Chapter 21

Electric Charge andElectric Field

Physics for Scientists & Engineers, 3rd EditionDouglas C. Giancoli

© Prentice Hall

7P01 -

Why Study Physics?

• Understand/Appreciate Nature• Understand Technology• Learn to Solve Difficult Problems• It’s Required

19P01 -

Physics is not Math…

20P01 -

…but we use concepts from 18.02

V= −∇E•Gradients

•Path Integrals

•Surface Integrals

•Volume Integrals

B

AV d∆ ≡ − ⋅∫ E s

0

in

S

Qdε

⋅ =∫∫ E A

Q dVρ= ∫∫∫

25P01 -

8.02: Electricity and MagnetismAlso new way of thinking…

How do objects interact at a distance? Fields We will learn about E & M Fields:

how they are created & what they effectBig Picture Summary:

0

0 0 00

in B

S C

Eenc

S C

Q dd ddt

dd d Idt

ε

µ µ ε

Φ⋅ = ⋅ = −

Φ⋅ = ⋅ = +

∫∫ ∫

∫∫ ∫

E A E s

B A B s

MaxwellEquations:

( )q= + ×F E v BLorentz Force:

46P01 -

Electric Charge (~Mass)Two types of electric charge: positive and negativeUnit of charge is the coulomb [C]

Charge of electron (negative) or proton (positive) is

Charge is quantized

Charge is conserved

19, 1.602 10e e C−± = ×

Q Ne= ±

n p e ν−→ + + e e γ γ+ −+ → +

Figure 21-3

Figure 21-4

Figure 21-5 (a)

Figure 21-5 (b)

Figure 21-5 (c)

Figure 21-2

Figure 21-6

47P01 -

Electric Force (~Gravity)The electric force between charges q1 and q2 is

(a) repulsive if charges have same signs (b) attractive if charges have opposite signs

Like charges repel and opposites attract !!

Figure 21-14

48P01 -

Coulomb's Law

1 212 2

ˆeq qkr

=F rCoulomb’s Law: Force by q1 on q2

9 2 2

0

1 8.9875 10 N m /C4ekπε

= = ×

ˆ :r unit vector from q1 to q2

rrr =ˆ 1 2

12 3eq qkr

⇒ =F r

Figure 21-15

Figure 21-16

50P01 -

The Superposition PrincipleMany Charges Present:Net force on any charge is vector sum of forces from other individual charges

3 13 23= +F F FExample:

In general:

1

N

j iji=

=∑F F

Figure 21-17

Figure 21-18

49P01 -

Coulomb's Law: Example

( )3132 2 2

ˆ ˆ m

1mr

= −

=

r i j?32 =F

a = 1 m

q1 = 6 C

q3 = 3 C

q2 = 3 C

32r

( )( )( )( )( )

129 2 2

3

ˆ ˆ3 m9 10 N m C 3C 3C

1m

−= ×

i j32 3 2 3ek q q

r=

rF

( )981 10 ˆ ˆ3 N

2×

= −i j

Figure 21-19

Figure 21-20

Figure 21-32

Figure 21-33 (a)

Figure 21-33 (b)

35P01 -

Visualizing Vector Fields:Three Methods

Vector Field DiagramArrows (different colors or length) in direction of field

on uniform grid.Field Lines

Lines tangent to field at every point along lineGrass Seeds

Textures with streaks parallel to field direction

All methods illustrated inhttp://ocw.mit.edu/ans7870/8/8.02T/f04/visualization

s/electrostatics/39-pcharges/39-twocharges320.html

36P01 -

Vector Fields – Field Lines

• Direction of field line at any point is tangent to field at that point

• Field lines never cross each other

38P01 -

Vector Fields – “Grass Seeds”

Source/Sink Circulating

Although we don’t know absolute direction, we can determine relative direction

42P01 -

2ˆg

MmGr

= −F r

Example Of Vector Field: Gravitation

Gravitational Force:

Gravitational Field:

2

2

/ ˆ ˆg GMm r MGm m r

= = − = −F

g r r

M : Mass of Earth

43P01 -

Example Of Vector Field: Gravitation

Gravitational Field:

