25.1

Chapter 25: Electric Potential The concept of work, potential energy, and conservation of energy proved to be extremely useful in our study of mechanics (Phys 4A) and will also be in 4B.

Review 1. When a force F acts on a particle that moves from point a to b, the work done is

b

a b aconstant force

variable force

W F d or F ds→ = ⋅ ⋅∫

2. If the force is conservative, then • work only depends on the end points (path independent) • a potential function U(x) can be defined:

a b b aW ΔU (U U )→ = − = − − DEMO drop or thrown a ball up

Example: when a baseball is drop, the positive work of gravity (Fg and y point in the same direction) increases the KE at the cost of decreasing the GPE. If instead the baseball is thrown straight up, the negative work of gravity reduces the KE were the GPE increases.

In the context of a “field”, when a mass moves in the direction of the g-field, ∆U decreases whereas moving against the field ∆U increases.

One characteristics of potential energy is moving with the field decreases potential whereas moving against the field increases the potential.

3. Using the work-energy theorem, one derives conservation of energy:

total 1 1 2 2 1 2W K U K U K U or E E= ∆ = −∆ → + = + = Applying these concepts to electricity, we ask the question what is EPE? First, we start with spring PE and then make the leap to electrical PE ≡ EPE. When work (Whand) is done on the spring by the hand, this work does not disappear but is transferred into the spring as PE according to the work-energy theorem (Whand = −∆US). Physically, one can interpret this storing of spring PE by looking at the spring rungs changing (“distorting”) shape via the compression of the spring.

In a very analogous manner, as the test charge approaches the source charge, the field due to the source charge gets more and more “distorted”. It is in this distortion of the field where EPE is stored. That is, if there is more distortion of the field, there is more EPE stored; however, if there is less distortion then less EPE is stored. In summary, when work is done on a test charge in the presence of the source charge, the work is transferred into EPE (Whand = −∆UE).

Now, we applying these concepts to the point charge.

Calculating EPE for a point charge Gravitational and electric forces are both (i) conservative forces and have (ii) inverse square law forces: Fg = Gm1m2/r2 and FE = kq1q2/r2. Therefore, the work done by both

25.2

forces change the PE of the system and are (iii) path independent and only depend on the ends points. Let’s calculate the EPE function of a point charge.

Suppose that a test charge q is placed near a source point charge Q. If the test charge has negative work done on it, there is an increase in its EPE because it moves against the field lines of the source. Since the work done is path independent, the easiest path to consider is a straight-line path where the E-field (∝ FE) is antiparallel to the displacement dr. The work is

test

0

b

a

test2

0

test

0

rb b b

ab E testa a ar

b a

qq

4

qqdrr 4

qq4

1W F dr q Edrr

1 1 0r r

−

π

−

π

+

π

= ⋅ = − = = −

= − >

∫ ∫ ∫

It is customary to rewrite this expression by setting ra ≡ rref and rb ≡ r:

( )test

0ref point

ref

qq4

1 1W U(r) U Ur r

−

π

= − ≡ − − = −∆

According to the definition of PE, only differences in energy can be measured, and therefore, the reference point is defined that is most useful to the user of ∆Upoint. Analogy: The GPE has its reference potential set to zero at the most convenient location. For many problems in Phys 4A, the surface of the earth has its reference point set to yref = 0. For example, if I drop a ball on the table, its most convenient to set Ug = 0 at the table height. In electrostatics, the exact situation occurs where the reference potential is set to zero (Uref = 0) at its most convenient location, which in our case rref = ∞:

b aU U U∆ = − refU(r) U≡ −

or

ref

test test

0 0 0

testpoint point

ref r

Qq Qq 14 4 4

Qq1 1 1U 0 Ur r r r

=∞

± ±

π π π

± ∆ = − = − = = ∆

where ∆Upoint > 0 (positive energy) and ∆Upoint < 0 (negative energy). Remarks 1. Don’t confuse FE = kqq0/r2 with ∆UE = kqq0/r. They look similar but they mean

different things and have different units: [FE] = Newtons and [∆UE] = Joules. 2. Only difference in EPE are meaningful and only differences can be measured in a

lab situation. This difference depends on the location of the reference point where U(rref) = 0. As we will see, different charge distributions will require different reference locations. For a point or sphere charge, setting rref = ∞ to get Uref = 0 seems “obvious”. However, doing this for a charged wire or cylinder will require a different reference location since setting rref = ∞ gets one Uref = ∞.

3. What is the physical meaning of EPE? EPE is the energy required to move an electric charge from one location to infinity. In terms of atoms, this is called the ionization energy or the binding energy, which is the work required to remove the electron from the atom to infinity.

4. EPE depends both on distance and charge:

0

testpoint

14

QqU (r) 0rπ

∆ = −

25.3

Distance dependence Moving a charge from r = ∞ to a • distance farther from the source charge (weaker force) requires doing less work

& therefore less EPE stored in the E-field • distance closer to the source charge (stronger force) requires doing more work &

therefore more EPE stored in the E-field

0 0

test testpoint

rpoint1 1

4 4Qq QqU 0 vs. 0r U

π π∆ = − = −∆

Charge dependence Moving a charge from r = ∞ to the same distance from the source charge requires • more work for a larger source charge and so stores more EPE in the E-field • less work for a smaller source charge and so stores less EPE in the E-field

0 0

test testpoint

Q

point1 1

4 4q qU 0 vs. 0r r

QUπ π

∆ = − = −∆

ELECTRIC POTENTIAL

DEMO wood block and power supply with a lightbulb

In order for some force to do work there needs to be at least two objects – the object (source) that provides the work and the object that will have work done on it.

