr⊥

, (36-29)

where the linear charge density λ is given by

λ =

Q

. (36-30)

Electric Fields In and Near Charged Conductors in Equilibrium:A conductor carries an electric charge Q and is in equilibrium. The electric field inside the conductor will be zero. (Whenever there is an electric field inside a conductor, a current will be established--valence electrons will move.) Very close to theoutside surface of the conductor, the electric field will have a magnitude given by

E =σεo

, (36-31)

where σ is the surface charge density and given by

σ =Q

Asurface

. (36-32)

Chapter 37

THE WORK DONE IN MOVING ELECTRIC CHARGES

A REVIEW OF WORK AND ENERGY

In our study of classical mechanics, the work done by a force F in the displacement of a point

mass M from some initial point A to some final point B was defined by

WA→B =

F •

ro

r f∫ dr , (37-1)

where dr represents an infinitesimal part of the path taken in going from A to B. Also, recall

Newton’s second law states that for point-like objects subjected to N forces simultaneously,

Fi

i=1

N

∑ =Fnet = M a . (37-2)

If one now places Ma into equation (37-1) and solves the integral, one will get the net or total

work done by all N forces. It turns out that no matter how many forces act on a point-like thing, or what path is taken in going from A to B, one always gets the same answer

Wnet = Wii=1

N

∑ =12

Mv f2 −

12

Mvo2 . (37-3)

The total work done depends only on the mass of the object displaced and the squares of the starting speed and the ending speed. This relationship of mass to speed squared is so important that we give it a name. It is called the kinetic energy and it is calculated using

K =

12

M v •v( ) =

12

Mv2 . (37-4)

So, for a point mass, the net or total work is simply equal to the change in kinetic energy, that is

Wnet = Wii=1

N

∑ = K f − Ko = ΔK . (37-5)

We also found out that the work done by a force of constant magnitude and constant

37-2

orientation to the path is given

Wconstant =

F • Δr = F( ) Δr( ) cos∠bet( ) , (37-6)

where, recall, ∠ bet is the measure of the angle between the direction of the force and the direction of the displacement.

Implicit in the definition of work is the notion of a displacement. In actually calculating the work done by a force, we usually must pay close attention to the path taken in the displacement. It turns out, however, that there is a group of forces for which the amount of work done in displacing a point-like physical thing from an initial point A to a final point B is exactly the same regardless of the path taken. Such forces are said to be conservative forces.

In classical mechanics, we worked with two conservative forces. The first conservative force of interest was the gravitational force between two masses. Recall that the gravitational force exerted on M 2 by M1 is given by

F21

G = −GM 1M 2

r122 r12 . (37-7)

When we put the gravitational force into equation (37-1) and solve for the work done in moving mass M2 from an initial point A to a final point B, we get

WG = −GM1M 2

r12 ,o

⎡

⎣⎢

⎤

⎦⎥ − −

GM1M 2

r12 , f

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (37-8)

The form of the expression inside the brackets above is so important we give it a name. We call this the gravitational potential energy, and define it by

U G = −GM1M2

r12

. (37-9)

So, we now know that the work done by the gravitational force in displacing mass M2 from point

A to point B is always given by

WG = UoG −U f

G = −ΔUG . (37-10)

When close to the surface of the Earth, the gravitational force can be approximated by

FG = −Mg j , (37-11)

where we have assigned the y -axis as positive vertically upward. When we place this definition of the gravitational force into equation (37-1) and solve for the work done in displacing a mass M from an initial point A to a final point B we get

37-3

WG = Mgyo[ ] − Mgyf⎡⎣ ⎤⎦ . (37-12)

Again, the terms in the brackets are gravitational potential energies of the formU G = Mgh , (37-13)

where h is the vertical distance from some arbitrarily defined zero. So, in displacing masses closeto the surface of the Earth, the work done by the gravitational force is always given by

WG = UoG −U f

G = −ΔUG . (37-14)

The second conservative force was that of the elastic spring. For a mass M attached to an elastic spring that is displaced

x from the natural equilibrium position of the spring, the force exerted on the mass by the spring is given by

Fsp = − ksp

x , (37-15)

where ksp is the so-called elastic spring constant and gives us information about the relative

stiffness of the spring. When we place the spring force into equation (37-1) and solve for the work done in

displacing the attached mass M from an initial point A to a final point B we get

Wsp =12

kspxo2⎡

⎣⎢⎤⎦⎥−

12

ksp x f2⎡

⎣⎢⎤⎦⎥

. (37-16)

The expression in the brackets are elastic spring potential energies. The spring potential energy is defined by

U sp = 12

ksp x2 . (37-17)

So, the work done by an elastic spring is always given by

Wsp = Uosp −U f

sp = −ΔUsp . (37-18)

There is a pattern here! The work done by a conservative force is always given byWconservative = −ΔU , (37-19)

where U is the appropriately associated potential energy.Finally, recall that if the only forces that do work are conservative, then from equations (37-5)

and (37-19) we can write−ΔU = ΔK , (37-20)

andUE −UL = K L − KE . (37-21)

37-4

Therefore,UE + KE = UL + K L . (37-22)

This is, of course, an expression of the conservation of the total mechanical energy.

ELECTRIC POTENTIAL ENERGY

We are interested in the amount of work done by the electrical force in the displacement of some electric charge Q2 from an initial point A to a final point B. What to do? It is simple. Take the electric force and put it into equation (37-1) and solve the integral. In Figure 37-1 below, we have a pictorial representation of two point-like electric charges. Please pay close attention to the vectors representing the positions of the charges and the line connecting the charges. The electrical force exerted on charge Q2 by charge Q1 is given by

F21

E = Q2

E1 = Q2

kQ1

r122 r12

⎡

⎣⎢

⎤

⎦⎥ =

kQ1Q2

r122 r12 . (37-23)

(Note that the electrical force has the same form as the gravitational force. It also is conservative.)Solving equation (37-1) with the force of equation (37-23), we find

WE =kQ1Q2

r12 , o

⎡

⎣⎢

⎤

⎦⎥ −

kQ1Q2

r12 , f

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (37-24)

This is illustrated below in Figure 37-2.

Figure 37-1

x ˆ i ( ) y ˆ j ( )

z ˆ k ( ) Q1

Q2

r1

r2

r12 = r12 r12 = r2 −r1

ˆ r 12

37-5

The expressions in the brackets of equation (37-24) are electrical potential energies. The electrical potential energy between two point-like charges is given by

U E = kQ1Q2

r12

. (37-25)

The work done by the electrical force in the displacement of charge Q2 from point A to point B is always given by

WE = UoE −U f

E = −ΔUE . (37-26)

You might be wondering what it is that the electric potential energy tells us. We have assumed that electric charge Q1 is fixed at the position

r1 , as represented below in Figure 37-2.

If we also assume that both charges are positive, then we would expect the repulsive nature of the force to increase the distance between the two charges, i.e. r12 , o < r12 , f . As there is a component of

the force in the same direction as the displacement, the work done should be positive as UoG > U f

G .

We are going to see later that if we want to move a positive charge closer to another positive charge, then as this can not happen spontaneously, we would have to hire an outfit like FP&L to forcibly move the charge closer. The electric potential energy serves as a measure of how much energy FP&L would have to expend to make the move. (Maybe just a tad abstract?) .

Figure 37-2

x ˆ i ( ) y ˆ j ( )

z ˆ k ( )Q1

r1

r12 ,o

r2

Q2

r12

Final Position

Path Taken

Initial Position

Arbitrary PositionQ2

r12 , f

Q2

37-6

THE ELECTRIC POTENTIAL

In Figure 37-2 above, we have represented the displacement of charge Q2 from an initial

point A to a final point B while charge Q1 remains fixed. The work done by the electric force in this process can be written as

WE = Q2kQ1

r12 , o

−kQ1

r12 , f

⎡

⎣⎢⎢

⎤

⎦⎥⎥

, (37-27)

where all that I have done is factor out the charge Q2 , the charge being moved around. Careful inspection of what remains in the bracket should convince you that the two terms have the same form. The quantity represented by these terms is called the electric potential. For point-like charges, the electric potential is defined by

V = kQr

. (37-28)

We have already learned that electric charges give rise to electric fields. In essence, an electric field is a mapping of two quantities at every point in space. The two quantities are the magnitude of the electric field and the direction of the electric field. We can think of the electric potential in a similar manner. Electric charges not only give rise to the vector field we call the electric field, but at the same time, they also give rise to another field, a scalar field, we call the electric potential. So, to map the electric potential at every point in space, we only need to specify its value at each point; the electric potential is a scalar. The electric potential for a point charge is given by equation (37-28) where Q is the source charge, the charge giving rise to the electric potential. The distance from the charge to the point in space where we want to know the electric potential is signified by r . The MKS standard unit for the electric potential is the Volt.

We can use the definition of the electric potential to rewrite equation (37-27) as

WE = Q2 Vo −Vf⎡⎣ ⎤⎦ = Q2 −ΔV( ) = −ΔUE . (37-29)

So, in general, the electric potential and electric potential energy are related byΔU E = Q ΔV( ) . (37-30)

(You must excuse physicists for giving these quantities such similar sounding names. Please pay close attention to the problems you are trying to solve to see if you are to find the electric potential, V, or the electric potential energy, U E .

You may also be wondering why we need this quantity, and how does it differ from the electric potential energy. We can think of the electric potential as a shorthand assessment of the energy needed to move exactly one Coulomb of charge from infinity to a point in space where the

37-7

electric potential is V. For example, if the electric potential at some point A is one hundred Volts, then it would require one hundred Joules of energy to move one Coulomb of positive electric charge from infinity to point A. (Note that equation (37-28) is zero when r is infinitely large. So, for point-like electric charges, the electric potential is zero at infinity.)

THE ELECTRIC POTENTIAL AND THE ELECTRIC FIELD

Consider the situation illustrated in Figure 37-3 below. Electric charge Q2 is displaced from

an initial point A to a final point B. Everywhere along the path taken by charge Q2 there is an

external electric field produced by charge Q1. (This electric field is shown at an arbitrary point on the path marked by an X.)

The work done by this electric force in the displacement of charge Q2 from point A to point

B , is given by

WE = Q2kQ1

r122   r12

⎡

⎣⎢

⎤

⎦⎥ro

r f∫ • dr12 = Q2

E1ro

rf∫ • dr12 = −ΔU E = −Q2 ΔV( ) . (37-31)

Figure 37-3

x ˆ i ( ) y ˆ j ( )

z ˆ k ( )Q1

r1

r12 ,o

r2

Q2

r12

Final Position

Path Taken

Initial Position

r12 , f

Q2

E1

37-8

Next, if we take equation (37-31) by the charge that is moved around, Q2 , we have

WE

Q2

= Q2kQ1

r122   r12

⎡

⎣⎢

⎤

⎦⎥ro

rf∫ • dr12 = Q2

E1ro

rf∫ • dr12

= −ΔUE

Q2

= − ΔV( ) = Vo −Vf . (37-32)

This is how, using the calculus, the electric potential is defined. Equation (37-32) gives us

ΔV = Vf −Vo = −

E

ro

rf∫ • dr . (37-33)

If it happens that the electric field is uniform in the displacement, then equation (32-33) reduces to

ΔV = Vf −Vo = −

E • Δr , (37-34)

where Δr is the displacement from the initial position to the final position.

Equation (37-33) has some important consequences for our investigations. First, note that if the potential difference between two points is zero, that is, if ΔV = 0 , then at every point on the

path E • dr = 0 . This requires that

E • dr = E dr E • dr

∧= E dr cos∠bet = 0 . (37-35)

Figure 37-4

Q2

path taken

E

r

Q2

Q1

A

B

∠ bet

dr

dr∧

37-9

Remember what we are doing with these integrals. First, dr represents an infinitesimal part

of the path taken by Q2 as it moves from an initial point A to a final point B . E is the magnitude

of the electric field on the path , and ∠ bet is the angle between the direction of the electric field and

the direction of dr , as illustrated in Figure 37-4 above.

This quantity, E dr cos ∠bet , is calculated at each infinitesimal path part. The total is found by adding up all of the infinitesimal products. (Remember, this is what the integral sign means, “add em up partner.”) I hope it is obvious that E is not zero anywhere along the path. The same is true for dr. The only way for the sum to be zero is if each infinitesimal product is zero! That can happen only if cos ∠bet = 0 at each infinitesimal path part. The cosine is zero when the angle is ninety degrees. We conclude then that the only way the change in the electric potential between two points A and B can be zero is if the two points are at the same potential, and the path taken in our integral is everywhere perpendicular to the electric field.

Consider the following, a positive point-like charge q is fixed at the origin of a Cartesian coordinate system. Notice that at every point on the surface of a sphere of radius r centered on the charge, the electric potential is the same, as represented below in Figure 37-5. Our analysis of equation (37-35) tells us that one could move a charge all over the equipotential surface without having to do any work. Also, the electric field lines are everywhere perpendicular to the equipotential surface of the sphere.

Another important consequence of equation (37-33) is that if there is a potential difference between two points in space, then there must be an electric field between the points. The electric field will always point from the point at higher potential to the point at lower potential.

Figure 37-5

V = kqr

E

EquipotentialSurface

E

E

E

E

E

E

Eq

r

37-10

Example One:An Electric Dipole

Two point charges are related by Q1 = −Q2 = q . (Such a charge distribution is called an

electric dipole.) They are fixed at positions given by

r1 = a i , and

r2 = − a i , respectively, as

represented in the diagram below. We want to find an expression for the total electric potential at the following points:

a)

rP1= x i , where x > a .

b)

rP2= y j . where y > 0 .

For point P1 , we have

V x( ) = V1 x( ) + V2 x( )

=kq

x − a−

kqx + a

= kq1

x − a−

1x + a

⎡⎣⎢

⎤⎦⎥

= kqx + a( ) − x − a( )x − a( ) x + a( )

⎡

⎣⎢

⎤

⎦⎥

=

kq 2a( )x2 − a2 =

kqx2 − a2 =

kpx2 − a2 , (1)

where, p is the magnitude of the electric dipole moment,

p = q . (2)

x ˆ i ( )

y ˆ j ( )

aa x − ax

y y 2 + a2y 2 + a2

P1

P2

q−q

p

37-11

For point P2 , we have

V y( ) = V1 y( ) + V2 y( ) =kq

y2 + a2−

kqy2 + a2

= 0 . (3)

Example Two:A proton is fixed at a point in space. A second proton is initially a very great distance from

the first. We want to find out with what initial speed vo one must launch this second proton toward

the first so that the closest approach of the second proton to the fixed proton is rc , as represented in the diagram below.

The only force that would act on the second proton after it is launched is the electrical force. As this force is conservative, the total mechanical energy will be conserved in this process. So, we can write

Ko + UoE =

12

MPvo2 + 0 = 0 +

ke2

rc

= K L + ULE , (1)

vo = 2ke 2

rc Mp

. (2)

To overcome the so-called Coulomb barrier and give this second proton enough speed to actually get close enough to “feel” the strong nuclear force, then we would have

rc ≈ 2.0 × 10− 15 m . For such a value, we would have to launch with a speed of

vo =2 8.99 × 109  N ⋅m2 ⋅C−2( ) 1.602 × 10−19  C( )2

2.0 ×10−15 m( ) 1.673× 10−27  kg( ) = 1.17 × 107 ms

. (3)

Temperatures greater than about eight million Kelvin are required to get a significant number of protons moving this fast; fast enough to get close enough to undergo fusion. Such extreme temperatures are rare. However, such temperatures are not uncommon in the cores of stars.

e e

vo

vL = 0p1 p1

effectively ∞ rc

p1

37-12

Example Three:We want to determine how much energy would have to be expended to fix four identical

positive electric charges of magnitude q at the vertices of a square of side length s, as represented in the diagram below. Such a configuration would not occur spontaneously, so we must assume the intervention of another agency, like FP&L. (You may assume that the charges were initially infinitely far apart.) We write, using work-energy considerations,

WFP&L + WE = ΔK = 0 , (1)

and

WFP&L = −WE = − −ΔU E⎡⎣ ⎤⎦ = U fE −Uo

E . (2)

As the charges were initially far apart,Uo

E = 0 . (3)For the final configuration, we have

WFP&L = U fE =

kq1q2

r12

+kq1q3

r13

+kq1q4

r14

+kq2q3

r23

+kq2q4

r24

+kq3q4

r34

=kq2

s+

kq2

s 2+

kq2

s+

kq2

s+

kq2

s 2+

kq2

s

=kq2

s4 +

22

⎡⎣⎢

⎤⎦⎥

= 4 + 2⎡⎣ ⎤⎦kq2

s= 5.414214( ) kq2

s . (4)

q1

s

s

s

s

q4q3

q2

Final Configuration

s2

37-13

Example Four:The Electric Potential Due a Continuous Charge

Distribution on a Circle of Radius RSo far, we have only dealt with point-like electric charges. Assume that a positive electric

charge Q is uniformly distributed along a thin, circular line of radius R , as represented in the

diagram below. We want to find the electric potential at an arbitrary point P on an axis that passes through the center of the circle and is perpendicular to the plane of the circle.

Since the charge is “smeared out” over space, we must modify equation (37-28). First, we divide the circular charge distribution into an infinite number of infinitesimal charges dq . One such infinitesimal charge is represented below. It is assumed that this charge is so small that it can be treated like a point charge. It is also assumed that it will produce an infinitesimal electric potential at point P on the z-axis, given by

dV =k dq

r . (1)

The electric charge is uniformly “smeared” over a circular line of radius R . So the linear charge density is given by

λ = Q / 2πR( ) . (2)

The infinitesimal charge dq is “smeared” over infinitesimal circular arc of length

d = R dϕ , (3)

z k( )

x i( )y j( )

P

R

dq = λ dϕ dϕ

z

r = z2 + R2

37-14

where, recall, dϕ is measured in radians.So, we can write

dq = λ d =Q

2π R⎡

⎣⎢

⎤

⎦⎥ R dϕ[ ] =

Q2π

  dϕ . (4)

The infinitesimal electric potential is given by

dV = k Q2π z2 + R 2

dϕ . (5)

To find the total electric potential, we add up all of the contributions of all of the infinitesimal charges. We have

V = dV∫ = kQ2π z 2 + R2

dϕϕο =0

ϕ f =2π

∫ = kQ2π z2 + R 2

2π[ ] = kQz2 + R2

. (6)

(I trust it is reasonable to the reader that if we add up all of the infinitesimal dϕ ’s around the entire

circle that we will get 2π .

Example Five:Assume that we have a positive electric charge Q uniformly spread over the surface of a

metal sphere of radius R , as represented in the diagram below. We want to find the electric potential at point on the surface of the sphere, and at the center of the sphere.

First, we want to look at that region of space where r > R . Recall that spheres have point symmetry so that when we are outside the sphere, it acts as if all of the charge is point-like and located at the center of the sphere. So, outside of the sphere, the electric potential, relative to infinity where it is zero, is given by

V =kQr

, when r > R . (1)

This, in turn, implies that the electric potential at the surface of the sphere, where r = R , is

Vsurface =kQR

≡ Rσεo

. (2)

To find out what is going on inside of the sphere, we use equation (37-33). We have

Vcenter −Vsurface = −

E

R

0

∫ • dr . (3)

37-15

From Gauss’ law, we know that inside the sphere, the electric field is zero everywhere. Therefore, the right-hand side of equation (3) is zero and

Vcenter −Vsurface = −

E

R

0

∫ • dr = 0 , (4)

and

Vcenter = Vsurface =kQR

, when r ≤ R (5)

Everywhere interior to a spherical surface charge distribution, including the center, the electric potential is the same as that at the surface. This is the case for any conductor in equilibrium! (Note, σ is the surface charge density on the sphere.)

R

r

σ = Qtotal

Atotal

= Q4π R2

rQ

Pinside

Poutside

37-16

Example Six:A Parallel-Plate Capacitor

Assume we have two metal plates which carry equal and opposite charges, as represented in edge-view in the diagram below. Such a device is called a parallel-plate capacitor. Capacitors are very important devices used to store electrical potential energy.

We want to calculate the potential difference between the plates. For applications to capacitors in general, it is useful for our answer to be positive. So, we are going to choose an integration path that goes from the negative to the positive plate. We can write

E = −E   j , (1)

while

dr = dy  j . (2)

We substitute these values into our equation for the electric potential and we get

ΔV = V+ −V− = −

Eidr

ro

r f∫ = − −E   j⎡⎣ ⎤⎦i dy  j  ( )o

d

∫ = E dyo

d

∫ = Ed . (3)

The value of the potential difference between parallel plates is the product of the the magnitude of the electric field between the plates and the plate separation distance.

A Parallel-Plate Capacitor

E

E

E

E

Ed

x i( )

y j( )

37-17

Example Seven:A Cathode-Ray Oscilloscope

The cathode-ray oscilloscope is an important electrical device used to analyze circuits that change very rapidly. The tube has parallel-plate capacitors that are used to deflect electrons. In the diagram below, we have a schematic representation of a typical cathode-ray tube used in oscilloscopes.

Much of the air has been removed from the tube. The heater raises the temperature of the cathode enough to “boil” electrons from the surface. (Before this process was well understood, these were called cathode-rays.) The accelerating anode is kept at a high electric potential relative to the cathode. This gives rise to an electric field that runs from the accelerating anode to the cathode. This field accelerates the electrons to a very great speed. The horizontal speed of the electron is essentially unchanged from the anode until it strikes the fluorescent screen. The control grid determines the brightness of the spot on the screen by controlling the number of electrons that get to the anode. The focusing anode ensures that electrons, initially on slightly different paths, reach the same spot on the screen.

When the electrons pass through the first set of deflecting plates, an electric field between the

A Cathode-Ray Tube for an Oscilloscope

Heater

Cathode

Control Grid

Electron Gun

Focusing Anode

Accelerating Anode

Horizontal DeflectingPlates

Vertical DeflectingPlates

Metallic Coating

Fluorescent Screen

37-18

plates deflects the electrons left or right. The electric field between the second pair of plates deflectsthe electrons up or down. If, of course, there were no electric fields between the plates, the electrons would pass through undeflected.

In the accelerating phase of the electrons, the only force that does work is the conservative electric force and, therefore, the mechanical energy is conserved. So, we can write

12

Me− vC2 − eVC =

12

Me− vA2 − eVA , (1)

where Me− is the mass of the electron, vC and vA are the speeds of the electron at the cathode and

accelerating anode, respectively. The electric potential at the cathode and the accelerating anode, respectively, are signified by VC and VA . The speed of the electron at the accelerating anode is so

much larger that its speed at the cathode we can ignore its initial speed. So, the horizontal speed of the electron as it emerges from the accelerating anode is given by

vA =2e VA −VC( )

Me−

=2e ΔVa

Me−

, (2)

where ΔVa signifies the accelerating potential difference between the accelerating anode and the

cathode.It is not uncommon for the accelerating potentials to be thousands of volts. So, if

ΔVa = 2,000 V , the speed of the electron would be

vA =2 1.602 × 10−19  C( )  2,000 V( )

9.109 ×10−31  kg( ) = 2.65 ×107  m / s , (3)

This speed is very large and is approaching a tenth the speed of light. At a tenth the speed of light, speeds are said to be relativistic and we need to modify our equations using Einstein’s special theory of relativity. The classical equations do not do a good job of correctly describing what happens to physical things that move at speeds close to the speed of light.

Now, I want to investigate how the deflection plates work. First, we will assume that there is no electric field between the horizontal deflection plates and an electric field of magnitude E between the vertical deflection plates. In the diagram below, we have a representation of this state of affairs.

Application of Newton’s second law gives us

FE = q

E = −e −E  j⎡⎣ ⎤⎦ = eE   j = Me−

a , (4)

37-19

and the acceleration of the electron between the plates is given by

a =eE  M −e

  j . (5)

The speed of the electron in the x-direction is constant, so we can use that to calculate the time it will take the electron to move through the plates. We find

′t =

vo

. (6)

We are now in a position to calculate the y-coordinate at the instant it emerges from the plates. We have

A Measure of the Deflection of Electrons in an Electric Field

x i( )

y j( )

d

vo− e

E

E

E

y

′yθ

θ ′v

•A

D

37-20

′y =12

a ′t 2 =12

eEM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

vo

⎡

⎣⎢

⎤

⎦⎥

2

=eE2

2Me−

vo2 . (7)

The electron emerges from the parallel plates with a speed ′v about which we can write

′v = vo  i + ′vy , (8)

where

′vy = ′vy   j = a ′t   j =eEM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

vo

⎡

⎣⎢

⎤

⎦⎥  j =

eEM

e−vo

  j . (9)

Also, the geometry of the path, as the electron emerges from between the plates, requires

tanθ =′vy

vo

=y − ′y

D . (10)

So, we have

y = D′vy

vo

⎡

⎣⎢

⎤

⎦⎥ + ′y = D

eE / Me−

vo

vo

⎡

⎣⎢

⎤

⎦⎥ +

eE2

2Me−

vo2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= DeE

Me−

vo2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+eE2

2Me−

vo2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= DeE

Me−

vo2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+eE2

2Me−

vo2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=eE

Me−

vo2 D +

2⎡⎣⎢

⎤⎦⎥

. (11)

Using equation (2), we can also write equation (11) as

y =E

2 ΔVa

D +

2⎡⎣⎢

⎤⎦⎥

=dd

E2 ΔVa

⎡

⎣⎢

⎤

⎦⎥ D +

2⎡⎣⎢

⎤⎦⎥

=

dEd

2 ΔVa

⎡

⎣⎢

⎤

⎦⎥ D +

2⎡⎣⎢

⎤⎦⎥

=

2dD +

2⎡⎣⎢

⎤⎦⎥ΔVcap

ΔVa

, (12)

where ΔVcap is the electric potential difference across the parallel plates. Later, we will show that

one can also use external magnetic fields to deflect moving charges.

37-21

Example Eight:The Millikan Oil Drop Experiment

We are now in a position to describe one of the most important experiments in the history of physics. In an absolutely remarkable series of investigations at the University of Chicago from 1909 to 1913, Robert A. Millikan was able to prove that the electron has a discrete charge and he was also able to measure the magnitude of that charge. In the diagram below, we have represented a schematic of the apparatus Millikan used to perform his experiment. We have also included force diagrams needed to understand his analysis.

Oil drops are introduced by means of an atomizer through a very small opening in the top plate of the Millikan apparatus. In collisions with air molecules, some of the drops become electrically charged. When charged, they interact with the electric field. If the plates are set up as represented below, positively charged drops will accelerate rapidly downward. For a negatively charged oil drop, however, it is possible to adjust the electric field strength until the drop is in equilibrium and not moving. (The person doing the experiment is looking through a telescope into the chamber between circular parallel plates that have a hole in the top to allow in the oil drops. Light coming into the chamber from another port is reflected off of the drops and they appear as very tiny points of light. While watching the proceedings, the experimenter has his hand on the voltage adjust knob that allows him to control the strength of the electric field, and, thereby, the potential difference between the two plates.)

In equilibrium, there are three forces acting on the oil drop: 1) an electric force; 2) a buoyancy force; 3) and the gravitational force. If q is the magnitude of the negative charge on the oil drop, then the magnitude of the electric force is given by

FE = qE . (1)

The Millikan Apparatus

••

••Battery

E

E

E

E

FG

FE

F B

FG

F fF B

vT

In Equilibrium

Constant Motion

ΔV d

37-22

The magnitude of the gravitational force is given by

FG = mog = ρoVvol ,og . (2)

where mo is the mass of the oil drop, ρo is average mass density of the oil used, while Vvol ,o is the

volume of the oil drop, and, of course, g is the magnitude of the Earth’s gravitational field at the location of the experiment.

The buoyancy force is, of course equal to the weight of the fluid that has been displaced by the oil drop. In this case, the displaced fluid is air. So, we can write

F B = ma g = ρaVvol ,a g = ρaVvol ,og , (3)

where ma is the mass of the air displaced, ρa is average mass density of the air, while Vvol ,a is the

volume of the displaced air. The volume of the displaced air is the same as that of the oil drop.Equilibrium requires

qE + ρaVol ,og = ρoVol ,og . (4)

Solving for the charge magnitude we find

q =ρo − ρa( )Vol,og

E=

ρo − ρa( ) 4 / 3( )πRo3⎡⎣ ⎤⎦ g

ΔVcap / d( ) =4πRo

3gd ρo − ρa( )3 ΔVcap

, (5)

where we have assumed the oil drop is spherical with a volume given by

Vol ,o = 4 / 3( )πRo3 . (6)

In experiments, equations do not do us much good if we can not measure the quantities given. So, we review the physical quantities expressed in equation (5). The magnitude of the Earth’s gravitational field g is well known as are the average mass densities of the oil ρo and the air ρa .

The plated separation distance d and the electric potential difference between the plates are easily

measured. However, Ro , the radius of the oil drop, is extremely difficult to measure. So, what to

do.If the electric field is turned off, the electron will no longer be in equilibrium and it will begin

to fall. A new force comes into play. It is a frictional force exerted on the electron which is proportional to the speed of the electron. Eventually, the frictional force is large enough to bring the electron back in equilibrium. The speed of the electron when equilibrium is reached is called the terminal speed and is signified by vT .

In 1851, George Stokes derived an equation to quantify the frictional force exerted on small spherical objects as they move through viscous fluids. This equation is called Stokes’ law and is

37-23

given by

F f = 6πηRvT , (7)

where R is the radius of the sphere and η is a constant that represents the viscosity of the fluid. At terminal speed, equilibrium requires

6πηRovT + ρaVol,og = ρoVol,o g . (8)

Solving for the radius of the sphere we find

6πηRovT = ρo − ρa( )Vol ,og = ρo − ρa( ) 4 / 3( )πRo3⎡⎣ ⎤⎦ g , (9)

and

Ro =9ηvT

2 ρo − ρa( ) g⎛

⎝⎜⎞

⎠⎟

1/2

. (10)

Substitution of the results of equation (10) into equation (5) gives us

q =

4π 9ηvT

2 ρo − ρa( ) g⎛

⎝⎜⎞

⎠⎟

1 /2⎡

⎣⎢⎢

⎤

⎦⎥⎥

3

gd ρo − ρa( )

3 ΔVcap

=4π

9ηvT

2 ρo − ρa( ) g⎛

⎝⎜⎞

⎠⎟

3/2

gd ρo − ρa( )

3 ΔVcap

= 18πd

ΔVcap

⎛

⎝⎜⎞

⎠⎟ηvT( )3

2 ρo − ρa( ) g . (11)

The terminal speed is found by measuring the time it takes the electron to move between two parallel lines on the eyepiece of the telescope through which the experimenter observes the process. The distance between the lines is measured with a traveling microscope. A statistical analysis is used to get from the charge on the oil drop to the charge on an electron. Where

q = ne , (12)

and n is an integer, such that n = 1, 2, 3,  . Millikan had to make a slight correction to Stokes’ law as the air is not quite a continuous fluid. For his experimental work on the charge of the electron and his experimental work on the photoelectric effect, Millikan won a Nobel Prize.

