Chapter 16: Electric Forces and Fields

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Chapter 16: Electric Forces and Fields

16.1 Electric Charge

16.2 Electric Force

16.3 The Electric Field

Chapter 16



Section 1 Electric ChargeChapter 16

Objectives for Section 1 Electric Charge• Understand the basic properties of electric charge.

• Differentiate between conductors and insulators.

• Distinguish between charging by contact, charging by induction, and charging by polarization.



Chapter 16Section 1 Electric Charge

Properties of Electric Charge• Atoms are the source of electric charge.

– Positively charged particles are called protons.– Uncharged particles are called neutrons.– Negatively charged particles are called electrons.

• Friction Rods• Electrons from animal fur are transferred to

atoms in ebonite (hard rubber). Ebonite acquires a net excess of electrons.

• Electrons from a glass rod will transfer to a silk cloth and give rise to an excess of electrons on the silk.

• Your hair becomes positively charged when you rub a balloon across it.

• Electric charge is conserved. The amount of positive charge acquired by your hair equals the amount of negative charge acquired by the balloon.



Chapter 16

Electric Charge

Section 1 Electric Charge




Properties of Electric Charge• There are two kinds of electric

charge.– like charges repel– unlike charges attract

• The magnitude of electrical forces between two charged bodies often exceeds the gravitational attraction between the bodies.

• Electric charge is conserved.




Properties of Electric Charge

• Electric charge is quantized. That is, when an object is charged, its charge is always a multiple of a fundamental unit of charge. +e, +2e, +3e, …

• The fundamental unit of charge, e, is the magnitude of the charge of a single electron or proton.

e = 1.602 176 x 10–19 C

• Charge is measured in coulombs (C).

• -1.0 C contains 6.2 x1018 electrons.



Chapter 16

The Milikan Experiment (1909)




Chapter 16

Milikan’s Oil Drop Experiment





Transfer of Electric Charge• An electrical conductor is a material in which

charges can move freely. Most metals are conductors.

• An electrical insulator, or nonconductor, is a material in which charges cannot move freely. (Electrons are tightly bound to the atom.) Nonmetallic materials, such as glass, rubber, and wood are good insulators.




Transfer of Electric Charge• Insulators and conductors can be

charged by contact.• Conductors can be charged by

induction.• Induction is a process of

charging a conductor by bringing it near another charged object and grounding the conductor.



Chapter 16

Charging by Induction





Transfer of Electric Charge

• A surface charge can be induced on insulators by polarization.

• With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation.



Section 2 Electric ForceChapter 16

Section 2 Electric Force Objectives• Calculate electric force using Coulomb’s law.

• Compare electric force with gravitational force.

• Apply the superposition principle to find the resultant force on a charge and to find the position at which the net force on a charge is zero.



Chapter 16

Coulomb’s Law (Charles Coulomb, 1780’s)• Two charges near one another exert a

force on one another called the electric force. Coulomb’s law describes this force.

• Coulomb’s law states that the electric force is proportional to the magnitude of each charge and inversely proportional to the square of the distance between them.

Section 2 Electric Force

1 22

2

charge 1 charge 2electric force = Coulomb constant

distance

electric Cq qF kr

kc = 8.99 x 109 N.m2/C2 when F in N, r in m, and q in C



Chapter 16

Coulomb’s Law• Electric force is a vector. When the charges (q1 and q2) are

alike they repel each other, and when they are opposite they attract each other.

• The resultant force on a charge is the vector sum of the individual forces on that charge. Adding forces this way is an example of the principle of superposition.

• When a body is in equilibrium, the net external force acting on that body is zero. A charged particle can be positioned such that the net electric force on the charge is zero. Set Coulomb’s Law forces equal and solve for the distance between either charge and the equilibrium position. (Practice C)




Chapter 16

Superposition Principle




Chapter 16

Sample Problem

The Superposition PrincipleConsider three point charges at the corners of a triangle, as shown at right, where q1 = 6.00 10–9 C, q2 = –2.00 10–9 C, and q3 = 5.00 10–9 C. Find the magnitude and direction of the resultant force on q3.