Suppose Carlos does positive work on a block (applies a force over a distance) and lefts it straight up. What do you call “Carlos” before he does the work? The potential to do work!

Question: does the earth have the ability to do work on the block before Carlos? Of course, it does. The earth’s “potential to do work” is technically called Earth’s gravitational potential. If some test mass (say, a block) is placed in Earth’s g-field, work will be done. There is a very important distinguish between GPE and the new concept of “POTENTIAL to do work”, doing work requires two objects while the “potential to do work” only involves one object:

Potential Energy is stored work involves two objectsPotential to do work involves only one object

→→

For electrical systems, the work done by a field involves two charges: a test charge qtest and the source charge’s qsource-field (that will do the work). If one was to isolate the source of the work or the “potential to do work”, this type of thinking has already been seen before between the E-force and the E-field. In a very similar way, one can isolate the “source of the EPE” called the Electric Potential:

test

2

0

test

0

test

2test 0

test

source source

isolated the cause of the force and is indep of q

source

isolated

?

q q Force 4 r charge

q q PE4 r charge

qF 1F E

q 4 r

UU V

q

π

π

= = = =π

∆∆ = = ∆

→

→ →

testthe cause of PE and is indep of q

25.4

Language: Electric Potential is usually called POTENTIAL. In circuit language, it is known as a Node VOLTAGE.

Definition The potential function V(r) at any point is the PE per unit charge:

testor

test

UV U q V

q∆

∆ ≡ ∆ = ∆←→

The potential function represents the concept of a field in a completely new way and it is a paradigm shift in thinking. The potential function is a topological (topo) parameter and it represents the electrical “landscape” produced by a source charge. Unfortunately, the vast majority of textbooks will not describe the potential function as a topographical map.

Remarks 1. The name. The word “potential” is a hideous misnomer because it inevitably reminds

you of potential energy. This is particularly confusing, because there is a connection between “potential” and “potential energy”. I am sorry that it is impossible to escape this word. The best I can do is to insist once and for all that “potential” and “potential energy” is completely different and should, by all rights, have completely different names.

2. V(r) is a scalar quantity not a vector quantity. In fact, it is a potential scalar field which is the true “force field” that is thought of in Star Wars.

3. Units of measurement: [∆V] = [∆UE]/[q] = J/C = 1 Volt = 1V 4. In terms of work, the work done by an electric field during a displacement is

testref

test

Uq

W V (V V )q

∆= − ≡ −∆ = − −

The Gravitational Potential In Phys 4A you were introduced to GPE, however, its more basic entity or cause of GPE is the “gravitational potential Vg”, which is completely different from GPE. Using the definition of gravitational potential, we write

2 2 g12 g g g1 1

test

UUW F dr mg dy mg y g y or V g y

m∆

∆ ∆ ∆= ⋅ = − = − ∆ = − → = ∆ =∫ ∫

Note that the gravitational potential is mass independent and only depends on position. Plotting Vg vs. y, there is “gravitational ramp or incline” and this is the gravitational landscape when in a constant gravitational field. Even though you cannot see this ramp, before a ball is thrown upwards the mass “sees/senses” this gravitational ramp. This is very analogous when you are riding your bike and see a steep hill ahead; you are likely to suddenly pedaling faster. Question, what told you to suddenly start speeding up? The physics answer is you saw the “gravitational landscape” setup by the earth, which in turn from personal experience, told you to increase your kinetic energy in order to work less when you are on the hill. The gravitational landscape (my language) is the gravitational potential setup by the hill. This is a pretty dramatic change in view point in potential energy. In summary, there are two viewpoints when throwing up a ball into the air: g-field picture vs. the potential field picture.

All constant vector fields have potentials that are ramps. So, the E-field between two large parallel plates (which is also a constant field), means it’s electric potential function is linear with distance and an electric ramp. Mathematically, we write

25.5

2 2

12 E E1 1

Eg

test

U

Ur

W F dr qE dr

V Eq

∆

∆∆ ∆

= ⋅ = − = −

→ = =

∫ ∫

DEMO Parallel plates and increase in voltage as plates are separated Analogous to that of gravitational case, the E-field pictures has a charge moving against the E-field stores electrical energy in the distortion of the E-field of the plates. On the other hand, the potential picture paints a radically different image: the test charge moves up an electrical ramp. As the charge increases in electrical height, it stores more EPE. The potential picture clearly demonstrates a good physical picture to apply conservation of energy. DEMO power supply and lightbulb Electrical Potential of a point charge As we already calculated, the change in electric potential is the amount of “work” to move a charge from r = ∞ to a distance r from the point charge:

test 0 0po int

0

test0

Qq14 r Q Q

q 4 4

0U V 0q r r

π∆∆

π π

−= → = − ≡

Plotting the electric potential for a point charge, the “electrical landscape” is an electrical hill where its peak height is at infinity. Under what physical circumstances does the electric potential increase or decrease?

⇒ ∆

⇒ ∆

Charges moving the field move in the direction of V Charges moving the field move in the direction of V

with decreasingagainst increasing

Locations where the potential is always a constant along some “path” is called an EQUIPOTENTIAL SUFACES or equipotentials. They are surfaces of equal potential values similar to how Gaussian surface are constant electric field surfaces.

why does the work increase the electrical potential energy of the plates? The electric potential sets up the “electrical landscape or elevation map” that the source of the field sets up. Immediately, one sees that this is similar to throwing a ball up.

