12-2 Coulomb's Law

Slide 1

2014 Stan Eisenstein

Physics Clarified Algebra-Based Physics with Extra Topics for Calculus-Based Physics

Physics

E = mc2

F = ma

V = IR

K = mv2

P = mv

L = Iw

t = RFsinq

x = v0t + at2

F = ma

P = mv

x = v0t + at2

K = mv2

V = IR

t = RFsinq

L = Iw

E = mc2

1

Chapter 12 Electrostatics:Getting Zapped by Your Doorknob and Sticking Balloons to Walls

-

+

+

Power

Supply

+

-

e-

Variable Power

Supply

+

+

-

-

2

Lesson 12.2 Coulombs Law

+q

+q

d

q

d

4q

1/3 d

9q

3

Introduction

The Coulomb Experiment

Coulombs Law

vector nature

in 2-D

Review

Closing

Return to Main Table of Contents

Lesson 12.2 Table of Contents

4

Introduction

Electrostatics is the study of the electric force.

Like charges repel and unlike charges attract.

-

-

-

+

5

Introduction

Neutral objects consist of an equal number of

positive and negative charges.

Neutral

+ - + - + - + - + - + -

+ - + - + - + - + - + -

+ - + - + - + - + - + -

+ - + - + - + - + - + -

6

Introduction

Neutral objects and charged objects attract. When a

charged object approaches a neutral object, the

neutral object is polarized. The unlike charges are

closer, so attraction is stronger than repulsion.

+ - - -

- - + -

+ - + - -

- + - + -

+ - -

+ + - + - + - + - + - -

+ + - + - + - + - + - -

+ + - + - + - + - + - -

+ + - + - + - + - + - -

Negative

Neutral but Polarized

+

-

7

Introduction

There are three methods of charging objects.

1) In the triboelectric effect, electrons are transferred from one object to the other as they come in contact often through rubbing.

+ - + - + - + -

+ - + -

+ - + -

-

Neutral

+ + - + + -

+ - + -

+ - + -

-

-

-

+

8

Introduction

2) In conduction, a charged object shares charge with another object through contact.

6+

+ + - +

+ + +

+ +

0

- + - +

+ - +

+ -

+ + - +

+ + +

4+

+ +

2+

-

-

-

+ + - +

+ + +

+ +

+ - +

+ - +

+ -

-

-

9

Introduction

3) In induction, a charged object is held near two neutral conductors in contact. The like charges move away from the charged object. When the conductors are separated, they are charged equally and opposite.

- + - +

+ - +

- + - +

0

0

- + - +

+ - +

- + - +

3+

3-

-

-

-

-

- + +

- + +

- + +

- - + - +

- + - +

- - + - +

-

-

-

-

3+

- + +

- + +

- + +

3-

- - + - +

- + - +

- - + - +

10

Introduction

In all these methods of charging, the law of

conservation of charge holds. The total amount of

charge stays constant.

6+

+ + - +

+ + +

+ +

0

- + - +

+ - +

+ -

+ + - +

+ + +

4+

+ +

2+

-

-

-

+ + - +

+ + +

+ +

+ - +

+ - +

+ -

-

-

11

Introduction

So far we have studied electric force and charge

qualitatively, without the use of the numbers and

equations. In this lesson we analyze the electric

force quantitatively.

+q

F = kq1q2/d2

+q

12


In 1784, Charles-Augustin de Coulomb invented

the torsion balance, the same balance that

Cavendish used to determine the value of G in

Newtons Law of Universal Gravitation.

+q

+q

Return to Table of Contents

13


Coulomb showed that the amount of force (torque,

actually) needed to maintain a certain amount of

twist of the supporting wire was directly

proportional to the angle of twist.

Side View

Top View

0 F

1 F

2 F

3 F

14


In 1785, Coulomb used his invention to study the

proportionality relationship between the electrical

force between two spheres, the amount of charge

on the spheres, and the distance between the

spheres.

