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
r
θ
d
x
r
-Q
Ey =
r = x2 + d 2 / 4
d /2
sin θ =
r
−2kQ
r2
d /2
− kQd
⎛ − kQ ⎞
sin θ = 2 ⎜ 2
=
⎟ 2
2
⎝ x + d /4 ⎠ x + d2 /4
x2 + d 2 / 4
Ey ≈ −
(
kp
x
3
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
+Q
d
x
x
Ex =
kQ
( x − d / 2)
Ex ≈
2kp
x
3
2
−
xd
kQ
( x + d / 2)
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
dE y
‹ Express dq, r, sinθ in terms of θ
‹ Top, bottom give same contribution
‹
=−
kdq
r
2
sin θ = −
π
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
Axis of Uniform Charged Ring (+Q)
ÎPoint
is distance z above center of charged ring, radius R
charged (ind. of angle φ)
‹ Horizontal components cancel
‹ Uniformly
x
dEz =
r
2
z
sin θ = =
r
z
θ
kdq sin θ
R
Ez = ∫
2π
0
Q
λ=
2π R
dq = λ Rdφ
Ez =
(z
k λ zRdφ
2
+R
2
)
3/ 2
=
k ( λ Rdφ )
z +R
2
sin θ
2
z
z 2 + R2
=
(z
kz 2πλ R
2
+R
2
)
3/ 2
kQz
(
z 2 + R2
PHY2049: Chapter 22
)
3/ 2
Check this!
15
Behavior at Large and Small z
ÎExact
ÎFor
ÎFor
Ez =
z small
kQz
(z
+R
2
Ez →
z large Ez →
kQz
R
kQ
z2
)
2 3/ 2
3
Becomes 0 when z = 0
Coulomb’s law
As expected
PHY2049: Chapter 22
16
Uniformly Charged Line
ÎDistance
y above midpoint of charged line of length L
‹ Uniformly
charged
P
‹Ex components cancel
λ = Q/ L
y
θ
L
Ey = ∫
Ey =
kdq
r
2
y y 2 + L2 / 4
dq = λ dx
L/2
L/2
k ( λ dx )
y
−L / 2
y +x
y +x
sin θ = ∫
kλ L
x
2
2
2
2
=∫
−L / 2
ky ( λ dx )
(y
2
+x
)
2 3/ 2
Check this!
PHY2049: Chapter 22
17
Special Cases for Charged Line
Î
Exact expression
Ey =
kλ L
y y 2 + L2 / 4
2k λ
ÎInfinite line (L →∞) E y =
y
Î
Zero length (point) E y =
kλ L
y2
Falls as inverse distance
=
kQ
y2
PHY2049: Chapter 22
Coulomb’s law
As expected
18
Uniformly Charged Disk
ÎPoint
is distance z above axis of charged disk, radius R
Ez = ∫
z
θ
ρ
σ=
R
Q
π R2
dq = σ dA
= σ ( ρ d ρ dφ )
kdq
r
2
sin θ
=∫
R 2π
=∫
Rk
∫
k (σρ d ρ dφ )
z
z2 + ρ 2
z2 + ρ 2
0 0
0
( 2π zσρ d ρ )
(z
2
+ρ
2
)
3/ 2
⎛
z
Ez = 2π kσ ⎜1 −
⎜
z 2 + R2
⎝
PHY2049: Chapter 22
⎞
⎟⎟
⎠
19
Charged Disk (cont)
⎛
z
ÎExact expression E z = 2π kσ ⎜ 1 −
⎜
z 2 + R2
⎝
Îz
Îz
<< R
>> R
σ
Ez → 2π kσ =
2ε 0
Ez →
kπ R 2σ
z
2
=
kQ
z2
PHY2049: Chapter 22
⎞
⎟⎟
⎠
Independent of z!
See next chapter
Coulomb’s law
As expected
20