25.6

Experimentally, the way one measures potential differences between two points is using a voltmeter. Mathematically, the potential function is similar in nature to calculating electric fields:

0 0

nnet net

n

1 14 4

q dqV Vr rπ π

= =∑ ∫

Example 25.1 An electric dipole consists of two-point charges, q1 = −q2 = 12nC, placed 10 cm apart. (a) Compute the potentials at points A, B, and C. Are these meaningful values? Explain. (b) Plot V(r) vs. r and interpret the plot. (c) A positive electron (positron) moves from B to A. What is the kinetic energy gained by the positron? Explain.

SOLUTION The electrical landscape set up by the dipole charges (q+ = 1, q− = 2) creates a potential field that is the superposition between two: Vnet = V1 + V2. a. The electric height change from V(r = ∞) = 0 to point V(r = A) is the superposition of

the two fields at point A:

( )

( )

99 2 2

0

99

0

11A

1

22A

2

A 1A 2A A

1 12 10 C9 10 N m /C 1800 N m/C 1800 J/C 1800 V

4 0.06 m

1 12 109 10 V

4 0.04

V V

qVrqV 2700r

V (V V ) V( ) 1800 2700 900

−

−

×× ⋅ = = =

π

− ×× =

π

=

= = ⋅

= = −

→ ∆ = + − ∞ = − = −

Is VA meaningful? Absolutely NOT! VA gives the location of a single point in the potential field. What is meaningful is change in potential ∆VA from the zero point V(∞) = 0. Similar calculations for points B and C lead to

( )

( )1B 2B

1C 2C

B 1B 2B Br 14cm; r 4cm

C 1C 2C Cr r 13cm

V V

V V

V V V V( ) 2700 770 1930

V V V V( ) 830 830 0= =

= =

=

=

∆ = + − ∞ = − = ∆

∆ = + − ∞ = − = ∆

What does the voltage 0V mean? You are most likely interpreting this as a vector field instead of a scalar field. Application of Electric Potentials – EKGs In terms of the heart, instead of calling them electric potentials, they are called Action Potentials. However, they mean exactly the same thing. An EKG is a mapping of a slightly distorted electric dipole on the left while on the right are the equipotentials of the heart.

25.7

Simulated heart attack; Computer simulation of the electrical activity of the heart during a heart attack.

Top image: This image shows the start of a pulse of electrical activity (red area at upper left). This will spread in a wave over the inactivated heart surface (blue) towards lower right. The wave should be smooth and cause the contraction of the heart muscle to send blood around the body. Disruption to this process during a heart attack (myocardial infarction) may cause ventricular fibrillation.

Bottom image: This image shows a chaotic pattern of electrical activity called ventricular fibrillation. The normal smooth wave of electrical activity (red) should have passed from upper left to lower right, causing inactivated heart muscle (blue) to contract and send blood around the body. Equipotential Program

The important types of Electric Potentials The whole point of developing the idea of the electric potential is to visualize the electrical topological map that a charge will experience as it moves through these electrical landscapes. By doing this, we can “easily” visualize what the motion of a charge in the potential field is going to do instead of relying on Newton’s Second Law with the E-force (or E-field picture). There are four main types of potential that we will deal with in this course:

1. Parallel Plate potential: Vparallel plates = Er 2. Point charge potential: Vpoint = Q/4πϵ0r 3. Charged solid conductors and insulators 4. Charged concentric spheres and shells

How will be the way I organized the rest of the chapter: 1. Knowing the potential fields are known, one applies conservation of energy to

interpret the motion of the electric charge.

2 21 11 1 2 2 1 1 2 22 2

1 2E E

K U K U mv qV mv qV=

+ = + + = +→

with the conversion factor 1 eV = 1.602 × 10−19 J. 2. “Coulomb’s Method” to determine the potential function (V(r) = ∫dq/4πϵ0r) is

challenging, as we saw in Chapter 23. For the type of charge distributions that we will deal with, Coulomb’s Method is “inefficient” in determining the potential function. An alternate method, which I call “Gauss’s method” (no Google search will find these words) is much more efficient because it is assumed that one can immediately write down the potential functions for spheres. If the E-field is known, then integrating along a straight line (remember that potential functions only depend on the end points), the potential can be determined via

25.8

2

1V E ds∆ = − ⋅∫

Once again, Gauss’s method will be the preferred method for calculating potentials.

3. With a more in depth knowledge of electrical potential, we will have a second look at dielectric breakdown and lightning.

Units of energy in terms of electron-volts It is common in electrostatic situations to use units of energy that are not in Joules since Joules are units of everyday situations. At the atomic level there is a preferred unit that makes calculations of energy considerably easier. One-electron volt ≡ eV is the energy gain by a particle with a proton charge to gain energy when “falling” through a 1-volt height:

19 19U q V e (1 V) (1.6 10 C)(1 V) 1.602 10 J 1 eV− −∆ = ∆ = ⋅ = × = × =

Since this is such a small number, it is usually measured in keV (Van de Graaff), MeV (large Van de Graaff), GeV and TeV (accelerators). CONSERVATION OF ENERGY By looking at a potential curve, one can interpret the motion via conservation of energy where (conversion factor 1 eV = 1.602 × 10−19 J)

2 21 11 1 2 2 1 1 2 22 2

1 2E E

K U K U mv qV mv qV=

+ = + + = +→

Example 25.2 The figure shows electric potential V(x) along an x axis. A proton moves inside the potential well. (a) Interpret the motion of the proton and calculate the force (magnitude and direction) as it moves (i) left of x = 3.0 cm and (ii) right of x = 5.0 cm. (b) The proton is released at x = 3.5 cm with initial kinetic energy 4.00 eV going left. What is the x coordinate and potential value of its turning point? (c) If instead the proton moves to the right with the same kinetic energy, what is the speed at x = 6.0 cm?