+q

+q

15


For instance, Coulomb charged two spheres with

like charge (q) so that they would repel, causing

the balance to twist through an angle q when the

charges were separated by a distance d. If the

charges were brought closer,

to a distance d, then the

angle of twist would be

multiplied by 4. For a

separation distance of d/3,

the angle of twist was 9q.

(click to animate)

Top View

+q

+q

d

q

d

4q

1/3 d

9q

16


Question 1a: What proportionality relationship

between electric force and separation distance did

Coulomb find?
Force (angle)q4q9qSeparation Distancedd/2d/3

17


Answer 1a: Force is inversely proportional to the

square of the separation distance.

F a 1/d2

If d is multiplied by , then F is multiplied by

1/(1/2)2 or 4.

If d is multiplied by 1/3, then F is multiplied by

1/(1/3)2 or 9.

18

In order to determine the effect of the size of the

charges on the force, Coulomb needed to be able

to change the size of the charge in a predictable

way. He did this by touching a charged conductor

to an uncharged conductor of the same size.

Although he did not know how much charge was

on the original conductor, he knew that the charge

must be cut in half. (click to animate)


+q

+q/2

+q/4

+q/4

+q/2

19

d

q/2

With a charge +q on each of the spheres, the

balance rotates through an angle q when the

sphere are held a distance d apart. When the

charge on one sphere is reduced to q/2, the angle

is halved. When the charge

is reduced to q/4, the

angle is reduced to q/4.

(click to animate)


Top View

+q

+q

d

q

+q/2

+q/2

+q/4

+q/4

d

q/4

20


Question 1b: What proportionality relationship

between electric force and charge did

Coulomb find?
Force (angle)qq/2q/4Chargeqq/2q/4

21


Answer 1b: Force is directly proportional to the

charge.

F a q

If q is multiplied by , then F is multiplied by

.

If q is multiplied by 1/4, then F is multiplied by

1/4.

22


Answer 1b:

F a q

It doesnt matter which sphere changes charge, so

actually, the force is directly proportional to the

charge on either sphere.

F a q1q2

23

Coulombs Law

F a 1/d2F a q1q2

Putting these proportionalities together yields

Coulombs Law:

F = kq1q2/d2

where q1 and q2 are the charges, d is the

separation distance between the charges, and k is

the proportionality constant.


24

Coulombs Law

F = kq1q2/d2

The unit for charge q is the coulomb. A coulomb

of charge is equal to 6.25 x 1018 elementary

charges. An elementary charge is the charge on

an electron or a proton. Conversely, one

elementary charge equals 1.6 x 10-19 coulombs.

e-

6.25 x 1018

= 1 coulomb

25

Coulombs Law

F = kq1q2/d2

A coulomb is a very large amount of charge, about

the amount of charge delivered in a small lightning

bolt. A typical lightning bolt delivers about

5 coulombs of charge.

A few coulombs

26

Coulombs Law

F = kq1q2/d2

A typical amount of charge that you might

experience from rubbing your feet on a carpet is

between 10-8 and 10-5 coulombs. A common unit

for measuring these charges is the microcoulomb

(mC). One mC equals 10-6 coulombs.

10-8 10-5 C

27

Coulombs Law

F = kq1q2/d2

The constant k has a value of

9.0 x 109 newtonmeter2/coulomb2. This large

value is indicative of the strength of the electric

force.

p+

e-

5 x 10-11 m

28

Coulombs Law

F = kq1q2/d2 F = Gm1m2/d2

To get an idea of the relative strength of the

electric force, lets compare the electric force to

the gravitational force between the electron and

the proton in a hydrogen atom. Recall that the

force of gravity is given by the

equation shown above in blue.

G = 6.67 x 10-11 nm2/kg2

p+

e-

5.3 x 10-11 m

29

Coulombs Law


Question 2:

An electron has a mass of 9.1 x 10-31 kg. A proton

has a mass of 1.7 x 10-27 kg. Both have a charge

of 1.6 x 10-19 coulombs. The separation distance

is 5.3 x 10-11 m. Determine the

sizes of the gravitational and

electric forces between the

proton and the electron.

p+

e-

5.3 x 10-11 m

30

Coulombs Law


Answer 2: The gravitational force is 3.7 x 10-47 n.

