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?
‹
(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 C
‹ r = 10 -15 m
E
= kq r
2
=
(
× 9
)(
( )
×
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
‹
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

Answer to Question #1
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

Answer to Question #2
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 r
2 sin
θ =
2
⎛
⎝ x
2
− kQ
+ d
2
/ 4
⎞
⎠ d / 2 x
2 + d
2
/ 4
=
(
− kQd x
2 + 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
=
(
2 kQxd x
2 − 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
General helpful rules
‹
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 =
π q r
= charge / unit length dq
= λ ds
= λ θ 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
( λ θ ) sin
θ r
2
θ dq
E y
=
=
2
∫
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
= λ θ
17 PHY2049: Chapter 22