Chapter 16Section 2 Electric Force

Sample Problem, continuedThe Superposition Principle1. Define the problem, and identify the known

variables.Given:

q1 = +6.00 10–9 C r2,1 = 3.00 mq2 = –2.00 10–9 C r3,2 = 4.00 mq3 = +5.00 10–9 C r3,1

= 5.00 mq = 37.0º

Unknown: F3,tot = ? Diagram:

Fel = kCq1q2

r2




Sample Problem, continuedThe Superposition PrincipleTip: According to the superposition principle, the resultant

force on the charge q3 is the vector sum of the forces exerted by q1 and q2 on q3. First, find the force exerted on q3 by each, and then add these two forces together vectorially to get the resultant force on q3.

2. Determine the direction of the forces by analyzing the charges.The force F3,1 is repulsive because q1 and q3 have

the same sign.The force F3,2 is attractive because q2 and q3 have

opposite signs.




Sample Problem3. Calculate the magnitudes

of the forces with Coulomb’s law.

–9 –9293 1

3,1 22 2

–83,1

–9 –9293 2

3,2 22 2

–93,1

5.00 10 C 6.00 10 CN m 8.99 10( 3,1) C 5.00 m

1.08 10 N

5.00 10 C 2.00 10 CN m8.99 10 ( 3,2) C 4.00m

5.62 10 N

C

C

q qF kr

F

q qF kr

F

2




Sample Problem, 4. Find the x and y components of each force.

At this point, the direction each component must be taken into account.

F3,1: Fx = (F3,1)(cos 37.0º) = (1.08 10–8 N)(cos 37.0º) Fx = 8.63 10–9 NFy = (F3,1)(sin 37.0º) = (1.08 10–8 N)(sin 37.0º) Fy = 6.50 10–9 N

F3,2: Fx = –F3,2 = –5.62 10–9 NFy = 0 N




Sample Problem 5. Calculate the magnitude of the total force acting

in both directions.

Fx,tot = 8.63 10–9 N – 5.62 10–9 N = 3.01 10–9 NFy,tot = 6.50 10–9 N + 0 N = 6.50 10–9 N




Sample Problem 6. Use the Pythagorean theorem to find the magni-

tude of the resultant force.

2 2 9 2 9 23, , ,

–93,

( ) ( ) (3.01 10 N) (6.50 10 N)

7.16 10 N

tot x tot y tot

tot

F F F

F




Sample Problem 7. Use a suitable trigonometric function to find the

direction of the resultant force.In this case, you can use the inverse tangent function:

–9,

–9,

6.50 10 Ntan3.01 10 N

65.2º

y tot

x tot

FF



Coulomb’s Law• The Coulomb force is a field force.• A field force is a force that is exerted by

one object on another even though there is no physical contact between the two objects.

• Gravitational attraction is also a field force.

FG = G


m1m2

r2

Fel = kCq1q2

r2



Section 3 The Electric FieldChapter 16

Objectives• Calculate electric field strength.

• Draw and interpret electric field lines.

• Identify the four properties associated with a conductor in electrostatic equilibrium.



Chapter 16Section 3 The Electric Field

Electric Field Strength (Intensity)• An electric field is a region where an

electric force on a test charge (small positive charge) can be detected. Q exerts an electric field on q0.

• The SI units of the electric field, E, are newtons per coulomb (N/C).

• The direction of the electric field vector, E, is in the direction of the electric force that would be exerted on a small positive test charge (the direction a + test charge accelerates).

E = Felec

q0

Q q0

Q in the figure above is (+) because it is exerting a force of repulsion on the test charge.



Chapter 16

Electric Fields and Test Charges

Section 3 The Electric Field




Electric Field Strength • Electric field strength (intensity) depends on charge and distance.

An electric field exists in the region around a charged object.

• Electric Field Strength Due to a Point Charge

2

2

charge producing the fieldelectric field strength = Coulomb constant

distance

CqE kr

Felec = kC

qq0

r2E = Felec

q0 and

Q q0Q




Electric Field Strength Direction • When q is (+) the electric field is directed outward radially from q. • When q is (-) the electric field is directed toward q.




