Chapter 20

Electric Forces and Fields: Coulomb’s Law

Reading: Chapter 20

1

Electric Charges

There are two kinds of electric charges- Called positive and negative

Negative charges are the type possessed byNegative charges are the type possessed by electrons

Positive charges are the type possessed by protons

Charges of the same sign repel one another and charges with opposite signs attract one another

El i h i l d i i l dElectric charge is always conserved in isolated system

Neutral – equal number of positive and negative chargesand negative charges

P iti l h d

2

Positively charged

Electric Charges: Conductors and Isolators

Electrical conductors are materials in which some of the electrons are free electrons

fThese electrons can move relatively freely through the material

Examples of good conductors include copper, aluminum and silver

Electrical insulators are materials in which all of the electrons are bound to atoms

These electrons can not move relatively freely through the materialthrough the material

Examples of good insulators include glass, rubber and wood

3Semiconductors are somewhere between

insulators and conductors

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Electric Charges

Coulomb’s Law

• Mathematically, the force between two electric charges:

= 1 2q qF k

• The SI unit of charge is the coulomb (C)

=12 2eF kr

• ke is called the Coulomb constant– ke = 8.9875 x 109 N.m2/C2 = 1/(4πeo)

e is the permittivity of free space– eo is the permittivity of free space– eo = 8.8542 x 10-12 C2 / N.m2

Electric charge:Electric charge:electron e = -1.6 x 10-19 Cproton e = 1.6 x 10-19 C

16

Coulomb’s Law

1 212 21 2

| || |e

q qF F kr

= =

21++

21

21F 12FDirection depends on the sign of the product

1 2q q 1 2 0q q >

21 12F F= − opposite directions, the same magnitude

−

1 2

1 2 0q q >−

21F 12F

1 2 0q q >

−+

1 2

0q q21F 12F

1 2 0q q <

The force is attractive if the charges are of opposite signThe force is repulsive if the charges are of like sign

17

The force is repulsive if the charges are of like sign

Magnitude: 1 212 21 2

| || |e

q qF F kr

= =

Coulomb’s Law: Superposition Principle

• The force exerted by q1 on q3 is F13

• The force exerted by q2 on q3 is F23y q2 q3 23

• The resultant force exerted on q3 is the vector sum of F and Fsum of F13 and F23

18

Coulomb’s Law

q q 5 6= 1 2

21 2eq qF kr

++

1 2 13F−

23F 33 21 31F F F= +

1 1q мC= 2 2q мC= −

5m 6m Resultant force:

3F1 1q мC 2 2q мC

3 3q мC=Magnitude:

6 6| || | 2 10 3 10

13F

23F

6 6| || | 10 3 10q q − −

6 69 43 2

23 2 2

| || | 2 10 3 108.9875 10 15 106e

q qF k N Nr

− −−⋅ ⋅ ⋅

= = ⋅ = ⋅

6 69 43 1

13 2 2

| || | 10 3 108.9875 10 2.2 1011e

q qF k N Nr

−⋅ ⋅= = ⋅ = ⋅

4 4 4(15 10 2 2 10 ) 12 8 10F F F N N− − −4 4 43 23 13 (15 10 2.2 10 ) 12.8 10F F F N N= − = ⋅ − ⋅ = ⋅

19

Coulomb’s Law

q q 5 61 221 212

€e

q qF k rr

= ++

1 2 32F−

12F 32 12 32F F F= +

F1 1q мC= 2 2q мC= −

5m 6m

F

Resultant force:

2F1 1q мC 2 2q мC

3 3q мC=Magnitude:

6 6| || | 2 10 3 10

32F

12F

6 6| || | 10 2 10q q − −

6 69 43 2

32 2 2

| || | 2 10 3 108.9875 10 15 106e

q qF k N Nr

− −−⋅ ⋅ ⋅

= = ⋅ = ⋅

6 69 41 2

12 2 2

| || | 10 2 108.9875 10 7.2 105e

q qF k N Nr

−⋅ ⋅= = ⋅ = ⋅

4 4 4(15 10 7 2 10 ) 7 8 10F F F N N− − −4 4 42 32 12 (15 10 7.2 10 ) 7.8 10F F F N N= − = ⋅ − ⋅ = ⋅

20

Coulomb’s Law

q q 5 61 221 212

€e

q qF k rr

= ++

1 221F−

31F 31 21 31F F F= +

F1 1q мC= 2 2q мC= −

5m 6m

F

Resultant force:

1F1 1q мC 2 2q мC

3 3q мC=Magnitude: 21F

31F

6 6| || | 10 3 10

6 6| || | 10 2 10q q − −

6 69 43 1

31 2 2

| || | 10 3 108.9875 10 2.2 1011e

q qF k N Nr

− −−⋅ ⋅

= = ⋅ = ⋅

6 69 41 2

21 2 2

| || | 10 2 108.9875 10 7.2 105e

q qF k N Nr

−⋅ ⋅= = ⋅ = ⋅

4 4 4(7 2 10 2 2 10 ) 5 10F F F N N− − −4 4 41 21 31 (7.2 10 2.2 10 ) 5 10F F F N N= − = ⋅ − ⋅ = ⋅

21

Coulomb’s Law

q q= 1 2

21 2eq qF kr +

131F

21F1 21 31F F F= +

F F1 1q мC= 2 2q мC= −5m 6m

Resultant force:

