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Solutions to Problem Set 1
1. The potential on the axis
z
is
Φ(
z
) =
1
4
π±
0
Z
S
σdS
r
=
1
4
π±
0
Z
a
0
2
πρdρ
p
z
2
+
ρ
2
=
σ
2
±
0
h
p
z
2
+
a
2
 
z

i
By symmetry,
E
points along the axis away from the disc (for
σ >
0),
E
=
E
z
e
z
, thus
E
z
=

∂
Φ(
z
)
∂z
=
±
σ
2
±
0
±
1


z

√
z
2
+
a
2
²
where the plus sign corresponds to
z >
0 while the minus sign for
z <
0.
2. The conductor and the metal plate have the same potential when they are contacted. After
the ﬁrst contact,
q/C
1
= (
Q

q
)
/C
2
. After the ﬁnal contact, there is no charge transfer, so
q
f
/C
1
=
Q/C
2
. The ultimate charge is
q
f
=
qQ
Q

q
.
3. We start with the Poisson equation,
∇
2
Φ =

ρ
±
0
. The potential has no angular dependence,
therefore,
ρ
=

±
0
1
r
d
2
dr
2
(
r
Φ) =

q
4
πr
d
2
dr
2
h
e

αr
³
1 +
αr
2
´i
=

qα
3
8
π
e

αr
This is valid for
r >
0. For
r
→
0, the potential Φ diverges and also
ρ
diverges. Note that in
this limit
lim
r
→
0
Φ(
r
) =
q
4
π±
0
r
Because
∇
2 1
r
=

4
πδ
(
r
) we have to add this piece to the solution. Finally, we get
ρ
(
r
) =
±
qδ
(
r
)

qα
3
8
π
e

αr
²
4. a) The potential satisﬁes the Poisson equation Φ
00
(
x
) =

ρ/±
0
. The boundary conditions are
Φ(0) = 0 and Φ
0
(
d
) =
σ
0
/±
0
. The solution is
Φ(
x
) =

ρ
2
±
0
x
2
+
ρd
+
σ
0
±
0
x.
b) The threshold shift is Φ(
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This note was uploaded on 02/04/2012 for the course PHY 501 at Princeton.
 '11
 MARLOW
 Magnetism

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