37-24

Summary of Chapter 37

Recall that work done by a force F in the displacement of a point mass M from some point

A to another point B is defined by

WA→B =

F • dr

rA

rB∫ . (37-36)

The net or total work done by N forces acting on a true point mass is:

Wnet = Wii =1

N

∑ = ΔK , (37-37)

where K is the kinetic energy and given by

K =

12

M v •v( ) =

12

Mv2 . (37-38)

The work done by a force of constant magnitude and constant orientation to the path is:

Wconstant =

F • Δr = F( ) Δr( ) cos∠between( ) . (37-39)

The work done by a conservative force is:Wconservative = Uo −U f = −ΔU , (37-40)

where U represents the associated potential energy.The gravitational potential energy is given by:

U G =− GM1M 2

r12

 , in general

Mgh  , near Earth's surface

⎧⎨⎪

⎩⎪ . (37-41)

The potential energy associated with an elastic spring, within its elastic limit, stretched or compressed a distance x from its natural equilibrium position is

U sp = 12

ksp x 2 . (37-42)

The work done by the electric force isWE = Uo

E −U fE = −ΔUE . (37-43)

For two point-like electric charges Q1 and Q2 separated by a distance r12 the electric potential energy is given by

U E = kQ1Q2

r12

. (37-44)

The electric potential is defined by

ΔV = Vf −Vo = −

E • dr

rA

rB∫ . (37-45)

37-25

If the electric field between the starting point and the ending point is uniform, equation (37-45) reduces to

ΔV = −

E • Δr = − E( ) Δr( ) cos∠bet( ) . (37-46)

A single point-like electric charge Q produces an electric potential at any point a distance r from Q is given by

V = kQr

. (37-47)

This equation implies that at infinity, the electric potential is zero. The zero value is, of course, the reference value.

The electric potential is related to the electric potential energy by′ q ΔV = ΔU E , (37-48)

where ′ q is the charge that is being moved to some final point B from some initial point A through a change in electric potential ΔV .

The total mechanical energy of a point-like physical system is always conserved when the only forces doing work on the system are conservative. In such situations, we have

Ko + Uo∑ = KL + UL∑ . (37-49)

PART TWO

Electrodynamics

Chapter 38

ELECTRIC CURRENT AND CONDUCTORS

Up to this point, we have been dealing with electric charges that are at rest. We found that a static electric charge gives rise to a vector field called the electric field. A static electric charge also gives rise to a scalar field called the electric potential. Static electric charges interact with each other by means of the electric force, what we call the Coulomb force. Also, the displacement of electric charges involves the expenditure of electrical potential energy.

We are now ready to begin our discussion of electrodynamics. When an electrically charged object moves, two things of physical significance emerge: a) First, a moving electric charge produces a second vector field called the magnetic

field. b) Second, a moving electric charge produces an electric current.

To understand the magnetic force, we begin by first looking at electric charges in motion.

ELECTRIC CURRENT (DEFINED)

An electric current exists wherever there is a net flow of electric charge through an area.

To quantify an electric current, we first choose some kind of planar area such as a circle. Second, we assign a positive direction. Third, we simply count charges that pass through this area and measure the amount of time taken for this process to unfold. For positive charges moving in the assigned direction you add the charge. For positive charges moving through the area opposite the assigned direction you subtract the charge. For negative charges moving opposite the assigned direction you add the charge magnitude, while for negative charges going in the assigned direction, you subtract the charge magnitude. Well, that is simple enough; we count charges passing through a planar area. The net charge passing through is Δq . The average current then is defined as

Iave = ΔqΔt

. (38-1)

Of course, the instantaneous current is defined by the limiting process. We have

I = limitΔt→0

ΔqΔt

= dqdt

. (38-2)

38-2

Example One:In the diagram below, we have represented a small circle through which, in a time interval of

one second, six protons move through the circle to the right while two protons move through the circle to the left. In the same time interval, one electron moves through the circle to the right while four electrons move through to the left. The average current then is given by

Iave =ΔqΔt

=6e( ) − e( ) − 2e( ) + 4e( )

1s( ) =7e1s( )

=7 1.602 × 10−19  C( )

1s( ) = 1.12 ×10−18 A , (1)

where A represents the MKS standard unit for current called the Ampere in honor of the French physicist André Marie Ampère . I Amp is equivalent to a transfer across a reference area of one Coulomb of electric charge per second. Remember that the direction of the current is determined by the direction positive charges would move if subjected to a reference electric field.

A Hypothetical Current

Assigned Direction

38-3

Example Two:Assume that a single electron orbits a fixed, solitary proton with a constant speed v along a

circular path of radius R , as represented in the diagram below. Use this information to find:

a) The speed v of the electron if R = 5.29 × 10−11m , the so-called Bohr radius.b) The period of this orbit.c) The magnitude of the average current represented by this electron motion.

The force acting on the electron is electrical. As the electron is moving on a circular path, then we also know that the magnitude of this electrical force must equal the magnitude of the net radial force. So, we have

ke2

R 2 =M

e−v2

R , (1)

and, therefore,

v =ke2

RMe−

=8.99 × 109  N ⋅m2 ⋅C−2( ) 1.602 × 10−19  C( )2

5.29 × 10−11 m( ) 9.109 × 10−31  kg( )

= 2.19 × 106 m / s . (2)

For circular motion, we know that

v = Rω = R 2π f( ) = 2πR / τ . (3)

An Electron Orbiting a Proton (Monatomic Hydrogen)

v

FE

R reference area

38-4

Therefore,

τ =2πR

v=

2π 5.29 × 10−11m( )2.19 × 106 m / s( ) = 1.52 × 10−16 s . (4)

For the average current, we have

Iave =eτ

=1.602 × 10−19  C

1.52 ×10−16 s= 1.05 × 10− 3 Amp ≈ 1 mA . (5)

CURRENTS IN METALLIC CONDUCTORS

An electron moving in free space produces a current. As we have seen in Example Two, as an electron “orbits” the nucleus of an atom, it produces a small current. Further, since electrons and protons can be modeled as small spheres with electric charge uniformly distributed over the spheres, then when electrons or protons spin, they also produce a current. Currents can arise under a myriad of circumstances. However established, all currents have in common a net flow of electric charge through a reference area.

In our everyday experience, the most common form of electric current is that found in metallic wires. As you probably know, metals are good conductors of electric charge. As such, they make a good material out of which to fabricate conductors. One useful function of wire conductors is that they provide a well defined path for the net flow of electric charge, a well defined path for electric current.

Figure 38-1Model of a Metallic Conductor

IonIonIonIon

Ion Ion Ion Ion

38-5

In metals, there are electrons that are not bound to any specific atom even though they are bound to the metal itself. These electrons are called conduction or free electrons. The nucleus of a metal atom is bound tightly with most of its electrons. However, usually one electron per atom isessentially free to move around in the metal. (This state of affairs is represented above in Figure 38-1.) The positively charged metallic ions form a crystal lattice that remains fixed in the conductor. The free electrons can be treated as an electron gas. (In copper, there are about

8 × 1022 free electrons per cubic centimeter.) This entire collection of free electrons moves as a group whenever an electric field is established in the wire. In the absence of a net electric field, the

electrons move randomly in all directions at speeds on the order of 106  m / s .

Assume we have a copper wire of length in electrostatic equilibrium. The wire itself is electrically neutral as there are as many electrons as there are protons. Also, as there is no net motion of the electrons in the metal, we have no electrical current. How can we get these free electrons to move? We need to establish an electric field in the conductor.

To that end, assume that we place simultaneously a positive charge of magnitude Q and a negative charge of the same magnitude at opposite ends of our conductor, as represented below in Figure 38-2. These source charges will produce an electric field in the conductor and the free electrons will move under the influence of the electric field. The problem with this setup, however, is that we have connected two reservoirs of opposite charge with a conductor. So in a very short time interval, the charges will neutralize and equilibrium will be reestablished. Unfortunately, this current will be transient unless we can keep supplying charge to the reservoirs. Enter the battery.

Figure 38-2Using Source Charges to Establish an Electric Field

+Q −Q

E

insulatorconductor

38-6

The battery is a device that, at least for a useful amount of time, will supply the “reservoirs” with the needed supply of charges to maintain the electric field and keep the free electrons moving, thereby producing a steady state current. (This process can not go on forever, there is a limit to the useful life of a battery. We will have more to say about batteries in the next chapter.) We represent this situation with a schematic diagram shown below in Figure 38-3.

Note please that in the wire itself, the electrons move toward the positive terminal of the battery. However, by convention, the direction of the current I is always opposite that of the electrons. In the academic scientific community, the direction of any current as the direction in which positive charge carriers would move, in other words, in the direction of the electric field that drives the process. (In the battery itself, the electrons must move from the positive to the negative terminal of the battery. Opposite of the direction of the electric field that moved them in the wire. Understanding how and why this must happen is crucial to understanding the function of a battery. We will have more to say about this later.)

So, whenever there is a net flow of charges in a conductor, it implies that there is an electric field present in the conductor with which the charges interact and “feel” the electric force that drives their motion. The presence of an electric field also indicates that there is a potential difference established along the conductor. In Figure 38-4 below, we represent the local electric field at five locations in the wire. Remember, electrons are negatively charged and so “feel” a force in a direction opposite the local electric field. The question naturally arises as to how the electric field takes on this pattern.

Figure 38-3A Schematic of a Battery and Conducting Wire

WireBattery

Ivd

e−

38-7

E

Wire

Battery

E

E

E

E

σ1σ 2

Enet

E2

E1

Figure 38 - 5A Nonuniform Ring Charge Density and the Net Electric Field Produced

Figure 38 - 4The Electric Field in a Conductor Connected to a Battery

38-8

Recall that in chapter thirty-five, Example Seven, we found that electric charge distributed in the shape of a circular ring produces an electric field along the axis of symmetry given by

E =

2π kλaz

z 2 + a2( )3/2 k , (38-3)

where λ is the charge per unit length, and where a is the radius of the ring, and z is the distance along the axis of symmetry from the center of the ring. In Figure 38-5 above, I have represented how a ring-like charge density that varies along the conductor might give rise to the local electric fields that point along the conductor. It is a small nonuniform surface charge density that is created along the wire when it is connected to the battery that produces the electric field that drives the motion of the electrons.

RESISTANCE AND RESISTIVITY

We an electric field is established in a metal conductor, the free electrons accelerate under the

influence of an electric force. However, each electron suffers on the order of 1014 collisions with the ions of the crystal lattice every second. As a result, the electron gas moves with a drift speed on the order of 10 mm / s . Since the net speed is so small compared to the random speed, the time interval between collisions, τ , is independent of the electric field and depends only on the type of metal.

In order to determine the drift speed of an electron, we analyze the changes in its linear momentum. In a collision, we would expect the electron to lose virtually all of its linear momentum. So, the loss of linear momentum can be written as

ΔpΔt

⎛⎝⎜

⎞⎠⎟ loss

=M

e−vd

τ . (38-4)

On the other hand, the electron gains linear momentum from the electrical force. So, we can writeΔpΔt

⎛⎝⎜

⎞⎠⎟ loss

= −eE . (38-5)

If the current in the conductor is steady-state, then the loss and the gain must be the same, and we have

−eE =M

e−vd

τ , (38-6)

and the drift velocity is

38-9

vd =−eEτM

e− . (38-7)

(Physically, the negative sign is indicating that the direction of the motion of the electron gas is opposite the direction of the electric field.)

The number of free electrons per unit volume depends on the type of material out of which the conductor is fabricated. If we assume the cross-sectional area of the wire is uniform with a value A , and that the conductor has a length , then the total charge represented by the free electrons is given by Δq = −enA . (38-8)The total time needed for all of the electrons to pass through the conductor is given by

Δt =

vd

. (38-9)

Therefore, the magnitude of the current is given by

I =ΔqΔt

=enA / vd

= enA vd = enAeEτM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=e2nτM

e− AE . (38-10)

(We have ignored the negative sign on the electric charge because of the convention that the direction of the current is the same as the electric field, the direction in which positive charges would move.)

The electric potential difference that must be maintained across the conductor to establish an electric field is given by

ΔV = E . (38-11)Substitution of equation (38-11) into equation (38-10) yields

I =e2nτ AM

e−  E( ) =

e2nτ AM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥ ΔV . (38-12)

Next, we use the term in the bracket to define a new quantity we will signify by R . We have

R =e2nτ AM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1

=M

e−

e2nτ A⎡

⎣⎢

⎤

⎦⎥ , (38-13)

and equation (38-12) becomes

38-10

I =ΔVR

. (38-14)

Equation (38-14) is called Ohm’s Law, and equation (38-13) gives us a theoretical definition of the resistance in terms of the parameters of the electron gas. The SI unit for resistance is the Ohm which we signify with the upper case Greek letter Ω (Omega). Unfortunately, the definition given in (38-13) is not very helpful as measuring τ directly is not practicable.

Currents in a metal conductor are the result of the motion of the free electrons in the metal when an electric field is established in the conductor. If we take the ratio of the current to the area through which it passes, we have what is called the current density. In Figure 38-6 below, we represent a conductor in cross section through which a current I passes. An infinitesimal area dA

has an infinitesimal current dI passing through it. We define a new quantity called the current density by

J =dIdA

. (38-15)

In terms of the current density, the current can be defined by

I =

J  i d

A∫ . (38-16)

So, it turns out that the current is also a kind of flux. With the current, we have charge carriers “popping” through an area.

The current density J in a conductor depends on the electric field E and the kind of material out of which the conductor is fabricated. For metals, the ratio of the electric field to the current density is virtually constant. We define still another new quantity called the resistivity in terms of this ratio, and signify it with the lower case Greek letter ρ (Rho). So, we have

ρ =EJ

. (38-17)

Figure 38-6

I

A

dAdI

38-11

Resistivity values are found experimentally. In Table 38-1 below, we give the measured resistivity values for some common materials at room temperature. On the following, page in Figure 38-7, we sketch some graphs to give one a sense of how these values depend on temperature.

Table 38-1Resistivities at Room Temperature

Conductors Semiconductors Insulators

Material ρ Ωm( ) Material ρ Ωm( ) Material ρ Ωm( )

Silver

Copper

Gold

Aluminum

Tungsten

Steel

Lead

Mercury

Manganin

Constantan

Nichrome

1.5 × 10−8

1.7 × 10−8

2.4 × 10−8

2.6 ×10−8

5.5 ×10−8

20 × 10−8

22 × 10−8

95 × 10−8

44 ×10−8

49× 10−8

100 × 10−8

Carbon 3.5 × 10−5

Germanium

Silicon

0.6

2300

Amber

Glass

Lucite

Mica

Fused Quartz

Sulfur

Teflon

Wood

5 × 1014

1010 −1014

> 1013

1011 − 1015

75× 1016

1015

> 1013

108 − 1011

38-12

Figure 38-7Resistivities as a Function of Temperature

Metalsρ

T

Superconductorsρ

T

Semiconductorsρ

T

Tc

38-13

At a specific point in a conductor, the electric field and the current density are related byE = ρJ . (38-18)

It is usually not easy to measure E and J directly. It would be useful if we could express this relationship in terms of quantities much easier to measure. Assume we have a conductor of length

and uniform cross-sectional area A , as represented in Figure 38-8 below. If we also assume a constant current density over the cross section and a uniform electric field along the conductor, then the current can be written as

I =

JidA∫ = JA . (38-19)

Also, as the electric field is uniform along the conductor, then

ΔV = E . (38-20)

Solving equation (38-20) for E and equation (38-19) for J and using equation (38-18), we have

ΔV

=ρIA

. (38-21)

Rearranging terms we have

I =ΔV

ρ / A[ ] =ΔVR

. (38-22)

If we compare this with Ohm’s Law given in equation (38-14), we can see that the term in the bracket is simply the resistance we defined in equation (38-13). The resistance and the resistivity are related by

Figure 38-8

A I

E

ΔV

38-14

R = ρ / A( ) . (38-23)

The resistivity of a material depends on its temperature. Therefore, the resistance also depends on the temperature. For a narrow temperature range that is not too far from room temperature, we can approximate the temperature dependence of the resistance by

RT = RO 1 + α T −TO( )⎡⎣ ⎤⎦ , (38-24)

where RO is the reference resistance at some reference temperature TO , and α is the so-called

temperature coefficient of resistivity. Below in Table 38-2, we have listed some values of α for some common materials. (All of the qualifiers that attend our description of equation (38-30) should make red lights flash in your brain. What it really means is that the physics is being glossed because it is just too complex! In fact, classical physics can not give an adequate description of these matters. Only quantum mechanics can provide a unified, coherent explication!)

Table 38-2 Temperature Coefficients of Resistivity Near Room Temperatures

Material

Aluminum

Brass

Carbon

Constantan (Cu60, Ni40)

Copper

Iron

Lead

Manganin (Cu84, Mn12 , Ni4)

Mercury

Nichrome

Silver

Tungsten

α   1 / C°( )0.0039

0.0020

−  0.0005

0.000002

0.00393

0.0050

0.0043

0.000000

0.00088

0.0004

0.0038

0.0045

38-15

Example Three:Number 10 copper wire is commonly used in homes for electrical wiring. The radius of this

wire is 0.129  cm . We want to find the resistance of a piece of this wire that is 35.25 m long. We also want to find the electric potential difference across this wire if it carries a steady-state current of 12.75 A .

To find the resistance, we write

R =ρA

=ρπ r2 =

1.7 ×10−8Ωm( ) 35.25 m( )π 0.129 0.01 m( )( )2 = 0.115  Ω . (1)

The electric potential difference needed to sustain this current is

ΔV = IR = 12.75  A( ) 0.115 Ω( ) = 1.47 V . (2)

38-16

Summary of Chapter 38

Any time we have a net flow of electric charge through a reference area, we have an electric current. The charge could be moving freely or in a conductor. To quantify the electric current, we must count the amount of charge that pass through a reference cross-sectional area in a specified direction in a specific amount of time. The average current is defined by

Iave = ΔqΔt

, (38-25)

where Δq represents the net amount of electric charge that passes through the cross-sectional area in the time interval Δt . We can define the instantaneous current in terms of our limiting process.

I = limΔt→0

ΔqΔt

= dqdt

. (38-26)

For charge carriers to move through conductors, an electric field must be established in the conductor. This process is very complicated in its analysis. Suffice it to say that when a conductor is connected to a source of electrical potential energy, such as a battery, a very complicated mosaic of varying surface charge densities is established on the conductor, and this gives rise to the net electric field in the wire that drives the charge carriers through the conductor.

The actual charges that are moving in a metallic conductor are the free electrons of the conducting material itself. These free electrons can be modeled as an electron gas. As the electrons move through the conductor, they scatter off of the positive ions which make up a crystal lattice--we call this resistance. In the collisions of the free electrons and the positive ions, the kinetic energy of the electrons is transferred to the lattice causing it vibrate more. These vibrations increase the temperature of the conductor. In terms of the parameters of the metal from which the conductor is fabricated, this resistance is defined by

R =e2nτ AM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−1

=M

e−

e2nτ A⎡

⎣⎢

⎤

⎦⎥ , (38-27)

where Me− is the mass of an electron, is the length of the conductor, e is the magnitude of the

charge of the electron, n is the number of electrons per unit volume for the specific metal of the conductor, τ the average time interval between collisions, and A is the cross-sectional area of the conductor. A more useful definition of the resistance is Ohm’s law. Here, the resistance is defined as

R =ΔVI

. (38-28)

38-17

The ratio of the current to the cross-sectional area is called the current density. We have

J =dIdA

. (38-29)

This notion enables us to define current as a kind of flux; namely, the flux of charge carriers through an area. We have

I =

J∫ idA . (38-30)

At a specific point in a conductor, the resistivity ρ is defined as the ratio of the electric field to the current density. We have

ρ =EJ

. (38-31)

For a uniform cross-sectional area and uniform electric field, the resistance and resistivity are related by

R = ρ

A . (38-32)

Within the proper limits, the temperature dependence of the resistance of a conductor is given by

RT = RO 1 + α T −TO( )⎡⎣ ⎤⎦ , (38-33)

where RO is the reference resistance at some reference temperature TO , and α is a number that

depends on the kind of material and is called the temperature coefficient of resistivity.(It is important to remember that this analysis has been done in terms of a classical model of

the motion of the charge carriers and there interactions--the so-called Drude model of conduction. A more precise view of the conduction and resistance of charge carriers requires a quantum mechanical analysis.)

Chapter 39

ELECTRICAL CIRCUIT ELEMENTS

THE BATTERY AND ELECTROMOTIVE FORCE

In the last chapter, we briefly discussed some of the properties of metallic conductors. We found out that the charge carriers of a metal conductor are the free electrons of the metal. In order to get these carriers to move, we need to establish an electric field in the conductor. One way to do this is with a battery.

Assume we have a copper wire of length in electrostatic equilibrium. If we now connect our copper wire to a battery, an electric potential difference is established across the wire and an electric field is established in the wire. The free electrons in the wire interact with this electric field and “feel” an electric force that drives some of the charges through the wire and produces a steady-state current in the wire for a useful period of time. (How long, of course, depends on the battery.) In Figure 39-1 below, we have a representation of our wire connected to a battery.

Figure 39-1Electric Field in a Conductor and a Battery

Battery

wire

vd

vd

E

E

E

Enc

P ′P

E

39-2

Assume a free positive charge is initially at point P , the positive terminal of the battery. The

carrier moves around the wire at an average speed vd that is very small. Its many collisions with

the positive lattice charges prevents it from gaining any appreciable kinetic energy by the time it reaches point ′P . However, the electric potential energy of the charge decreases as it moves

around the wire. In order for the charge to get back to point P , the battery is going to have to supply the charge with the necessary electric potential energy. Most batteries transform chemical energy into electrical potential energy. To measure the “strength” of the source of electrical potential energy, we introduce the notion of electromotive force, or simply EMF. We will signify the EMF with a funky script “E”: E . The EMF of our battery is defined as the amount of electric potential energy delivered by the battery to every Coulomb of charge that passes through the battery from the low-potential terminal to the high-potential terminal. The SI unit is the Volt. (It is an electric potential energy per unit charge; a Joule/Coulomb; a Volt.)

The EMF is very poorly named as it is not a force at all. It was so named before physicists had a clear understanding of the difference between force and energy. (We have seen this kind of thing before. Potential energy is a very bad description for what is obviously positional energy. We are not going to be able to change that name either!)

If our battery were an ideal battery, had no internal resistance to charge flow, then for a steady-state situation, the energy supplied by the battery is equal to the energy lost in the wire. For an ideal battery, we can write

E + ΔV = 0 . (39-1)In the last chapter, we argued that a nonuniform surface charge distribution on the conductor

is what produces the electric field in the conductor that drives charge carriers. This electric field is conservative and like those we have discussed earlier. However, note that in the battery itself, this electric field is in the wrong direction to ferry charge carriers across the battery. (This state of affairs is represented below in Figure 39-2.) So, inside the battery, at least, there must be another kind of electric field. We will call this electric field non-conservative, and signify it with an Enc .

Example One:

A new flashlight battery has an EMF of 1.50 Volts . The battery can deliver a current of

1.25 Amps for about 1.125 hrs before running down. We want to determine how much energy is provided by the battery. As the charge carriers are electrons, can write

ΔUE = Δq( ) ΔV( ) = I Δt( ) ΔV( ) = 1.25 A( ) 1.125 3600 s( )⎡⎣ ⎤⎦ 1.50 V( )

= 7,594 Joules . (1)

39-3

The conservative electric field E established in the wire can move positive charge carriers in

the wire from the positive terminal of the battery through the wire to the negative terminal of the battery. However, to complete the circuit, the positive charge carriers must go through the battery from the negative terminal to the positive terminal. For this part of the trip, the conservative electric field can not help. We conclude that the chemical reactions that go on in the battery must establish

a non-conservative electric field

Enc in the battery to move the positive charge carriers across the

battery to complete the circuit. The battery “pumps” the positive charge carriers. The battery provides them with the energy to make the trip and complete the circuit.

(You may be wondering how the electrons know the difference between a conservative and a non-conservative electric field? They don’t. This suggests that the distinction is simply a construct of human thought to help us make sense of this matter. Einstein reasoned that if the charges themselves do not know the difference, there is not any and it means we have an incomplete understanding of the matter. One of the great triumphs of his relativity theory was it allowed him to construct a more complete theory of the electromagnetic interaction. However, as you may have guessed, this more unified theory is much more complicated and requires a level of mathematical sophistication way beyond the level of this discourse.)

For our purposes, suffice it to say that if we are going to have current moving in a complete circuit, then somewhere in the circuit there must be a device that provides energy to the charge carriers to allow them to complete the trip! A battery is only one such possible source of energy. A generator is another device that uses mechanical energy, rather than chemical energy, to accomplish the task.

Returning to the battery, a more realistic view of the battery will include the fact that charges also encounter resistance when moving through the battery. If we signify the resistance of the

Figure 39-2A Non-Conservative Electric Field in the Battery

Battery

E

Enc

E

E

E

E

E

39-4

battery with a lower case r , then we can correct equation (39-1) withEgained = Elost = Elost  wire + Elost battery . (39-2)

The language I have used can be misleading. Energy is not created or lost. Energy is merely transformed. The energy gained is the chemical energy supplied by the battery that is transformed into electrical potential energy. The energy lost is thermal; thanks to resistance, things heat up! The energy “lost” or “gained” per Coulomb of charge is given by

E = ΔV+− + Ir , (39-3)

whereΔV+− = V+ −V− , (39-4)

V+ is the electric potential at the positive terminal and V− is the electric potential at the negative

terminal. If the resistance of the wire is R , then, using Ohm’s law, we could write equation (39-3)as

E = IR + Ir = I R + r( ) . (39-5)

We will be using schematic diagrams to represent electrical circuits. The symbol for an ideal battery is represented below in Figure 39-3 a). The two common ways to signify a battery with an internal resistance is represented below in Figure 39-3 b). Note the longer line always signifies the positive terminal.

Figure 39-3Schematic Symbols for an Ideal Battery

And a Battery with an Internal Resistance

+ −

−+

+ −

a)

b)

39-5

RESISTORS

All circuit elements are used to control the amount of energy in a part of the circuit, or they are used to control the amount of current in a specific part of the circuit, or to perform some other useful purpose.

As we have seen, when a current passes through a conductor, the electrons are scattered off of the positive ions which make up the crystal lattice of the metal. These interactions give rise to the resistance of the conductor. We can also design a specific circuit element called a resistor. In general, a resistor is made of a material with a much larger resistivity than that of a conductor. For example, tungsten is used in an incandescent light bulb. This material offers so much resistance to the current that it becomes hot enough to gives off visible light. Other kinds of resistors make very good heating elements in toasters and ovens, for example.

Now, we want to look at a simple circuit that is made of a battery, conductors and a single resistor. Such a circuit is represented below in Figure 39-4. A steady-state current I is established

in circuit for which the resistance outside the battery is R . The electric potential difference across

the battery is ΔV . The electric potential difference represents the amount of energy per Coulomb of charge that is available to the circuit for work. If we assume that the energy lost in the conductors is very small compared to R , then, using Ohm’s law, we can write

ΔV = IR . (39-6)Next, I want to see what would happen if we connect, using conductors, our resistor to a

second resistor. It turns out that there are essentially two different ways to do this. In Figure 39-5 below, I represent the two ways. The resistors in circuit a) are said to be in series, while the resistors in circuit b) are said to be in parallel.

Figure 39-4A Simple Circuit with a Single Resistor

ΔV R

I

39-6

Figure 39-5The Two Possible Ways to Connect Two Resistors

R1 R2ΔV

I

I

I1 I2

ΔV

I

R1

R2

P

′P

a) Resistors in series

b) Resistors in parallel

I

39-7

For the resistors in series, the conservation of charge requires that the current going through resistor one is identical to that of resistor two. So using Ohm’s law, we can write

ΔV1 = IR1 , (39-7)

and in a similar fashion,ΔV2 = IR2 . (39-8)

If we add equation (39-7) to equation (39-8), we get

ΔV1 + ΔV2 = ΔV = IR1 + IR2 = I R1 + R2( ) = IRtot . (39-9)

I hope the reader agrees that a power supply does not know the configuration to which it is attached. It is possible to find a single resistor that can be used to replace a combination of resistors without changing the current delivered to the circuit by the given potential difference ΔV .That single resistor has a resistance called the equivalent resistance. Equation (39-9) give us

ΔVI

= R1 + R2 = Rtot = Req . (39-10)

So, for N resistors in series, the equivalent resistance is given by

Req, series = Rii=1

N

∑ = R1 + R2 ++ RN . (39-11)

Look carefully at the circuit where the resistors are in parallel; Figure 39-5 b) above. Point P in the circuit is called a branch point. The conservation of charge requires that

I = I1 + I2 . (39-12)

Point ′P is also a branch point. There are two paths from point P to point ′P . The potential difference between the two points must be the same for each path. Therefore,

ΔV1 = I1R1 = I2 R2 = ΔV2 = ΔV . (39-13)

So,

I1 =ΔVR1

, (39-14)

while

I2 =ΔVR2

. (39-15)

Adding equations (39-14) and (39-15) we have

39-8

I1 + I2 = I =ΔVR1

+ΔVR2

= ΔV1R1

+1R2

⎡

⎣⎢

⎤

⎦⎥ . (39-16)

Finally,

IΔV

=1

Req

=ΔV 1

R1

+ 1R2

⎡

⎣⎢

⎤

⎦⎥

ΔV=

1R1

+1R2

. (39-17)

We conclude that for N resistors in parallel, the equivalent resistance is given by

1Req, parallel

=1Rii =1

N

∑ =1R1

+1R2

++1

RN

. (39-18)

CAPACITORS

Parallel-Plate Capacitors

Capacitors are electrical devices with a wide range of uses in electrical circuits. We are going to limit our discussion to capacitors that have two conductors separated by an insulator. We are going to further restrict our attention to capacitors where the conductors have equal and opposite charge.

