Electrostatic pressure on a conducting sphere
A conducting sphere of radius a has a net charge Q and it is electrically isolated. Find the pressure
at the surface of the sphere
a) directly from the knowledge of the surface charge density and the electric field on the sphere,
b) by calculating the variation of the electrostatic energy with respect to a.
c) Now calculate again the pressure on the sphere assuming the latter not to be isolated, but
connected to an ideal generator keeping the sphere at the constant voltage V .
1
Solution
a) The surface charge is σ = Q/S where S = 4πa2 is the surface of the sphere. The electric field at
the surface is E = 4πk0 σ and the pressure is
1
Q2
P = σE = 2πk0 σ 2 = k0
.
2
8πa4
b) Using Gauss’s theorem the electric field of the sphere is
0
(r < a)
E(r) =
2
k0 Q/r (r > a)
and thus the electrostatic energy is
2
Z ∞
Z
1
k0 Q
Q2
2
3
2
.
E (r)d r =
4πr
dr
=
k
Ues =
0
8πk0
8π r2
2a
a
(1)
(2)
(3)
The derivative of Ues , which has the dimension of a force, can be interpreted as the integral of the
electrostatic pressure over the surface of the sphere. Since for symmetry reasons the pressure is
uniform, we can write
dUes
Q2
1 k0 Q2
1
−
=
k
,
(4)
=
P =
0
4πa2
da
4πa2 2a2
8πa4
as we previously obtained.
c) The electric potential of the sphere is
V (r) =
V
(r < a)
V (a/r) (r > a)
(5)
from which we obtain the charge on the sphere as Q = aV /k0 . By substituting for Q in Eq.(1) we
obtain
P =
V2
.
8πk0 a2
(6)
Alternatively, we can write the electrostatic energy (3) as a function of V ,
Ues =
aV 2
,
2k0
(7)
and use the fact that for a variation of a at constant voltage, the corresponding variation in the total
energy of the system (the sphere plus the voltage generator) is ∆Utot = ∆Ugen + ∆Ues = −∆Ues since
∆Ugen = −2∆Ues . Thus
1
dUtot
V2
P =
−
=
,
(8)
4πa2
da
8πk0 a2
which is again the correct result.
2