Chapter 22: Electric Field
PHY2049: Chapter 22
1
Electric Field of Single Point Charge
G
kq
E = − 2 rˆ
r
G kq
E = 2 rˆ
r
PHY2049: Chapter 22
2
Example: Electric Field on Proton
ÎAt
surface of proton
‹q
= e = 1.6 x 10-19 C
‹ r = 10-15 m
E=
ÎE
kq
r
2
(
=
)(
9 ×109 1.6 ×10−19
(
10−15
)
2
) = 1.44 ×10
21
N/C
points radially outward for + charge
PHY2049: Chapter 22
3
E Field of Two Equal, Positive Point Charges
PHY2049: Chapter 22
4
E Field of Two Equal, Unlike Point Charges
PHY2049: Chapter 22
5
Field Between Two Charged Parallel Plates
ÎAssume
plates are much larger
than separation
‹E
is approx. constant between plates
‹ E is zero outside the plates
‹ This is a capacitor!
ÎE
points from + plate to – plate
ÎWe
will calculate E in Chap. 23
‹ Gauss’
law
‹ Proportional to surface charge density
PHY2049: Chapter 22
6
1. Rank magnitude of E at P1, P2, P3.
Assume charges on rings are +q and +q
PHY2049: Chapter 22
7
Answer to Question #1
Î P1
has E = 0 since it is equidistant from ring A and B and
they are same sign.
Î P3
has largest E because it has contributions from ring A
and B
Î P2
has no contribution from ring B because it is at the
center, thus it is only affected by ring A.
ÎSo
the order (smallest E to largest E) is P1, P2, P3
PHY2049: Chapter 22
8
2. Rank magnitude of E at P1, P2, P3.
Assume charges on rings are +q and −q
PHY2049: Chapter 22
9
Answer to Question #2
Î P1
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.
ÎHard
to rank E field of P3 and P2, in my opinion. Relative
distances from the two rings are different and there is a
cancellation in P3.
PHY2049: Chapter 22
10
Calculate E of Dipole (⊥ axis)
ÎAt
point x, Ex = 0 and Ey < 0. Why?
+Q
d
x
-Q
d /2
− kQd
⎛ − kQ ⎞
Ey = 2⎜ 2
=
⎟ 2
2
⎝ x + d /4 ⎠ x + d2 /4
x2 + d 2 / 4
(
Ey ≈ −
kQd
x3
xd
)
3/ 2
p = Qd (dipole moment)
PHY2049: Chapter 22
11
Calculate E of Dipole (along axis)
ÎAt
point x, Ex > 0 and Ey = 0. Why?
-Q
Ex =
Ex ≈
+Q
d
x
kQ
( x − d / 2)
2kQd
x3
2
−
kQ
( x + d / 2)
xd
2
=
2kQxd
(x
2
−d /4
2
)
2
p = Qd (dipole moment)
PHY2049: Chapter 22
12
Finding E Field from Charge Distribution
ÎPerform
‹ Each
integral over charge distribution
component must be calculated separately (vector addition)
dE y =
ÎGeneral
kdq
r
2
( sin θ or cosθ )
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
13
Center of Uniformly Charged Circle
ÎE
field is down. Why?
Uniform distribution of charge
‹ Express dq, r, sinθ in terms of θ
‹ Top, bottom give same contribution
‹
dE y = −
kdq
r
2
π
sin θ = −
Ey = 2 × ∫ −
0
k λ rdθ
r
2
k λ rdθ
r
2
sin θ
sin θ
−2k λ
π
−
cos
θ
(
)0
r
−4k λ
=
r
dq
=
θ
q
λ=
πr
Ey = −
4kq
π r2
dq = λ ds = λ rdθ
PHY2049: Chapter 22
14