In Figure 39-6 below, we have a representation of a very common type of capacitor called a parallel-plate capacitor. Each conductor has a surface area A over which electric charge Q is

distributed. Positive charge Q is distributed on the bottom surface of the top plate, while negative

charge of magnitude Q is distributed on the top surface of the bottom plate. The plates are

separated by a distance d , and we assume a vacuum between the plates. The charge distributions give rise to an electric field that is everywhere uniform, except at the edges of the plate where there is some fringing, as represented below in Figure 39-7.

There is a limit to the amount of charge that can be put on the plates. This limit turns out to depend on the geometry of the plates, and the insulator between the plates. To quantify this limit, we introduce a concept called the capacitance, and signify it with C . The capacitance is defined by

C = Q / ΔV , (39-19)

39-9

where Q is the magnitude of charge on each plate, and ΔV is the potential difference between the two plates. The potential difference in this equation is always assumed positive as the capacitance can not be negative. So, technically, the potential difference in equation (39-19) is defined by

ΔV = V+ −V− . (39-20)

Using Gauss’s law, we can write for the magnitude of the electric field between the plates as

E =σεo

=Q

Aεo

. (39-21)

As the electric field is uniform through out most of the volume between the plates, the potential difference between the two plates is approximated by

ΔV = Ed =Q

Aεo

⎡

⎣⎢

⎤

⎦⎥d . (39-22)

So, the capacitance of a parallel-plate capacitor is given by

C =QΔV

=Q

Q / Aε o[ ] d= εo

Ad

. (39-23)

As equation (39-23) indicates, the capacitance of a parallel-plate capacitor depends on the area and the plate separation--geometric factors--and the permittivity of free space, εo . Free space, is of

course, telling us that we have the best insulator between the plates, namely, a vacuum.

Example Two:

The SI unit for the capacitance is the Farad F( ) , named in honor of Michael Faraday.

Assume we have a parallel-plate capacitor where the plates are separated by a distance of one millimeter. We also assume a vacuum between the plates. Determine the area of the plates if the capacitance is to be one Farad. Using equation (39-23), we can write

A =Cdεo

= 4πkCd

= 4π 8.99 × 109  N ⋅m2 ⋅C−2( ) 1 F( ) 0.001 m( ) = 1.13 × 108 m2 . (1)

This is a very large number. It represents a square of about 10,600 m on a side! More typical

values for capacitors are microfarads, µF ≡ 10−6 F , or picofarads, pF ≡ 10−12 F .

3-10

E

+Q

−Q

d

A

E

Figure 39 - 7Electric Field Lines

Figure 39 - 6Parallel - Plate Capacitor

39-11

Capacitors Connected in Series

Assume we have two capacitors, initially uncharged, connected in series, as represented below in Figure 39-8. Any charge sent to this combination must reside on the outside plates, the +Q1 and

−Q2 plates. In equilibrium, when the current stops, these must have the same magnitude. So, we

can write

+Q1 = −Q2 = Q . (39-25)

In turn, these charges will induce equal charges on their opposite plates, So, we can also write

− Q1 = Q2 = Q . (39-26)

The definition of the capacitance allows us to write

C1 =QΔV1

, (39-27)

and

C2 =QΔV2

. (39-28)

This implies that

ΔV1 + ΔV2 = ΔV =Q

Ceq

=QC1

+QC2

= Q1

C1

+1

C2

⎡

⎣⎢

⎤

⎦⎥ . (39-29)

Figure 39-8

ΔV

C1

C2

+Q1−Q1

+Q2

−Q2

39-12

Equation (39-29) tells us that the equivalent capacitance for this system of capacitors is given by1

Ceq

=1

C1

+1

C2

. (39-30)

We can generalize this for N capacitors in series. We have

1Ceq, series

=1Cii=1

N

∑ =1C1

+1

C2

++1

CN

. (39-31)

Note that this is the same form as the equivalent resistance for resistors in parallel--see equation (39-18).

Capacitors Connected in Parallel

Assume we have two capacitors, initially uncharged, connected in parallel, as represented below in Figure 39-9. Once the charging has stopped and equilibrium is reached, we have

ΔV1 =Q1

C1

= ΔV2 =Q2

C2

= ΔV =Q

Ceq

. (39-32)

Therefore,

C1 =Q1

ΔV , (39-33)

and

Figure 39-9

ΔV

P

′P

C1 C2

39-13

C2 =Q2

ΔV . (39-34)

If we add the two capacitances, we have

C1 + C2 =Q1 + Q2

ΔV=

QΔV

= Ceq . (39-35)

According to equation (39-35), the equivalent capacitance of our two capacitors in parallel isCeq = C1 + C2 . (39-36)

Again we can generalize this for N capacitors in parallel. We have

Ceq,parallel = Cii=1

N

∑ = C1 + C2 ++ CN . (39-37)

Note that the form of this equation is the same as the equivalent resistance of N resistors in series--see equation (39-11).

Example Three:Using the circuit schematically represented in the diagram below, we want to do the

following:a) Find an expression for the equivalent resistance of the circuit.b) If Rn = nR , where R = 1250 Ω , find a value for the equivalent resistance.

c) Using the same configuration, replace the resistors with capacitors and find an expression for the equivalent capacitance of the circuit.d) If Cn = nC , where C = 1250  pF , find a value for the equivalent

capacitance.Resistors:

We begin by replacing series resistors two and four with their equivalent. We haveReq,1 = R2 + R4 = 2R + 4 R = 6R . (1)

Now, with resistor three, in parallel with our first equivalent resistor, we find our second equivalentresistor. We have

1Req,2

=1R3

+1

Req,1

=1

3R+

16R

=3

6 R=

12R

, (2)

andReq,2 = 2R . (3)

39-14

Finally, resistor one and five are in series with the second equivalent resistance. So, the total circuit equivalent resistance is given by

Req = R1 + R5 + Req ,2 = R + 5R + 2R = 8R = 8 1250 Ω( ) = 10,000 Ω . (4)

Capacitors:We begin by replacing series capacitors two and four with their equivalent. We have

1Ceq,1

=1

C2

+1

C4

=1

2C+

14C

=3

4C , (5)

and

Ceq,1 =43

C . (6)

Now, with capacitor three, in parallel with our first equivalent capacitor, we find our second equivalent capacitance. We have

Ceq,2 = C3 + Ceq,1 = 3C + 4 / 3( ) C = 13 / 3( )C , (7)

Finally, capacitor one and five are in series with the second equivalent capacitance. So, the total circuit equivalent capacitance is given by

R2

R4

ΔV

R1

R5

R3

39-15

1Ceq

=1

C1

+1

C5

+1

Ceq ,2

=1C

+1

5C+

113 / 3( ) C

=65 + 13 + 15

65C=

9365C

, (8)

and

Ceq =6593

C =6593

1250  pF( ) = 873.7  pF . (9)

Capacitors with Dielectrics

If an electric charge is fixed in free space, then the electric field it produces propagates in free space and is described by Coulomb’s law. Free space means the absence of other electric charges; in essence, a vacuum. (Also, we have made no effort to distinguish between a vacuum and air.) In reality, however, the material through which the electric field propagates will affect the net electric field.

The space between the conducting plates of a real world capacitor is usually filled with an insulator, or dielectric. The dielectric material significantly reduces--attenuates--the strength of the net electric field between the plates. We want now to discuss how this might happen. To help, note carefully Figure 39-10 below. In the diagram, we have represented a side view of the conducting plates and the dielectric material between. (The gaps between the plates and the dielectric are simply for purposes of clarifying the geometry of the situation.) The dashed line rectangles are intended to draw your attention to the “effective” bound charges on the surfaces of the dielectric. The ovals with plus and minus signs represent polar molecules.

Many materials are comprised of polar molecules. For many of these, the orientation of the molecules is random. However, for some materials, if they are placed in an external electric field, a significant fraction of the molecules will align themselves with the external field. The free charges

on the conductors produce this external electric field that we signify with

Ef . This induces a

smaller bound charge density on the parallel surfaces of the dielectric. Thus, the dielectric produces

an effective electric field

Eb . The net electric field will then be given by the vectorial sum of these

fields. So, we can write

Enet =

E f +

Eb . (39-38)

In most dielectric materials, the response of the polar molecules is proportional to the electric field and they are said to be linear. The dielectrics are characterized by the dielectric constant κ ,where κ ≥1 . In Table 39-1 below, is a list of dielectric constants for some materials at standard

39-16

temperature and pressure. For linear dielectrics, the net electric field is proportional to the electric field produced by the free charges and we can write

Enet =

1κE f . (39-39)

The potential difference across the capacitor with a dielectric is given by

ΔV =1κ

E f d( ) =1κ

ΔVf , (39-40)

and, therefore, the capacitance becomes

C =QΔV

=Q

1 /κ( )ΔVf

=κQΔVf

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= κC f . (39-41)

In terms of the geometry of the capacitor, we have

C = κC f = κεoAd

. (39-42)

Figure 39-10

Ef

Eb

Ef

Eb

Ef

Eb

Ef

Eb

Ef

Eb

σb−

σb +

σ f −

σ f +

39-17

We can use equations (39-38) and (39-39) to calculate the surface density of the bound charges of the dielectric. We have

Enet =

1κE f =

E f +

Eb , (39-43)

and

Eb =

1κEf −

Ef =

1κ− 1⎡

⎣⎢⎤⎦⎥Ef = −

κ − 1κ

⎡⎣⎢

⎤⎦⎥Ef . (39-44)

Figure 39-1Dielectric Constants of Some Materials

Values at Room Temperature 20° C and 1 atm

Material   At 20°C and 1 atm ( )Vacuum

HeliumAir

Carbon DioxideCarbon Tetracloride

ParaffinPolyethyleneHard Rubber

Transformer OilPlexiglas

NylonEpoxy Resin

Paper

BakelitePyrex Glass

GlassPorcelain

Distilled WaterMetals

κ1.000000001.0000641.000551.00092

2.2 22.32.8 33.43.53.6 4 5 5 6 7 80∞

39-18

The surface charge density due to the bound charges of the dielectric is given by

σb = −κ −1κ

⎡⎣⎢

⎤⎦⎥σ f . (39-45)

This charge density associated with the dielectric arises from polarized molecules. Physicists call this the bound charge density.

Example Four:A parallel-plate capacitor is partially filled by a dielectric with dielectric constant κ , as

represented in the diagram below. The charged surface area of the plates is A . The plates are

separated by a distance d , while the dielectric has a thickness s , where s < d . We want to find the capacitance of this contraption.

In the gaps, the net electric field is simply that produced by the free charge density. Interior to the dielectric, the net electric field is, as we have seen, proportional to the free charge electric field. So, the potential difference across the capacitor is given by

ΔV = E f d − s( ) +1κ

E f s = 1−sd

+s

dκ⎡⎣⎢

⎤⎦⎥

E f d = 1−sd

+s

dκ⎡⎣⎢

⎤⎦⎥ΔVf . (1)

The capacitance, then, is given by

C =QΔV

=Q

1− sd

+ sdκ

⎡⎣⎢

⎤⎦⎥ΔVf

=1

1− sd

+ sdκ

⎡⎣⎢

⎤⎦⎥

QΔVf

=Cf

1− sd

+ sdκ

⎡⎣⎢

⎤⎦⎥

=C f

1− s / d( ) 1− 1 /κ( )⎡⎣ ⎤⎦. (2)

d s

κ

39-19

The Energy in a Capacitor

Assume that we begin with a capacitor without charge on either plate. When we have finished charging the capacitor we will have

Q = C  ΔV . (39-46)Recall that for a charge that is being moved, we have

U E = qmovedΔV , (39-47)

and, therefore,

dUcapE = dq ΔV( ) = dq 

qC

, (39-48)

where we have used

ΔV =qC

. (39-49)

(The potential difference between the plates changes as charge is moved.) Solving equation (39-48) requires using calculus. We have

UcapE = dUcap

E∫ =1C

dq q0

Q

∫ =1C

12

q2⎡⎣⎢

⎤⎦⎥ q =0

q =Q

=Q2

2C . (39-50)

Using equation (39-46). we can also write equation (39-50) as

Q2

2C=

12

C   ΔV( )2 =12

Q ΔV( ) . (39-51)

You may be wondering just how one stores electric potential energy. Note that we can also write equation (39-51) as

UcapE =

12

C ΔV( )2 =12

εoAd

⎡⎣⎢

⎤⎦⎥

Ed[ ]2 =12εoE 2 Ad[ ] . (39-52)

The term Ad is, of course, simply the volume of space between the plates of our capacitor. This allows us to identify a quantity called the energy density, the energy per unit volume, by

U E

Volume

= uE =12εo E2 =

12εo

EiE . (39-53)

Although we derived this result with capacitors, it turns out that it is completely general for regions of empty space where there is an electric field. So, electric potential energy is stored in the electric

3-20

field. For a region of space that is not empty but filled with a dielectric material, the energy density then is given by

uE =

12κεoE 2 =

12κεo

EiE . (39-54)

Gauss’ Law in Dielectrics

We developed Gauss’ law assuming that the source charges of the electric fields were in vacuum. We have seen that the introduction of dielectric material attenuates the electric field. We can modify Gauss’ law to incorporate this state of affairs and give us a more general expression. We have

κEnet id

A∫ =

Q free, enclosed

εo

. (39-55)

39-21

Summary of Chapter 39

In order to sustain a steady current in an electric circuit, a device such as a battery or a generator is needed to provide electrical potential energy to the charges carriers. The energy provided per Coulomb of charge is called the EMF--electromotive force--and signified by E .

For an ideal battery, a battery with no internal resistance, the electrical potential energy provided by the battery is equal to the energy “lost” in the external part of the circuit. We can write

E + ΔV = 0 . (39-56)For a battery with internal resistance r , we have

E = IR + Ir = I R + r( ) . (39-57)

The resistance of a circuit element is defined by

R =ΔVI

. (39-58)

For resistors operating under normal conditions, the voltage and current characteristics of the resistor is described by Ohm’s:

ΔV = IR . (39-59)

For N resistors connected in series, the equivalent resistance is given by

Req, series = Rii=1

N

∑ = R1 + R2 ++ RN . (39-60)

For N resistors connected in parallel, the equivalent resistance is given by

1Req, parallel

=1Rii =1

N

∑ =1R1

+1R2

++1

RN

. (39-61)

Capacitors are devices that store electrical potential energy. We limit our discussion to capacitors with two conductors carrying equal and opposite charges. The capacitance is defined by

C =QΔV

. (39-62)

For a parallel-plate capacitor with vacuum between the plates, the capacitance is given by

C =QΔV

=Q

Q / Aε o[ ] d= εo

Ad

. (39-63)

39-22

For N capacitors connected in series, the equivalent capacitance is given by

1Ceq,series

=1Cii=1

N

∑ =1C1

+1

C2

++1

CN

. (39-64)

For N capacitors connected in parallel, the equivalent capacitance is given by

Ceq, parallel = Cii=1

N

∑ = C1 + C2 ++ CN . (39-65)

One can increase the capacitance of a parallel-capacitor by introducing a dielectric material between the plates. These materials attenuate the net electric field between the conducting plates. These materials can be characterized by the constant κ , called the dielectric constant. For linear dielectric materials, the capacitance becomes

C = κC f , (39-66)

where C f is the capacitance it device would have if there were no dielectric material between the

plates.A capacitor stores electrical potential energy in the electric field between the plates. This

energy is given by

Q2

2C=

12

C   ΔV( )2 =12

Q ΔV( ) . (39-67)

The energy density, the energy per unit volume, in a region of free space where there is an electric field is given by

U E

Volume

= uE =12εo E2 =

12εo

EiE . (39-68)

If the electric field is established in a linear dielectric material, the energy density is

uE =

12κεoE 2 =

12κεo

EiE . (39-69)

In a region of space filled with a dielectric material, Gauss’ law has to be modified to

κEnet id

A∫ =

Qfree,enclosed

εo

. (39-70)

Chapter 40

DIRECT CURRENT CIRCUITS

KIRCHHOFF’S RULES

In this chapter, we want to discuss some techniques for analyzing more complicated direct current circuits. We limit ourselves to circuits with sources of EMF, resistors and capacitors. To aide us in this analysis, we are going to rely heavily on research done by the German physicist Gustav Robert Kirchhoff (1824-1887). Kirchhoff’s rules consist of two statements:

Point Rule:(1) The algebraic sum of the currents toward any branch point is zero.

I∑ = 0 . (40-1)

Loop Rule:(2) The algebraic sum of EMF values around any loop is equal to the sum of

IR products around the loop.

E∑ = IR∑ . (40-2)

The point rule is an expression of the conservation of charge. The loop rule is an expression of the conservation of energy. These rules are fairly straightforward. However, initially, the implementation of the rules can be confusing. Here are some important points to remember in applying these rules.

a) A current is positive if it is moving toward the branch point.b) If there are n branch points, apply the the point rule only n − 1 times.c) For a loop, assign a sense around the loop; clockwise, for example. Currents

moving through resistors in this sense produce positive IR values. Currents moving through a battery from the negative terminal to the positive terminal are

considered positive.Maybe some examples will help.

40-2

Example One:Consider the simple circuit represented in the diagram below. In the diagram on the left, the

sense around the loop is clockwise. The current goes through the battery and the resistor in the same direction as the clockwise sense around the loop. Therefore, E , Ir , and IR are all positive values in our loop rule. In the diagram on the right, we have simply changed the sense in which we go around the loop and all this does is change the sign on the E , Ir , and IR values.

E

r

R

a E

r

R

aI I

E = I r + R( ) −E = −I r + R( )

E1

E2

r1

r2

R1 R2

a I

Example Two :

40-3

Two batteries are connected in series with two resistors, as represented in the diagram above.

The EMF values are related by E1 = 3 / 2( )E2 = 20.00 V , and the internal resistances are related

by r1 = r2 = 2.50 Ω , and the resistor values are related by R1 = 3 / 4( ) R2 = 1,000 Ω . We

wish to find the current in the circuit.We apply the loop rule starting at point a and going in the sense indicated. We find

E2 − E1 = I r2 + R2( ) + I r1 + R1( ) = I r2 + r1 + R2 + R1( ) , (1)

and

I =

E2 − E1

r2 + r1 + R2 + R1( )

=13.33 V( ) − 20.00 V( )

2.50 Ω( ) + 2.50 Ω( ) + 1,333 Ω( ) + 1,000 Ω( )⎡⎣ ⎤⎦=− 0.00285 A . (2)

Since E1 > E2 , then the current must go in a direction opposite of that chosen for the loop

analysis. That is, of course, the significance of the negative sign in our answer.

Example Three:Consider the circuit represented in the diagram below. I have assigned current directions

arbitrarily. Positive loop directions are indicated. If we apply the point rule to branch point a, we can write

I3 − I2 − I1 = 0 . (1)

If we now apply the loop rule to loop one, we can write

E1 − E2 = I1 r1 + R1( ) − I2 R2 + r2( ) . (2)

Next we apply the loop rule to loop two. We have

E2 = I 2 r2 + R2( ) + I 3R3 . (3)

Using equation (1), we noteI3 = I1 + I2 (4)

Substitution of equation (4) into equation (3) gives us

E2 = I2 r2 + R2( ) + I1R3 + I2 R3 . (5)

Now, we solve equation (5) for I2 and get

40-4

I2 =

E2 − I1R3

r2 + R2 + R3( ) . (6)

Substitution of equation (6) into equation (2) gives us

E1 − E2 = I1 r1 + R1( ) − E2 − I1R3

r2 + R2 + R3( )⎡

⎣⎢

⎤

⎦⎥ R2 + r2( ) . (7)

We now solve this equation for I1 . First, we have

E1 − E2 = I1 r1 + R1( ) − E2 R2 + r2( )

r2 + R2 + R3( ) +I1R3 R2 + r2( )r2 + R2 + R3( ) . (8)

Next, we have

E1 − E2 +

E2 R2 + r2( )r2 + R2 + R3( ) = I1 r1 + R1( ) +

R3 R2 + r2( )r2 + R2 + R3( )

⎡

⎣⎢

⎤

⎦⎥ . (9)

E1

E2

r1

r2

R1

R2I1

I3

I2

L1

L2

a b

R3

40-5

And, finally, we have

I1 =

E1 r2 + R2 + R3( ) − E2R3

r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( ) . (10)

To find I2 , we substitute equation (10) into equation (6) and find

I2 =

E2 −E1 r2 + R2 + R3( ) − E2 R3

r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( )⎡

⎣⎢

⎤

⎦⎥R3

r2 + R2 + R3( ) . (11)

Now, we attempt to simplify. We find

I2 r2 + R2 + R3( ) = E2 −

E1R3 r2 + R2 + R3( ) − E2R32

r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( ) , (12)

and then

I2 r2 + R2 + R3( ) =

E2 r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( ) + R32⎡⎣ ⎤⎦

r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( )

−

E1R3 r2 + R2 + R3( )r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( ) , (13)

and again

I2 r2 + R2 + R3( ) =

E2 r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2 + R3( )⎡⎣ ⎤⎦r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( )

−

E1R3 r2 + R2 + R3( )r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( ) , (14)

and, at long last, we can write

I2 =

E2 r1 + R1 + R3( ) − E1R3

r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( ) . (15)

Finally, to find I3 , we add I1 and I2 . We have

40-6

I3 =

E1 r2 + R2 + R3( ) − E2R3⎡⎣ ⎤⎦ + E2 r1 + R1 + R3( ) − E1R3⎡⎣ ⎤⎦r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( )

=

E1 r2 + R2( ) + E2 r1 + R1( )r1 + R1( ) r2 + R2 + R3( ) + R3 R2 + r2( ) . (16)

This problem illustrates just how complicated circuit analysis can be even for fairly simple circuits. To solve more complicated networks, physicists and engineers must use sophisticated matrix programs to keep track of everything.

THE R-C SERIES CIRCUIT

The simple circuits we have looked at so far have all had constant currents in them. This is not the case in a circuit where a resistor and capacitor are connected in series. Below, in Figure 40-1, we have represented an R-C series circuit with an open switch. We assume that the capacitor is initially uncharged. When we close the switch, the capacitor will begin to charge and acurrent will be established in the circuit. Initially, as there is no charge on the capacitor, then we canwrite

E − Io r = ΔV = I oR , (40-3)

Figure 40-1

E

R

r

C

S

40-7

and the initial current is

Io =

E

R + r=

E

Rtot

. (40-4)

At some time t , the current in the circuit is I and the charge on the capacitor is q and we can write

E − Ir = ΔV = IR +

qC

, (40-5)

and, therefore,

E = I r + R( ) +

qC

= IRtot +qC

=dqdt

Rtot +qC

. (40-6)

Equation (40-6) is a differential equation, the solution of which requires calculus. I know some calculus, so I will solve this equation for you. First, we rearrange terms so that we have

EC =

dqdt

Rtot C + q , (40-7)

and then

EC − q =

dqdt

Rtot C . (40-8)

Next, we can write

 

dtRtot C

=dq

EC − q . (40-9)

So solve equation (40-9), we must integrate. We have

dqEC − qqo = 0

qL =Q

∫ =1

Rtot C  dt

to =0

tL =t

∫ . (40-10)

The integral on the right hand side of equation (40-10) is very simple and gives us1

RtotC  dt

to =0

tL =t

∫ =t

Rtot C   . (40-11)

A useful substitution will greatly simplify the integral on the left hand side of equation (40-10).So, we let

x = EC − q , (40-12)

40-8

so thatdx = −dq , (40-13)

and the limits of integration become

qo → EC , (40-14)

and

qL →EC −Q . (40-15)

Substitution of these values gives us

dqEC − qqo = 0

qL =Q

∫ = −dxxEC

EC−Q

∫ = − ln EC − Q( )⎡⎣ ⎤⎦ − − ln EC( )⎡⎣ ⎤⎦ = − lnEC − Q

EC⎡⎣⎢

⎤⎦⎥

. (40-16)

Equating equation (40-11) and (40-16) we have

ln

EC −QEC

⎡⎣⎢

⎤⎦⎥

= −t

RtotC   . (40-17)

Operating on both sides of equation (40-35) with the inverse natural logarithm, we find

EC −QEC

= e− t / RtotC . (40-18)

Solving for the charge, we first have

EC −Q = EC( ) e− t /RtotC , (40-19)

and

Q = EC − EC( ) e− t /RtotC = EC 1− e− t /RtotC⎡⎣ ⎤⎦ . (40-20)

The current is given by

I =

dQdt

=ddt

E C 1− e− t / RtotC( )⎡⎣ ⎤⎦ = E C1

RtotCe− t /RtotC⎡

⎣⎢

⎤

⎦⎥

=

E

Rtot

e− t /RtotC = I oe− t /RtotC . (40-21)

In Figure 40-2 below, we have a graphic representation of the current in the circuit and the charge build up on the capacitor after we close the switch.

40-9

Figure 40-2Charge and Current Values During the Charging Phase

I Q

t00

E

Rtot

E

2Rtot

4RtotC 8Rtot C

EC

EC / 2

Q = EC 1− e−t /RtotC( ) Qfinal = EC

I = Ioe−t / RtotC Io = E / Rtot

charge on the capacitor

c urrent   in  the circuit

40-10

Summary of Chapter 40

The analysis of direct current circuits is not a trivial exercise. There are however, two basic principles that at least provide us with a method. These are called Kirchhoff's laws. The first is called the point rule and follows from the conservation of charge.

The Point Rule:

(1) The algebraic sum of the currents toward any branch point is zero.

I∑ = 0 . (40-22)

For a circuit with n branch points, you can apply this rule n − 1 times. A current is treated as positive if going toward a branch point. A branch point is a point where three or more conductors intersect.

The second rule is called the loop rule and follows from the conservation of energy.

The Loop Rule:

(2) The algebraic sum of EMF values around any loop is equal to the sum of IR products around the loop.

E∑ = IR∑ . (40-23)

To use the loop rule, first assign a sense around a closed conducting path. If the current goes through a battery from the negative to the positive terminal in the same sense as you assigned, the EMF value is positive. A current going through a resistor in the same sense as you assigned has a positive IR . Currents going opposite the assigned sense yield negative values.

A capacitor C is an initially uncharged and connected in series with a resistor R . The

circuit is powered by a battery with EMF E and internal resistance r, the charge on the capacitor at time t after charging begins is given by

Q = EC 1− e− t / RtotC⎡⎣ ⎤⎦ . (40-24)

while the current in the circuit at that time is given by

I =

E

Rtot

e− t /RtotC , (40-25)

where Rtot = r + R .

PART THREE

Magnetism

Chapter 41

THE MAGNETIC FORCE

THE MAGNETIC FORCE: AN INTERACTION BETWEEN MOVING ELECTRIC CHARGES

Two moving electric point charges are represented below in Figure 41-1. The positions and velocities of both charges are defined in the coordinate system shown. Each charge “feels” a force that is magnetic. When measured in the laboratory, the magnetic force exerted on q2 by q1 is

given by

F21

M =′k q2q1v2v1

r212 v2 × v1 × r21( )⎡⎣ ⎤⎦ . (41-1)

Figure 41-1The Magnetic Force on Moving Point-Like Charges

x i( )y j( )

z k( )

q1

v1

r1

q2

v2

r2

r21 = r21   r21 = r

2 − r1r21

41-2

In equation (41-1), ′k is the magnetic constant given by

′k =µo

4π= 10− 7 N ⋅s2 ⋅C-2 , (41-2)

and µo is another constant called the permeability of free space. The vector that runs from

charge q1 to charge q2 , is defined by

r21 = r21 r21 = r2 −r1 . (41-3)

There is also a magnetic force exerted on charge q1 that would be given by

F12

M =′k q1q2v1v2

r122 v1 × v2 × r12( )⎡⎣ ⎤⎦ . (41-4)

One startling fact that emerges is that there is no necessity for these two forces to be equal in magnitude and oppositely directed. Newton’s third law breaks down for rapidly moving elementary charged particles. This breakdown is mathematically represented by the kind of multiplication going on here, namely, the cross product. As we have seen before, the cross product is not commutative!

Now, I want to rearrange equation (41-1) factoring out all of the terms that are explicitly connected with moving charge q2 . We have

F21

M = q2v2 ×

′k q1

r212v1 × r21( )⎡

⎣⎢

⎤

⎦⎥ . (41-5)

Let us make sure we understand the physical significance of the terms in the bracket on the right-hand side of equation (41-5). First, the unit vector r21 signifies a direction from charge q1 toward

charge q2 . The distance from charge q1 to charge q2 is represented by r21 . So, the physical

quantity in the bracket is a quantity to be evaluated at the location of q2 even though it is moving

charge q1 that produces it. The term in the bracket is what we call the magnetic field produced by

q1 at that point in space where q2 is located! We are going to signify the magnetic field with B .

(This symbol is often called the magnetic induction rather than the magnetic field.) The MKS unit for the magnetic field is the Tesla, signified by T . So, we have

B21 =

′k q1

r212v1 × r21( ) , (41-6)

41-3

where the subscripts are indicating that the magnetic field is produced by charge q1 at the location

of charge q2 . Equation (41-5) can be written as

F21

M = q2v2 ×B21⎡⎣ ⎤⎦ . (41-7)

THE MAGNETIC FORCE EXERTED ON A POINT-LIKE ELECTRIC CHARGE

We can use equation (41-7) to define a general equation for the magnetic force exerted on a point charge q moving with an instantaneous velocity

v in the presence of an external magnetic

field

Bext . We have

FM = q v ×

Bext⎡⎣ ⎤⎦ . (41-8)

(The magnetic field in equation (41-8) is called an external magnetic field because it is not the magnetic field produced by moving charge q . To determine the magnetic field produced by moving charge q , we would have to use an appropriately modified equation (41-6).)

