# Chapter 22: Electric Field PHY2049: Chapter 22 1 Chapter 22: Electric Field

PHY2049: Chapter 22

1

Concept Quiz

Î

For which situations can we put a charge q’ on the left side of the two charges so that it is in equilibrium?

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(1) (a) only



(2) (a) and (b)



(3) (c) and (d)



(4) (b) and (d)



(5) (c) only

Should be near the smaller charge because if it is near the larger charge, there can be no total cancellation

q

q

q

q

PHY2049: Chapter 22

2

Concept Quiz

Î

Which situation has the largest force acting on q?

q q q q

All E fields are in the same direction in (3). Note that (4) has E = 0.

PHY2049: Chapter 22

3

Electric Field of Single Point Charge

q

>

0

G

E

=

kq r

ˆ

r

2

PHY2049: Chapter 22

q

<

0

G

E

=

kq r

ˆ

r

2

4

Example: Electric Field on Proton

Î

At surface of proton

 q = e = 1.6 x 10

-19

 r = 10

-15 m

C

E

=

kq r

2

=

(

×

9

10

)(

15

2

19

)

=

21

N/C

Î

E points radially outward for + charge

PHY2049: Chapter 22

5

E Field of Two Equal, Positive Point Charges

PHY2049: Chapter 22

6

E Field of Two Equal, Unlike Point Charges

PHY2049: Chapter 22

7

Field Between Two Charged Parallel Plates

Î

Assume plates are much larger than separation



E is approx. constant between plates



E is zero outside the plates



This is a capacitor!

Î

E points from + plate to – plate

Î

We will calculate E in Chap. 23

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Gauss’ law



Proportional to surface charge density

PHY2049: Chapter 22

8

1. Rank magnitude of E at P

1

, P

2

, P .

Assume charges on rings are +q and +q

PHY2049: Chapter 22

9

Î

P

1 has E = 0 (equidistant from ring A and B and they are same sign)

Î

P

3 has largest E (contributions from ring A and B)

Î

P

2 has no contribution from ring B because it is at the center, thus it is only affected by ring A.

Î

So the order (smallest E to largest E) is P

1

, P

2

, P

3

PHY2049: Chapter 22

10

2. Which point has largest E?

Assume charges on rings are +q and − q

PHY2049: Chapter 22

11

Î

P

1 has largest E field since it is equidistant from ring A and

B and their E contributions add, rather than cancel, as in the first question.

PHY2049: Chapter 22

12

Calculate E of Dipole ( ⊥ axis)

Î

At point x, E x

= 0 and E y

< 0. Why?

+Q

r r

=

x

2

+

d

2

/ 4 sin

θ

=

d

/ 2

r

θ

d

x

r

-Q

E y

=

2

kQ

sin

θ

r

2

=

2

x

2

kQ

+

d

2

/ 4

d

/ 2

x

2

+

d

2

/ 4

=

(

x

2

+

kQd d

2

/ 4

)

3/ 2

E y

≈ −

kp x

3

x d p

=

Qd

(dipole moment)

13

PHY2049: Chapter 22

Calculate E of Dipole (along axis)

Î

At point x, E x

> 0 and E y

= 0. Why?

Q

d

+

Q

x

x

E x

=

(

kQ

/ 2

kQ x

+

d

/ 2

)

2

=

(

x

2

2

kQxd

d

2

/ 4

)

2

Show this yourself

(algebra)

E x

2

kp x

3

x d p

=

Qd

(dipole moment)

PHY2049: Chapter 22

14

Finding E Field from Charge Distribution

Î

Perform integral over charge distribution



Each component must be calculated separately (vector addition)

dE y

=

kdq

(

r

2

θ

)

Î



Use symmetry to see if any component must be zero



Use symmetry to see if any component is doubled, etc.



Express dq, r and trig functions in terms of “natural” variables defined by the problem



Then we can integrate!

PHY2049: Chapter 22

15

Find E at Center of Uniformly Charged Circle

+q on top half

− q on the bottom half

Symmetry:

E = 0: E x

E x y doubled: E y

(right) = − E x

(top) = +E y

(left)

(bottom)

λ

dq

=

=

π

q r

=

λ

ds

charge / unit length

=

r

= const

16

PHY2049: Chapter 22

Center of Uniformly Charged Circle

Î

Find E





Express dq, r, sin θ in terms of θ

Top, bottom give same contribution

dE y

= −

kdq r

2 sin

θ

= −

k

(

r

2

) sin

θ

θ

dq

E y

=

=

π

0

r

2

2

k

λ

4

r k

λ

(

− cos

θ

)

π

0 sin

θ

r

λ

=

q

π

r

E y

= −

4

kq

π

r

2

dq

=

λ

ds

=

17

PHY2049: Chapter 22