Solution a. The electrical landscape produced by the source charge allows us to visualize what the proton’s motion would be if inserted in this world. As it moves to the left, it would go up an electrical incline that increases it EPE while decreasing its KE. If below the 5.0V line, you could visualize an oscillatory motion between the two inclines. However, if the proton had enough KE it would be able to “roll out” of this potential well and roll away (behaving as a free particle). Although the potential landscape is easy to visualize, seeing the E-field picture is not easy at all.

(i) For a proton moving to the left, if below the 9V potential line, we see a negative sloped potential that will repel the proton with a constant E-field, which in turn, applies a positive E-force to the proton in the direction opposite to its motion:

field points to the right

19 17x x

V 300 900E 3 V/cm 300 V/m Er 3 1

F eE 1.6 10 C 300 V/m 4.8 10 N F− −

∆ − = − = − = = + = ∆ −

= = × ⋅ = × =

25.9

(ii) For a proton moving to the right, if below the 5V potential line, will see a positive sloped potential that repels the proton with a negative E-force:

−∆ − = − = − = − = − = → = − × ∆ −

17E

field points to the left

V 500 300E 2 V/cm 200 V/m E F 3.2 10 Nr 6 5

b. When the proton is released and moves to the left, its total energy (derived from the graph) is K + U = 4.0 eV + 3.0 eV = 7 eV. At the 7-eV height, the horizontal line shows that the proton will intersect the potential wall and from the graph, its turning point is x ≈ 1.7 cm.

We can mathematically calculate this by solving the triangle problem of the right electrical incline. Using the slope, the value of x is

4 Vslope E 3 V/cm x 1.33 cmx

= = = → =

However, measured from the origin, the proton is turned at the location of turning pointx 3 cm 1.33 cm 1.67 cm= − =

c. When the proton moves to the right out of the well, there is an increase in electrical height and consequently, it must slow down. The left over KE is given by conservation of energy and its speed is

−

−

×= = → = =

×

= =

1921

2 272(2 eV) 2(2 eV)(1.6 10 J/eV)K 2eV mv v

m 1.67 10 kg

20 km/s v

Application: Ionic Bounding of NaCl GAUSS’ METHOD FOR CALCULATING POTENTIALS USING

b

aV E ds∆ = − ⋅∫

We will only focus on charge distributions that are continuous charge distributions for spheres and cylinders, and this requires us to integrate to get the potential function. Unfortunately, using “Coulomb’s Method” is really the wrong tool to determine the potential function because it is so challenging to use. However, “Gauss’ Method” is another approach that is considerably easier but it requires that we already know the functional form of the E-field. That is, in some cases (spheres and cylinders) in which the E-field is known, it is easier to determine V(r) from E(r). To derive an equation for the potential function in terms of the electric field, we start from the definition of work:

b b b

ab 0a a a

Gauss' Method

dVW F ds q E ds V V E ds or Eds

= ⋅ = ⋅ = −∆ → ∆ = − ⋅ = −∫ ∫ ∫

From this integral equation, the customary units for the electric field are

[ ] [ ] [ ] [ ]V E s E V/m= ⋅ → =

To see how this integral equation is used, we can calculate the potential of a point charge since we already know the E-field of a point charge. If a test charge is brought from infinity to a point a distance r from the point charge (∞→r), we integrate Epoint along a straight radial line:

0 0 0

rr r

point 2q q 1

4 4 4dr 1 qV E dsr r r∞ ∞

∞

− −

π π π∆ = − ⋅ = = − =∫ ∫

We now want to calculate the potential functions for

25.10

1. Charged solid conductors 2. Charged solid insulators 3. Charged concentric spheres and shells

A Charged Conducting Sphere Consider a solid conducting charged sphere of radius R and total charge Q. What does the electrical landscape of a charged conducting sphere look like? At a point outside of the sphere, having the same E-field of a point, the potential field must also look like a point charge too. Since we just calculated the potential field of a point charge, we already possess it solutions.

0 0outside surface

Q Q4 4

V and Vr Rπ π

∆ = ∆ =

Key point: when determining the potential function inside of a charge distribution, remember to always integrate from infinity and work your way to the location where you want to determine this potential at. For inside the sphere, we start at infinity and work our way into the sphere to get Vinside(r < R).

R r r

R R0 0 work to go through

work to get to the conductor the surface

0

E ds E ds (0) drQ QV( inside) 0

4 R 4 R

Q constant V( inside4 R

∞∆ ⋅ ⋅ ⋅ =

= ∆

∞ → = − − = − +π π

= = ∞ →π

∫ ∫ ∫

inside) V (r)∆=

For inside the sphere, the potential field Vinside is a constant even though the E-field is zero. Mathematically, this make sense because the only way a derivative of something can be zero is if that thing is a constant according to E =−∇V=0. So the potential field inside a conductor must be a constant. Remember Lab 3 when we mapped the equipotentials to get the E-field inside a circle conductor? Why was the potential inside the conductor a constant, so that the superposition of E-fields inside summed to zero (Enet = Eext – Eind = 0).