It is hoped by the author that you recall that we can break down all of this cross product stuff into the following:

FM =

FM FM = F M FM = q v v( ) × Bext Bext( )⎡⎣ ⎤⎦

= qvBext v × Bext⎡⎣ ⎤⎦ = qvBext sin∠bet( ) C , (41-9)

where the unit vector ˆ C is perpendicular to both ˆ v and ˆ B ext in the right-hand sense. Note that if q

is a positive electric charge, then ˆ C = ˆ F M . However, if q is a negative electric charge, then

ˆ F M = − ˆ C . We can write this as

FM =C , if  q is positive

−C , if q isnegative

⎧⎨⎪

⎩⎪ . (41-10)

Example One:

An electron is moving instantaneously due north with a speed of v = 6.26 × 106 m / s in a

region of space where there is a uniform magnetic field of magnitude B = 0.500 Tesla directed

41-4

vertically downward, as represented in the diagram below. We wish to determine the magnitude and the direction of the magnetic force exerted on the electron at the instant described.

Using the information given, we can write

v = v j , (1)

B = −B k , (2)

while q = − e . (3)Therefore,

FM = − e( ) v j( ) × −B k( )⎡

⎣⎤⎦ = evB( ) − j × − k( )⎡

⎣⎤⎦

= evB( ) j × k⎡⎣ ⎤⎦ = evB i

= 1.602 ×10−19  C( ) 6.26 × 106 m / s( ) 0.500 T( ) i = 5.01× 10− 13 N i . (4)

Note that the force is directed due east and is perpendicular to both the velocity and the magnetic field. Also, recall that if a force is directed perpendicular to the direction of motion, it will change the direction of motion. This charge would, if this field were uniform over a large enough area, move in a horizontal circle. In this case, the magnetic force would also be radially directed, and act as a centripetal force.

East ˆ i ( )

North ˆ j ( )Up ˆ k ( )

B

vC = v × B( )

FM

e−

−e

41-5

THE MAGNETIC FORCE EXERTED ON A CURRENT CARRYING CONDUCTOR

One of the most common forms of moving electric charge is electric charge constrained to move through a conducting wire. Let us look at equation (41-8) again. For an infinitesimal charge dq moving through a conducting wire in the presence of an external magnetic field, then we would

expect this charge to “feel” an infinitesimal amount of magnetic force dF M . For such a case, we

could write equation (41-8) as

dF M = dq

d

dt×Bext

⎡

⎣⎢

⎤

⎦⎥ , (41-11)

where I have taken the liberty of writing the velocity of infinitesimal charge as

v =d

dt , (41-12)

where d signifies an infinitesimal part of the conductor through which the electric charge is

passing as it interacts with the external magnetic field. The direction of this vector, d∧

, is the same

as the direction of the current; the direction positive charge carriers would move through the wire. We can rearrange equation (41-11) getting

dFcond

M =dqdt

d ×Bext⎡⎣ ⎤⎦ = I d

×Bext⎡⎣ ⎤⎦ . (41-13)

Of course, to use this form of the magnetic force equation, one must be able to do integral calculus. We are, however, in luck. I can do some integral calculus.

In a region of space where the magnetic field is everywhere uniform, equation (41-13) becomes

Fcond

M = dFcond

M∫ = I d∫⎡⎣ ⎤⎦×Bext = I

×Bext⎡⎣ ⎤⎦ , (41-14)

where the vector is given by

= , (41-15)

while is the length of the conductor, and is the direction in which the current moves through the conductor.

41-6

Example Two:Suppose we have a very long, straight conductor of length

= 20.00  m through which a

steady-state current I = 2.750 A is moving due north. The conductor is located in a region of

space where there a uniform magnetic field directed due east with a magnitude B = 0.784 T , as

represented in the diagram below. We want to find the magnitude and direction of the magnetic force exerted on the conductor. We can write

B = B i , (1)

and

= j . (2)

Using equation (41-14), we have

Fcond

M = I ×Bext⎡⎣ ⎤⎦

= I j( ) × B i( )⎡⎣ ⎤⎦ = IB j × i( ) = − IB k

= − 2.750 A( ) 20 m( ) 0.784 T( ) k = − 43.12 N k . (3)

East ˆ i ( )

North ˆ j ( )

Up ˆ k ( )

B

I

Fcond

M

41-7

Example Three:The outer core of the Earth is metallic and electrically conducting. It is believed that this

material, coupled with the Earth’s rotation, gives rise to a bipolar magnetic field that surrounds the Earth and the axis of which is inclined approximately twelve degrees with respect to the Earth’s axis of rotation. (The magnitude of this magnetic field, the magnitude of the angle and the polarity of the field change over time!)

The north magnetic pole is close to the south geographic pole, while the south magnetic pole is close to the north geographic pole. The Earth’s magnetic field lines emerge from the north magnetic pole and reenter the Earth at the south magnetic pole. (Please see the crude diagram below.) So, in the southern hemisphere of the Earth the magnetic field lines have a component due north and a component directed vertically upward. In the northern hemisphere, the vertical component is directed vertically downward. The angle by which the magnetic field lines diverge from the north-south line depends on the latitude; being zero at the magnetic equator. The magnitude of the Earth’s magnetic field at the surface of the Earth is approximately

B⊕ = 4.0 ×10− 5 T . Also represented below in a diagram is an energetic proton moving due

south with a speed of v = 1.250 × 107 m / s . The proton is at a northern latitude where the

Earth’s magnetic field line dips below the north-south line by an angle of θ = 15° . We wish to determine the following:

a) The magnitude and direction of the magnetic force exerted on the proton due to its interaction with the Earth’s magnetic field. b) The magnitude and direction of the magnetic force exerted on the proton due to its interaction with the Earth’s magnetic field if it is moving with the same speed due east rather than due south.

If the proton is moving south, we can write

B⊕ = B⊕ cosθ j − sinθ k( ) , (1)

and

v = −v j . (2)

So,

Fp ,S

M = e v ×B⊕⎡⎣ ⎤⎦ = e −v j( ) × B⊕ cosθ j − sinθ k( )( )⎡

⎣⎤⎦

= evB⊕ cosθ − j × j( ) + sinθ − j × − k( )⎡⎣

⎤⎦ = evB⊕ sinθ   i

= 1.602 × 10−19  C( ) 1.250 ×107 m / s( ) 4.0 × 10− 5 T( ) sin15°( )  i

= 2.07 × 10−17 N i ≡ 2.07 × 10−17 N east( ) . (3)

41-8

If, however, the proton is moving east, we can write

B⊕ = B⊕ cosθ j − sinθ k( ) , (4)

and

v = v i . (5)

So,

Fp ,E

M = e v ×B⊕⎡⎣ ⎤⎦ = e v i( ) × B⊕ cosθ j − sinθ k( )( )⎡

⎣⎤⎦

= evB⊕ cosθ i × j( ) + sinθ i × − k( )⎡⎣

⎤⎦ = evB⊕ sinθ j + cosθ k( ) . (6)

NG

NM

SG

SM

B⊕

B⊕

East ˆ i ( )North ˆ j ( ) Up ˆ k ( )

B⊕

v

B⊕ ,N

B⊕ ,Down θ

p

e

41-9

So, Fp,EM = evB⊕ = 1.602 × 10−19  C( ) 1.250 ×107 m / s( ) 4.0 × 10− 5 T( )

= 8.01×10−17 N . (7)The unit vector that represents the direction of this magnetic force is given by

Fp ,EM = sinθ   j + cosθ   k . (8)

This implies that the magnetic force makes an angle ′θ to the north branch of the north-south line, where

   ′θ = tan−1 cosθsinθ

⎡⎣⎢

⎤⎦⎥

= 75° . (9)

Up k( )

North j( ) East i( ) is into the page as is v

B⊕

Fp ,E

Mθ

θ

p

e′θ

Example Four:

A steady-state current Io passes through a rectangular loop of wire of length and width w ,

as represented in the first diagram below. The loop is free to pivot with negligible friction about a horizontal axis. The loop is in a region of space where there is a uniform magnetic field directed vertically upward. We wish to determine the net torque acting on the loop. The net torque will cause the loop to spin. (This is the basic idea behind the electric motor.)

41-10

In the second diagram below, we look down the horizontal axis in the ˆ j direction. This

affords a better view of the forces exerted on the various segments of the rectangular loop. The force exerted on the leftmost wire of the conducting loop is given by

FL

M = Io

×B⎡⎣ ⎤⎦ = Io − j( ) × B k( )⎡

⎣⎤⎦ = − IoB i . (1)

x ˆ i ( )

z ˆ k ( )

ϕϕ

Io

B

FL

M

FR

M

rL

rR

ϕ

A

w2

x ˆ i ( )

y ˆ j ( )

z ˆ k ( )

B

A

ϕ

w

Io

Io

ϕ

Io

41-10

This force produces a torque on the leftmost wire given by

Γ L = rL ×

FL

M =w2

− cosϕ i + sinϕ k( )⎛⎝⎜

⎞⎠⎟ × −IoB i( )

= −

12

IowB sinϕ j . (2)

In a similar fashion, we can write for the force exerted on the rightmost conductor

FR

M = Io

×B⎡⎣ ⎤⎦ = Io j( ) × B k( )⎡

⎣⎤⎦ = I oB i , (3)

while the torque is given by

Γ R = rR ×

FR

M =w2

cosϕ i − sinϕ k( )⎛⎝⎜

⎞⎠⎟ × IoB i( )

= −

12

IowB sinϕ j . (4)

The torque exerted on the front most and back most conductors are equal in magnitude and in opposite directions so that they cancel out. The net torque exerted on the loop is given by

Γ net = −IowB sinϕ j . (5)

We now introduce a new quantity that we call the magnetic moment. The magnetic moment of a current loop is given by

µ = NIA A , (6)

where N is the number of windings in the loop, I is the current passing through the loop, A is the

cross-sectional area of the loop, an ˆ A is a unit vector that points in a direction perpendicular to the plane of the loop in the right-hand sense. (The sense being determined by the circulation direction of the current around the loop.) Finally then,, we can write for the net torque on a current loop as

Γ net =

µ ×Bext . (7)

For this situation, N = 1 and we have

Γ net = wIo   A( ) × B  B( ) = IowB A× B( ) = −I owB sinϕ   j . (8)

41-11

Example Five:Consider a point-like physical thing with electric charge q , mass M , and instantaneous

velocity v that moves into a region of space where there is both an external electric field and an

external magnetic field, as represented in the diagram below. The electric charge q will interact

with the electric field

Eext and “feel” an electric force. The moving electric charge q will

interact with the magnetic field

Bext and be subjected to a magnetic force. The total or net force

acting on the electric charge is the vectorial sum of these two forces. This sum is called the Lorentz force, and is written as

FLorentz = q

Eext + v ×

Bext( )⎡⎣ ⎤⎦ . (1)

We want to find the conditions under which the charge will pass through undeflected.For our example, we have

Eext = −E j , (2)

while

Bext = −B k . (3)

The electric force is

FE = − qE j , (4)

and the magnetic force is given by

FM = q v i( ) × −B k( )⎡

⎣⎤⎦ = qvB i × − k( ) = qvB j . (5)

So, the Lorentz force or the total force acting on the charge is

Ftot = − qE j + qvB j . (6)

q

M

v

E

B

y ˆ j ( )

x ˆ i ( )

E

E

z k( ) out of the page

B

41-12

Please note that if qE > qvB , (7)the charge will be deflected toward the lower plate. However, if

qE < qvB , (8)the charge will be deflected toward the upper plate. The charge will pass through undeflected only if the magnitudes are equal. So, if

qE = qvB , (9)then the charge will pass through if and only if

v = EB

. (10)

This kind of device is called a velocity selector.

41-13

Summary of Chapter 41

A point-like electric q moving with an instantaneous velocity v in a region of space where

there is an external magnetic field

Bext , is subjected to a magnetic force given by

FM = q v ×

Bext⎡⎣ ⎤⎦ . (41-16)

If there is also an external electric field

Eext in this region of space, then the total force exerted on

the point-like electric charge is called the Lorentz force and given by

FLorentz = q

Eext + v ×

Bext( )⎡⎣ ⎤⎦ . (41-17)

If a steady-state current I passes through a conductor of length in a region of space where

there is a uniform external magnetic field

Bext , then the magnetic force exerted on the conductor is

given by

Fconductor

M = I×Bext⎡⎣ ⎤⎦ , (41-18)

where the vector is given by

= , (41-19)

and the unit vector is in the direction of the current.

A conducting loop of N turns with a steady-state current I that is oriented appropriately in an

external magnetic field

Bext will be subjected to a net external torque that will cause the loop to

spin. This torque is given by

Γ net =

µ ×Bext , (41-20)

where the vector µ is the so-called magnetic moment and given by

µ = µ µ , (41-21)

where µ = NIA , (41-22)

and A is the area of the loop, and

ˆ µ = ˆ A , (41-23)

and A is perpendicular to the plane of the loop in the right hand sense--the sense of the current circulation about the loop.

Chapter 42

THE MAGNETIC FIELD

Moving electric charges also give rise to magnetic fields.

THE MAGNETIC FIELD OF A MOVING POINT CHARGE

A static electric charge gives rise to an electric field. A moving electric charge also produces an electric field as well as a magnetic field. In Figure 42-1 below, we have represented the magnetic field at an arbitrary point P produced by a positive point charge q moving with

instantaneous velocity v . The magnetic field is given by

B =

′k qr2v × r( ) , (42-1)

where r is a unit vector pointing from the charge to point P .

Figure 42-1The Magnetic Field Produced by a Moving Point Charge

x ˆ i ( )

z ˆ k ( )

y ˆ j ( )

r

v

B

ˆ r ˆ v

ˆ B P

q

42-2

We can now modify equation (42-1) to describe the magnetic field produced by charges moving in a conductor. The idea is an infinitesimal moving charge dq will produce an

infinitesimal magnetic field dB at an arbitrary point P in space given by

dB =

′k  dqr 2

v × r[ ] . (42-2)

The charges are constrained to move inside a conductor. The path is determined by the orientation of the conductor. So, for a charge moving along a conducting path, the velocity of the charge can be written as

v =d

dt , (42-3)

where d represents an infinitesimal segment of the conductor. If we substitute equation (42-3)

into equation (42-2), we have

dB =

′k  dqr 2

d

dt× r

⎡

⎣⎢

⎤

⎦⎥ =

′kr2

dqdt

d × r⎡⎣ ⎤⎦ =

′k Ir2 d

× r⎡⎣ ⎤⎦ . (42-4)

A steady-state current I passing through an infinitesimal segment of a conductor will give rise to an infinitesimal magnetic field. The magnitude and direction of that magnetic field is given by equation (42-4). Now, let us apply this to an infinitely long straight wire through which a steady current I passes.

Example One:The Magnetic Field Produced by and Infinitely Long

Conductor Carrying a Steady-state CurrentA steady-state current I passes through an infinitely long straight wire as represented in the

diagram Figure below. We want to find the magnetic field at an arbitrary point P on an axis that is a perpendicular bisector of the wire. In this instance, the z-axis.

First, we note that the infinitesimal conductor has an infinitesimal length dy and is directed in

the direction of the current, ˆ j . So, we have

d = d  d

∧= dy  j . (1)

Next, we want to write an expression for r , the unit vector that points from d to point P . We

have

42-3

r = − cosθ j + sinθ k . (2)

So, we can write

d × r = dy  j⎡⎣ ⎤⎦ × −cosθ j + sinθ k⎡⎣ ⎤⎦

= dy − cosθ j × j( ) + sinθ j × k( )⎡⎣

⎤⎦ = dy sinθ   i . (3)

Also, note that

sinθ =z

y2 + z2   and    cosθ =

yy2 + z2

. (4)

So, for this specific configuration, equation (42-4) becomes

dB =

′k Ir 2 d

× r⎡⎣ ⎤⎦ =

′k Ir2 sinθ  dy i =

′k Ir2

zr

dy i =′k Izr3 dy i

= ′k Iz dy

y2 + z2( )3/2

⎡

⎣⎢⎢

⎤

⎦⎥⎥ i . (5)

To find the total magnetic field at the point P , we must add up the contributions of all of the infinitesimal segments of current. To that end, we write

x ˆ i ( )

z ˆ k ( )

y ˆ j ( )

dB

ˆ r

I

r = y2 + z2

y

z

d

θ

P

dy

42-4

B = d

B∫ = ′k Iz

dy

y2 + z2( )3/2   i−∞

∞

∫ . (6)

The rules for adding up infinitesimal quantities is the subject matter for integral calculus. However, this level of math is not required for this class. So, I will simply tell you what the sum turns out to be. The magnetic field produced by a steady-state current I passing through an infinitely long straight conductor at a point P a distance z along an axis perpendicular to the conductor is given by

BLSC =

2 ′k Iz

i . (7)

Equation (7), though correct, does not give us the entire sense of the matter. To better understand what is going on, I am going to redraw Figure 42-2 from a perspective where we look directly along the y-axis into the current. (That is, the current will be coming out of the page. ( See the diagram below.) For any circle of radius z for which the conductor passes through its center and for which the conductor is perpendicular to the plane of the circle, then the magnitude of the magnetic field is the same at every point on the circumference of the circle. The direction of the magnetic field at every point on the circle is tangent to the circle in the right-hand sense. This

x ˆ i ( )

z ˆ k ( )

y j( ) is out of the page

B

B

B

B

B

B

B

B

z

I

r⊥

P

42-5

suggests that we can rewrite equation (7) as

BLSC =

2 ′k Ir⊥

ϕ , (42-5)

where r⊥ is the perpendicular distance of the point of interest from the conductor, and ϕ is the unit

vector used to signify that a vector is tangent to a circle.

Example Two: The Magnetic Field Produced by a Current in a Circular Conductor

At a Point on an Axis Passing Through the Center of the CircleAnd Perpendicular to the Plane of the Circle

A steady-state current I passes through a circular conductor of radius R in the sense indicated in the diagram below. We want to find the magnetic field produced by this current at a point P on an line that passes through the center of the circle and is perpendicular to the plane of the circle.

First, we note that

d = Rdϕ  ϕ , (1)

while r = − cosθ  r⊥ + sinθ   k , (2)

where cosθ = R / r    and    sinθ = z / r , (3)

x ˆ i ( )

z ˆ k ( )

y ˆ j ( )

dBcircle

I

R

z r = z2 + R2

ˆ r

r⊥

dϕ

ˆ ϕ

r⊥′

θ

dBr⊥

dBz

P

Rdϕ

42-6

and

r = z2 + R2 . (4)

We also note

d × r = R dϕ  ϕ[ ] × − cosθ r⊥ + sinθ k⎡⎣ ⎤⎦

= R dϕ − cosθ ϕ × r⊥( ) + sinθ ϕ × k⎡⎣ ⎤⎦ = R dϕ cosθ   k + sinθ  r⊥⎡⎣ ⎤⎦ . (5)

Using the diagram above, note that for every infinitesimal segment d , there is another one on the

opposite side of the circle for which r⊥′ is pointed opposite to ˆ r ⊥ . Physically, this means that the

r⊥ term in equation (5) above vanishes by symmetry; in each pairing, one cancels the other out. So,

the direction of the magnetic field will be along the positive branch of the z-axis, k . So, we have

dBcircle =

′k Ir 2 d

× r⎡⎣ ⎤⎦ =

′k Ir2 Rcosθ  dϕ  k⎡⎣ ⎤⎦ =

′k IR2

r 3 dϕ  k =′k IR2

z 2 + R2( )3/ 2 dϕ   k .(6)

Again, to find the total magnetic field, we must add up the contributions of all of the infinitesimal segments. We find

Bcircle = d

Bcircle∫ =

′k IR 2

z2 + R2( )3/2 dϕ0

2π

∫⎡⎣⎢⎤⎦⎥ k =

2π ′k IR2

z2 + R2( )3/ 2 k . (42-6)

Example Three:A long, straight conductor carries a steady-state current I = 8.750 A along the positive y

axis, as represented in the diagram below. We want to find the magnetic field produced by this

current at a point P in space the coordinates of which are P 6 m , 0,− 3m( ) .

To calculate the magnetic field for a single long straight conductor, it is almost always best to draw a diagram of the physical states of affairs from a point looking down the conductor so that the current is coming directly toward the observer; as we have done in the diagram below. (Currents

coming out of the page will always be signified by • , currents into the page by .)

42-7

To solve this problem, we are going to use equation 42-5,

BLSC = BLSC BLSC =

2 ′k Ir⊥

ϕ . (42-5)

Once we have a clear diagram of the conductor, the current direction, the point of interest and an appropriate coordinate system, then we can draw a line segment from the conductor to the point of interest. The line segment must be perpendicular to the conductor, hence the use of r⊥ . Using

the Pythagorean theorem, we can write

r⊥ = x2 + z2 = 6m( )2 + −3m( )2 = 45  m ≈ 6.7 m . (1)

So, the magnitude of the magnetic field at point P is given by

BLSC =2 ′k Ir⊥

=2 × 10−7  N ⋅ s2 ⋅C−2( ) 8.750 A( )

45  m= 2.61× 10−7  Tesla . (2)

The direction of the magnetic field at point P is given by

ϕ = −cosθ  i − sinθ   j = −3 / 45   i − 6 / 45  k =− 0.4472  i − 0.8944  k , (3)

where θ = cos−1 3 / 45⎡⎣ ⎤⎦ = 63.4° . (4)

x ˆ i ( )

z ˆ k ( )

y j( ) is out of the page

B

r⊥

P

I•

θϕ

ϕ

ϕθ

42-8

Example Four:Two long, straight conductors are parallel to each other. They carry currents in the same

direction and are related by I1 = 2I 2 = 24.00 A . The conductors separated by a distance

d = 1.5 m , as represented in in the diagram below. We wish to find the magnetic field at point P

the coordinates of which are P 6.0m ,3.0m ,0( ) .

Again, we use equation (42-5). For the first current, we can write

B1 =

2 ′k I1

r⊥1

sinϕ1  i − cosϕ1  j⎡⎣ ⎤⎦ =2 ′k I1

d 20220 i −

420  j⎡

⎣⎢⎤⎦⎥

. (1)

For the second current, we have

B2 =

2 ′k I 2

r⊥2

sinϕ2  i − cosϕ2   j⎡⎣ ⎤⎦ =2 ′k I1 / 2( )

d 17117  i −

417  j⎡

⎣⎢⎤⎦⎥

. (2)

So, the total magnetic field at point P is given by

Btot =

B1 +

B2 =

′k I1

d4

20+

117

⎛⎝⎜

⎞⎠⎟  i −

820

+4

17⎛⎝⎜

⎞⎠⎟   j⎡

⎣⎢⎤⎦⎥

=′k I1

d0.2588  i − 0.6353  j⎡⎣ ⎤⎦ . (3)

So, the magnitude of the total magnetic field is given by

Btot =′k I1

d0.2588( )2 + 0.6353( )2 = 0.6860( ) ′k I1

d , (4)

x i( )

y j( )

I1 ϕ1

ϕ 2

B2

r⊥ , 2 = d 17

dr⊥ ,1

= d 20I2

B1

4d

2d

Btot

ϕ tot

P

42-9

and

Btot = 0.6860( )10−7  N ⋅ s2 ⋅C−2( ) 24 A( )

1.5 m( ) = 1.10 ×10−6  T . (5)

The unit vector that represents the direction of the total magnetic field at point P is given by

Btot =Btot

Btot

=

′k I1

d0.2588  i − 0.6353  j⎡⎣ ⎤⎦

0.6860( ) ′k I1

d

= 0.3773  i − 0.9268   j . (6)

The angle ϕ tot can be found may ways. We can write

ϕ tot = cos−1 0.3773( ) = 67.8° = tan−1 0.92680.3773⎡⎣⎢

⎤⎦⎥

. (7)

Example Five:A conducting coil is made up of N closely spaced circular windings of radius R through

which a steady-state current I passes. This current configuration produces an axial magnetic field. The magnitude of the magnetic field produced by such a coil at a point on the axis of symmetry a distance z from the center of the windings is given by

B = N2π ′k IR2

z 2 + R2( )3/ 2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (42-7)

We wish to find the magnitude and the direction of the magnetic field that would be produced at point equidistant from two such circular windings each of which carries the same amount of current in the same sense, as represented in the diagram below. Such windings are called Helmholtz coils.

By symmetry we can write

Btop = Bbot =2π ′k NIR2

d / 2( )2 + R2( )3/2 =12

Btot , (1)

from which we have

42-10

Btot =4π ′k NIR2

d / 2( )2 + R2( )3/2 =4π ′k NIR2

d2 / 4( ) + R2( )3/ 2 =4π ′k NIR 2

d2 + 4R2( ) / 4( )3/2

=4 4( )3 /2 π ′k NIa2

d 2 + 4 R2( )3/ 2 =32π ′k NIR2

d2 + 4R2( )3/2 . (2)

For that special case when R = d / 2 , equation (2) becomes

Btot =32π ′k NIR2

2R( )2 + 4 R2( )3 /2 =32π ′k NIR2

8R2( )3/2 =32π ′k83 /2

⎛⎝⎜

⎞⎠⎟

NIR

= 4.4429 × 10−7 T ⋅mA

⎛⎝⎜

⎞⎠⎟

NIR

. (3)

d

d / 2

P

N

N

I

I

z k( )

x i( )y j( )

Btot

R

Top

Bottom

42-11

Example Six: Moving Charges in External Magnetic Field

There are many important technological applications which employ magnetic fields. One of the more important is the cathode-ray tube. J. J. Thomson used such a device in 1897 to measure the ratio of the charge on an electron to its mass. In the diagram below, we have a schematic representation of the cathode-ray tube used by Thomson in his famous experiment.

The glass tube has had air pumped out and several metal electrodes are sealed in the tube. Electrode C is the cathode from which the electrons emerge. Electrode A is the anode, and maintained at a potential much higher than the cathode. The electrons are accelerated from the cathode to the anode. A small hole in the anode allows some of the electrons to pass through The second hole in electrode ′A ensures that a narrow beam of electrons will enter the region between

the parallel plates P and ′P . The electrons continue until they strike a fluorescent screen at S . The fluorescent material glows wherever the electrons strike. This is the basic idea behind the cathode-ray tubes used in oscilloscopes, computer monitors, and television picture tubes. (Newer technologies have replaced the cathode-ray tube in many applications.)

Assume an electron of mass Me− is moving with an instantaneous speed vo parallel to the

plates when it enters the plates, as represented below in the diagram below. Also, assume that in this region there is only a uniform magnetic field of magnitude B between the plates. We wish to derive an expression for how much the particle will have been deflected when it strikes the screen.

The magnetic force exerted on the electron is given by Newton’s second law as

FM = −  e vo   i( ) × −B  k( )⎡

⎣⎤⎦ = −evo B  j = Me−

ay . (1)

A Representation of Thomson’s Cathode-Ray Tube

C

S

′PP

′AA

D

B

j

ik  out of page

42-12

and the acceleration is

ay = − evoBM

e−  j . (2)

The time needed for the particle to move through the plates is given by

tP = / vo . (3)

The y-coordinate of the particle as it emerges from the plates is

yP = −12

evo BM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥t p

2   j = −12

evoBM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

vo

⎡

⎣⎢

⎤

⎦⎥

2

  j = −eB2

2Me−

vo

  j . (4)

As the electron emerges from the plates, it has an instantaneous velocity given by

vP = vo   i + vP ,y   j = vo  i + −evo BM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

vo

⎡

⎣⎢

⎤

⎦⎥  j = vo   i −

eBM

e−  j . (5)

Between the plates, the electron was moving on a circular path. Once out of the magnetic field, its path is a straight line. The geometry of the straight path is given by

− e

vo

R

D

′d

ϕ ySc

x

y

B

vP

yP

42-13

tanϕ =vP ,y

vo

=y − yP

′d , (6)

and

y = ′dvP ,y

vo

+ yP = ′d− eB

Me−

vo

+eB2

2Me−

vo

= ′deB

Me−

vo

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+eB2

2Me−

vo

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=eB

Me−

vo

′d +

2⎡⎣⎢

⎤⎦⎥

. (7)

Example Seven:The Mass Spectrometer

Every atom of a specific element has the same number of protons in its nucleus. However, the number of neutrons does not have to be the same. Atomic nuclei with the same number of protons but different numbers of neutrons are called isotopes of that element. Assume that a source gives off atomic nuclei with atomic number A and with varying speeds. The nuclei pass through slits S1

and S2 , as represented in the diagram below. The nuclei then move through a region of space

where there is an electric field of magnitude E produced by charged parallel plates. In this region

of space, there is also a uniform magnetic field of magnitude B directed perpendicular to the

electric field. Slit S3 ensures that the only nuclei to get through are the ones that are undeflected

and, therefore, have the same speed given by

v =EB

. (1)

(See Example Five, Chapter 41.) The undeflected nuclei next move into another region of space where there is another magnetic field of magnitude ′B directed perpendicular to the initial direction of the nuclei in this region. The charged nuclei will interact with the magnetic field and “feel” a magnetic force. As this force is always perpendicular to the direction of motion, the nuclei will move on a circular path of radius R .

42-14

The magnitude of the magnetic force exerted on a nucleus of charge Ae and mass MA in this

region is given by

F M = qvBext sin∠bet = Ae( ) EB

⎛⎝⎜

⎞⎠⎟ ′B( ) sin 90°( ) =

AeE ′BB

. (2)

This force keeps the nucleus moving on a circular path, so it must also be equal to the net radial force. We have, then,

F M = Fnet ,rad =AeE ′B

B=

M Av2

R=

MA E / B( )2

R=

MAE 2

RB2 . (3)

The radius is given by

R =MAE 2

B2

BAeE ′B

=MAE

AeB ′B=

M Avq ′B

, (4)

A Model of a Mass Spectrometer

B

E

E

Ae

MA

Source

PC

Nucleiof

v

S3

′B

Rx i( )

z k( )y j( )

is directed into the page

Photographic Plate

S1 S2

42-15

while the mass would be given by

MA =Ae ′B BR

E=

qR ′Bv

. (5)

In the diagram above, point PC represents where the nucleus strikes a photographic plate

leaving an “image.” Isotopes of the element with atomic number A would have different masses corresponding to their having a different number of neutrons. So each specific isotope would strike the plate at a specific radius. Counting the number of strikes at each specific radius would allow one to determine the relative abundance of the elemental isotopes. Measuring the radii of the various strikes would allow us to calculate the mass of the specific isotope.

