Electrically connected spheres
Two conducting spheres of radii a and b, respectively (with b < a), are connected by a thin metal
wire of negligible capacitance. The centers of the
b
two spheres are at a distance d ≫ a > b. A total
a
net charge Q is located on the system (spheres and
wire).
d
Assuming at first that the surface charge on each
sphere is uniformly distributed, find
a) how the charge Q is partitioned between the two spheres,
b) the electrostatic potential V of the system and the capacitance C = Q/V ,
c) the electric field at the surface of each sphere, comparing the intensities and discussing the limit
b → 0.
Now consider electrostatic induction effects, which lead to a non-uniform distribution of the
surface charge on each sphere.
d) Using the method of image charges, improve the preceding results to the lowest order in a/d and
b/d.
1
Solution
a)
If we assume (as an approximation) the surface charge to be uniformly distributed, the electrostatic potential and field generated by each sphere are those of point charges located in the center
of the spheres. Let Qa and Qb be the charge on each sphere, with Qa + Qb = Q. The electrostatic
potentials of the spheres are
Qb
Qa
,
Vb ≃ k0
,
(1)
a
b
respectively. Since the spheres are electrically connected, Va = Vb ≡ V . Solving for the charges we
obtain
Q
Q
Qa ≃
,
Qb ≃
,
(2)
1 + b/a
1 + a/b
Va ≃ k0
from which Qa > Qb follows.
b) From the results of point a) it immediately follows
V ≃ k0
Q
,
a+b
C≃
a+b
.
k0
(3)
c) For the electric field we obtain
Ea ≃ k0
Q
Qa
=
k
,
0
a2
a(a + b)
Eb ≃ k0
Qb
Q
=
k
,
0
b2
b(a + b)
from which we obtain Eb > Ea . In the limit b → 0, Ea = k0 Q/a2 while Eb → ∞.
d) To lowest order we consider the field of each
sphere as that due to a point charge at its cenqa
q’a
ter. Let qa e qb the values of such charges (yet to
be determined). To next order we consider that
each charge induces on the opposite sphere a surface charge distribution whose field is equivalent to
the field of an image charge of value
(4)
q’b qb
a
b
,
qb′ = −qa ,
(5)
d
d
at a distance from the center of the corresponding sphere equal to a2 /d and b2 /d, respectively. (In
turn, these image charges also lead to further induction effects which we may take into account by
adding other images; this would lead to higher order terms in a/d and b/d.)
Since the total charge is Q, the relation qa + qb + qa′ + qb′ = Q holds. Moreover, the values of the
potential for each sphere are
qa
qb
Va ≃ k0 ,
Vb ≃ k0 ,
(6)
a
b
since at the surface of each sphere the sum of the potentials generated by the opposite central charge
and the corresponding image is zero. By posing Va = Vb we eventually obtain
qa′ = −qb
qa ≃
Q
,
1 + b/a − 2b/d
qb ≃
2
Q
.
1 + a/b − 2a/d
(7)