2ˆMG

r= −g r g m=F g

Created by M Felt by m

ˆ :r unit vector from M to m

rrr =ˆ

M : Mass of Earth

3

MGr

⇒ = −g r

45P01 -

From Gravitational toElectric Fields

51P01 -

Electric Field (~g)The electric field at a point is the force acting on a test charge q0 at that point, divided by the charge q0 :

0q≡

FE

2ˆe

qkr

=E rFor a point charge q:

http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/04-MovingChargePosElec/04-MovChrgPosElec_f223_320.html

52P01 -

Superposition Principle

The electric field due to a collection of N point charges is the vector sum of the individual electric fields due to each charge

1 21

.. . . .N

total ii=

= + + =∑E E E E

53P01 -

Summary Thus Far

Mass M Charge q (±)

2ˆMG

r= −g r 2

ˆeqkr

=E rCREATE:

g m=F g E q=F E

This is easiest way to picture field

FEEL:

Figure 21-33 (c)

Figure 21-21

Figure 21-23

Figure 21-25

Figure 21-26

6P02 -

In-Class Problem

d

s

q− q+

P

ij

Consider two point charges of equal magnitude but opposite signs, separated by a distance d. Point Plies along the perpendicular bisector of the line joining the charges, a distance s above that line. What is the E field at P?

24P02 -

Continuous Charge Distributions

25P02 -

Continuous Charge Distributions

( ) ?P =E

Vi

iQ q

Break distribution into parts:

= ∆∑

2ˆe

qkr∆

∆ =E r

E field at P due to ∆q

Superposition:

= ∆∑E E

V

dq→ ∫

d→ ∫ E

2ˆe

dqd kr

→ =E r

26P02 -

Continuous Sources: Charge DensitydVdQ ρ=

R

L

2Volume V R Lπ= = QV

ρ =

LQ

=λ

QA

σ =

dAdQ σ=w

L

Area A wL= =

dLdQ λ=Length L=

L

27P02 -

Examples of Continuous Sources: Line of charge

LQ

=λ

Length L=

L

dLdQ λ=

http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect

rostatics/07-LineIntegration/07-

LineInt320.html

28P02 -

Examples of Continuous Sources: Line of charge

LQ

=λ

Length L=

L

dLdQ λ=

http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect

rostatics/08-LineField/08-

LineField320.html

36P02 -

r

2L

−2L

+

s

P

j

i

In-Class: Line of Charge

Point P lies on perpendicular bisector of uniformly charged line of length L, a distance s away. The charge on the line is Q. What is E at P?

37P02 -

r

θ

θ

2L

−2L

+

xd ′

x′

xddq ′= λs

22 xsr ′+=

P

j

i

Hint: http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/07-LineIntegration/07-LineInt320.html

Typically give the integration variable (x’) a “primed” variable name.

32P02 -

Ring of Charge

Symmetry!0E⊥ =1) Think about it

http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect

rostatics/09-RingIntegration/09-

ringInt320.html

2) Define Variables

dq dlλ=22 xar +=

( )a dλ ϕ=

38P02 -

E Field from Line of Charge

2 2 1/ 2ˆ

( / 4)eQk

s s L=

+E j

Limits:

2ˆlim e

s L

Qks>>

→E j Point charge

ˆ ˆ2 2lim e es L

Qk kLs s

λ<<

→ =E j j Infinite charged line

29P02 -

Examples of Continuous Sources: Ring of Charge

2Q

Rλ

π=dLdQ λ=

http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect

rostatics/09-RingIntegration/09-

ringInt320.html

30P02 -

Examples of Continuous Sources: Ring of Charge

2Q

Rλ

π=dLdQ λ=

http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/elect

rostatics/10-RingField/10-

ringField320.html

31P02 -

Example: Ring of Charge

P on axis of ring of charge, x from centerRadius a, charge density λ.