42-16

Summary of Chapter 42

Moving electric charges give rise to magnetic fields. A point-like electric charge q moving with velocity

v produces a magnetic field at some point P that is given by

B =

′k qvr2 v × r[ ] , (42-8)

where r is the distance from the charge q to point P and ˆ r is a unit vector that points from the charge q to point P , as illustrated in Figure 42-11 below. Also, recall that the magnetic constant ′ k is, in MKS units, is measured to be

′k =µo

4π= 10−7  N ⋅s2 ⋅C−2 , (42-9)

and µo is the so-called permeability of free space. Recall that in this case, free space means that

there are no other charges between q and P.

Figure 42-2Magnetic Field Produced by a Moving Point-Like Charge

q

v

ˆ r

B

v × rr

B if q is negative

P

42-17

For a steady-state current I passing through a very long straight conductor, the magnetic field produced at some point P is given by

BLSC =

2 ′k Ir⊥

ϕ , (42-10)

as represented below in Figure 42-3.

Figure 42-3 The Magnetic Field Produced by a Current in a Very Long Straight Conductor

I

B

r⊥

ˆ ϕ

ϕ Signifies that B is tangent to thecircle.

P

42-18

Represented in Figure 42-4 below is a steady-state current I passing through a circular conductor of radius R. The magnetic field produced by the current at some point P a distance z from the center of the circle on an axis that passes through the center of the circle and is perpendicular to the plane of the circle is given by

Bcircle =

2π ′k IR 2

z2 + R2( ) k . (42-11)

Figure 42-4 Magnetic Field Produced by a Current in a Circular Conductor

x ˆ i ( ) y ˆ j ( )

z ˆ k ( )

I

I

Bcircle

 Bcircle for a  current in theoppositedirection

P

z

Chapter 43

MAGNETIC FLUX AND AMPERE’S LAW

MAGNETIC FLUX

In chapter thirty-six, you were introduced to the idea of an electric flux. Remember that flux is the word used to describe situations where vectors are “popping through” some specific area. You will also recall that we talked about representing an area as a vector. As it is meaningful to talk about the magnetic field as “popping through” an area, it is meaningful to speak of a magnetic flux.

In general, the magnetic flux is defined by

ΦM =

B • d

A∫ . (43-1)

In Figure 43-1 below, we have represented a region of space where a uniform magnetic field is “popping through” a rectangle.

Figure 43-1Magnetic Field Lines “Popping Through” a Rectangle

B

B

B

43-2

The magnetic flux is defined in terms of an integral. Since this is a non-calculus based course, we will be limited to problems for which the magnetic field is constant over the area. When the magnetic field is constant over the area, equation (43-1) reduces to

ΦM =

BiA = BA cos∠bet . (43-2)

GAUSS’ LAW FOR MAGNETIC FLUX

We know now that electric charges give rise to electric fields. We also know that electric charges are discrete. So, electric fields begin and end on these electric charges. However, even though we know that moving electric charges give rise to magnetic fields, the magnetic field does not begin and end on discrete charges. Rather, magnetic fields circulate around moving charges. There is not a magnetic analog for electric charges. We do not have something like a “magnetic charge.” (These analogous entities are called magnetic monopoles.)

Whenever we encounter a magnetic north pole, it is always paired with a magnetic south pole. If we divide a bar magnet into two parts, we do not isolate a north and south magnetic pole, but, rather, wind up with two smaller magnets each having a north and a south magnetic pole. (See Figure 43-2 below.)

As a result, there is no net magnetic flux with respect to a closed volume. As magnetic field lines circulate, then for an arbitrarily shaped closed surface, there is as much flux from field lines “popping” into the volume as there are field lines “popping” out of the volume. Therefore, the

Figure 43-2Dividing a Bar Magnet into Two Smaller Magnets

N

S

N

S

N

S

43-3

net magnetic flux over a closed surface is zero. This result is know as Gauss’ Law for the magnetic field, and is written as

ΦM , closed surface =

B • d

A∫ = 0 . (43-3)

I have attempted to represent this state of affairs with the diagram in Figure 43-3. (Recall that field lines penetrating into a closed surface produce negative flux, while field lines exiting the closed surface produce positive magnetic flux.)

Figure 43-3Gauss’ Law for the Magnetic Field

B

B

43-4

AMPÈRE'S  LAW

Consider the following situation. We have a steady-state current I passing through a very long straight wire in the positive z direction. If we look along the wire, the magnetic field produced “encircles” the wire, as illustrated below in Figure 43-4. Now, imagine that we walk around the wire in a counterclockwise sense along the circle that is pictured. Notice that as we walked around the circle, we would always be moving tangent to the circle, in the same direction as the magnetic field. Also, the magnitude of the magnetic field at each point along the circle would be the same. This suggests that we could do something formally similar to a work calculation except we will not be using a force. We will, however, be dealing with a path integral.

An infinitesimal path part d along our circular walk is expressed quantitatively as

d = dd

∧= r⊥ dϕ ϕ , (43-4)

Figure 43-4

x ˆ i ( )

y j( )

z k( ) out of the page

B

B

B

B

I

r⊥ϕ

dϕ d = r⊥dϕ

43-5

as illustrated above in Figure 43-4. We know that the magnetic field at d is given by

B =

2 ′k Ir⊥

ϕ . (43-5)

Now, if we use the dot product to multiply B and d

we will get an infinitesimal quantity. It

represents the product of that part of the magnetic field that is parallel to the path and the path length itself. We have,

Bid =

2 ′k Ir⊥

ϕ⎡

⎣⎢

⎤

⎦⎥i r⊥ dϕ ϕ[ ] = 2 ′k I dϕ ϕ iϕ = 2 ′k I dϕ . (43-6)

To find the total of all of all the contributions made by all of the infinitesimal segments, we must add them up all the way around the circle. We get

Bid∫ = 2 ′k I dϕ∫ = 2 ′k I 2π[ ] = 4π ′k I = µo I . (43-7)

I hope it seems reasonable that if we add up all of the infinitesimal angles all the way around a circle we get 2π radians.

In walking around the circular path in the exercise above, we used a closed path. This path bounded a planar area; in this specific example, a circle. Along this specific path, the magnitude of the magnetic field was constant. Finally, notice the current I is “enclosed” by the path and “pops through” the area bounded by the path.

This result is not a mere coincidence. For any planar area bounded by a closed path, if one adds up all of the infinitesimal dot products of the magnetic field and infinitesimal path segments around the closed path, then the sum is always equal to µo Ienclosed , where Ienclosed is the total, or net current “popping” through the planar area bounded by the closed path. This is called Ampère's law. The closed path that bounds the planar area is called an Amperian path. Mathematically, we write Ampère's law as

Bid∫ = µ oI enclosed . (43-8)

Just as we use Gauss’ law to find electric fields when we have symmetrical charge distributions, we use Ampère's law to find magnetic fields when we have certain kinds of patterned current distributions. Perhaps some examples will help make this powerful idea a bit clearer.

43-6

Example One:The Magnetic Field Interior to a Cylindrical Conductor

Assume that a steady-state current I is uniformly distributed over the cross-sectional area of a long, straight cylindrical conductor of radius R , as represented in the diagram below. We want to use Ampère's law to find the magnetic field in the following regimes:

a) When r⊥ < R . Point P1 is a representative point in this regime.

b) When r⊥ > R . Point P2 is a representative point in this regime.

Ampère's law is, among other things, an extremely useful tool for finding the magnetic field. The key to using Ampère's law is in choosing a closed path, continuous or segmented, along which the magnitude of the magnetic field is constant and/or zero. Since the magnetic field “circles” a long, straight conductor, the perfect choice for an Amperian path of a long, straight conductor is a circle. When we have chosen well, the left-hand side of equation (43-8) always reduces to something like

Bid∫ = B , (43-9)

where is the length of the path.

Point P1 is inside the conductor where r⊥ < R. We construct an Amperian path of radius

r⊥ on which we find point P1 . Using equations (43-8) and (43-9), we have

r⊥

R

r⊥I

Closed Amperian Paths

The current I is coming out of the page.

P1

P2

43-7

B 2π r⊥( ) = µ oIenclosed = µ oI π r⊥2

π R2⎡ ⎣ ⎢

⎤ ⎦ ⎥ , (1)

and, B =µo I

2π r⊥

π r⊥2

π R 2

⎡

⎣⎢

⎤

⎦⎥ =

µo

2πI

R2⎡⎣⎢

⎤⎦⎥

r⊥ =2 ′k IR2

⎡⎣⎢

⎤⎦⎥

r⊥ . (2)

Note that if r⊥ = R , then the magnetic field goes to 2 ′ k I / R as we have already found.

Point P2 is outside the conductor where r⊥ > R . We construct an Amperian path of radius

r⊥ on which we find point P2 . Using equations (43-8) and (43-9), we have

B 2π r⊥( ) = µ oI , (3)

and, therefore, B = 2 ′ k Ir⊥

. (4)

Recall that the direction of the magnetic field for a long straight conductor is always tangent to the Amperian path in the right-hand sense.

Example Two:The Magnetic Field of a Solenoid

A solenoid is a winding of wire. Represented in the first diagram below is a solenoid of NS

circular windings of radius rS . The total length of the solenoid is S . When a steady-state current

IS is established in the solenoid, a magnetic field is also established. Inside the solenoid, the

magnetic field is parallel to the major axis of symmetry, we say the field is axial. If we take a cross-section of the solenoid, we get something that looks like the representation

in second diagram below. I have taken the liberty to include two rectangular Amperian paths that will allow us to use Ampère's law to calculate the magnetic field inside and outside of the solenoid. The in Figure 43-8 is used to signify that the current is going into the page, while the signifies that the current is coming out of the page.

I want now to use Ampère's law to determine the magnetic field outside of the solenoid. To do this I will use the larger of the two rectangular Amperian paths. We will be going in theclockwise sense around the path, and I will start at the top left-hand corner. Recall that Ampère's

43-8

law states

Bid∫ = µ oI enclosed . (1)

A Solenoid

S

NS

IS

IS

Bsol

RS

Binside

Amperian paths

y

x

Boutside

Boutside

The Solenoid in Cross - Section

43-9

Above and below the solenoid, we would expect the magnetic field to point to the right, while inside the solenoid, we expect it to point to the left. Finally, since the top part of the Amperian path is at a different distance from the major axis of symmetry we would expect the magnitude of the magnetic field above to be different from the one below. So, we have

Bout, top i

top +

Bout ,right i

right +

Bout ,bot i

bot +

Bout ,left i

left = 0 . (2)

First, note that the sum must vanish as there is no net current “popping” through the planar area bounded by the path. As much current leaves the page as goes into it. Next, I want to look at each term on the left-hand side of equation (2). We have

Bout, top i

top = Bout, top i( )i top i( ) = Bout, top top( ) ii i( ) = Bout , top top . (3)

Bout, right i

right = ± Bout , right i( )i − right j( ) = Bout , rightright( ) ± ii− j( ) = 0 . (4)

Bout, bot i

bot = Bout ,bot i( ) i − bot i( ) = − Bout ,botbot( ) ii i( ) = −Bout , botbot . (5)

Bout, left i

left = ± Bout , left i( )i left j( ) = Bout , left left( ) ± ii j( ) = 0 . (6)

Substitution of equations (3) through (6) into equation (2) yields

Bout , top top + 0 − Bout, bot top + 0 = 0 , (7)

and, therefore, Bout , top top = Bout , botbot . (8)

As top = bot , (9)

then Bout , top = Bout , bot . (10)

The results of equation (10) are curious. As the bottom part of the Amperian path is closer to the solenoid that is the top part of the path, then we should have

Bout , top < Bout ,bot , (11)

and the sum not vanish. The only way that equation (10) can be true at any point outside the solenoid is if the magnetic field is zero outside of the solenoid. Outside the solenoid, the magnetic field is so small that it can be ignored.

Now, we want to do this procedure again with the smaller of the two Amperian paths represented in the second diagram above. We wish to use Ampère's Law to find out what the magnitude of the magnetic field is inside of the solenoid. We have

Bout, top i

top +

Bout, right i

right +

Binside , bot i

bot +

Bout, left i

left = µ oI enclosed . (12)

First, note that the sum this time does not vanish as there is net current “popping” through

43-10

the planar area bounded by the path. Next, I want to look at each term on the left-hand side of equation (12). We have

Bout, top i

top = 0 i( )i top i( ) = 0 . (13)

Bout, right i

right = ± Bout, right i( )i − right j( ) = Bout, rightright( ) ±i i− j( ) = 0 . (14)

Binsidei

bot = −Binside i( )i − i( ) = Binside( ) ii i( ) = Binside . (15)

Bout, left i

left = ± Bout, left i( )i left j( ) = Bout, left left( ) ±i i j( ) = 0 . (16)

Substitution of equations (13) through (16) into equation (12) gives us

0 + 0 + Binside + 0 = µ oI enclosed , (17)

and

Binside = Bsol =µo Ienclosed

=

µ o

S

NS IS⎡

⎣⎢

⎤

⎦⎥ = µo

NS

S

⎡

⎣⎢

⎤

⎦⎥ I S . (18)

We often define a quantity called the turn density that is signified by n and defined by

n =

N

. (19)

Therefore, in terms of the turn density, the magnitude of the axial magnetic field of a solenoid can be written as

Bsol = µ onS IS . (20)

43-11

Summary of Chapter 43

The magnetic flux is defined by

ΦM =

BidA∫ . (43-10)

If the magnetic field is constant over the defined area, then the magnetic flux is given by

ΦM =

BiA = BA cos∠bet . (43-11)

To date, physicists have not been able to find magnetic monopoles. So, the magnetic field does not originate on something like a magnetic charge. For a closed surface, there is no net magnetic flux through the closed surface. This result is expressed in Gauss’ Law for the magnetic field and is given by

B • d

A = 0∫ . (43-12)

For a planar area bounded by a closed path, the dot product of the magnetic field and an infinitesimal path segment, when summed all the way around the closed path, yields

B • d

= µ o∫ I enclosed . (43-13)

The net current that “pops” through the bounded area is signified by Ienclosed . Equation (43-13) is

called Ampère's Law. We can use Ampère's Law as a tool for finding the magnetic field. Using

Ampère's law, we found that for a solenoid of NS windings, and length S through which a

constant current IS passes, there is produced inside the solenoid, a magnetic field the magnitude of

which is given by

Bsol = µ oNS

S

⎡

⎣⎢

⎤

⎦⎥ IS . (43-14)

The magnetic field is directed along the major axis of symmetry of the solenoid in the right-hand sense. We treat the magnetic field outside the solenoid as negligible.

Chapter 44

CHANGING MAGNETIC FLUX ANDINDUCED ELECTRIC FIELDS

MOTIONAL EMF AND FARADAY’S LAW

A horizontal conductor of length is located in a region of space where there is a uniform

magnetic field B that is directed vertically downward, as represented in Figure 44-1 below. A

positive charge carrier is highlighted for purposes of developing the theory. (Recall that in metal conductors, electrons are the charge carriers that actually move.) If this conductor is moved horizontally with a constant velocity

v that is perpendicular both to the conductor and to the magnetic field, then the positive charge carrier with “feel” a magnetic force. This magnetic force will induce a current in the conductor.

Figure 44-1

B

v

FM

East i( )

North j( )

Up k( )out of page

44-2

After a time, the top of the conductor will become positively charged while the bottom of the conductor will be negatively charged. This charge distribution will set up an electric field directed toward the bottom of the conductor, as represented below in Figure 44-2. Eventually, the electrical force exerted on our positive charge carrier will be as large as the magnetic force and the current will cease.

Equilibrium conditions on a single positive charge carrier requireqE = qvB , (44-1)

orE = vB . (44-2)

The electric field established in the conductor is not unlike that of a small capacitor and will have a magnitude given by

E = ΔV / , (44-3)where ΔV is the electric potential between the ends of the conductor. Using equation (44-3) and equation (44-2) gives us

ΔV

= vB , (44-4)

Figure 44-2Equilibrium Established

East i( )

North j( )

Up k( )out of page

E

E

B

v

FM

FE

44-3

and, therefore,

ΔV = Bv . (44-5)Note that this potential difference here is the direct result of a static charge distribution, and the electric field established is conservative.

Next, let us look at a variation of this scenario. This time the conductor is placed across a

horizontal set of conducting rails. A uniform magnetic field B is directed vertically downward, as

represented in Figure 44-3 below. Now, when the conductor is moved to the right with a constant velocity

v , the magnetic force exerted on positive charge carriers in the conductor will establish an induced current Iin that can move around a closed path. In this situation, electric charges will not build up at the top or the bottom of the moving conductor.

Next, let us look at the magnetic flux through the area bounded by the closed path of the induced current. The magnetic field is constant over the bounded area so we can use

ΦM =

B •A . (44-6)

At time to , the magnetic flux is given by

ΦM ,o = − B k( ) • Ao k( ) = − BAo k • k = −BAo , (44-7)

while at a later time tL . the magnetic flux is given by

ΦM ,L = −B k( ) • AL k( ) = −BAL k • k = −BAL . (44-8)

Figure 44-3A Variation on a Theme

to tL

B

v

FM

Iinduced

xo v ΔtEast i( )

North j( )

Ao ΔA

44-4

You may be wondering how I knew the direction of the area. The convention is right-handed. If the current is circulating in a counterclockwise sense, the unit vector representing the direction of the area comes out of the page. If the current is in a clockwise sense, the unit vector representing the direction of the area goes into the page, as represented below in Figure 44-4.

The change in the magnetic flux over this time interval is given by

ΔΦM = ΦM ,L −ΦM ,o = − B AL − Ao[ ] = −BΔA . (44-9)

Inspection of Figure 44-3 should convince one that the change in the area is given by

ΔA = vΔt . (44-10)

So, the change in magnetic flux is given by

ΔΦM = − BvΔt . (44-11)

Finally, note that the time rate of change of the magnetic flux is

ΔΦM

Δt=−Bv Δt

Δt= − Bv . (44-12)

The magnitude of equation (44-12) is precisely the same as equation (44-5.) If a conductor is placed in a region of space where there is a changing magnetic flux, a current will be induced in the conductor just as surely as if it had been connected to a battery with an E = Bv . The average EMF of this “battery” is given by Faraday’s law of electromagnetic induction:

Eave = −ΔΦM / Δt . (44-13)

Equation (44-13) was first expressed by Michael Faraday, a brilliant British experimental physicist, as a result of his extensive investigations. It’s validity is not dependent on the motion of the conductor. All that needs to happen for there to be an induced current in the conductor is for the magnetic flux to change! How can a magnetic flux be changed? There are essentially three ways. Again, if we assume the magnetic field is constant over a well defined area, we can write

Figure 44-4Determining the Direction of an Area Bounded by a Current

I

A

A

I

44-5

Δ ΦΜ( ) = Δ

BiA⎡⎣ ⎤⎦ = Δ BA cos∠bet[ ]

= ΔB Acos∠bet( ) + Bcos∠bet( ) ΔA + BA Δ cos∠bet( )( ) . (44-14)

Equation (44-14) implies that the magnetic flux can be changed if:1) The magnetic field over the area is not constant in time.2) The area is not constant in time.3) The orientation of the area to the magnetic field is not constant in time.

So, the induced EMF can be calculated by

Eave = −

ΔΦM

Δt

= −ΔBΔt

A cos∠bet( ) + B cos∠bet( ) ΔAΔt

+ BAΔ cos∠bet( )

Δt⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (14-15)

The negative sign in equation (44-15) is not a mere triviality. The negative sign is indicating that the induced current must always go in the direction which would oppose the very change that brought it into existence. This is known as Lenz’s law. To illustrate, consider again the positive charge carrier depicted in Figure 44-3, above. In Figure 44-5 below, we show the magnetic forces exerted on the carrier. As the conductor is pulled to the right with velocity

v , this motion of a

charged particle in the presence of an external magnetic field gives rise to a magnetic force FM that

is directed along the conductor. This force causes the charge carrier to also move along the conductor with some velocity

′v . The second component of motion then gives rise to another

magnetic force FM ′ that is directed to the left. In this situation, the magnetic flux is changing

Figure 44-5The Magnetic Forces Exerted on a Positive Charge Carrier

v

′v

FM

FM ′

B

44-6

because the enclosed area is increasing. The induced current goes up the conductor giving rise to a

magnetic force FM ′ that has the effect of opposing this increase in area.

Magnetic flux requires a magnetic field passing through an area. An area requires a boundary. If the boundary is a single closed conductor, then a current will be induced in the conductor for as long as the magnetic flux changes. Faraday’s law would then take on the form

Eave = −

ΔΦΔt

= I in R , (44-16)

where R is the total resistance of the conductor. If the conductor had N closely wound turns, then the induced EMF would be given by

Eave = − N

ΔΦM ,sc

Δt= Iin R , (44-17)

where ΦM ,sc is the magnetic flux through a single conductor.

INDUCED ELECTRIC FIELDS

I wish to return to Faraday’s law. We can use the limit process to ge the instantaneous induced EMF. The instantaneous EMF is given by

E = −

dΦM

dt . (44-18)

In the last chapter, the magnetic flux was defined by

ΦM =

BidA∫ . (44-19)

For conservative electric fields, we have seen that the electrical potential energy per unit charge in the displacement of charge ′q from some initial point to some final point is given by

ΔUE

′q= −

FE idr∫

′q= −

Eidr∫ = ΔV , (44-20)

where ΔV is the electric potential difference between the final point and the starting point and signified by

ΔV = Vf −Vo . (44-21)

If we now displace charge ′q from some initial point to some intermediate point and then back to

44-7

the starting point, we have displaced the charge on what is called a closed path. It is hoped by the author that equation (44-21) convinces you that the potential difference in this displacement would be zero. We can write the result for a displacement around a closed path mathematically as

ΔV = −

Eidr∫ = 0 . (44-22)

As we saw in our discussion of the battery, in order to move charges completely around a closed circuit we must have a source of EMF and this source can not be conservative, it can not be the result of a static charge distribution only! It requires a mechanism that gives rise to an induced, non-conservative electric field. In terms of such a field, the EMF can be defined as

E =

Enc id

r∫ . (44-23)

Substitution of equations (44-23) and (44-19) into (44-18) gives us

Enc id

r∫ = −ddtBidA∫ = E . (44-24)

The closed path indicated by the integral on the left-hand side of equation (44-24) bounds the area of the integral in the middle term of equation (44-24). If the magnetic flux changes in the area, then a non-conservative electric field is produced along the closed path. If there are any charge carriers on that closed path, they will feel an electric force. The amount of electrical energy per unit charge available to these carriers is E . Example One:

A single, circular conductor of radius R spins with a constant angular speed ω about an east-west axis in a region of space where there is a uniform magnetic field directed vertically downward, as represented in the diagram below. If the plane of the conductor is, at time to , oriented horizontally, we wish to determine the average induced EMF in the conductor over a time interval Δt , where the later time is given by tL = to + τ / 6( ) .

The axis of rotation runs east-west. I have assume that the induced current in the conductor will “flow” in the sense indicated in the diagram. At a later time tL , the circular loop will have rotated through an angle

τΔt

=360°Δϕ

→τ

τ / 6=

360°Δϕ

→ Δϕ =360°

6= 60° . (1)

The magnetic flux is given by

ΦM ,o = B B( ) • A Ao( ) = BA B • Ao = BA cos 0° = BA , (2)

44-8

while

ΦM ,L = B B( ) • A AL( ) = BA B • AL = BA cosθL . (3)

So, the change in magnetic flux is

ΔΦM = ΦM ,L −ΦM ,o = BA cosθL − cosθo[ ] = BA cos 60° − cos0°[ ]

= −12

B πR2( ) . (4)

Therefore, the average induced EMF is given by

Eave = −

ΔΦM

Δt= −

− 1 / 2( ) BπR2

τ / 6⎡

⎣⎢

⎤

⎦⎥ =

3BπR2

2π /ω=

32

BR2ω . (5)

The Loop at Times to and t( Looking Down the Axis of Rotation )

Up k( )

N j( )

B

to

t

ω

Ao

A

R

Iin

Iin

θ

44-9

Example Two:A solenoid of length

S has NS circular turns of radius RS . A circular coil with NC turns

of radius RC , where RC > RS , is coaxial with the solenoid. Both are represented in cross-section

in the first diagram below. A steady-state current IS is established in the solenoid and gives rise to

an axially directed magnetic field Bsol . We are model the solenoid as having a negligible magnetic field outside the solenoid and a uniform axial magnetic field inside the solenoid. This model provides us with a very good approximation.

I want first to calculate the amount of magnetic flux through a single turn of the coil. To facilitate this calculation, a second diagram below is provided where we look at the configuration down the major axis of symmetry. From this perspective, one easily can see that the magnetic field is “popping” through only a fraction of the cross-sectional area of the coil. This is, of course, the only area associated with the coil through which there is a magnetic flux. So, for a single turn of the coil, the magnetic flux is given by

ΦM , cs =

Bsol •

AS = Bsol i⎡⎣ ⎤⎦• ± AS i⎡⎣ ⎤⎦ = ± Bsol AS . (1)

Bsol

RS

RC

NS

NC

IS

S

x ˆ i ( )

y ˆ j ( )

44-10

The total magnetic flux through the coil is given byΦM , C = NC ΦM , sc = ± NC Bsol AS

= ± NC µoNS

S

⎛⎝⎜

⎞⎠⎟

I S

⎡

⎣⎢

⎤

⎦⎥ πRS

2⎡⎣ ⎤⎦ = ±4π 2 ′k NC N S

S

⎡

⎣⎢

⎤

⎦⎥ I S , (2)

where we have used µo = 4π ′ k . The plus-minus sign is a result of not knowing how to assign the direction of the area; other than arbitrarily. Note also that all of the terms in the bracket are constants. Only the current can be changed. So, as long as the current is constant, there can be no

Bsol

NS

NC

ΔIS < 0

Bsol

NS

NC

ΔIS > 0

IS

IS

RC RS

Magnetic Field

No Magnetic Field

44-11

be no induced EMF and no induced current.There are two ways to change the current: increase it or decrease it. We now want to calculate

the average induced EMF in the coil if the current in the solenoid changes by ΔIS in a time interval

Δt . We have

Eave = −

ΔΦM

Δt= −

± 4π 2 ′k NC NS / S⎡⎣ ⎤⎦ ΔIS

Δt=

4π 2 ′k NC NS

S

⎡

⎣⎢

⎤

⎦⎥ΔIS

Δt . (3)

The induced EMF must be a positive number. The negative sign in Faraday’s law is telling us the induced current moves through the conductor in a sense so as to oppose the change that brought it into existence. If the current decreases, the magnetic field of the solenoid will decrease and the induced current in the coil will circulate in a sense so that it produces an induced magnetic field in the same direction as that of the magnetic field of the solenoid itself. If the current increases, the directions are reversed. These physical states of affairs are represented in the bottom diagram above.

GENERATORS

The rectangular loop is rotating with constant angular speed ω about the y-axis, as

represented below in Figure 44-6. A uniform magnetic field B is directed vertically upward. The

magnetic flux through the rectangular loop is given by

ΦM =

B •A = B  B⎡⎣ ⎤⎦ • A  A⎡⎣ ⎤⎦ = B ab( ) cosϕ . (44-25)

The change in the magnetic flux arises from the change in the angle ϕ . So, we have

ΔΦM = B ab( ) Δ cosϕ( )⎡⎣ ⎤⎦ = B ab( ) cosϕ L − cosϕ E[ ]

= B ab( ) cos ϕ + Δϕ( ) − cosϕ⎡⎣ ⎤⎦ . (44-26)

Using a trig identity, we can write

cos ϕ + Δϕ( ) = cosϕ cosΔϕ − sinϕ sinΔϕ , (44-27)

and substitution of this into equation (44-26) gives us

ΔΦM = B ab( ) cosϕ cosΔϕ − sinϕ sinΔϕ − cosϕ[ ] . (44-28)

Recall, we are trying to find the instantaneous EMF. So, we need to determine what happens to equation (44-28) as Δϕ gets very close to zero. To that end, we note that as Δϕ→ 0 , then

44-12

cosΔϕ → 1, and sinΔϕ→ dϕ , while ΔΦΜ → dΦΜ . In the limit, then, equation (44-28)

becomes

dΦM = B ab( ) cosϕ 1( ) − sinϕ dϕ( ) − cosϕ⎡⎣ ⎤⎦ = −B ab( ) sinϕ( ) dϕ . (44-29)

Therefore, the instantaneous EMF value is given by

E = −

dΦM

dt= B ab( ) sinϕ( )  dϕ

dt⎡⎣⎢

⎤⎦⎥

= B ab( ) sinϕ( )ω . (44-30)

Note that ab( ) is the area of the loop. Also, if there were N loops rather than one loop, equation

Figure 44-6Rectangular Conducting Loop Rotating with

Constant Angular Velocity ω in a Uniform Magnetic Field

x i( )

z k( )

y j( )

b / 2

a

B

B

B

ω

ϕ

v

ϕ

Enc

Enc

A

ϕ

Enc

Enc

44-13

(44-30) would become

E = NBA sinϕ( )ω , (44-31)

where we have used the definition of the instantaneous angular speed

ω =dϕdt

. (44-32)

Also, as a point of interest, equation (44-32) impliesdϕ = ω dt . (44-33)

If the angular speed ω is constant over some time intervalΔt = t − to = t , then we have

dϕ0

ϕ

∫ = ω dt0

t

∫ , (44-34)

and ϕ = ωt . (44-35)

INDUCTANCESelf-Inductance

Figure 44-7

I

I

I

I

B

B

′B

B

B

B

w

B

44-14

Assume we have an isolated rectangular circuit through which a steady-state current I passes,

as represented in Figure 44-7 above. This current will give rise to a magnetic field B that

“encircles” the conducting wire. However, there is also magnetic field which “pops through” the area bounded by the circuit and produces a magnetic flux ΦM . If this current were to change, the

magnetic flux through the circuit would also change and an EMF E would be induced in the circuit. In this case, the changing current would also result in a change in the magnetic field!