+

2−

3

31F 21F

1F

1 1q мC 2 2q мC

3 3q мC= 7mMagnitude:ag tude

6 69 41 2

21 2 2

| || | 10 2 108.9875 10 5 106e

q qF k N Nr

− −−⋅ ⋅

= = ⋅ = ⋅

6 69 33 1

31 2 2

| || | 10 3 108.9875 10 1.1 105e

q qF k N Nr

− −−⋅ ⋅

= = ⋅ = ⋅

13F+

1

F

5m 6m

+

112F

5m6m

13F

23F3F

12F1F

32F

22

+

213F −

23F3 7m+

232F−

3 7m

Conservation of Charge Electric charge is always conserved in isolated system

1 1q мC= 2q мC

g y yTwo identical sphere

1 1q мC2 2q мC= −

Th t d b d ti i Wh t i th l t iThey are connected by conducting wire. What is the electric charge of each sphere?

The same charge q. Then the g qconservation of charge means that :

1 2 1 2 0.52 2

q qq мC мC+ −= = = −1 22q q q= +

2 2

For three spheres: 1 1q мC= 2 2q мC= − 3 3q мC=

23

1 2 33q q q q= + +

1 2 3 1 2 3 13 3

q q qq мC мC+ + − += = =

Chapter 20

El i Fi ldElectric Field

24

Electric Field

An electric field is said to exist in the region of space around a charged object

This charged object is the source chargeThis charged object is the source chargeWhen another charged object, the test charge, enters this electric

field, an electric force acts on it.The electric field is defined as the electric force on the test charge

per unit chargeFE =

If you know the electric field you can find the force

0

Eq

=

If you know the electric field you can find the force

If i iti F d E i th di ti

F qE=

25

If q is positive, F and E are in the same directionIf q is negative, F and E are in opposite directions

Electric Field

The direction of E is that of the force on a positive test chargeThe SI units of E are N/C

0

FEq

=

= 02e

qqF kr

Coulomb’s Law:

20

eF qE kq r

= =

Then

0q r

26

Electric Field

q is positive, F is directed away from qfrom q

The direction of E is also away from the positive source chargep g

q is negative, F is directed toward qE is also toward the negative

source charge

= 02e

qqF kr 2

0e

F qE kq r

= =

27

Electric Field: Superposition Principle

• At any point P, the total electric field due to a group of source charges equals the vector sum of electric g qfields of all the charges

= ∑ ii

E Ei

28

Electric Field

5 6+

1 2 1E−

2E

Electric Field:

1 2E E E= +

1 1q мC= 2 2q мC= −

5m 6m2eqE kr

=

E E1 1q мC 2 2q мC

Magnitude:6| | 2 10

2E

1E E

6| | 10q −

69 22

2 2 2

| | 2 108.9875 10 / 5 10 /6e

qE k N C N Cr

−⋅= = ⋅ = ⋅

69 21

1 2 2

| | 108.9875 10 / 0.7 10 /11e

qE k N C N Cr

= = ⋅ = ⋅

2 2 22 2 22 1 (5 10 0.7 10 ) / 4.3 10 /E E E N C N C= − = ⋅ − ⋅ = ⋅

29

Electric Field

1E

2E

Electric field

1 1q мC= 2 2q мC= −5m 6m

1 2E E E= +

EE

2eqE kr

=

+

1 2−

1 1q мC 2 2q мC

7m

2E1E

E

Magnitude:

69 22| | 2 108 9875 10 / 5 10 /qE k N C N C

−⋅

69 21| | 108 9875 10 / 0 37 10 /qE k N C N C

−

= = ⋅ = ⋅

9 222 2 2

| | 8.9875 10 / 5 10 /6e

qE k N C N Cr

= = ⋅ = ⋅

1 2 28.9875 10 / 0.37 10 /5eE k N C N C

r= = ⋅ = ⋅

30

Electric Field

1 10q мC= 2 10q мC=

2eqE kr

= z

1 10q мC 2 10q мC

E Direction of electric field?

