Alternatively, this problem can be solved by finding the solution to Poisson’s (Laplace’s)
equation for
r < R
(
r > R
). In this method, we would find the solution to the differential
equation
∇
2
ϕ
(
r
) =

ρ
(
r
)
/ε
0
. The electric field would then computed using
E
=
∇
ϕ
.
(b) The electric field and the potential are sketched in the figure.
1
(c) The stable equilibrium position is at the origin
r
= 0, at which the force due to the
field of the charged sphere vanishes.
(d) In equilibrium the total force vanishes
F
sphere
+
F
ext
= 0
.
or
qE
(
r
)ˆ
r
+
qE
0
ˆ
x
= 0
.
Using the expression for
E
(
r
) from part (a) inside the sphere, we find for the new equi
librium position
d
=
E
0
4
πε
0
R
3
q
ˆ
x .
We can also minimize the total potential energy obtained by adding

E
0
x
to the elec
trostatic potential inside the sphere.
(e) Outside the sphere, we can take its charge

q
to be at the origin. Hence, the superpo
sition this negative charge with the positive point charge (+
q
) gives the system a dipole
moment of
p
=
q
d
= (4
π ε
0
R
3
)
E
0
ˆ
x .
2
(f) Comparing the expression in (e) with with
p
=
α
E
, we find
α
= 4
π ε
0
R
3
.
Interestingly, the polarizability depends only on the radius of our classical atom; the larger
the radius the larger the polarizability, meaning that our classical atom exhibits a larger
response to a given applied field.
3
Problem 2: Electromagnetism II
(a) Using BiotSavart Law (in SI units):
d
~
B
=
μ
0
I
d
~
l
×
ˆ
r
4
πr
2
,
where d
~
l
is the differential directed length of the current, and
r
is the distance from
the differential current to the observation point.
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 Winter '16