Solutions to Homework Set 9
1) Conducting strip
:
A
(
k
) =
Z
+
∞
∞
V
(
x
)
e

ikx
dx
=
V
0
Z
a

a
e

ikx
dx
=
2
V
0
sin
ka
k
From this we have
V
(
x,
0
, y
) = 2
V
0
Z
∞
∞
dk
2
π
sin(
ka
)
k
e
ikx
e

k

y

.
Taking the
y
gradients to get
E
y
, and then getting
σ
=
0
(
E
y

y
=+
ε

E
y

y
=

ε
) gives us
σ
(
x
) = 4
V
0 0
Z
∞
∞
dk
2
π
sgn(
k
) sin(
ka
)
e
ikx
e

ε

k

.
The integral is elementary, and gives
σ
(
x
) =
4
V
0 0
4
π
1
x
+
a

iε

1
x

a

iε
+
1
x
+
a
+
iε

1
x

a
+
iε
We can take safely take the
ε
regulator to zero. We end up with
σ
(
x
) =
2
0
V
0
π
1
a
+
x
+
1
a

x
=
4
0
V
0
π
a
a
2

x
2
.
σ
x
a
a
A sketch of the charge distribution on the strips.
2) Qual Problem
:
a) We have, in general
b
l
=
2
l
+ 1
2
R
l
+1
Z
1

1
d
(cos
θ
)
V
(
θ
)
P
l
(cos
θ
)
.
1
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Plugging in the explicit expression for the relevent
P
l
gives
b
1
=
3
4
R
2
(
V
1

V
2
)
b
2
=
0
(1)
b
3
=
7
16
R
4
(
V
2

V
1
)
b) If
V
(
r, θ, φ
) =
1
4
π
0
Q
r
+
d
·
ˆ
r
r
2
+
· · ·
,
then
Q
is the total charge, and
d
is the dipole moment. Thus our divided sphere has

d

= 3
π
0
R
2
(
V
1

V
2
)
.
c) For the sphere immersed in the external field we have
V
(
r, θ, φ
) =

E

r
cos
θ
+
V
sphere
(
r, θ, φ
)
,
and so
V
(
r, θ, φ
) =

E

r
cos
θ
+
1
4
π
0
Q
r
+

d

cos
θ
r
2
+
· · ·
.
The surface of the sphere at
r
=
R
must be an equipotential, and so the coefficients of
the cos
θ
terms must add to zero:

E

R
+

d

4
π
0
R
2
= 0
.
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 Fall '07
 Stone
 Centrifugal Force, Force, Work, Rotation, Cos, Frame of reference, Inertial frame of reference

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