Electric Field Lines• Electric field lines are used

to analyze electric fields and show strength and direction.– The number of electric

field lines is proportional to the electric field strength.

– Electric field lines are tangent to the electric field vector at any point.



Chapter 16

Rules for Drawing Electric Field Lines




Chapter 16

Rules for Sketching Fields Created by Several Charges





4 Properties of Conductors in Electrostatic Equilibrium (Distribution of Static Charges)

1. The electric field is zero everywhere inside the conductor (otherwise charges would move and it would not be at equilibrium).

2. Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface (because the excess charges repel each other).

http://sol.sci.uop.edu/~jfalward/electricforcesfields/electricfieldofchargedconductor.jpg




4 Properties of Conductors in Electrostatic Equilibrium (Distribution of Static Charges)3. The electric field just outside a charged conductor is perpendicular

to the conductor’s surface.

4. On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.




Discharging Effects of Points• The charge density is greatest at the point of greatest curvature.

St. Elmo’s Fire

A corona discharge or brush is a slow leakage of charge that occurs when the electric field intensity is great enough to produce ionization at sharp projections or corners.

Spark Discharge Dry air can ionize at atmospheric pressure when a difference of 30,000 J/C/cm (V/cm) exists between two charged surfaces.



Chapter 16

Calculating Net Electric Field





Sample Problem

Electric Field StrengthA charge q1 = +7.00 µC is at the origin, and a charge q2 = –5.00 µC is on the x-axis 0.300 m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis 0.400 m from the origin.



Chapter 16

Sample Problem, continuedElectric Field Strength1. Define the problem, and identify the known

variables.Given:

q1 = +7.00 µC = 7.00 10–6 C r1 = 0.400 m

q2 = –5.00 µC = –5.00 10–6 C r2 = 0.500 mq = 53.1º

Unknown:E at P (y = 0.400 m)

Tip: Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.




Chapter 16

Sample Problem, continuedElectric Field Strength2. Calculate the electric field strength produced by

each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge.

–69 2 2 51

1 2 21

–69 2 2 52

2 2 22

7.00 10 C8.99 10 N m /C 3.93 10 N/C(0.400 m)

5.00 10 C8.99 10 N m /C 1.80 10 N/C(0.500 m)

C

C

qE kr

qE kr




Chapter 16

Sample Problem, continuedElectric Field Strength3. Analyze the signs of the

charges.The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2 at P due to q2 is directed toward q2 because q2 is negative.




Chapter 16

Sample Problem, continuedElectric Field Strength4. Find the x and y components of each electric field

vector.

For E1: Ex,1 = 0 N/C Ey,1 = 3.93 105 N/C

For E2: Ex,2= (1.80 105 N/C)(cos 53.1º) = 1.08 105 N/C

Ey,1= (1.80 105 N/C)(sin 53.1º)= –1.44 105 N/C




Chapter 16

Sample Problem, continuedElectric Field Strength5. Calculate the total electric field strength in both

directions.

Ex,tot = Ex,1 + Ex,2 = 0 N/C + 1.08 105 N/C = 1.08 105 N/C

Ey,tot = Ey,1 + Ey,2 = 3.93 105 N/C – 1.44 105 N/C = 2.49 105 N/C




Chapter 16

Sample Problem, continuedElectric Field Strength6. Use the Pythagorean theorem to find the

magnitude of the resultant electric field strength vector.

22

, ,

2 25 5

5

1.08 10 N/C 2.49 10 N/C

2.71 10 N/C

tot x tot y tot

tot

tot

E E E

E

E




Chapter 16

Sample Problem, continuedElectric Field Strength7. Use a suitable trigonometric function to find the

direction of the resultant electric field strength vector.In this case, you can use the inverse tangent function:

5,

5,

2.49 10 N/Ctan1.08 10 N/C

66.0

y tot

x tot

EE




Chapter 16

Sample Problem, continuedElectric Field Strength8. Evaluate your answer.

The electric field at point P is pointing away from the charge q1, as expected, because q1 is a positive charge and is larger than the negative charge q2.





Charging By Induction




Transfer of Electric Charge




Electric Field Lines

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Chapter 16: Electric Forces and Fields