The electric force is 8.2 x 10-8 n. The electric force

is almost 1040 times stronger than the gravitational

force.

p+

e-

5.3 x 10-11 m

31

Coulombs Law

F = kq1q2/d2

The constant k is also stated as 1/4pe0.

e0 is called the constant of permittivity and has a

value of 8.85 x 10-12 coulomb2/newtonmeter2.

1/4pe0. = 9.0 x 109 nm2/c2

k = 1/4pe0

32

Coulombs Law

F = kq1q2/d2

Coulombs Law applies only to spherical charges

or point charges. Point charges have sizes that

are much smaller than the separation distance.

Coulombs Law applies to Spherical and Point Charges

33

Coulombs Law

F = kq1q2/d2

For instance, suppose the two rectangles shown

below are both charged. Coulombs law does not

apply to these charges. The sizes of the charged

bodies are comparable to the separation distance.

These bodies are not point charges.

d size F kq1q2/d2

34

Coulombs Law

F = kq1q2/d2

As the rectangles are moved further apart, they

approach becoming point charges. As the

separation distance increases, it becomes

substantially larger than the size of the charged

objects. Coulombs Law applies better as the

bodies move further apart.

d >> size F kq1q2/d2

35

Vector Nature of Coulombs Law

F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Question 3a: A point charge of +1.0 mC is placed

10 meters to the left of a point charge of -1.0 mC.

What is the force acting on each charge?

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC


36


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Answer 3a: (-)9 x 10-5 n or (-)90 mn

Dont forget to convert 1.0 mC into Coulombs.

F = (9.0 x 109 nm2/C2)(+1.0 x 10-6 C)(-1.0 x 10-6 C)

(10 m)2

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

37


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Question 3b: (-)9 x 10-5 n or (-)90 mn

What does the negative sign for force mean?

Does it tell you the direction the charge is forced?

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

38


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Answer 3b: (-)9 x 10-5 n or (-)90 mn

The negative sign tells you that the charges

attract. If the charges had been the same sign

(either both positive or both negative), then the

force would be positive, meaning that the charges

repel.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

-90 mn

-90 mn

Attraction

39


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Answer 3b: The negative sign does not tell you the

absolute direction the charge is forced. In this

case, both charges experience a negative force.

This means that one charge is force to the left and

the other is forced to the right.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

-90 mn

-90 mn

Attraction

40


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Physics Trick: When solving electricity problems

that involve vector quantities, such as force, do

not include the sign of the charge in the

calculation. Make all numbers positive, and then

use your reasoning to determine the sign.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

-90 mn

-90 mn

Attraction

41


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Question 4a: A +2.0 mC charge is placed 5 m to

the right of a +1.0 mC charge and 5 m to the left of

a 1.0 mC charge. What is the size and direction of

the net force on the +2.0 mC charge?

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

42


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Answer 4a: 0.00144 n to the right.

Each charge exerts a force of 0.00072 n

F = (9.0 x 109 nm2/C2)(2.0 x 10-6 C)(1.0 x 10-6 C)

(5 m)2

Notice that no negative signs are used.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

43


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0


Now we note that the +1.0 mC charge repels the

+2.0 mC charge to the right. The -1.0 mC charge

attracts the +2.0 mC charge, also to the right.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

0.00072 n

0.00072 n

44


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0


Because the forces are in the same direction, they

are added.

Fnet = F+ + F- = (0.00072 n) + (0.00072 n) = 0.00144 n

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

0.00072 n

0.00072 n

45


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0


Notice that if we had used the signs of the charges

in the equation, we would have arrived at an

erroneous result of zero.

Fnet = F+ + F- = (0.00072 n) + (-0.00072 n) = 0 n

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

46


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Question 4b: What is the size and direction of the

force acting on the +1.0 mC charge?

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

47


Answer 4b: 6.3 x 10-4 n to the left.