In summary,

0

0

outside

conductingsphere

inside surface

Q4

Q4

V , r Rr

VV V , r R

R

π

π

∆ = >∆ = ∆ = ∆ = ≤

Physical interpretation When we calculate the potential field of a charge distribution, we are calculate the electrical height change of a test charge moves through this electrical landscape. With a test charge located at infinity, it sees a flat potential field. As we move the test charge closer to the charged conductor it feels an electrical repulsion, which we interpret as moving up the electrical hill. When we’ve pushed the test charge to the surface, the amount of electrical height achieved as well as the work done is

0 0height to surface nonelectrical surface

Q qQ4 4

V W q VR R

−

π π∆ = → = − ∆ =

This is how high or how much work had to be done to “overcome” the electrical repulsion to get to the surfce.

25.11

Once we get to the surface, to move the test charge into and across the conductor, there is no electrical height change and therefore, no work is required since the potential field a flat. That is, the potential field everywhere inside the conductor has the same value at every point. Mathematically, we see this as

inside inside0 no height change

through a conductorheight to surface

QV( inside) 0 W q V 04 R

∆ ∞ → = + → = ∆ =π

To get a physical sense of what is happening, image that you are climbing up a mountain and as you get to the top, you suddenly stumble upon a frictionless icy plateau that you enter at some initial speed. What would happen to you? Since there is no friction, you would continue to move at a constant velocity across this icy plateau. In a very similar manner, as a test charge “stumbling” into a conductor, which has a flat electrical potential field there. From the work-energy theorem, no work implies that there is no change in the kinetic energy (W = ∆K = 0) and the test charge moves at a constant speed across this flat landscape. In some sense, the test charge is having the ultimate free lunch because it gets a “free ride” across the conductor.

As a check, compute E V= −∇

to make sure you get the correct E-field value:

0 02

q q4 4

E Vr rπ π

= −∇ = −∇ =

A Charged Insulating Sphere Consider a solid charged insulating sphere of radius R and total charge Q. What does the electrical landscape look like? As before, at a point outside of the sphere, the potential field must also look like a point charge and we write

0 0outside surface

Q Q4 4

V and Vr Rπ π

∆ = ∆ =

For inside the charged insulating sphere, we start at infinity and work our way inwards.

( )

R r r

3R R0 0

height to the surface

22 21

23 30 0 0 0 0

height to height from the surfathe surface

E ds E ds r drQ QV( inside)

4 R 4 R

Q Q Q Q Qr r R4 R 4 R 4 R 8 R 8 R

∞′ ′∆ ⋅ ⋅

= − =

∞ → = − − = −π π

⋅ − + −π π π π π

∫ ∫ ∫

ce

to somewhere inside

2

total height inside3 change0 0

3Q Qr V (r) V (r)8 R 8 R

− ∆ ∆= = ≡π π

In summary,

0outside

insulating sphere 2

inside 30 0

Q4

V , r Rr

V3Q QrV , r R

8 R 8 R

π

−

∆ = ≥∆ = ∆ = < π π

As a check, compute E = −∇V to make sure you get the correct E-field function inside the insulating sphere:

0

2

3 30 0

Qr4

3Q QrE V8 R 8 R R

−π

= −∇ = −∇ = π π

25.12

Physical Interpretation of crossing the boundary condition If we continue to interpret the potential field as the change in electrical height, by looking at the potential function we see that

There are three different electrical height changes: (i) the height change to the surface, (ii) the maximum height change from the surface to the center of the insulating sphere, and (iii) the total height change from infinity to the center of the insulating sphere:

height to surface height from surface total heightto center of sphere0 0 0

Q Q 3QV V V4 R 8 R 8 R

∆ ∆ ∆= = =π π π

So what about the r-dependence of the potential field inside the insulating sphere? 2

inside 30 0

total height change

3Q QrV (r)8 R 8 R

∆ −=π π

According to Gauss’ law applied to the insulating sphere, the smaller the enclosed charge, the smaller the E-field. Because of the E-field is decreasing inside, the slope of the electrical hill also decreases too, as shown on the plot. In other words, it requires less work to continue to push the charge up the hill.

Key point: it is tempting to suppose that you could figure out the potential inside the sphere on the basis of the field there alone, but this false. The potential inside the sphere is sensitive to what's going on outside the sphere as well. If I placed a second uniformly charged shell out at some radius r > R, the potential inside R would change, even though the field would still be zero. Gauss' law guarantees that charges exterior to a given point (that is, at larger r) produces no net field at that point, provided it is spherically or cylindrically symmetric. But there is no such rule for potentials, when infinity is used as the reference. Let’s now apply

b

aV E ds∆ = − ⋅∫

to a very important example

Example 25.3 Two isolated, concentric, conducting spherical shells of negligible thicknesses have radii and charge as shown. (a) Plot E(r) and V(r) before you start your calculations where V (r = ∞) = 0. (b) Calculate the potentials V(r) at radial distances r = 4.00m, 1.00m, 0.70m, 0.50m, 0.20m and r = 0?