Find E at P

Figure 21-27

33P02 -

Ring of Charge3) Write Equation dq a dλ ϕ=

2

ˆe

rd k dqr

=E

a) My way

3x exdE k dqr

=

b) Another way

22 xar +=

3erk dqr

=

cos( )xdE d θ= E 2 3

1e e

x xk dq k dqr r r

= ⋅ =

34P02 -

Ring of Charge4) Integrate

3x x exE dE k dqr

= =∫ ∫22 xar +=

dq a dλ ϕ=

3exk dqr

= ∫

Very special case: everything except dq is constant

2aλ π=dq∫2 2

0 0a d a d

π πλ ϕ λ ϕ= =∫ ∫

Q=

35P02 -

Ring of Charge5) Clean Up

3x exE k Qr

=

( )3/ 22 2x exE k Q

a x=

+

0a →

( )3/ 22 2ˆ

exk Q

a x=

+E i

6) Check Limit

( )3/ 2 22

ex e

k QxE k Qxx

→ =

39P02 -

In-Class: Uniformly Charged Disk

P on axis of disk of charge, x from centerRadius R, charge density σ.

Find E at P

( 0 )x >

40P02 -

Disk: Two Important Limits

( )1/ 22 2ˆ1

2disko

x

x R

⎡ ⎤⎢ ⎥= −⎢ ⎥+⎣ ⎦

E iσε

Limits:

2

1 ˆlim 4diskx R o

Qx>>

→E iπε

*** Point charge

ˆlim 2diskx R o<<

→E iσε

Infinite charged plane

41P02 -

E for Plane is Constant????

1) Dipole: E falls off like 1/r3

2) Point charge: E falls off like 1/r2

3) Line of charge:E falls off like 1/r4) Plane of charge: E constant

Figure 21-30

Figure 21-33 (d)

Figure 21-35

Figure 21-36

Figure 21-37

Figure 21-39

Figure 21-40

12P02 -

Electric DipoleTwo equal but opposite charges +q and –q,

separated by a distance 2a

q

-q

2a charge×displacementˆ ˆ×2 2q a qa

≡

= =

p

j j

Dipole Moment

p

p points from negative to positive charge

Figure 21-41

13P02 -

Why Dipoles?

Nature Likes To Make Dipoles!

http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/20-Molecules2d/20-mole2d320.html

14P02 -

Dipoles make Fields

15P02 -

Electric Field Created by DipoleThou shalt use components!

3 3x ex xE k q

r r+ −

⎛ ⎞∆ ∆= −⎜ ⎟

⎝ ⎠

3 3y ey yE k q

r r+ −

+ −

⎛ ⎞∆ ∆= −⎜ ⎟

⎝ ⎠

3/2 3/22 2 2 2( ) ( )e

x xk qx y a x y a

⎛ ⎞⎜ ⎟= −⎜ ⎟⎡ ⎤ ⎡ ⎤+ − + +⎣ ⎦ ⎣ ⎦⎝ ⎠

3/2 3/22 2 2 2( ) ( )e

y a y ak qx y a x y a

⎛ ⎞− +⎜ ⎟= −⎜ ⎟⎡ ⎤ ⎡ ⎤+ − + +⎣ ⎦ ⎣ ⎦⎝ ⎠

2 3 3 3

ˆ ˆ ˆx yr r r r

∆ ∆= = +

r r i j

19P02 -

Dipoles feel Fields

21P02 -

Dipole in Uniform FieldˆE=E i

ˆ ˆ2 (cos sin )qa= +p i jθ θ

( ) 0net q q+ −= + = + − =F F F E E

tends to align with the electric field p

Total Net Force:

Torque on Dipole: = ×τ r F( )( )2 sin( )a qE θ=

= ×p Esin( )rFτ θ+= sin( )pE θ=

Figure 21-43

Figure 21-47

Figure 21-50

Figure 21-51

Figure 21-52

Figure 21-53

Figure 21-55

Figure 21-56

Figure 21-57

Figure 21-59

Figure 21-60

Figure 21-61

Figure 21-62

Figure 21-63

Figure 21-66

Figure 21-67

Figure 21-68

Figure 21-69

Figure 21-70

Figure 21-71

Figure 21-72