We know that the magnetic field produced by a current in a conductor is proportional to that current. We also know that the magnetic flux through the area bounded by the circuit is proportional to the magnetic field, and, therefore, is also proportional to the current in the circuit; that is, ΦM ∝ I . To change this proportionality into an equation, we introduce a

proportionality constant L , called the self-inductance. The self-inductance is defined by

ΦM = LI . (44-36)

The change in the magnetic flux through the circuit is then given byΔΦM = L  ΔI . (44-37)

So, we can write Faraday’s law as

E = −

dΦM

dt= −L 

dIdt

. (44-38)

This EMF is called back EMF and it attempts to keep the current constant! The MKS unit for

inductance is called the henry, in honor of Joseph Henry. 1 H ≡ 1 Volt / Amp ≡ V ⋅s ⋅C−1 .

Example Three:

Represented in the diagram below, is a toroidal coil with N turns, a mean radius Rm and

cross-sectional area A . I want to find the self-inductance of this configuration. (For simplicity, we assume the flux is a result of the magnetic field produced by the current in the windings only and that there are no extraneous magnetic fields.)

Using Ampère's law, we can write

B 2πRm( ) = N µoI , (1)

and the magnitude of the magnetic field is given by

B =µo NI2πRm

=2 ′k NRm

I . (2)

44-15

The magnetic flux through one of the windings would be given by

ΦM ,sc =2 ′k NA

Rm

I , (3)

and the total through the entire toroid would be

ΦM ,tot = N ΦM ,sc =2 ′k N 2 A

Rm

I . (4)

The change in the flux would be given by

ΔΦM ,tot =2 ′k N 2A

Rm

 ΔI , (5)

I

I

N  Turns

Amperian Path

Rm

Cross − Sectional Area A

B

B

B

44-16

and the self-inductance would be

L =ΔΦM ,tot

ΔI=

2 ′k N 2 ARm

 ΔIΔI⎡⎣⎢

⎤⎦⎥

=2 ′k N 2 A

Rm

. (6)

Mutual Inductance

With the self-inductance, we were dealing with a single isolated circuit. It is possible for there to be magnetic linkage between more than one circuit. Now, we direct our attention to the mutual inductance of two circuits.

Consider carefully the coaxial coils represented in cross section below in Figure 44-8. A

current I1 in coil one produces a magnetic

B1 . Some of the field lines of

B1 intercept coil two

and establishes a total magnetic flux ΦM ,21 . If current I1 varies, then there will be a change in the

magnetic flux in coil two and an EMF, E2 , will be induced in coil two.

Figure 44-8Flux Linkage between Two Circuits

Coil  1 Coil  2

N1

N2

B1

}ΦM ,21

I1

I1

44-17

Again, we use the fact that the magnetic field

B1 produced by I1 is proportional to I1 ; that is,

B1 ∝ I1 . Also, the magnetic flux through coil two ΦM ,21 is proportional to

B1 , and, therefore,

also proportional to I1 ; that is, ΦM ,21 ∝ I1 . To make this proportionality an equation, we

introduce a new constant of proportionality M 21 called the mutual inductance and defined by

ΦM ,21 = M 21I1 . (44-39)

The change in magnetic flux in coil two would be given byΔΦM ,21 = M 21 ΔI1 , (44-40)

while the induced EMF in coil two would be

E2 = −

dΦM ,21

dt= −M 21  

dI1

dt . (44-41)

Example Four:

A long solenoid of length S and cross-sectional area A has NS closely wound turns. A

small coil of NC turns is wound closely around the solenoid at its center, as represented in the

diagram below. A steady-state current IS is established in the solenoid. We wish to calculate the

mutual inductance MCS of the coil due to the solenoid.

The current IS produces a magnetic field Bsol given by

Bsol = µ o NS / S[ ] IS . (1)

The total magnetic flux through the coil due to the magnetic field of the solenoid is given by

ΦM ,C = NC Bsol A = NC µ o NS / S( ) I S⎡⎣ ⎤⎦ A = µo NC NS AIS / S . (2)

S ISNS NC

B

44-18

Therefore,

MCS =ΔΦM ,CS

ΔI=

µo NC NS AS

. (3)

MAGNETIC ENERGY

We saw earlier that capacitors store electric potential energy in the electric field. It is also the case that inductors store energy in the magnetic field. To calculate the amount of magnetic energy stored, we recall that the back EMF is given by

E = −L

dIdt

. (44-42)

If we now multiply equation (44-42) by I , we have the power supplied by the power supply

IE = −

dU M

dt= −LI

dIdt

, (44-43)

and

 dU M = L   I  dI . (44-44)So, the energy is

  dU M

U E =0

U E

∫ = L   I  dII = 0

I

∫ , (44-45)

and

U M =12

LI 2 . (44-46)

Example Five

A solenoid of length has N circular turns of radius r . A steady-state current I is established in the solenoid. We wish to calculate the magnetic energy per unit length in the solenoid.

First, we note that the magnetic flux through a single turn of the solenoid is given by

ΦM ,sc =BiA = µo

N

⎛⎝⎜

⎞⎠⎟ I⎡

⎣⎢⎤⎦⎥π r2⎡⎣ ⎤⎦ = µonI[ ] πr 2⎡⎣ ⎤⎦ = µoπ nr2 I , (1)

44-19

where we have used the so-called turn density defined by

n =

N

. (2)

The total magnetic flux is given by

ΦM = NΦM ,sc = µoπNnr 2 I = LI . (3)

So, the magnetic energy is given by

U M =12

LI 2 =12

µoπNnr 2 I 2 , (4)

Magnetic Energy as a Function of the Magnetic Field

We can write the the magnetic energy per unit length of the solenoid would be

U M

=

12

µoπNnr 2 I 2

=

12

µ oπn2r2 I 2 , (44-47)

and

U M =12

µoπ n2r 2I 2 =

12

µon2I 2( ) πr2( ) =

B2Volume

2µ o

. (44-48)

Therefore, the magnetic energy density in the solenoid is, then, given by

uM ,sol =U M

Volume

=Bsol

2

2µo

. (44-49)

44-20

Summary of Chapter 44

In the diagram below, we have represented a planar area bounded by a closed path. If the magnetic flux through this area is changing in time, then a non-conservative electric field will be induced along the boundary. The relationship between the changing magnetic flux and the induced electric field is given by

Enc id

r∫ = −ddtBidA∫ . (44-50)

If the boundary is formed by a conducting material, then there will also be induced EMF given by

E =

Enc id

r∫ = −ddtBidA∫ , (44-51)

and an induced current given by

Iin =

E

Rtot

, (44-52)

where Rtot is the total resistance in the conductor.

A Bounded Planar Area

boundary

planar  area

44-21

If a current moves in a rigid circuit, it will give rise to a magnetic field that is proportional to the current. A magnetic flux through the circuit will also be established and it too will be proportional to the current. The self-inductance L is a constant of proportionality and defined in terms of

ΦM = LI . (44-53)

If this current changes in time, then a back EMF will be induced in the circuit given by

E = −

dΦM

dt= L

dIdt

. (44-54)

For two rigid circuits that have magnetic flux linkage, the magnetic flux in one circuit is proportional to the current in the other circuit. The mutual inductance L21 is defined in terms of

ΦM ,2 = L21 I1 . (44-55)

If current one changes in time, then there will be induced an EMF in circuit two given by

E2 = −

dΦM ,2

dt= L21

dI1

dt . (44-56)

It can be proven that these mutual inductances are equal, that is, thatL21 = L12 . (44-57)

Inductors are circuit elements made up of windings of wire. They store magnetic energy in the magnetic field. The amount of energy stored is given by

U M =12

LI 2 . (44-91)

The energy density of an inductor is given by

uM =B 2

2µ o

. (44-58)

(All of our values have assumed that the core of the inductor is a vacuum, or air.)

Chapter 45

MAGNETIC MATERIALS

We have seen how magnetic fields can be produced by moving electric charges. An isolated charge moving in empty space or charges moving in a conductor produce magnetic fields. There is, however, still the mystery of how there can be something like a permanent magnet.

You have probably at some time in your life seen a “bar magnet.” The bar magnet produces a magnetic field even though there is no apparent moving charge to cause it. It turns out that to explain how this happens we must resort to looking at the atomic structure of matter. Unfortunately, that can not be done completely or correctly without using quantum mechanics. Even though quantum mechanics is beyond the scope of this course, I want to begin our discussion of magnetic materials by looking at a model of an electron orbiting a fixed proton.

A MODEL OF AN ELECTRON ORBITING A FIXED PROTON

Imagine a solitary electron in a circular orbit around a fixed proton, as represented below in Figure 45-1. The electron is held in its circular orbit by the electrical force. The moving electron establishes an atomic current and produces a small magnetic field. The current produced is given

Figure 45-1A Model of an Orbiting Electron

v e−

R

FE

p

B

45-2

by

I =eτ

=e

2πR / v=

ev2πR

. (45-1)

Treating the orbit as circular and using equation (42-17), we can write the magnitude of the magnetic field at a point on the axis of symmetry a distance z from the proton as

B =2π ′k ev / 2πR( ) R2

z2 + R2( )3 /2 =′k evR

z2 + R2( )3 /2 . (45-2)

At a point in space where z >> R , equation (45-2) reduces to

B z >> R( ) =′k evRz3 . (45-3)

Also, using equation (41-53), we can write the orbital magnetic moment of the electron as

µorb = IA =ev

2πR⎡⎣⎢

⎤⎦⎥πR2⎡⎣ ⎤⎦ =

12

evR . (45-4)

Using equation (45-4) in equation (45-3), we can write

B z >> R( ) =2 ′k µorb

z3 . (45-5)

The classical angular momentum of the electron would by given by

Lorb = rorb ×

p = RM e− v Lorb . (45-6)

In terms of the classical angular momentum, the orbital magnetic moment would be expressed as

µorb = − 

12

eM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥ Lorb , (45-7)

where the negative sign is telling us that the orbital magnetic moment and the orbital angular momentum are in opposite directions. It turns out that one can only measure a component of the orbital angular momentum of an electron. These values are assigned a quantum number signified by ml . So, a component of the orbital angular momentum of an electron can take on only the

following possible values:

Lorb = ml , for

ml = 0, ±1, ±2,  limit . (45-8)

(The possible values have a limit that is defined by the quantum rules.)

45-3

Electrons also behave as if they spin. This, in turn, sets up another “current” and has its own magnetic moment. The magnitude of the spin magnetic moment of an electron is always

µspin =12

eM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥ , (45-9)

the value of which is

µspin =12

eM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥ =

12

1.602 × 10−19 C9.109 × 10−31 kg

⎡

⎣⎢⎤

⎦⎥1.0546 × 10−34 J ⋅s( )

= 9.274 × 10−24 J / T . (45-10)This value is called the Bohr magneton.

Protons and neutrons also have “spin” magnetic moment values but they are about three orders of magnitude smaller. So, the electron magnetic moments tend to dominate the magnetic contributions of individual atoms. Listed below in Table 45-1 are the magnetic moments for some atoms and ions.

Table 45-1Some Typical Magnetic Moment Values

Atom Magnetic Moment

HHe

Li

ONe

Na

C+++

Yb++

9.27 ×10−24 A ⋅m2

0

9.27 ×10−24 A ⋅m2

13.9 × 10−24 A ⋅m2

0

9.27 ×10−24 A ⋅m2

19.8 × 10−24 A ⋅m2

37.1×10−24 A ⋅m2

45-4

MAGNETIC MATERIALS IN AN EXTERNAL FIELD

For most materials, the magnetic moments of their constituent atoms are randomly oriented. However, if a material is placed in a region of space where there is a net external magnetic field

Bo , then the magnetic moments of the atoms in the material will respond to the external field and

give rise to a magnetic field of their own, signified by

BM . The field produced by the material is

proportional to the original external field such that

BM = χM

Bo . (45-11)

The constant of proportionality χM is called the magnetic susceptibility. The total magnetic

field then would be given by

B =Bo +

BM =

Bo + χM

Bo = 1+ χM( ) Bo

= κ M

Bo , (45-12)

where we have introduced another value κM called the relative permeability that is defined by

κM = 1 + χM =BBo

. (45-13)

There are three basic categories of magnetic material. The categories are based on how the magnetic moments of the atoms in a specific material respond to an applied external magnetic field

Bo .

PARAMAGNETISM

The atoms of a paramagnetic material have permanent magnetic dipole moments. When there is no external magnetic field, the atoms are randomly oriented and, therefore, make zero contribution to a magnetic field. When an external magnetic field is established, a fraction of the atoms become aligned with the external field and the material becomes magnetized. The aligned magnetic moments produce a magnetic field that enhances the existing external magnetic field.

Consider carefully Figure 45-2 below, where I have attempted to illustrate how such a state of affairs can come about. In Figure 45-2 (a), we have represented a cube of paramagnetic material when there is no external magnetic field. The small arrows represent atomic magnetic dipole moments and are intended to convey the fact that they are randomly oriented and, as such, do not give rise to a magnetic field.

45-5

Figure 45-2A Representation of the Behavior of Magnetic Dipole Moments

In a Paramagnetic Material

Bo

Bo

Bo

a( ) b( )

c( ) d( )

IS

BM

45-6

In Figure 45-2 (b) above, we represent the fact that a uniform, external magnetic field

Bo has

been established in the paramagnetic cube. This results a net alignment of magnetic dipole moments with the external field. (In reality, of course, this alignment is not perfect; only partial.)

Also, in Figure 45-2 (c), we represent the orientation of the “atomic currents” which produce the magnetic dipole moments. Note that at any point interior to the cube where the currents are in close proximity, the currents are in opposite direction. However, on the surfaces of the cube, there is in effect a net surface current, represented by IS in Figure 45-2 (d). This produces an additional

magnetic field

BM interior to the cube in the same direction as the external field

Bo . The total

magnetic field inside the cube is the vector sum of

Bo and

BM , as given in equation (45-12)

above. This result also suggests that we must revisit Ampère's law. In the presence of a paramagnetic material, we have

1κM∫

Bidr = µo I free . (45-14)

Listed in Table 45-2 below are some typical permeability values for some paramagnetic materials. All values are for room temperature unless otherwise noted.

Table 45-2 Some Typical Permeability Values for Some Paramagnetic Materials

Material κM

AirOxygenOxygen  −190°C , Liquid( )Manganese ChlorideNickel MonoxideManganesePlatinumAluminum

1.0003041.001331.003271.001341.0006751.0001241.00001381.00000817

45-7

FERROMAGNETISM

As the table above indicates, paramagnetic materials do not enhance the magnetic field by very much. Ferromagnetic materials, however, vastly increase the magnetic field. Also, once a ferromagnetic material is magnetized, it remains so. Ferromagnetic materials can be used to make permanent magnets.

What causes this increase? When ferromagnetic materials are placed in an external magnetic field, there is a very strong alignment of magnetic dipole moments. This favorable configuration is the result of a very complicated interaction that takes place between the electron spin states of adjacent atoms. Unfortunately, quantum mechanics is needed to describe this coupling and, therefore, its quantitative description is beyond the scope of this text; not to mention the expertise of this author.

It turns out that in a specimen of typical iron, the crystal structure is made up of a large number of small domains within which the magnetic moments are all parallel. However, on a large scale, the individual domains are randomly oriented For a very crude representation of this kind of structure, see Figure 45-3 below.

Figure 45-3Magnetic Domains

45-8

When the iron is placed in an external magnetic field, the domains change. The domains that were already oriented in the same direction as the external field “grow” in size compared to their less favorably oriented neighbors. Also, some initially unaligned domains will rotate to align themselves with the field. The net effect of these processes is that a significant fraction of magnetic dipole moments aligns with the external field. This can give rise to fields in the iron the magnitudes

Figure 45-4

The Magnetic Field

BM in Annealed Iron as a Function of

Bo

Bo

2 4 6 8 10

10−4 T( )

1.4

1.2

1.0

0.8

0.6

0.4

0.2

T( )

BM

45-9

of which are over a thousand times the magnitude of the external field. The graph in Figure 25-4 above illustrates this extraordinary increase for annealed iron that had never been magnetized.

Hysteresis

One of the extraordinary qualities of ferromagnetic materials is that they do not lose all of

their magnetism once the external magnetic field

Bo is removed--provided the material is below the

so-called Curie temperature. Exposure to an external magnetic field transforms a ferromagnetic material into a permanent magnet. It is as if the material retains some “memory” of the initializing field. The dependence of a physical system on its “history” is called hysteresis.

It turns out that for ferromagnetic materials, magnetization rearranged the domains into favorable energetic states. After such, some domains do not return to their original configurations and a permanent alignment of magnetic dipole moments is established. Recall that this, in turn, gives rise to an effective surface current which can, for very strong bar magnets, amount to several hundred amperes per centimeter.

As alluded to above, the amount of magnetization retained by a ferromagnetic material is dependent on the temperature of the material. The higher the temperature the less magnetism retained. At temperatures above the Curie temperature, the material will not retain any magnetization. For iron, the Curie temperature is 1,043°C .

DIAMAGNETISM

To recapitulate, when paramagnetic and ferromagnetic materials are placed in an external magnetic field, the intrinsic magnetic dipole moments of the electrons are aligned, more or less, with the external magnetic field. Diamagnetism, however, arises from induced magnetic dipoles.

Imagine that an electron is orbiting a nucleus on a circular orbit between the poles of a large electromagnet that is initially switched off, as represented in Figure 45-5 below. As we discussed above, the moving electron constitutes a current loop. The current gives rise to a magnetic field

signified as

Be− in the diagram below. A magnetic dipole moment

µ is also produced; what we

might call the intrinsic magnetic dipole moment.

When the electromagnet is turned on, a magnetic field

BE is established. This field is

directed downward. However, now there is also a radially directed magnetic force exerted on the electron. For a circular orbit, we can write

eE + evBE =M

e−v2

R . (45-15)

45-10

In terms of the angular speed ω , we can rewrite equation (45-15) as

eE + eRωBE = Me− Rω 2 . (45-16)

Before the electromagnet was turned on, only the electrical force would have acted on the electron and we could write, in terms of the the initial angular speed ω o ,

eE = Me− Rω o2 . (45-17)

Figure 45-5Strong Magnetic Field Linking an Electron Orbit

v

FE

Be−

BE

FM

N

S

BE

BE

−

+R

45-11

Substitution of equation (45-17) into (45-16) gives us

Me− Rω o2 + eRωBE = Me− Rω 2 . (45-18)

Rearranging the terms in equation (45-18) we have

eωBE = Me− ω 2 −ω o2( ) . (45-19)

The change in angular speed, recall, is given byΔω = ω −ω o . (45-20)

As long as BE is not too large, then then

Δω << ω o , (45-21)

and we can write

ω 2 −ω o2 = ω o + Δω( )2 −ωo

2 = ωo2 + 2ω o Δω( ) + Δω( )2 −ωo

2

= 2ω o Δω( ) + Δω( )2 ≈ 2ω o Δω( ) . (45-22)

So, for a not too excessive BE , equation (45-19) becomes

eBE = 2Me− Δω( ) , (45-23)

where we have used the fact that ω ≈ωo . Therefore, the change in the angular speed, or angular

frequency of the electron is approximately given by

Δω =eBE

2Me−

. (45-24)

This change Δω is called the Larmor frequency, and tells how much faster the electron will be

moving as a result of an external magnetic field BE .

A change in the orbital speed of the electron will also produce a corresponding change in the magnetic moment. Using equation (45-4) we can write

µo =12

evR =12

eR2ω o , (45-25)

and, therefore,

Δµ =12

eR2 Δω( ) . (45-26)

The fractional change can be written as

45-12

Δµµ o

=eR 2 Δω( ) / 2

eR2ω o / 2=Δωω o

. (45-27)

Example One:

Assume we have an electron orbiting a solitary proton on a circular path the radius of which is the Bohr radius. Let us calculate the angular speed of the electron. We can write using Newton’s second law of motion

ke2

rBohr2 = Me− rBohrω

2 , (1)

and, therefore,

ω =ke2

Me−

rBohr3 =

8.99 × 109 N ⋅m2 / C2( ) 1.602 ×10−19 C( )2

9.11 ×10−31 kg( ) 5.29 ×10−11 m( )3

= 4.14 ×1016  rad / s . (2)Now, let us calculate what would be the change in the angular frequency of the electron if a

strong external magnetic field of one Tesla were established in the space of the orbiting electron. Using equation (45-24) we can write

Δω =eBE

2Me−

=1.602 × 10−19 C( ) 1 T( )

2 9.11×10−31 kg( ) = 8.79 ×1010  rad / s . (3)

This represents a fractional change of

Δωωo

=8.79 × 1010 s−1

4.14 ×1016 s−1 ≈ 2 × 10−6 . (4)

So, impressing a very strong magnetic field on the atom does not change the angular speed of the electron much. This indicates how small the diamagnetic effect is.

Paramagnetic and Ferromagnetic materials also have diamagnetic responses but diamagnetic response is smaller than either the paramagnetic or ferromagnetic effects.

In Table 45-3 below we have listed some typical permeability values for some diamagnetic

materials. Unless otherwise indicated, the materials are at room temperature 20°C( ) and subject

45-13

to a pressure of 1 atm .

Table 45-3 Some Typical Permeability Values for Some Diamagnetic Materials

Material κM

BismuthBerylliumMethane

EthyleneAmmoniaCarbon Dioxide

Glass heavy flint( )

0.9999810.9999890.9999690.9999800.9999860.99999470.999985

45-14

Summary of Chapter 45

If we model electron orbits in atoms as circular, and if we use a classical approach, we find that the moving electron establishes an “atomic current” given by

I = ev / 2πR , (45-28)

where v is the instantaneous speed of the electron and R is the radius of its orbit. The magnetic field produced by this current loop is given by

B =′k evR

z2 + R2( ) , (45-29)

where z is the distance from the center of the loop on an axis perpendicular to the plane of the loop. When we are a large distance from the loop, where z >> R , the magnetic field strength of the atomic current loop is given by

B z >> R( ) =′k evRz3 . (45-30)

At such a distance, the magnetic field looks like it is produced by a small dipole magnet.This current loop has a magnetic dipole moment given by

µorb =12

evR . (45-31)

In terms of the angular momentum, the orbital magnetic moment is given by

µorb = −

12

eM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Lorb , (45-32)

and the magnitude of the angular momentum is given by quantum mechanics as

Lorb = ml  , for ml = 0, ±1, ±2 , limiting value . (45-33)

The electron behaves as if it “spins” about its axis. This spin state gives rise to another magnetic moment given by

µspin =12

eM

e−

⎡

⎣⎢⎢

⎤

⎦⎥⎥ = 9.274 × 10−24 J / T. (45-34)

The value of the electron’s spin magnetic moment is called the Bohr magneton. Even though protons and neutrons also have spin states, the magnetic moment of atoms is dominated by the

45-15

electrons.

When a magnetic material is placed in an external magnetic field

Bo , another magnetic field

is produced in the material due to the response of the magnetic moments of its atomic structure.

This magnetic field

BM is proportional to the external magnetic field

Bo and given by

BM = χM

Bo , (45-35)

where χM is the constant of proportionality and called the magnetic susceptibility. The total

magnetic field in the material becomes

B =Bo +

BM = 1 + χM( ) Bo =κ M

Bo , (45-36)

where we have introduced a new constant κM called the relative permeability. This result requires a

modification of Ampère's law in magnetic materials. We have

1κM

Bidr∫ = µ oI free . (45-37)

Magnetic materials are divided then into three categories:Paramagnetic: κM > 1 ; Ferromagnetic: κM >> 1 ; Diamagnetic: κM < 1

Paramagnetic and ferromagnetic materials have responses dominated by the relative alignment of magnetic moments to an impressed external magnetic field. In paramagnetic materials, this alignment is not very strong. In ferromagnetic materials, however, there is a very large alignment of magnetic moments produces an increase in magnetic field strength that can be as much as three orders of magnitude larger than the external field. Ferromagnetic materials subjected external magnetic fields at temperatures less than the Curie temperature--1,043°C for iron--will become a permanent magnet.

The diamagnetic response is present in all materials and is the result of induced magnetic moments when the material is subjected to an external magnetic field. (In paramagnetic and ferromagnetic materials, the alignment of intrinsic magnetic dipole moments is of greater consequence than the small influence of induced magnetic moments.)

For magnetic fields that are not too large, the change in the angular frequency of electrons in orbits influenced by external magnetic fields is given by

Δω = eBE / 2Me− , (45-38)

where BE is the magnitude of the external magnetic field. This change in angular frequency is

called the Larmor frequency and gives us information on how large the diamagnetic response will be.

Chapter 46

ALTERNATING CURRENT CIRCUITS

As you know, the electrical power that is provided to your home is called AC. This is shorthand for alternating current. The typical AC outlet has a potential difference of ΔV = 120 Volts between the plates. This system is said to be alternating because the polarity of these plates alternates between positive and negative sixty times every second. The frequency of the electrical system in the United States of America is 60 Hz .

All of the wonderful electrical devices that we connect to this system must be designed to deal with alternating currents. In this chapter, we want to look briefly at some of the more important features of alternating current circuits. We want to begin by looking at how resistors, capacitors, and inductors behave when in an AC circuit.

A RESISTOR IN AN AC CIRCUIT

In chapter forty-four, we briefly described how a rectangular loop rotating is a region of space where there is a uniform magnetic field can generate an alternating EMF. Without attempting to describe the wonderfully complex electrical engineering involved in designing modern alternators, we are going to assume we have an electrical power supply capable of delivering an alternating potential difference to a circuit. We will signify such a device with the symbol shown below in Figure 46-1.

In general, we can describe such a potential difference mathematically using sine or cosine functions. The instantaneous voltage Vab across an AC power supply can be written as

Vab = Vm cos ωt( ) , (46-1)

where Vm is the maximum voltage or the voltage amplitude. The angular frequency of the AC

Figure 46-1The Symbol for an AC Voltage Source

46-2

source is signified by ω , it is not the angular speed, and related to the frequency f by

ω = 2π f . (46-2)

Figure 46-2 below, is a schematic diagram of a resistor of resistance R connected to an AC source. Using Ohm’s law, we can write

I =Vab

R=

Vm

Rcos ω t( ) . (46-3)

The maximum current, or the current amplitude, is given by

Im =Vm

R . (46-4)

For purposes of illustration, we assume that V = 5R = 10 , and, therefore, I = 5 . In Figure

46-3 (a) below, is a graph of both the instantaneous voltage V across the resistor and the

instantaneous current I through the resistor. As both quantities are directly proportional to the

cos ω t( ) , then both quantities have maximum and minimum values at the same time. The current

and the voltage are said to be in phase. In diagram (b) of Figure 46-3 below, we represent the voltage and the current as vectors in a phasor diagram; phase vector or phasor. The instantaneous values of V and I correspond to the horizontal components of the maximum values of these quantities, respectively. The quantity ω t is, of course, the angle the vectors make to the horizontal reference line.

Figure 46-2An Alternating Source and a Resistor

∼

a bR

46-3

Figure 46-3A Resistor in an AC Circuit

A Graph of the Current and Voltage as a Function of TimeAnd a Phasor Diagram

2

1

0

-1

- 2

I

76543210-1- 2t

I , V

V

I

a( )

VmIm

Vm

Im

b( )

VI

ω t

t

46-4

An interesting complication should be mentioned. Earlier, we saw that in a steady-state direct current the current density is constant over the cross-sectional area of the wire. This is not the case with an alternating current! There is a greater density near the surface of the conductor. This phenomenon is called the skin effect. The effective resistance of the wire is increased with AC current. The skin effect is caused by induced EMF’s. Even though the resistance is a function of the frequency, unless the frequency is very large, on the order of millions of Hz , this effect can be neglected.

The instantaneous power dissipated in the resistor is given by

P = I 2 R =Vm

Rcos ω t( )⎡

⎣⎢⎤⎦⎥

2

R =Vm

2

Rcos 2 ω t( ) . (46-5)

To find an average value, we use

Pave =12

Pf + Po⎡⎣ ⎤⎦ =12

Vm2

Rcos 2 2π( )⎛

⎝⎜⎞⎠⎟

+Vm

2

Rcos2 0°( )⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ =

Vm2

2R . (46-6)

Many values involving sine and cosine functions have averages of zero over one complete cycle--the voltage in this example is one such case. Zero is not very useful. So, the so-called root mean square value is often used. In this example, the maximum voltage value is Vm and its rms

value would be given by

Vrms =Vm

2≈ 0.7071( )Vm . (46-7)

Note that

Vrms2 =

Vm2

2 . (46-8)

So, we could also write the average power as

Pave =Vrms

2

R . (46-9)

A CAPACITOR IN AN AC CIRCUIT

Figure 46-4 below is a schematic diagram of a capacitor of capacitance C in an AC circuit. At some time t after the source is turned on, an initially uncharged capacitor would have an instantaneous charge q . This instantaneous charge is related to the capacitance and the potential

46-5

difference across the capacitor by

q = CVab = CVm cos ω t( ) . (46-10)

The instantaneous current is given by

I =dqdt

= CVmddt

cos ω t( )⎡⎣ ⎤⎦ . (46-11)

Since this requires calculus, I am going to simply state the correct result. (In the appendices, I have derived a few important calculus results using algebra, geometry, trigonometry and then applying the limit notion.)

The time rate of change of the cosine function in equation (46-11) is found to beddt

cos ω t( )⎡⎣ ⎤⎦ = −ω sin ωt( ) , (46-12)

This means the instantaneous current is

I = −  ωCVm sin ω t( ) . (46-13)

Using the following trig identity,

cos x + 90°( ) = − sin x , (46-14)

we can write the instantaneous current as

I = ωCVm cos ω t + 90°( ) . (46-15)

A graph of the voltage and current as a function of time is shown below in Figure 46-5 below. Note that the voltage lags the current by 90° .