SOLUTION a. Plot E(r) and V(r) and interpret the plots The E-field for a spherically symmetric charge distribution has already been calculated several now and is E = q/4πε0r2. The E-field inside the first shell is zero since there is no enclosed charge. Outside the first shell, the enclosed charge is twice that of the outer shell, so the first is stronger just outside the first shell and weaker just outside the outer shell. That is, I should see a dip in the E-field as I approach the outer shell, then because of the sudden increase in electric charge on the outer shell, there is a jump in E-field value but has as “high” as in the inner shell value.

25.13

From far away, I see a point charge instead of two concentric shells, so the electric potential is an electric hill. As I move a test charge towards the shells, I have to push harder and harder to get the test charge up the electrical hill. As we already know, Einside(conductor) = 0, so that means the FE(inside) = 0 and therefore, no work required to move a charge across a conductor. Because the inner shell is a conductor, no work is required to move a test charge from just outside the inner shell, through the inner shell and just to inside the inner shell. However, because there is no more enclosed charge, I don’t have to do any more work to push a test across the inside of the inner shell.

Electrical Field Calculations using ( ) enc 0 GaussianE r q / A= b. The E-field for a spherically symmetric charge distribution has already been

calculated several now and is E = q/4πϵ0r2. • For r = 4.00 m > R2 > R1: outside both shells

( )1 2

9 63

21 2

2r 4mq 2q 2 C

(8.99 10 )(2.00 1.00) 101.69 10 V/m .

(4.00)q qE r4 r

−

=0= = µ

× + ×= ×

+= =

π

• For R2 > r = 0.700 m > R1: in-between both shells

( )9 6

412 2

r 0.70m

(8.99 10 )(2.00 10 )3.67 10 V/m .

(0.700)qE r

4 r

−

0 =

× ×= ×= =

π

• For R2 > R1 > r = 0.200 m: inside the inner shell has qenc = 0, so E = 0.

c. Electrical Potential Calculations using ( ) ( )r

refrefV r V E r dr.− = −∫ • For r = 4.00 m > R2: outside both shells

( )2 refV r R V ( )> −∆ ∞ ( )r 4m

0

r 4m1 2 1 2 1 2

220 0

q drV r R

4q q q q q1

4 r 4 rr=

∞

=

∞

= − >π

+ + += − − = = ∆π π

∫

Substituting in the numbers, we get

( )9 6

31 22

r 4.00m

q q (8.99 10 )(2.00 1.00) 10V r 6.74 10 V V(r R .

4 r (4.00))

−

0 =

+ × + ×= = = × = >

π∆

• At the surface, r = R2 = 1.00 m. The potential at the surface must match the potential outside when r = R2:

( )2

41 2surface 2 outside 2 surface 2

2 R 1.00m

q qV V R 2.70 10 V V (r R .

4 R(R ) )

0 =

+= = × = =

π=

• For R2 > r = 0.700 m > R1 (in-between the shells). Remember that energies are always measured as difference. Since we are inside the second shell, we cannot reference the potential to be zero at r = ∞ because at the surface it is NOT zero.

Analogy: when a drop a ball above a table, instead of measuring GPE from the ground, I will measure from the table instead. The only meaningful statement is potential differences, not individual potential points.

25.14

Even though now I will be measuring from the shell surface because I am forced to because the shells have a nonzero potential, it doesn’t really matter where the reference potential is set only differences are meaningful. So our reference potential must now change to the surface of the outer shell. In more technical language, the potential function must be continuous at the boundary and this applies a boundary condition to the potential at the surface given by

1 2surface 2

2

q qV R

4 R( )

0

+

π∆ =

[Analogy: are we measuring the GPE from the ground or the table when I drop a ball?] The integral gives us

( )2

2

r 0.7m

1 R0 0 0

r 0.7m1 1 1 1

2 20 2R

q q qdrV R r R

4 4 4q 1

4 r r Rr=

=

> = −π π π

∆ > = − − = −π

∫

Putting this altogether without the numbers yet, we have

( ) 1 2

0 0 0 0

between 2 10 0

1 1 1 1ref 2

2 2 2

1 2

2

q q q q q qV r V

4 r 4 R 4 R 4 r 4 R

q qV (R r R )

4 r 4 R

(R )

0

+=

π π π π π

> >π π

+ − = + −

= + =

Important result: whenever we measure anything inside this shell, we have to reference the potential relative to the outside shell and is given by the above expression.

Numerically, we compute

( )

( )

6 691 2

2

4

between 2 1

between 2 1

q q1 2.00 10 1.00 10V R r R (8.99 10 )

4 r R 0.700 1.00

3.47 10 V V R r R

− −

0

× ×> = + = × +

π

= × = >

∆ >

>

• For R1 = r = 0.500 m (at the surface of the inside shell),

( )

( )

6 691 2

1 2

41

q q1 2.00 10 1.00 10V r (8.99 10 )

4 R R 0.500 1.00

4.50 10 V V R r

− −

0

× ×= + = × +

π

= × =

∆

>

• For R1 > r and r = 0 (inside the inner shell); we know that the potential inside must be the same value as that of the surface:

4inside surfaceV 4.50 10 VV = ×∆ = ∆

Physlet concentric shells of conductors Qconductor = −5Q, Qinsulator = Q

http://www.suu.edu/faculty/penny/Phsc2210/Physlets/PhysletsForWeb/Semester2/c06_spheres.html

