QUALIFYING EXAMINATION, Part 2
Solutions
Problem 1: Electromagnetism I
(a) Since our system has spherical symmetry the electric field and potential have, respec
tively, the form
E
=
E
(
r
) ˆ
r
and
ϕ
=
ϕ
(
r
).
We use the integral form of Gauss’s law to determine the electric field. For
r < R
, we
have
E
(
r
)(4
πr
2
) =
Q
enc
ε
0
=

q
ε
0
4
3
πr
3
4
3
πR
3
=

q
ε
0
r
R
3
,
or
E
(
r
) =

q
4
πε
0
R
3
r .
For
r > R
, we find
E
(
r
) =

q
4
πε
0
1
r
2
.
Since
E
=
∇
ϕ
, we can find the electrostatic potential by integrating the electric
field. Up to an integration constant, the potential is given by
ϕ
(
r
) =

R
E
r
(
r
)
dr
. For
r > R
, using the boundary condition
φ
(
r
)
→
0 for
r
→ ∞
, we find
ϕ
(
r
) =

q
4
πε
0
1
r
.
For
r < R
, we find
ϕ
(
r
) = +
1
2
q
4
πε
0
R
3
r
2
+
C ,
where
C
is an integration constant. We determine
C
using the continuity condition for the
electrostatic potential at
r
=
R
, and find
C
=

3
2
q
(4
πε
0
R
)

1
. Therefore, the potential
for
r < R
is
ϕ
(
r
) =

1
2
q
4
πε
0
R
"
3

r
R
2
#
.
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 Winter '16