Figure 46-4A Capacitor in an AC Circuit

∼

a bC

46-6

Figure 46-5A Capacitor in an AC Circuit

A Graph of the Current and Voltage as a Function of TimeAnd a Phase Diagram

3

2

1

0

-1

- 2

- 3

q

76543210-1- 2t

V ,I

I

V

t

Vm

Vm

Im

Vi

I

ω t

Im

46-7

So, for our “simple circuit” with a capacitor, we have found that the charge on the capacitor

is given by: q = CVm cos ω t( ) , (46-16)

with a maximum charge on the plates ofQ = CVm . (46-17)

The current in the circuit is given by

I = ωCVm cos ω t + 90°( ) , (46-18)

with the maximum current being: Im = ωCVm . (46-19)

For the resistor, Ohm’s law gives us

Im =Vm

R. (46-20)

For our capacitor, we have: Im =Vm

1 /ωC . (46-21)

This prompts us to introduce a new quantity called the capacitive reactance signified by XC and

defined by

XC =1ωC

. (46-22)

Then we have an “Ohmic” form for the capacitor, and

Im =Vm

Xc

. (46-23)

The instantaneous power delivered to the circuit is given by

P = IV = ωCVm cos ω t + 90°( )⎡⎣ ⎤⎦ Vm cos ωt( )⎡⎣ ⎤⎦

= ωCVm2 cos ω t + 90°( ) cos ω t( ) . (46-24)

The average value, however, is zero.

Example One:Assume we have a capacitor of capacitance C = 1 µF that is connected to an AC current

that with a frequency f = 60 s−1 ≡ 60 Hz . We want to find the reactance of this capacitor.

We have

46-8

XC =1

ωC= 2π 60 s−1( ) 1× 10−6 F( )⎡⎣ ⎤⎦

−1= 2.65 × 103  Ω . (1)

AN INDUCTOR IN AN AC CIRCUIT

Figure 46-6 below is a schematic diagram of an inductor of negligible resistance and with self-inductance L connected to an AC source. Earlier, in our discussion of magnetic energy, we found that the current in an inductor is related to the potential difference by

Vab = Vm cos ωt( ) = LdIdt

. (46-25)

We can rearrange equation (46-25) and get the following differential equation:

dI =Vm

Lcos ω t( ) dt . (46-26)

To solve this equation, we have to use integral calculus. The integral solution of equation (46-26) is given by

I = dI∫ =Vm

Lcos ω t( ) dt∫ =

Vm

ω Lsin ω t( ) . (46-27)

We now use the trig identity

sinx = cos x − 90°( ) (46-28)

to write

I =  Vm

ω Lcos ωt − 90°( ) . (46-29)

Figure 46-6An AC Source and an Inductor

∼

a b

L

46-9

The maximum current is given byIm = Vm / ωL , (46-30)

from which we can writeVm = ImωL . (46-31)

As we did with the capacitor, we introduce a new quantity called the inductive reactance signified by XL and defined by

XL = ωL . (46-32)

Then, again, we have an “Ohmic” formV = IXL . (46-33)

The instantaneous power is given by

P = IV =Vm

ω Lsin ω t( )⎡

⎣⎢⎤⎦⎥

Vm cos ω t( )⎡⎣ ⎤⎦

=Vm

2

ωLsin ωt( ) cos ω t( ) . (46-34)

As with the capacitor, the average power is zero.Below, in Figure 46-7, is a graph of the current and voltage as a function of time. There is

also a phasor diagram, note how the current lags the voltage.

Example Two:An inductor of self-inductance L = 1 H connected to an AC source at a frequency

f = 60 s−1 . We want to find the reactance of this inductor. We can write

XL = ωL = 2π 60 s−1( ) 1 H( ) = 377 Ω . (1)

46-10

Figure 46-7An Inductor in an AC Circuit

A Graph of the Current and Voltage as a Function of TimeAnd a Phasor Diagram

2

1

0

-1

- 2

q

76543210-1- 2

t

VmIm

Vm

VIω t

I

V

t

Im

V , I

ω t

46-11

THE R-L-C SERIES CIRCUIT

Many useful circuits incorporate resistors, capacitors and inductors together. We want to investigate one simple example of such a circuit. Figure 46-8, on the next page, is a schematic diagram of a resistor, capacitor and inductor connected in series in an AC circuit. For simplicity, we will also assume that the frequency of the AC current is not too high; this will ensure that the current is the same everywhere in the circuit.

The instantaneous potential difference across the AC source must equal the sum of the instantaneous potential differences across each circuit element. We can write

Vad,m = Vab,m + Vbc, m + Vcd ,m = VR ,m + VL,m + VC ,m . (46-35)

Earlier, we found out that the voltage across the resistor is in phase with the current. We have represented this on the next page in the phasor diagram in Figure 46-9. We know that the maximum voltage will be given by

VR ,m = ImR . (46-36)

We have also represented the voltage across the inductor which we found out leads the current by ninety degrees. The maximum voltage across the inductor will be

VL ,m = ImX L . (46-37)

And, finally, we represent the voltage across the capacitor which we found out that lags the current by ninety degrees. The maximum voltage value across the capacitor will be

VL ,m = ImX L . (46-38)

A final comment about the phasor diagram. The angle φ is called the phase angle. Since the vectorial sum of the potentials across the circuit elements must equal that of the AC

source, inspection of the phasor diagram, along with the Pythagorean theorem, should convince one that

Vad,m = VR, m2 + VL ,m −VC,m( )2

= ImR( )2 + Im XL − ImXC( )2

= Im R2 + XL − XC( )2 = Im R2 + ω L − 1 /ωC( )⎡⎣ ⎤⎦2

. (46-39)

The quantity XL − XC is called the reactance of the circuit and will be signified by X .

Now, we introduce a new quantity Z called the impedance of the circuit that is defined by

Z = R2 + X 2 = R2 + ωL − 1 /ωC( )⎡⎣ ⎤⎦2

, (46-40)

46-12

and we can write, in Ohmic form,Vad,m = ImZ . (46-41)

VL ,m

Im

VR ,m

∼

a bLR

cC

d

VC ,m

Figure 46 - 8An R - L - C Circuit

Figure 46 - 9A Phasor  Diagram

ω tφVL ,m −VC ,m

Vad,m

46-13

There is a special angular frequency value, ω r , at which the current in our R-L-C series

circuit is a maximum. When this condition exists, the circuit is said to be in series resonance. The maximum current implies that the impedance is a minimum. The impedance is a minimum when the reactance is zero. So, for series resonance, we require

ω r L =1

ω rC , (46-42)

and therefore,

ω r = 1 / LC . (46-43)

Example Three:

We want to find the following values for the R-L-C circuit shown below:a) The angular frequency at which there is series resonance.b) The inductive and capacitive reactance at resonance.c) The rms current at resonance.d) The rms potential difference across each circuit element at resonance.

a) At resonance, the angular frequency is given by

ω r = 1 / LC = 1 / 3.0 H( ) 1.0  µF( ) = 577  rad / s . (1)

b) XL = ωr L = 577 rad / s( ) 3 H( ) = 1,731 Ω , (2)

and XC = 1 /ω rC = 1 / 577 rad / s( ) 1.0 µF( )⎡⎣ ⎤⎦ = 1,733 Ω . (3)

L = 3 HR = 2000 ΩC = 1.0  µF

Vrms = 200 Volts

∼

a b c d

46-14

c) Irms = Vrms / R = 200 V( ) / 2000 Ω( ) = 0.10 A . (4)

d) VR ,rms = I rmsR = 0.10 A( ) 2000 Ω( ) = 200 V , (5)

VL ,rms = Irms X L = I rms ω r L( ) = 0.10  A( ) 577 rad / s( ) 3 H( )⎡⎣ ⎤⎦ = 173 V , (6)

VC ,rms = Irms XC = I rms1

ωr C⎛⎝⎜

⎞⎠⎟

= 0.10  A( ) 1577 rad / s( ) 1µF( )

⎡

⎣⎢

⎤

⎦⎥ = 173 V . (7)

THE R-L-C IN PARALLEL CIRCUIT

On the next page, in Figure 46-10, is a schematic diagram of a resistor, inductor, and capacitor in parallel with an AC power supply. In a parallel configuration, we know the potential difference across each element is the same.

For the resistor, we know the current and the voltage are in phase. We represent this on the next page in Figure 46-11. We know the maximum current through the resistor is related to the voltage across the resistor by

Vm = VR ,m = IR ,m R . (46-44)

For the inductor, we know the current lags the voltage by ninety degrees. We represent this in the phasor diagram below. The maximum current in the inductor is related to the voltage across the inductor by

Vm = VL,m = IL ,m XL . (46-45)

For the capacitor, we know the current leads the voltage by ninety degrees. This also is represented below in the phase diagram. The maximum current through the capacitor is related to the voltage across the capacitor by

Vm = VC ,m = IC ,m XC . (46-46)

Analysis of the phase diagram leads us to conclude that

Im = IR ,m2 + IC ,m − IL,m( )2

= Vm / R( )2 + VmωC( ) − Vm /ω L( )⎡⎣ ⎤⎦2

=Vm

2

R2 + Vm2 ωC −

1ω L

⎡⎣⎢

⎤⎦⎥

2

= Vm1

R2 + ωC −1

ω L⎡⎣⎢

⎤⎦⎥

2

. (46-47)

So, we can write: Vm = Im ′Z , (46-48)

46-15

where ′Z is the impedance for the parallel case and given by

′Z =1

1R2 + ωC − 1

ωL⎡⎣⎢

⎤⎦⎥

2 . (46-49)

IC ,m

IR, m

Vm

IL,m

Figure 46 - 11A Phase Diagram

ω tφIC ,m − IL ,m

LR C ∼

Figure 46 - 10An R - L - C Parallel Circuit

Im

46-16

The current will be minimized when the impedance is maximized. Although the proof involves calculus, it can be shown that the angular frequency that minimizes the current is

′ω =1LC

. (46-50)

THE TRANSFORMER

The long distance transmission of AC electrical energy is more efficient at high voltage and low current. To match the actual voltage needs at the point where the energy is to be used, one needs an electrical device that “transforms” the high voltage to a lower value. This is accomplished with a step-down transformer.

Figure 46-12 below is a schematic diagram of an ideal transformer with the secondary winding open. Basically, a transformer consists of two sets of windings that are electrically insulated from each other and wound on the same soft-iron core. An AC source sets up an alternating flux in the core and induces an EMF in the second winding.

The windings attached to the AC source is called the primary coil and the second set of windings is called the secondary. The schematic circuit symbol is shown below.

∼ N1V1N2 V2

Figure 46 - 12Schematic of a Transformer with an Open Secondary

46-17

In real world transformers, the output energy must be smaller than the energy input due to inescapable energy losses. However, in our ideal transformer, the ratio of the potential differences is the same as the ratio of the number of turns. So, we write

V2

V1

=N 2

N1

, (46-51)

from which we can write

V2 =N2

N1

V1 . (46-52)

If V2 > V1 , we have a step-up transformer, while if V1 > V2 , we have a step-down transformer.

46-18

Summary of Chapter 46

AC sources produce alternating EMF’s and can be quantified by using sine or cosine functions. For example, we could write the instantaneous voltage across an AC source as

v = V cos ωt( ) , (46-53)

where ω is the so-called angular frequency, it is not the angular speed. and it is related to the frequency f by

ω = 2π f , (46-54)and to the period τ by

ω = 2π / τ . (46-55)

For a resistor of resistance R connected to an AC source, the circuit current would be given by

I =Vm

Rcos ωt( ) , (46-56)

and the current would be in phase with the applied voltage. The average power dissipated in the resistor is given by

Pave = Vm2 / 2R , (46-57)

or, in terms of the root-mean-square voltage as

Pave = Vrms2 / R , (46-58)

where the rms voltage is defined by

Vrms = Vm / 2 . (46-59)

For a capacitor of capacitance C connected to an AC source, the current in the circuit is given by

I = ωCVm cos ω t + 90°( ) , (46-60)

where 90° is called the phase angle. The current leads the applied voltage by ninety degrees. In

order to write the current-voltage relationship in Ohmic form, we introduce a quantity XC called the

capacitive reactance that is defined byXC = 1 /ωC . (46-61)

So, we can writeV = I /ωC = IXC . (46-62)

46-19

The average power delivered to the circuit is zero.For an inductor of negligible resistance and self-inductance L connected to an AC source,

the current in the circuit is given by

I =Vm

ωLcos ω t − 90°( ) , (46-63)

where −90° is the phase angle. The current lags the voltage by ninety degrees. In order to write

the current-voltage relationship in Ohmic form, we introduce a quantity XL called the inductive

reactance that is defined byXL = ωL . (46-64)

So, we can writeVm = ImωL = Im XL . (46-65)

The average power delivered to the circuit is zero.If we now connect the resistor, capacitor and inductor is series with AC source, the voltage of

the AC source will be out of phase with the voltage across the resistor by an angle φ , where

φ = tan−1 VL ,m −VC,m

VR ,m

⎡

⎣⎢

⎤

⎦⎥ = tan−1 Im XL − Im XC

ImR⎡

⎣⎢

⎤

⎦⎥ = tan−1 X

R⎡⎣⎢

⎤⎦⎥

, (46-66)

where X is the reactance and is defined by

X = XL − XC . (46-67)

In order to write the current-voltage relationship in Ohmic form, we use Z called the “series” impedance and defined by

Z = R2 + ωL −1

ωC⎡⎣⎢

⎤⎦⎥

2

. (46-68)

So, for the R-L-C series circuit, the voltage is given byVm = ImZ . (46-69)

There is a specific angular frequency ω r at which the current will be a maximum and the circuit

will be in resonance. That value is given by

ω r = 1 / LC . (46-70)

The average power input to the circuit is given by

46-20

Pave = Vrms Irms cosφ , (46-71)

where cosφ is called the power factor.If the resistor, inductor and capacitor are connected in parallel with the AC source, then the

source current I will be our of phase by an angle φ with the source voltage. The phase angle is given by

φ = tan−1 IC − IL

IR

⎡

⎣⎢

⎤

⎦⎥ = tan−1 V / XC( ) − V / XC( )

V / R( )⎡

⎣⎢

⎤

⎦⎥ = tan−1 R ωC −

1ω L

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

. (46-72)

In an ideal transformer, the voltage across the secondary winding is related to the applied AC voltage across the primary by

V2 =N2

N1

V1 , (46-73)

where N1 is the number of turns in the primary coil and N2 is the number of turns in the

secondary. We have a step-up transformer if V2 > V1 , and a step-down transformer if V1 > V2 .

PART FOUR

Maxwell’s Equations, Electromagnetic Waves and Optics

Chapter 47

LIGHT AS AN ELECTROMAGNETIC WAVE

An electric charge at rest gives rise to an electric field. An electric charge in motion also gives rise to a magnetic field.

An accelerating electric charge also gives rise to electromagnetic radiation.

Maxwell’s equations, coupled with the Lorentz force, provide the theoretical basis for classical electromagnetism. It was James Clerk Maxwell who first understood the physical nature of light. That moment of supreme synthesis represents maybe the most significant moment in the history of the human quest to understand the nature of things.

From the perspective of the science of physics, it is difficult to overestimate how important light and the understanding of light is to our comprehension of the physical world. On the practical side of things, modern man is still being swept along by the tidal force of the electrical revolution; a revolution that dwarfs the mechanical revolution initiated by Isaac Newton.

It is, of course, one of the tragic ironies of history that Maxwell himself did not live to see his greatest intellectual insight verified by the experimental work of Heinrich Hertz. Maxwell died in 1879 at the age of forty-eight. Hertz’s famous experiment confirming the existence of electromagnetic waves was performed in 1888.

MAXWELL’S EQUATIONS

In our brief discussion of Maxwell’s equations, we write them in integral form and assume all electric and magnetic fields are propagated in vacuum. The first of the so-called equations of Maxwell is Gauss’ law for the electric field. We have

ΦE =EidA∫ =

1εo

ρq  dV∫ =Qenclosed

εo

. (47-1)

The integral on the left hand side of equation (47-1) is a surface integral. It tells us to add up all of the electric field lines popping through an arbitrarily shaped closed surface. (Yes, this is our old friend the electric flux.) That closed surface forms the boundary of the volume signified by the Vin the integral of the middle term of equation (47-1). The average charge per unit volume in this

47-2

enclosed space is signified by ρq . So, the integral is telling us to add up all of the electric charge

in this volume. When you have finished doing that, you will get the total amount of charge in the volume, the Qenclosed .

Recall that from a practical point of view, we used Gauss’ law not so much because we wanted to know the electric flux but as another tool for finding the electric field in a region of space. If we can find a symmetrical surface over which the electric field is uniform, then this equation reduces to

EA =Qenclosed

εo

, (47-2)

and we can find the electric field. From a theoretical point of view, Gauss’ law establishes the relationship between electric charge and the electric field.

The second of Maxwell’s equations is Gauss’ law for the magnetic field. We write

ΦM =

BidA∫ = 0 . (47-3)

This equation highlights the difference between the magnetic field and the electric field. The electric field is produced by electric charges that come in two forms; positive and negative. There are no such “magnetic charges” giving rise to the magnetic field. There is, however, something about equation (47-4) that theoretical physicists do not like. They do not like the fact that the right hand part of the equation does not have the same mathematical form as the right hand side of equation (47-1). They have this aesthetic sensibility that sees formal similarity on the left hand sides of the equations and desires the same on the right hand sides.

The modern cosmological model we lovingly call the Big Bang postulates that there was a time when there was an “abundance” of magnetic monopoles in the universe but a symmetry break put an end to that idyllic epoch. By some calculations, however, cosmologists speculate that there may be as many as five of these rascals left over. Finding them, however, is not a bit unlike the proverbial hunt for a needle in a haystack.

The third of Maxwell’s equations is Faraday’s law. We write Faraday’s law as

Eidr∫ = −

ddtBidA∫ . (47-4)

On the left hand side of equation (47-4) we have what is called a closed path integral. It asks us to calculate the dot product of the electric field and an infinitesimal path part at every path part all the way around a closed path. The dot product gives us that part of the electric field that is parallel to the path. The closed path constitutes the boundary of the area represented in the flux integral on the right hand side of equation (47-4). The most important point for our discussion is that a changing magnetic field gives rise to an electric field.

47-3

For physicists like Faraday and Maxwell, formal symmetry convinced them that a changing electric field should give rise to a changing magnetic field. It was, of course, Maxwell who demonstrated how that could happen. This notion provided the missing piece that allowed Maxwell to come up with a completely unified theory of electromagnetic waves. According to Maxwell’s notion, the mutual induction of electric and magnetic fields produces self-sustaining electromagnetic waves that can propagate in empty space. Visible light is an electromagnetic wave phenomenon. Visible light, it turns out, is only a small segment of a vast range of electromagnetic radiation.

Assume that we have an infinitely long straight wire carrying a constant current I , as

represented below in Figure 47-1. Imagine that we center a circle of radius r⊥ on the wire oriented

such that the plane of the circle is perpendicular to the wire. The current in the wire will produce a magnetic field at point P given by

B =

2 ′k Ir⊥

ϕ , (47-5)

where, recall, ϕ is a unit vector signifying that the direction of the magnetic field is tangent to the

circle. This result follows directly from integrating the definition of the magnetic field produced by a moving charge.

We also could approach this situation using Ampère's law. Recall that Ampère's law is given by

Bidr∫ = µoI enclosed . (47-6)

Figure 47-1

I

r⊥

B

P

47-4

We choose the circle itself as our Amperian path. We know that magnetic field encircle wires, so the magnitude of the magnetic field would be constant at every point on the wire. They are also directed tangentially. So, we can write

B = B  ϕ . (47-7)

The “infinitesimal bitty” path is a part of a circle and can be written as

dr = r⊥dϕ  ϕ . (47-8)

When we take the dot product we get

Bidr = B ϕ[ ]i r⊥dϕ  ϕ[ ] = Br⊥dϕ . (47-9)

Substituting this into equation (47-6) gives us

Br⊥dϕ∫ = Br⊥ dϕ∫ = B2π r⊥ = µo I , (47-10)

from which we get

B =µo I

2πr⊥=

2 ′k Ir⊥

. (47-11)

Figure 47-2

I

a( )

b( )

E

47-5

When we talk about the geometric object we call a circle, we know that it is is a planar figure

with a well know area of πR 2 . However, think for a moment of a strange bag like item, crudely represented above in Figure 47-2 (a). The opening is indeed a circle but the bounded area is certainly not planar.

You may be wondering what all of this babbling has to do with our discussion. Let me show you. Imagine, as Maxwell did, that we place this bag around one of the plates of a parallel plate capacitor, as represented above in Figure 47-2 (b). We assume the capacitor is connected to a variable power supply so that we can maintain a constant current I . As charge builds on the plates, the electric field increases and the is a change in the electric flux through the bag. However, Ampère's law can not account for the magnetic field that would be generated as there is no current “popping” through the bag.

Maxwell postulated that there is a quantity called the displacement current that can be used to modify Ampère's law. The displacement current is defined by

Id = εodΦE

dt , (47-12)

and Maxwell’s modification of Ampère's law--the last of the so-called Maxwell equations--is given by

Bidr∫ = µo I + Id[ ] = µo

JidA∫ + µ o

ddtEidA∫ . (47-13)

Along with its complete generality, this equation also has the advantage of indicating the connection between a changing electric field and the magnetic field to which it gives rise. Maxwell’s modification of Ampère's law allowed for a unified understanding of the connection between electric and magnetic fields and a coherent theory of electromagnetism. For this reason, the equations are called Maxwell’s equations.

Maxwell’s Equations

EidA∫ =

1εo

ρq  dV∫

BidA∫ = 0

Eidr∫ = −

ddtBidA∫

47-6

Bidr∫ = µo

JidA∫ + µ oεo

ddtEidA∫

The Lorentz Force

F = q

E + v ×

B( )⎡⎣ ⎤⎦

ELECTROMAGNETIC WAVES

We now know that changing magnetic fields can produce an electric field. Conversely, a changing electric field can generate a magnetic field. Electromagnetic waves consist of time-varying electric and magnetic fields that propagate through space. Some of the more common examples of electromagnetic waves are radio and television signals, X-rays, visible light, and micro waves. Now, it is time to take a more quantitative look at electromagnetic waves.

We begin with a simple example called a plane wave pulse, represented below in Figure 47-3 The electric field vectors are parallel to the y-axis, while the magnetic field vectors are parallel to the z-axis. So, the wave front lies in the y-z plane and is moving parallel to the x-axis with an unknown speed c. (We ignore the question as to how such a pulse is produced.)

Figure 47-3Propagation of a Plane Wave

wave  front

E

B

E

B x

y

z

tL

E

B

E

B

tE

a

b

47-7

We use Faraday’s law with respect to the rectangle that lies in the x-y plane, shown above in Figure 47-3. Recall that Faraday’s law states

Eidr∫ = −

ddtBidA∫ = −

dΦM

dt . (47-14)

For the magnetic flux, we have

ΦM =

BiA = B  k⎡⎣ ⎤⎦i −a k⎡⎣ ⎤⎦ = −Ba , (47-15)

while the change in magnetic flux is given by

ΔΦM = −Ba Δ . (47-16)

We have, then,

−  

ΔΦM

Δt= Ba  

ΔΔt

= Bac . (47-17)

We now want to look at the left hand side of equation (47-14). (We are going around the loop in a clockwise sense.) We have

Eidr∫

= E   j⎡⎣ ⎤⎦i dy  j⎡⎣ ⎤⎦0

a

∫ + E  j⎡⎣ ⎤⎦i dx i⎡⎣ ⎤⎦0

a

∫ + 0  j⎡⎣ ⎤⎦i dy  j⎡⎣ ⎤⎦a

0

∫ + E   j⎡⎣ ⎤⎦i dx  i⎡⎣ ⎤⎦

0

∫= Ea + 0 + 0 + 0 . (47-18)

The first integral of the sum of integrals deals with the left side of the rectangle. On the left hand side, the magnitude of the electric field is E and the length of this part of the path is a. Since the direction of the field and the path are the same, the dot product is the product of the magnitudes of the two vectors. The second and fourth integrals are zero because every point on the path is perpendicular to the electric field and, as such, the dot product vanishes. The third integral is zero because there is no electric field in this region of space as the plane wave has not yet reached that region.

We can now write

Eidr∫ = Ea = −

dΦM

dt= Bac , (47-19)

and, therefore, E = cB . (47-20)

Next, we use the modified form of Ampère's law to analyze the rectangle in the x-z plane

47-8

shown above in Figure 47-3. Since there is only a displacement current in this region of space, the modified form of Ampère's law is given by

Bidr∫ = µoεo

ddtEidA∫ = µoεo

dΦE

dt . (47-21)

We begin with the electric flux. We have

ΦE = E   j⎡⎣ ⎤⎦i b  j⎡⎣ ⎤⎦ = Eb . (47-22)

The change in the electric flux is

ΔΦE = Eb  Δ . (47-23)

The time rate of change in the electric flux is

ΔΦE

Δt= Eb  

ΔΔt

= Ebc . (47-24)

Now, we look at the path integral on the left hand side of equation (47-21). (We are going around the rectangle in a counterclockwise sense.) We have

Bidr∫

= 0  k⎡⎣ ⎤⎦b

0

∫ i dz k⎡⎣ ⎤⎦ + B k⎡⎣ ⎤⎦

0

∫ i dx  i⎡⎣ ⎤⎦ + B k⎡⎣ ⎤⎦0

b

∫ i dz  k⎡⎣ ⎤⎦ + B  k⎡⎣ ⎤⎦0

∫ i dx  i⎡⎣ ⎤⎦

= 0 + 0 + Bb + 0 . (47-25)The first integral in the sum of integrals deals with the right hand side of the rectangle. This integral is zero because there is no magnetic field in that region of space as the wave front has not yet reached that region. The second and fourth of integrals are zero because at every point on each path the magnetic field is perpendicular to the path and, therefore, the dot product vanishes. For the third integral, the direction of the path and the magnetic field are the same so the result is simply the product of the magnitudes of the two vectors; namely, the magnetic field magnitude B times b the magnitude of the path.

We can now write

Bidr∫ = Bb = µ oεo

dΦE

dt= µoεoEbc , (47-26)

from which we writeB = µ oεocE . (47-27)

Of course, equations (47-20) and (47-27) must both be true. So, we can write

47-9

B = µ oεocE =Ec

, (47-28)

and

c =1

µoεo

. (47-29)

Numerically, the speed at which the wave front propagates is given by

c =1

µoεo

=1

4π ′k( ) 1 / 4πk( )=

k′k

=8.99 × 109 N ⋅m2 / C2

10−7 N ⋅s2 / C2

= 3.00 × 108  m / s . (47-30)This is the speed of light in vacuum.

We chose the plane wave because it is much easier mathematically. It does, however, provide us with some insight into some universal characteristic of electromagnetic waves. We have found the following to be true:

1) The electromagnetic wave is a transverse wave. It propagates in a direction that is perpendicular to the direction of the oscillations that give rise to the wave.

2) The electric field is perpendicular to the magnetic field.3) The ratio of the magnitude of the electric field to the magnitude of the magnetic

field is equal to the speed of propagation. That isEB

= c .

4) The speed of electromagnetic waves in vacuum is constant and equal to c ,where

c = 2.99792458 × 108 m / s .

THE ENERGY OF AN ELECTROMAGNETIC WAVE

Earlier in the text, see equation (39-54), we found that the energy density of an electric field established in vacuum is given by

uE =12εo E2 . (47-31)

We also found, see equation (44-49), that for a magnetic field in vacuum, the energy density is given by

47-10

uM =B 2

2µ o

. (47-32)

For an electromagnetic wave, we have both a magnetic field and an electric field. So, we would expect the energy density for an electromagnetic wave to be given by

uEM =12εoE 2 +

B2

2µ o

. (47-33)

This can be written in terms of the electric field alone by using equations (47-27) and (47-29). We have

uEM =12εoE 2 +

µoεo 1 / µ oεo( ) E⎡⎣

⎤⎦

2

2µo

= εoE 2 . (47-34)

This shows us that the energy density in an electric field is the same as that in the magnetic field.As the electromagnetic wave moves in vacuum, energy is being transferred from one region of

space to another. We can, then, define a new quantity S that quantifies the energy transferred per unit time, per unit cross-sectional area; where the area A is assumed to be perpendicular to the direction of motion. This quantity is analogous to the current density. Using equation (47-34), we can write

uEM =U EM

Volume

= εoE 2 = εoE cB( ) = εocEB . (47-35)

If the wave moves for a time t , then the volume swept out will be given by

Volume = Act . (47-36)

Substitution of this value for the volume into equation (47-35) yields

U EM = εoE cB( ) = εocEBAct = εoc2EBAt . (47-37)

The energy transferred per unit of time, per unit of cross-sectional area would be

S =U EM

At= εoc

2 EB =εoEBεoµo

=EBµo

. (47-38)

The energy is propagated perpendicular to both the electric field and the magnetic field. The electric field and the magnetic field, in turn, are perpendicular to each other. This suggests that we can write equation (47-38) in vector form using the cross product. This vector quantity is called the Poynting vector and is defined by

47-11

S =

1µ o

E ×B . (47-39)

The power delivered through any surface would be given by

Power =

SidA∫ . (47-40)

The power delivered by electromagnetic waves is another kind of flux.

Example One:Using our plane wave illustrated above in Figure 47-3, if the magnitude of the electric field is

E = 100 N / C , then we wish to find the following:a) The magnitude of the magnetic field.b) The energy density of the wave.c) The magnitude of the Poynting vector.

The magnitude of the magnetic field is given by

B =Ec

=100 N / C

3.0 × 108 m / s= 3.33 ×10−7 Tesla . (1)

The energy density is given by

uEM = εoE 2 =E2

4πk=

100 N / C( )2

4π 8.99 ×109 N ⋅m2 / C2( ) = 8.85 × 10−8  J / m3 . (2)

The magnitude of the Poynting vector is given by

S = εocE 2 =cE 2

4πk=

3.0 × 108 m / s( ) 100 N / C( )2

4π 8.99 ×109 N ⋅m2 / C2( ) = 26.6  Watts / m2 . (3)

SOME OTHER PROPERTIES OF ELECTROMAGNETIC WAVES

It may seem strange that energy can be transported in wave form from one point to another without the need for a material medium. However, when you have been outside and been “burned” by the Sun, you have been the “victim” of solar energy transported through space by electromagnetic waves. (Think of how fortunate we are that this is so! Virtually all of the energy on this planet comes directly or indirectly from the Sun. Most of that energy transported by

47-12

electromagnetic waves.)It turns out that electromagnetic waves also have linear momentum. This is another amazing

phenomenon as the photon, the particle responsible for light, has no mass. I will let you ponder how it is that a massless field can have linear momentum. We must digress for a moment.