25.15

INFINITIE LINE CHARGE CONDUCTOR Let’s find the potential at a distance r from an infinite line charge with λ = constant. The easiest is to use the relationship

refb r

b a refa rV V V E ds V V E ds ∆ = − = − ⋅ ⇒ − = − ⋅∫ ∫

We have already found the electric field for a cylinder earlier:

0

12

E(r)rπ

λ=

Substituting this into the line integral and integrating along a straight line, we get

[ ]ref ref

0 0 0

0

r r

refr r

refref

12 2 2

2

drV dr ln(r ) ln(r)r r

r V V V lnr

−λ −λ

π π π

−λ

π

λ∆ = − ⋅ = = −

∆ = − =

∫ ∫

As we have been doing earlier, we set the potential to zero at rref = ∞ . However, this presents a problem since doing exactly that leads to an infinite potential function, which is not possible:

refV V∆ =refref0 0

ref

rr2 2rV ln V ln r r =∞=∞

−λ λ

π π

∞ − = ⇒ = = ∞

This shows that if we try to define the potential V to be zero at infinity, then V must be infinite at any finite distance from a line charge: not a useful way to define V for this problem! The problem is we are dealing with a charge distribution that extends to infinity and leads to this result. However, the potential of a very long wire has many applications in real life if r << L. So we need to be better in our choice of reference potential location.

To get around this difficulty, remember that we can define V to be zero at any point we like. The question now becomes when is the natural log zero (ln(r) = 0)?

It is customary to choose rref at the surface of the cylinder, so that rref = R results in

refV V∆ =ref0 0

ref

r R2 2r RV ln V(r) lnr r=

−λ λ

π π − = ⇒ =

such that at the surface, the potential function goes to zero:

25.16

02RV(r R) ln 0R

λ

π = = =

Since the cylinder is a conductor, Einside = 0 so the potential is a constant and equal to the value inside the surface. In summary, we get

0cylinder 2Rln , r > R

V r0 r < R

λ

π

=

Example Three long parallel lines of charge, with the linear charge densities shown, extend perpendicular to the page in both directions. Sketch some of the (a) equipotential surfaces and (b) electric field lines. Explain your reasoning.

Deeper insight into the electric field The E-field E

is not just any old function; it is a very special kind of vector function, one whose curl is always zero. For example, ˆE yx=

could not be an electrostatic field since E 0∇ × ≠

. Therefore, no set of charges, regardless of their sizes and positions could ever produce such a field. We now exploit this special property of E-fields to reduce a vector problem (finding E) down to a much simpler scalar problem. Since the curl of E is zero, then a scalar function must exist:

E 0 E V where V scalar∇ × = → = −∇ ≡

Mathematically is all ties together with the use of Stokes' theorem from vector calculus:

E da E ds 0 E V∇ × ⋅ = ⋅ = → = −∇∫ ∫

Comments on Potential 1. Advantage of potential formulation: (i) If you Know V, you can easily get E V= −∇

(ii) This is extra ordinary when you stop to think of it, for E

is a vector quantity (3 components), but the answer is that the 3 components of E

are not really as independent as they look. Why?

y x x y y z z y z x x zE 0 E E 0, E E 0, E E 0∇ × = → ∂ − ∂ = ∂ − ∂ = ∂ − ∂ =

(iii) This brings us back to the previous observation - E

is a very special kind of vector. What the potential function does is to exploit this feature to maximum advantage, reducing a vector down to a scalar problem.

Deeper look at the physics of Dielectric Breakdown The dielectric breakdown is when an insulating material suddenly becomes conductive. The maximum voltage and E-field of a charged conducting sphere is at its surface:

0

surface surface

q4

V RERπ

= =

The dielectric breakdown around a charged conductor depends on two things: type of material and the radius of the conducting sphere.

Material Dependence Typically, air is a very good insulating material but at the breakdown voltage, Emax(air) = 3 ×106 V/m, air becomes conductive. This breakdown voltage at the angstrom level is

25.17

= × = =6breakdownE (air) 3 10 V/m 30,000 V/cm 3 mV/nm

Oxygen (O2) has a lower BE than nitrogen (N2) and therefore, we focus on N2. We can estimate the binding energy of nitrogen gas since two nitrogen molecules covalently bond three electrons. To ionize nitrogen, one has to overcome its electrical potential energy. The distance between the nuclear charge and one of these electrons (covalent bound distance) is 0.11 nm. The potential energy for one of these electrons (this is approximate because we need to account for charge screen)

⋅≈ = = ≈

2

2 2ke 1.44 eV nmBE(N ) U(N ) 10 eVR 0.11 nm

Since this is the potential energy that binds the electron, we can estimate the E-field via

11breakdown 2breakdown

to ionize one electron

Ve BE(N ) 10 eV 100 eVE 100V/nm 10 V/mr r 0.11 nm nm

∆≈ = ≈ = → ≈

This is exactly what we obtained using E-field calculations. Clearly this breakdown field is strong enough to ionize one of nitrogen’s gas electrons (and oxygen molecules since it has a lower ionization energy at 5.1 eV), which then starts the avalanche. Coming back to the charged conductor in air, as soon as I try to increase the voltage of the conductor beyond the breakdown voltage, free electrons are accelerated in air in the vicinity of the conductor that will ionize the air around the conductor. As a consequence, the air around the conductor is now conducting allowing charges to “leak off” the conductor into the air. In the case of the van de Graaff, the van de Graaff has less charge, which in turn, lowers the voltage, and the van de Graaff has to recharge again up to the breakdown value.