The -1.0 mC charge attracts the +1.0 mC charge to

the right with a force of 9 x 10-5 n (see question 3)

The +2.0 mC charge repels the +1.0 mC charge to

the left with a force of 7.2 x 10-4 n (see question 4a)

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

7.2 x 10-4 n

9 x 10-5 n

48



Because the forces oppose each other, they

subtract.

Fnet = F+ + F- = (-0.00072 n) + (0.00009 n)

= (-0.00063 n)

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

7.2 x 10-4 n

9 x 10-5 n

49



Fnet = -0.00063 n

The negative sign means that the net force points

left.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

7.2 x 10-4 n

9 x 10-5 n

50


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Question 5a: Charge A of +1.0 mC charge is

placed at a position of +2 m. Charge B of +4.0 mC

charge is placed at a position of -1 m. At what

position (not including ) could a third positive

charge be placed and experience a net force of

zero?

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

+ 4.0 mC

A

B

51


Answer 5a: +1 m.

In order to experience a net force of zero, the third

charge must experience opposite forces. If placed

to the left of B, both repelling forces are to the left.

If placed to the right of A, both repelling forces are

to the right. Only between A and B do the repelling

forces oppose each other.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

52


Answer 5a: +1 m.

The forces must be equal as well.

FB = kq(4 x 10-6 C)/dB2 = kq(1 x 10-6 C)/dA2 = FA

Canceling terms and rearranging yields:

dB2 = 4 dA2 dB = 2 dA

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

dB

dA

53


Answer 5a: +1 m.

dB = 2 dA

We also know that dB + dA = 3 m

Substituting: 2dA + dA = 3 m dA = 1 m; dB = 2 m

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

dB

dA

54


Question 5b: How would your answer to question

5a have been different if the third charge were

negative instead of positive?

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

+ 1.0 mC

+ 4.0 mC

55


Answer 5b: No difference. The answer would be

the same. To the left of B, the attracting forces

would both point right. To the right of A, the

attracting forces would both point left. Only

between A and B do the attracting forces oppose

each other.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

56


F = kq1q2/d2k = 9.0 x 109 nm2/C2 = 1/4pe0

Question 5c: How would your answer to question

5a change if charge A were negative?

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

- 1.0 mC

+ 4.0 mC

A

B

57


Answer 5c: Now the position where the third

charge experiences no net force is +5 m.

The correct position cannot be between A and B

because now the attraction from A and the

repulsion from B point in the same direction.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

58


Answer 5c: +5 m.

The correct position cannot be to the left of B

either. Although the attraction from A to the right

opposes the repulsion from B to the left, B has a

larger charge and a lesser distance, so it will

always exert a larger force.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

59


Answer 5c: +5 m.

To the right of A, the attraction from A to the left

opposes the repulsion from B to the right. B has a

larger charge but a further distance, so the forces

can balance somewhere in this region.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

60


Answer 5c: +5 m.

Setting the forces equal again yields:

FB = kq(4 x 10-6 C)/dB2 = kq(1 x 10-6 C)/dA2 = FA

Canceling terms and rearranging yields:

dB2 = 4 dA2 dB = 2 dA

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

dB

dA

61


Answer 5c: +5 m.

dB = 2 dA

We also know that dA + 3 m = dB

Substituting: dA + 3 m = 2dA dA = 3 m; dB = 6 m

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

A

B

dB

dA

62

Coulombs Law in 2-D

Question 6a: A charge of +4q is placed a distance

2d to the west of a charge of +2q. A charge of + q

is placed a distance d to the south of the +2q

charge. What are the size and direction of the net

force acting on the +2q charge? Note: leave k as

k; do not substitute its numerical value.

+4q

+2q

+q

2d

d


63

Coulombs Law in 2-D

Answer 6a: (kq2/d2)22 45 N of E

F4q = k(4q)(2q)/(2d)2 = 2kq2/d2 East (repulsion)

Fq = k(q)(2q)/(d)2 = 2kq2/d2 North (repulsion)

Fnet = [(2kq2/d2)2 + (2kq2/d2)2] = (kq2/d2)22

q = tan-1[(2kq2/d2)/(2kq2/d2)] = 45

+4q

+q

2d

d

Fq

F4q

Fnet

q

64

Coulombs Law in 2-D

Question 6b: A charge of -4q is placed a distance

2d to the west of a charge of +2q. A charge of + q

is placed a distance d to the south of the +2q

charge. What are the size and direction of the net

force acting on the +2q charge? Note: leave k as

k; do not substitute its numerical value.