Albert Einstein in his famous special theory of relativity discovered that the energy associated with a particle is given by

E2 = pc( )2 + moc2( )2 , (47-41)

where p is the magnitude of the linear momentum of the particle, c the speed of light in vacuum,

and mo is the so-called rest mass of the particle. If a particle is at rest, it does not have linear

momentum and equation (47-41) reduces to the famous equation

E = moc2 . (47-42)

However, the particle of light, a photon, moves at the speed of light but has no rest mass. So, for the photon, equation (47-41) reduces to

Eγ = pγ c , (47-43)

where we have used γ to signify a photon. So, the momentum of the photon is given by

pγ =Eγ

c . (47-44)

One more interesting thing before we move back to our discussion of electromagnetic waves. The mathematicians have taught us that if we solve an equation like (47-43) for E , we will get

E = ± pc( )2 + moc2( )2 . (47-45)

Initially, physicist thought that the negative value was an artifact of the math and had no physical significance. It was the great English physicist Paul Adrien Maurice Dirac who speculated that the negative sign indicates an entirely new class of particles that we today call antiparticles. We now, of course, have extensive experimental confirmation of the existence of antiparticles.

Returning now to electromagnetic waves, we wish to define a new quantity called the linear momentum density, namely, the linear momentum per unit of volume. Equation (47-46) suggests that we would have

pγ

Vol

=U EM / c

Vol

=SAt / c

Act=

Sc2 . (47-46)

As we have seen earlier, particles that carry linear momentum are able to exert forces in collisions. If we use Newton’s definition of a force, we can write

47-13

F =dpγ

dt=

ddt

SAc

⎛⎝⎜

⎞⎠⎟  t

⎡⎣⎢

⎤⎦⎥

=SAc

. (47-47)

We can also now define the radiation pressure exerted on material surface of cross-sectional area A. We have

Pressure =FA

=Sc

=EB

µ oc . (47-48)

ELECTROMAGNETIC WAVES IN MATTER

So far, we have limited our discussion to electromagnetic waves propagating in vacuum. We now wish to discuss what happens to these waves in a dielectric material. When an electromagnetic wave enters a material from a vacuum, the wave propagates at a slower speed in the material. For the time being, we signify the speed of light in a matter as ′c .

The relationship between the electric field and the magnetic field now becomesE = ′c B . (47-49)

Furthermore, in the modified form of Ampère's law becomes

Bidr∫ = κM µ o( ) κεo( ) dΦE

dt , (47-50)

and we have

B = κ M µo( ) κεo( ) ′c E . (47-51)

Using equations (47-49) and (47-51), we have

B = κ M µo( ) κεo( ) ′c ′c B( ) , (47-52)

and

′c 2 =1

κ Mκ1

µoεo

=1

κMκc2 . (47-53)

So,

′c =1κ Mκ

c . (47-54)

We can now define a new quantity n called the index of refraction and defined by

47-14

c′c

= n = κ Mκ . (47-55)

Note that the refractive index for vacuum is one.The energy density would now be given by

uEM =12εE2 +

B2

2µ= εE 2 , (47-56)

while the Poynting vector becomes

S =

1µE ×B . (47-57)

THE ELECTROMAGNETIC SPECTRUM

Our eyes are sensitive to a specific band of electromagnetic radiation called visible light. In the universe, however, there is an immense spectrum of electromagnetic radiation to which our eyes are not attuned. (As an aside, your generation is the first generation of human kind to have surveyed the sky with detectors that are sensitive to the entire spectrum of electromagnetic radiation. And what a wondrous “sight” it is!)

Later when we discuss optics, we will make more extensive use of the fact that electromagnetic radiation deigns to show us two faces in the laboratory. There are experiments the results of which seem to make sense only if light is localized to a very same region of space. We say that it “behaves” as if it were a particle. As we noted above, we call this particle a photon. In other wide ranging experiments, the results seem to match a non-localized phenomenon. In these cases, we say that light appears to be a wave form.

Like all wave forms, electromagnetic radiation can be described by the following equation:

fλ =cn

, (47-58)

where f is the frequency of the wave form, λ is the wavelength of the wave form, n is the refractive index, and c is the speed of light in vacuum. It turns our that the energy of a photon is given by

Eγ = hf , (47-59)

where h is Planck’s constant. In vacuum, we can use equations (47-58) and (47-59) to write

Eγ = hf = h c / λ[ ] = hc / λ . (47-60)

47-15

The energy of a photon is directly proportional to the frequency and inversely proportional to the wavelength.

Figure 47-4The Electromagnetic Spectrum

1022

1021

1020

1019

1018

1017

1016

1015

1014

1013

1012

1011

1010

109

108

107

106

105

104

103

102

101

3×10−14

3×10−13

3×10−12

3×10−11

3×10−10

3×10−9

3×10−8

3×10−7

3×10−6

3×10−5

3×10−4

3×10−3

3×10−2

3×10−1

3×100

3×101

3×102

3×103

3×104

3×105

3×106

3×107

f Hz( ) λ m( )

Gamma Rays

X  − Rays

UltravioletVisible

Infrared

Short  Radio Waves

Long Radio  Waves

FM  TV

AM  Broadcast

47-16

Above in Figure 47-4, we represent the frequency range and the range of wavelengths for the major portions of the electromagnetic spectrum. The major regions are:

Gamma RaysX − Rays

UltravioletVisibleInfrared

Radio Waves

Below in Figure 47-5, we represent the major regions of the visible spectrum of light. (Note how little of the total electromagnetic spectrum is that portion to which our eyes are sensitive.)

Figure 47-5Visible Light

λ nm ≡ 10−9 m( )

400 500 600 700

VIOLET

BLUE

GREEN

YELLOW

OR

RED

ANGE

47-17

Summary for Chapter 47

Classical electromagnetism is theoretically founded in Maxwell’s equations and the Lorentz force. The form of Maxwell’s equations presented below assume that all fields are propagated in vacuum and the equations are given in integral form. We have

Maxwell’s Equations

EidA∫ =

1εo

ρq  dV∫ (47-61)

BidA∫ = 0 (47-62)

Eidr∫ = −

ddtBidA∫ (47-63)

Bidr∫ = µo

JidA∫ + µ oεo

ddtEidA∫ (47-64)

The Lorentz Force

F = q

E + v ×

B( )⎡⎣ ⎤⎦ (47-65)

We use plane waves as mathematically simple representations of electromagnetic waves. For electromagnetic waves propagating in vacuum, we found using Faraday’s law that

E = cB , (47-66)while using Maxwell’s generalization of Ampère's law to analyze the plane wave we found that

B = µ oεocE . (47-67)

So, the speed of light in vacuum is given by

c =1

µoεo

= 2.99792458 ×108  m / s . (47-68)

The energy density of the electromagnetic wave in vacuum is given by

uEM =12εoE 2 +

B2

2µ o

= εoE2 . (47-69)

The energy transferred through a cross-sectional area A per unit time is called the Poynting vector and given by

47-18

S =

1µ o

E ×B . (47-70)

The power delivered by an electromagnetic wave is given by

Power =

SidA∫ . (47-71)

The electromagnetic field not only has energy but it also has linear momentum given by

pγ =SAc

⎡⎣⎢

⎤⎦⎥ t . (47-72)

The radiation pressure on a material surface as a result of incident electromagnetic waves is given by

Pressure =Sc

=EBµ oc

. (47-73)

In matter, the speed of light, ′c , is less than in vacuum, c . They are related by

′c =1µε

=1κMκ

1µ oεo

=1κMκ

 c . (47-74)

We define the refractive index n by

n =c′c

= κ Mκ . (47-75)

APPENDICES

APPENDIX A

Mathematical Review

A.1Powers of Ten and Scientific Notation:

Some Prefixes for Powers of Ten:

POWER PREFIX ABBREVIATION

10−18 atto a10−15 femto f10−12 pico p10− 9 nano n10− 6 micro µ10−3 milli m10−1 deci d101 deka da103 kilo k106 mega M109 giga G1012 tera T1015

peta P1018 exa E

Some Examples:

100 = 1101 = 10102 = 10 ×10 = 100103 = 10 ×10 ×10 = 1,000104 = 10 ×10 × 10 ×10 = 10,000105 = 10 ×10 ×10 × 10 ×10 = 100,000106 = 10 ×10 × 10 ×10 ×10 × 10 = 1,000,000107 = 10,000,000108 = 100,000,000109 = 1,000,000,000

A-1

10−1 = 110

= 0.1

10− 2 = 110

× 110

= 0.01

10−3 = 110

× 110

× 110

= 0.001

10− 4 = 110

× 110

× 110

× 110

= 0.0001

10− 5 = 110

× 110

× 110

× 110

× 110

= 0.00001

10− 6 = 110

× 110

× 110

× 110

× 110

× 110

= 0.000001

Multiplication and Division of Powers of Ten:

10m × 10n = 10m +n

Example: 104 ×106 = 104 +6 = 1010

10m

10n = 10m−n

Example: 104

106 = 104−6 = 10− 2

10m( ) n= 10m×n

Example: 104( )2= 104×2 = 108

A.2Some Algebra Rules:

Factors and Expansion:a ± b( )2 = a2 ± 2ab + b2

a ± b( )3 = a3 ± 3a2b + 3ab2 ± b3

a ± b( )4 = a4 ± 4a3b + 6a2b2 ± 4ab3 + b4

a2 + b2 = a + bi( ) a − bi( ) , where i = −1a3 − b3 = a − b( ) a2 + ab + b2( )

A-2

a3 + b3 = a + b( ) a2 − ab + b2( )The Binomial Expansion:

a + b( )n = an + nan−1b + n n −1( )2!

an−2b2

+ n n −1( ) n − 2( )3!

an−3b3 + ...+ nabn−1 + bn , when n > 0

Powers and Roots:a x × ay = ax +y

a0 = 1, if a ≠ 0ab( ) x = ax b x

a x

a y = ax −y

a− x = 1ax

ab

⎛ ⎝

⎞ ⎠

x

= a x

b x

a1/ x = ax

abx = ax bx

ayx = axy

a x/y = axy

ab

x = ax

bx

Proportion:

If ab

= cd

, then

ad = cband

a + b

b= c + d

d

A-3

and

a − b

b= c − d

dand

a − ba + b

= c − dc + d

Approximations:

If x <<1 , then 1+ x( )n ≈ 1 + nx

and

1+ x( )1/2 ≈ 1 + x2

and

1

1+ x( ) = 1 + x( ) −1 ≈ 1− x

and

1

1+ x( )1/2 = 1 + x( ) −1 /2 ≈ 1− x2

and

1+ x( )n ≈ 1 + nx + n n −1( )2!

x 2 + n n − 1( ) n − 2( )3!

x3 + ...

Exponential and Logarithmic Functions:

The exponential function exp x( ) is defined by

exp x( ) = ex = 1 + x + x2

2!+ x 3

3!+ x 4

4!+ ...

The natural logarithm ln x( ) is the inverse of the exponential function, so ln ex( ) = x ,

and/or eln x = x .

Note that for the natural logarithm function ln xy( ) = ln x + ln y ,

and

ln xy

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = ln x − ln y ,

and ln y a( ) = a ln y

A-4

while ln e( ) = 1 ,

and e = 2.71828183...

The common logarithm uses base 10 and is designated with log . The common logarithm and the natural logarithm are related by

ln x = ln 10 log x( ) = log x( ) ln10( ) = 2.3026( ) log x( )

The Quadratic Formula:

Assume we have a quadratic equation of the forma x2 + b x + c = 0 . (1)

Dividing every term by a yields

x2 +ba

x +ca

= 0 , (2)

which we can write as

x2 +ba

x = −ca

. (3)

Now, note that

x +b

2a⎛⎝⎜

⎞⎠⎟

2

= x2 +ba

x +b

2a⎛⎝⎜

⎞⎠⎟

2

= −ca

+b

2a⎛⎝⎜

⎞⎠⎟

2

=b2 − 4ac

4a2 . (4)

Taking the square root of equation (4) gives us

x +b

2a= ±

b2 − 4ac4a2 . (5)

Solving for x we have

x = −b2a

±b2 − 4ac

4a2 . (6)

Finally, note that if a = 1 , then

x = − b2

⎛ ⎝

⎞ ⎠ ± b

2⎛ ⎝

⎞ ⎠

2

− c . (7)

Equation (7) suggests that it will be useful to algebraically manipulate a quadratic equation until the coefficient of the quadratic term is one. So, equation (7) is the form of the quadratic formula that I will be using in this book.

A-5

A.3Some Plane and Solid Geometry:

Circle of radius R

x i( )

y j( )

R

The circumference of a circle of radius R :C = 2πR

The area of a circle of radius R :

A = π R2

Sphere of radius R

x i( )

y j( )

R

Surface area of a sphere of radius R :

Asurf = 4π R2

Volume of a sphere of radius R :

Vsphere = 4 / 3( )π R3

A-6

Cylinder of radius R

RR

h

Lateral surface area of a right circular cylinder of radius R and height h :

Alat surf = 2πRh

Top and bottom areas of a right circular cylinder of radius R and height h :

Atop = πR2 = Abot

Volume of a right circular cylinder of radius R and height h :

Vcylinder = π R2h

Right Triangle

ϕ

θ

a

b

c

Pythagorean Theorem:c2 = a2 + b2

Area of a Scalene Triangle (see figure below)

A = 12

bh

A-7

Scalene Triangle

a

b

c

h

A.4Some Trigonometry:

Angles:

Δ = Rθθ

R

R

Two radii intersect at the center of the circle and form an angle θ . The length of the circular arc

Δ intercepted by the two radii is the same fractional part of the circumference as the angle θ

is to the central angle of the entire circle. Or, if you will,

Δ2πR

=θ

entirecentral ∠ . (12)

Mathematicians have defined the radian so that the entire central angle of a circle is simply2π radians. Equation (12) then becomes, in radian measure,

Δ2πR

=θ

2π , (13)

and, consequently, the arc length is given by Δ = Rθ , (14)

where θ is measured, remember, in radians. Radians and degrees are related byA-8

π radians ≡ 180° , (15)and

1 radian ≡180°π

= 57.29577951° , (16)

while

1° ≡π

180radian = 0.17453293 radian . (17)

The Trigonometric Functions in Terms of the Right Triangle:

A Right Triangle

θ

a

b

c

sinθ =ac≡

length of sideopposite the anglelength of the hypoteneuse

≡ sineθ

cosθ =bc≡

length of side adjacent theanglelength of thehypoteneuse

≡ cosineθ

tanθ =sinθcosθ

=ab≡

length of side opposite theanglelength of side adjacentthe angle

≡ tangentθ

Trigonometric Identities:

cotθ = 1tanθ

≡ cotangentθ

secθ = 1cosθ

≡ secantθ

cscθ = 1sinθ

≡ cosecantθ

A-9

sin2 θ + cos2 θ = 1sec 2θ = 1 + tan2 θcsc 2θ = 1 + cot2 θ

sin 2θ = 2 sinθ cosθcos 2θ = 2 cos2 θ −1

sin 12θ = 1− cosθ

2

cos 12θ = 1+ cosθ

2

sin α + β( ) = sinα cosβ + cosα sinβ

cos α + β( ) = cosα cos β − sinα sinβ

tan α + β( ) = tanα + tanβ1− tanα tanβ

sinα + sin β = 2 sin 12α + β( ) cos 1

2α − β( )

cosα + cos β = 2 cos 12α + β( ) cos 1

2α −β( )

Law of Cosines and Sines:

c2 = a2 + b2 − 2abcos γ ,

sinαa

= sin βb

= sinγc

a

b

c

α

β

γ

A-10

A.5Time Independent Equation of Motion With Constant Acceleration:

Recall that for constant acceleration in one dimension, the fundamental equations of motion are

vx = vox + ax t , (1)

Δx = x − xo = voxt +12

ax t2 . (2)

If we solve equation (1) for t , we find

t =vx − vox

ax

. (3)

Substitution of equation (3) into equation (2) gives us

Δx = x − xo = voxvx − vox

ax

⎛⎝⎜

⎞⎠⎟

+12

axvx − vox

ax

⎛⎝⎜

⎞⎠⎟

2

=voxvx − vox

2

ax

+vx

2 − 2voxvx + vox2

2ax

=2voxvx − 2vox

2 + vx2 − 2voxvx + vox

2

2ax

=vx

2 − vox2

2ax

. (4)

So, we can write

vx2 − vox

2 = 2ax Δx( ) , (5)

and

vx2 = vox

2 + 2ax Δx( ) , (6)

or, as I am more likely to write,

vx2 = vox

2 ± 2 ax Δx , (7)

where the plus sign indicates the object is speeding up, and minus sign that the object is slowing down.

A-11

A.6The Vector Cross Product:

There is a way to multiply two vectors where the result is also a vector. This form of vector multiplication is called the vector cross product. We want to see what this is all about.

Assume that we have two vectors A and

B such that

A = A A = Ax i + Ay j + Az k , (1)

and

B = B B = Bx i + By j + Bz k . (2)

The cross product is defined by

A ×B = A A⎡⎣ ⎤⎦ × B B⎡⎣ ⎤⎦ = AB A× B⎡⎣ ⎤⎦ . (3)

The interesting vector stuff takes place when we take the cross product of the unit vectors. So, remember this:

Whenever one takes the cross product of any two unit vectors, one always gets the sine of the angle between the unit vectors, and one also gets another unit vector that is perpendicular to both of the unit vectors being crossed. That is

A × B⎡⎣ ⎤⎦ = sin∠bet C , (4)

where ˆ C is perpendicular to both ˆ A and ˆ B . Careful inspection of the diagram below,

should convince you that there are two possible directions ˆ C and − ˆ C . The correct

choice is the right-handed choice, ˆ C . The cross product is not commutative, and it is right-handed.

A ×B =C = AB sin∠bet C , (5)

while

B ×A = −

C = −AB sin∠bet C . (6)

The Cartesian coordinate system is right-handed so thatˆ i × ˆ j = ˆ k , (7)

ˆ j × ˆ k = ˆ i , (8)

ˆ k × ˆ i = ˆ j . (9)A-12

Reversing the order of these, of course, produces negative values. That isˆ j × ˆ i = − ˆ k , (10)

ˆ k × ˆ j = − ˆ i , (11)

ˆ i × ˆ k = − ˆ j . (12)Furthermore, note that the cross product between any two vectors that are in the same

direction or in opposite directions must vanish. That isˆ i × ˆ i = ˆ i × − ˆ i = ˆ j × ˆ j = ˆ j × − ˆ j = ˆ k × ˆ k = ˆ k × − ˆ k = ˆ A × ˆ A = ˆ A × − ˆ A = 0 . (13)

Remember that the sin 0 ° = sin180 ° = 0 . (14)

Pictorial Representation of the Vector Cross Product

A-13

A

B

A ×B

B ×A

∠ bet

ˆ C

− ˆ C

We can also take the cross product using the components. We find

A ×B = Ax i + Ay j + Az k⎡⎣ ⎤⎦ × Bx i + By j + Bz k⎡⎣ ⎤⎦

= Ax Bx i × i

=0

+ Ax By i × j=k

+ Ax Bz i × k=− j

+ Ay Bx j × i

=− k

+ Ay By j × j= 0

+ Ay Bz j × k= i

+ Az Bx k × i

= j

+ Az By k × j=− i

+ Az Bz k × k=0

. (15)

Collecting terms with like unit vectors we can write

A ×B = i Ay Bz − Az By⎡⎣ ⎤⎦ + j Az Bx − Ax Bz⎡⎣ ⎤⎦ + k Ax By − Ay Bx⎡⎣ ⎤⎦ . (16)

Let us do an example. The angular momentum of an object about the origin is defined by

LO = r × p = r × M v = M r × v[ ] . (17)

If the mass of the object is M = 2.00 kg, and the position of the object is given by

r = 2.00 m i − 3.0 m j + 4.00 m k ,

and the instantaneous velocity of the object is given by

v = − 5.00ms

i − 10.0ms

j + 15.0ms

k ,

then we can write

r × v = i − 3 m( ) 15ms

⎛⎝⎜

⎞⎠⎟ − 4 m( ) −10

ms

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

+ j 4 m( ) − 5ms

⎛⎝⎜

⎞⎠⎟ − 2 m( ) 15

ms

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

+ k 2 m( ) −10ms

⎛⎝⎜

⎞⎠⎟ − − 3m( ) −5

ms

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

. (18)

A-14

So,

r × v = − 5.00m2

s i −  50.00

m2

s  j −  35.00

m2

s k , (19)

and, therefore,

LO = M r × v[ ] = 2 kg( ) − 5.00

m2

s i − 50.00

m2

s  j −  35.00

m2

s k

⎡

⎣⎢⎤

⎦⎥

= −10.0 kg m2

si −100.0  

kg m2

sj − 70.0  

kg m2

sk . (20)

A-15

A.8Some Calculus Results

Even though calculus is not required to understand this text, there are many results that we can use that are not too difficult to prove using only algebra and trigonometry. There are two main branches to calculus: differential and integral. The two operations are related and both operations rely on one of the most important ideas in the history of human thought, namely, the idea of a limit.

The most important concern of calculus from our perspective, is that it gives us a method by which we can analyze how a mathematical function changes is we change the value of the independent variable. As an example, assume that we have a function f which depends on an independent variable x . We signify this relationship of function to independent variable by f x( ) , which we read as “ f of x.”

Next, assume we have the specific functionf x( ) = ax2 , (1)

where a is a constant.If x = 1, then f = a . However, if we change the x value, then we expect a change the f value. For example, if x = 2 then f = 4a . So, we can reasonable ask what happens to our function if we change a variable on a little bit; not double it like we did in the example. Let us see what that might mean by looking more closely at equation (1).

Let us say that we change x by a little bit, say Δx . Then we first want to look and see what happens to the function f as a result of the change in x . We have

f x + Δx( ) = a x + Δx( )2 = a x2 + 2x Δx( ) + Δx( )2⎡⎣ ⎤⎦ . (2)I hope that it seems reasonable to you that if we take equation (2) and subtract equation (1) from it, what we will have left is simply how much the function changed. So, we define the change in a function that depends on x by

Δf = f x + Δx( ) − f x( )= a x2 + 2x Δx( ) + Δx( )2⎡⎣ ⎤⎦ − a x2⎡⎣ ⎤⎦

= 2ax Δx( ) + a Δx( )2 . (3)If we now divide equation (3) by Δx we will have a ratio of the change in the function to the

change in our variable. When we do this, we getΔfΔx

=2ax Δx( ) + a Δx( )2

Δx= 2ax + a Δx( ) . (4)

Notice that the smaller we make our change in Δx , the closer the ratio becomes to being simply 2ax . So, we define a new quantity called the first derivative by

dfdx

= limitΔx→0

ΔfΔx

. (5)

We take the ratio of the change in the function to the change in the independent variable and then we see what happens to this ratio as Δx gets very, very close to zero. The limit exists and is the first derivative if the ratio gets closer and closer to a specific value as Δx gets closer, and closer to zero.

If we have a function for which the first derivative exists, then in principle, we can use the A-16

simple process outlined above to calculate the first derivative. Let us do a couple more examples. f x( ) = ax−2 , (6)

where a is a constant. First, we write

f x + Δx( ) =a

x + Δx( )2 =2

x2 + 2x Δx( ) + Δx( )2 , (7)

and then

f x + Δx( ) − f x( ) =a

x2 + 2x Δx( ) + Δx( )2 −ax2 =

ax2 − a x2 + 2x Δx( ) + Δx( )2⎡⎣ ⎤⎦x2 x2 + 2x Δx( ) + Δx( )2⎡⎣ ⎤⎦

=−2ax Δx( ) − a Δx( )2

x2 x2 + 2x Δx( ) + Δx( )2⎡⎣ ⎤⎦=

−a 2x Δx( ) − Δx( )2⎡⎣ ⎤⎦x 4 + 2x3 Δx( ) + x2 Δx( )2⎡⎣ ⎤⎦

. (8)

For the ratio, we haveΔfΔx

=−a 2x − Δx( )⎡⎣ ⎤⎦

x4 + 2x3 Δx( ) + x 2 Δx( )2⎡⎣ ⎤⎦ . (9)

Finally, when we take the limit, we getddx

ax−2⎡⎣ ⎤⎦ = −2axx 4 = −2x−3 . (10)

Mathematicians are very, very clever! One important way they help us is by solving general forms of equations so that we do not have to do it for each individual case. For example, they have proven that for a monomial of the form

f x( ) = axn , (11)where a is a constant and n ≠ −1 , the first derivative is given by

ddx

axn⎡⎣ ⎤⎦ = anxn−1 . (12)

(See how useful mathematicians are?)Another function of use to us is the exponential. Assume we have

f x( ) = eax , (13)where a is a constant. We first write

f x + Δx( ) = ea x +Δx( ) = eaxea Δx( ) , (14)while

f x + Δx( ) − f x( ) = eaxea Δx( ) − eax = eax ea Δx( ) − 1⎡⎣ ⎤⎦ . (15)Recall from your precalculus math that

es = 1 + s +

s2

2!+

s3

3!+ . (16)

we use this to rewrite equation (15) as

A-17

f x + Δx( ) − f x( ) = eax a Δx( ) +a Δx( )( )2

2!+

a Δx( )( )3

3!+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (17)

The ratio becomes

ΔfΔx

= eax a +a2 Δx( )

2!+

a3 Δx( )2

3!+

⎡

⎣⎢⎢

⎤

⎦⎥⎥

. (18)

When we take the limit we get the first derivative. We findddx

eax⎡⎣ ⎤⎦ = aeax . (19)

I want to do two more functions that we will use frequently. Assume we havef t( ) = sin at( ) , (20)

where a is a constant. First, we writef t + Δt( ) = sin a t + Δt( )( )( ) = sin at + a Δt( )( ) . (21)

Using the trig identity sin c + d( ) = sin c( ) cos d( ) + sin d( ) cos c( ) . (22)

We rewrite equation (21) asf t + Δt( ) = sin at( ) cos a Δt( )( ) + sin a Δt( )( ) cos at( ) . (23)

Next, we writef t + Δt( ) − f t( ) = sin at( ) cos a Δt( )( ) + sin a Δt( )( ) cos at( ) − sin at( ) . (24)

The ratio becomesΔfΔt

=sin at( ) cos a Δt( )( ) + sin a Δt( )( ) cos at( ) − sin at( )

Δt . (25)

To take the limit, we need to know that as Δt→ 0 ,sin a Δt( )( ) → a Δt( )

cos a Δt( )( ) → 1 , (26)

Taking the limit we haveddt

sin at( )⎡⎣ ⎤⎦ =sin at( ) + a Δt( ) cos at( ) − sin at( )

Δt= acos at( ) . (26)

We can follow a similar argument to prove thatddt

cos at( )⎡⎣ ⎤⎦ = −a sin at( ) . (27)

In general, integral calculus is much more difficult than differential calculus. However, we have found that there are functions for which the first derivative exists. When such is the case, we can write

d f x( )dx

= F x( ) . (28)

This says simply that if we take the derivative of a function of x, we get yet another function of x.

A-18

We can rearrange equation (28) asd f x( ) = F x( ) dx . (29)

Equation (29) is what mathematicians call a differential equation. To solve this equation, one performs the same operation on both sides of the equation. The operator is signified by ∫ and is called an integral and the process is called integration. We have

f x( ) = d f x( )∫ = F x( ) dx∫ + C , (30)where C is an arbitrary constant.

This is an important result, and it is called the fundamental theorem of the calculus. It states simply that when we integrate a function F x( ) we get another function f x( ) for which F x( ) is the first derivative. Integration is the opposite of differentiation and, in earlier times, was sometimes called anti-differentiation.

A-19

APPENDIX BValues of Physical Constants

B.1Universal Physical Constants:

Quantity Symbol Value MKS Units

Speed of light in vacuum c 2.99792458×108 m / sPermittivity of free space εo 8.854187817 ×10−12 C2 / N ⋅m2

Permeability of free space µo 4π × 10−7 N ⋅s2 / C2

Gravitational Constant G 6.674 × 10−11 N ⋅m2 / kg2

Planck’s Constant h 6.62606896 ×10− 34 J ⋅ sCharge on an electron − e −1.602176487 × 10−19 CMass of an electron me− 9.10938215 ×10− 31 kgCharge on a proton e 1.602176487 × 10−19 CMass of the proton mp 1.672621637 × 10−27 kgMass of the neutron mn 1.674927211× 10−27 kg

B.2Some Other Useful Physical Values:

Quantity Symbol Value MKS Units

Mass of the Sun M 1.99 × 1030 kg

Mean Radius of the Sun R 6.96 ×108 m

Luminosity of the Sun L 3.9 × 1026 W

Mass density of the Sun ρ 1,410 kg / m3

Mass of the Earth M⊕ 5.97 ×1024 kgMean Radius of the Earth R⊕ 6.371× 106 mMass density of the Earth ρ⊕ 5,520 kg / m3

Mean Distance from the Sun 1 AU 1.50 × 1011 mMass of the Moon MM 7.35× 1022 kgRadius of the Moon RM 1.74 × 106 mMass density of the Moon ρM 3,340 kg / m3

Mean Distance from the Earth d⊕M 3.84 × 108 mA-20

Quantity Symbol Value MKS Units

Avogadro’s Number NA 6.02214179 ×1023 1 / moleAtomic mass unit u 1.660538782 × 10−27 kg

Boltzmann constant kB 1.3806504 ×10−23 J / KAtmospheric pressure atm 1.01325 × 105 N / m2

Density of dry air ρair 1.29 kg / m3

at 0°C,1atm( ) Molecular mass of air Mair,molecular 0.02898 kg / moleSpeed of sound in air vsound ,air 331 m / s at 0°C,1atm( )Density of water ρH2 O 1000 kg / m3

Earth’s surface gravity g 9.810 m / s2

B.3The Greek Alphabet:

Α α alpha Ν ν nuΒ β beta Ξ ξ xiΓ γ gamma Ο ο omicronΔ δ delta Π π piΕ ε epsilon Ρ ρ rhoΖ ζ zeta Σ σ ,ς sigmaΗ η eta Τ τ tauΘ θ theta Υ υ upsilonΙ ι iota Φ ϕ ,φ phiΚ κ kappa Χ χ chiΛ λ lambda Ψ ψ psiΜ µ mu Ω ω omega