5m ϕ

+ +

5m5m

ϕ

1 26m

31

Electric Field

Electric field

1 10q мC= 2 10q мC=1 2E E E= +

2eqE kr

=E

2E

ϕ1E2E

1 10q мC 2 10q мC

ϕ

ϕ

+ +

5m5m

ϕ

+

1 2+

6mMagnitude:

6| | 10 10q −69 32

2 1 2 2

| | 10 108.9875 10 / 3.6 10 /5e

qE E k N C N Cr

⋅= = = ⋅ = ⋅

2E E −2 25 3 4 8

32

= 12 cosφE E −= =

5 3 4cosφ5 5 1

85

E E=

Example

1 10q мC= 2 10q мC=1l m=

2ϕ1l m=

T1ϕ

+

1 2+

EF Electric field

1 10q мC 2 10q мC

r1 2m m m= =

0ET mg F+ + =mg

g

1 22E e

q qF k

r=

T iT F2cosT mgϕ = 2sin ET Fϕ =

1 21 2tan E

eF q q

kmg r mg

ϕ = =1 22 2tan E

eF q q

kϕ = =

33

mg r mg2 2emg r mgϕ

2 1ϕ ϕ=

0planeE

eσ=2πk σ=

2εplaneE

0Q >

σ>0

planeE

02ε

Q=σAplaneE

σ<0planeE= e

0

|σ|2πk |σ|=2εplaneE

< 0Q

σ<0

planeE Q=σA

34

Fi d l t i fi ld

Important Example

Find electric field σ>0

+ + + + ++ + + + +

+

-σ<0− − − − −

− − − − −−

σ>0E−E+

−σ<0

+ =σ

2εE

E+ E−− =

σ2ε

E

35

02ε 02ε

Fi d l t i fi ld

Important Example

σ>0+ + + + +Find electric field σ>0

−σ<0

+ + + + ++ + + + +

+

− − − − −+ =σ

2E − =

σ2

E σ<0− − − − − −

+02ε 02ε

E E E= +

σ>0E−E+

E E E+ −= +

0E E E+ −= − =

−σ<0E−E+ + −= + =

0

σε

E E E

E+E−

0E E E+ −= − =

36

Important Example

0у >+ + + + + 0у >

0у− <

+ + + + ++ + + + +

+

− − − − − 0у <− − − − − −

σ>00E =

E= =0 0

σε ε

QEA

−σ<00E =

37

Electric field due to a thin uniformly charged spherical shell

• When r < a, electric field is ZERO: E = 0

1A∆ 1 1q Aσ∆ = ∆ 2 2q Aσ∆ = ∆

2A∆ ∆Ω 2A∆ ∆Ω1r

the same

21 1A r∆ = ∆Ω 2

2 2A r∆ = ∆Ω

2E∆ 21 1 1

1 2 2 2e e e eq A r

E k k k kr r r

σ σσ

∆ ∆ ∆Ω∆ = = = = ∆Ωthe same

solid angle1E∆

1 1 1r r r

22 2 2

2 2 2 2e e e eq A r

E k k k kr r r

σ σσ

∆ ∆ ∆Ω∆ = = = = ∆Ω

2r2 2 2r r r

1 2E E∆ = ∆

1 2 0E E∆ + ∆ =

382A∆ 2

1Er

∝Only because in Coulomb law

M i f Ch d P i lMotion of Charged Particle

39

Motion of Charged Particle

• When a charged particle is placed in an electric field, it experiences an electrical force

• If this is the only force on the particle it must be the net force• If this is the only force on the particle, it must be the net force• The net force will cause the particle to accelerate according to

Newton’s second law

F qE= - Coulomb’s law

F ma= - Newton’s second law

qa Em

=

40

Motion of Charged Particle

What is the final velocity?

F qE= - Coulomb’s lawq

F ma= - Newton’s second law

| |q fv| |y

qa Em

= −

vMotion in x with constant velocity

l

Motion in x – with constant acceleration 0v | |

yqa Em

= −Motion in x – with constant velocity

0

ltv

= - travel time

After time t the velocity in y direction becomes0xv v=

41| |

y yqv a t Etm

= = −fvyvthen

220f

qv v Etm

= +

Chapter 20

C d t i El t i Fi ldConductors in Electric Field

42

Electric Charges: Conductors and Isolators

Electrical conductors are materials in which some of the electrons are free electrons

fThese electrons can move relatively freely through the material

Examples of good conductors include copper, aluminum and silver

Electrical insulators are materials in which all of the electrons are bound to atoms

These electrons can not move relatively freely through the materialthrough the material

Examples of good insulators include glass, rubber and wood

43Semiconductors are somewhere between

insulators and conductors

Electrostatic Equilibrium

Definition:when there is no net motion of charges within a conductor, the conductor is said to be in electrostatic equilibrium

Because the electrons can move freely through the material F qE

no motion means that there are no electric forcesno electric forces means that the electric field

inside the conductor is 0

F qE=

inside the conductor is 0

If electric field inside the conductor is not 0, then0E ≠

44

If electric field inside the conductor is not 0, then there is an electric force and, from the second Newton’s law, there is a motion of free electrons.

0E ≠F qE=

Conductor in Electrostatic Equilibrium

• The electric field is zero everywhere inside the conductor

• Before the external field is applied, free electrons are distributedfree electrons are distributed throughout the conductor

• When the external field is applied, the l t di t ib t til thelectrons redistribute until the

magnitude of the internal field equals the magnitude of the external field

• There is a net field of zero inside the conductor

45

Conductor in Electrostatic Equilibrium

• If an isolated conductor carries a charge, the charge resides on its surface

0E =

46