DEMO Van de Graaff with conductor has a spark in between them. The maximum field and maximum voltage in air is given by

πε= = = = × ⋅→

0

6surface surface max max

q4

V RE V RE (3 10 V/m) RR

For example, for a sphere with R = 1 cm, the maximum voltage is 6

max2V (3 10 V/m) (10 ) 30,000 Vm−= × ⋅ =

Suppose you wanted to increase the potential of the spherical conductor above the 30,000 V. NO AMOUNT of “charging” beyond this point could raise the potential in air higher than 30,000 V. The reason for this is ⇒ Attempting to raise the potential further by adding extra charge would cause the

surrounding air to become ionized and conductive. ⇒ Any extra charge added would leak charge into the air since the air at this is a

conductor. Note that these values are for air. I know that there are high-voltage machines that are placed in other gases that are not as conductive as air such as SF6 (Emax(SF6) >Emax(air)).

Plasma Balls: A plasma ball consists of a small metal ball charged to a potential of about 2000V inside a hollow glass sphere. The glass sphere is filled with a gas (either neon or argon because of the colors they produce) at a pressure of about 0.01 atm. The E-field of the high-voltage ball is sufficient to cause a gas breakdown at this pressure, creating “lightning bolts” between the ball the glass sphere. cool. DEMO Plasma ball

25.18

Radius Dependence According to Vmax = REmax, the radius depends linearly on the radius. The larger (smaller) the radius, the more (less) voltage the conductor can be brought to. DEMO Telsa coil; compare the copper wire to the sphere attachment

To obtain higher potentials, high-voltage machines (such as Van de Graaff) use spherical terminals with very large radii. Application: Car radio antennas have a ball on the end to help prevent corona effect that would cause static in the reception. At the other extreme is the effect produced by a surface of a very small radius, such as a sharp point, or the tip of a thin wire. Again, using our relationship between Vmax and Emax we find that

( )max maxR smallV

smaller values of potential will also lim RE decreases

cause the breakdown of airR

→→ ⇒

= ∝

Let’s convert the dielectric breakdown value into cm and mm:

maxE 3,000,000 V/m 30,000 V/cm 3,000 V/mm= = =

If a wire has a diameter of 1 mm, a field of 3,000 V/mm will cause the breakdown of air. For sharp points there is an associated decrease in the breakdown potential in air to produce sufficiently high fields just outside the point to ionize the surrounding air, making it a conductor. The result is a glow due to the air conducting called the CORNOA DISCHARGE. Along with this ionization comes ozone, which is unhealthful in large quantities.

Question: someone shows you this picture of standard household wires that do this and claim that it was done with 120V outlet voltage. Is this possible?

6

max max 6

5V RE (3 10 V/m) R 3 10 V/m

120 VR 4 10 m 0.04 mm−= = × ⋅

×= = × =→

Are these wires 0.04 mm in diameter? NO! So to achieve dielectric breakdown the voltage has to be higher and can be achieved by attaching a transformer to increase the voltage to a higher value. Application: Lightning rods – the Empire State building has a metal mast that acts as a lightning rod that averages about 500 strikes per year.

YOUTUBE Power line workers

Application: Kirlian photography of the "corona discharge". It obtains a photographic image due to interactions between the subject & an applied electric field. No external light source is used; light emitted as photons from the electrical interaction produces the image on film.

25.19

A source of Electric Potential Macroscopic View of Electric Circuits Potential Difference across a Battery What determine how much potential difference the motor of a “mechanical battery” can maintain across the battery? To answer this question, 1. First start with uncharged parallel plates, and transport negative charged from the left

plate to the right plate. The conveyor belt, driven by the motor, exerts what I call a nonelectrical force on the charges (FNE).

2. As the motor transport electrons to the right, the plates exert an electrical force on the electrons (FE) in the opposite direction.

3. Eventually there is enough charge on the plates to make FE = eEplates = FNE. At this point the motor cannot pump any more charge, and the plates charged up as much as they can.

We see that the function of a battery is to produce and maintain a charge separation (pull electrons out of the positive plate and push them onto the negative plate). The amount of charge separation is limited by and determined by the strength of the motor in a mechanical battery (or a car battery it is the chemical reactions in a chemical battery). So the voltage across a battery is given by

NE NEbattery platesV

F d F dE d where emfe e

∆⋅ ⋅

= ⋅ = ≡

The emf (electromotive force, which is bad name) of a battery is a property of the battery. Why is this a terrible name? It is not a force al all but is energy per unit charge. We will avoid this terminology by just using the abbreviation emf or ∆Vsource.

Source S emf V V≡ ≡ ∆ ∆=ε In a chemical battery the emf is a measure of the chemical energy per unit charge expended by the battery in moving charge through the battery. The emf of a flashlight battery is about 1.5 V, and the associated charge separation is due to chemical reactions in which electrons are reactants or products.

Role of a Battery: A battery maintains a potential difference across the battery. This potential difference is numerically equal to the battery’s emf.

There are different emf’s: 1. Van de Graaff generator – the nonelectrostatic force is mechanical which moves

charge using a belt from a low to high potential. 2. Battery or fuel cell - the nonelectrostatic force is associated with a diffusion process

and varying electrolyte concentrations resulting from chemical reactions that take charge from a low to high potential.

3. Generators – magnetic fields acting on moving charges. A source