-4q

+2q

+q

2d

d

65

Coulombs Law in 2-D

Answer 6b: (kq2/d2)22 45 N of W

F4q = k(4q)(2q)/(2d)2 = 2kq2/d2 West (attraction)

Fq = k(q)(2q)/(d)2 = 2kq2/d2 North (repulsion)

Fnet = [(2kq2/d2)2 + (2kq2/d2)2] = (kq2/d2)22

q = tan-1[(2kq2/d2)/(2kq2/d2)] = 45

-4q

+q

d

Fq

F4q

Fnet

q

66

Coulombs Law in 2-D

Question 7: Determine the net force acting on

charge C from charges A and B. Hint: First

determine the sizes of each of the forces. Then

determine the size and direction of x- and y-

components of each force.

A

0.50 m

0.50 m

37

+2.5 x 10-6 C

+2.5 x 10-6 C

+1.0 x 10-6 C

37

B

C

67

Coulombs Law in 2-D

Answer 7: 0.108 n in the +y direction.

F = (9.0 x 109 nm2/C2)(2.5 x 10-6 C)(1.0 x 10-6 C)

(0.5 m)2

FA = FB = 0.090 n

A

0.50 m

0.50 m

37

+2.5 x 10-6 C

+2.5 x 10-6 C

+1.0 x 10-6 C

37

B

C

68

Coulombs Law in 2-D


FAx = (0.09 n)cos(37) = 0.072 n

FBx = (0.09 n)cos(37) = -0.072 n

FAx + FBx = 0 n

A

0.50 m

0.50 m

37

+2.5 x 10-6 C

+2.5 x 10-6 C

37

B

C

FB

FBy

FBx

FA

FAy

FAx

37

37

69

Coulombs Law in 2-D


FAy = (0.09 n)sin(37) = 0.054 n

FBy = (0.09 n)sin(37) = 0.054 n

FAy + FBy = 0.108 n

A

0.50 m

0.50 m

37

+2.5 x 10-6 C

+2.5 x 10-6 C

37

B

C

FB

FBy

FBx

FA

FAy

FAx

37

37

70

Coulombs Law in 2-D


Notice that because of symmetry, the x-

components of the forces cancel and the y-

components add.

A

0.50 m

0.50 m

37

+2.5 x 10-6 C

+2.5 x 10-6 C

37

B

C

FB

FBy

FBx

FA

FAy

FAx

37

37

71

Review

Coulombs Law describes the force between two point charges and is given by the equation:

F = kq1q2/d2

in which q1 and q2 are the charges, d is the

separation distance between the charges, and k is

the proportionality constant equal to

9.0 x 109 nc2/m2.


72

Review

Electric Force is a vector. Forces in the same direction add and forces in opposite directions subtract. When using Coulombs Law, use the absolute value of the charges in the equation and then use reasoning regarding attractions and repulsions to determine the directions of the forces.

-

+

0

1

2

3

4

5

-5

-4

-3

-2

-1

X (m)

+ 1.0 mC

- 1.0 mC

+ 2.0 mC

0.00072 n

0.00072 n

73

Review

When working with Coulombs Law in two dimensions, we determine the x- and y-components of the net force separately. Again, attractions and repulsions are used to determine the direction of each force.

-4q

+q

d

Fq

F4q

Fnet

q

74

Review

Sometimes there is sufficient symmetry that one of the force components cancels while the other force component adds.

A

0.50 m

0.50 m

37

+2.5 x 10-6 C

+2.5 x 10-6 C

37

B

C

FB

FBy

FBx

FA

FAy

FAx

37

37

75

Closing

12-2 Coulomb's Law

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Transcript of 12-2 Coulomb's Law