# Energy of a charged sphere ```Energy of a charged sphere
A total charge Q is distributed uniformly into a spherical volume of radius R. Find the electrostatic
energy of this configuration by each of the following three methods:
a) Evaluate the work done to build up the charged sphere “layer after layer” by carrying the requiring
amount of charge from infinite distance.
b) Evaluate the volume integral of uE = ǫ0 |E|2 /2, where E is the electric field.
c) Evaluate the volume integral of ρV /2, where ρ is the charge density and V is the electrostatic
potential. Discute the differences with the approach used in b).
1
Solution
a) We build up the sphere by adding subsequent infinitesimal layers of charge (carried from infinite
distance). From Gauss’s theorem we know that, for an uniformly charged sphere having charge
density ρ, radius r, and total charge q = q(r) = ρ(4πr3 /3), the field and the potential outside
the sphere are those of a point charge q located in the center. Thus, in building the sphere, when
a new layer of charge dq = ρ4πr2 dr is added, its charge will be located at the potential V (r) =
k0 q(r)/r, where k0 = 1/4πǫ0 ; thus, the corresponding stored amount of electrostatic energy is V (r)dq.
Integrating over r up to the final radius R we find for the total energy U0
Z
Z R
q(r)
k0
U0 =
V (r)dq =
ρ4πr2 dr
r
0
Z R
Z
4π 3 1
4πρ2 R 4
4πρ2 R5
2
k0 ρ
=
r
ρ4πr dr =
r dr =
3
r
3ǫ0 0
15ǫ0
0
2
3k0 Q
,
(1)
=
5 R
being ρ = Q/(4πR3 /3).
b) The electric field is easily found from Gauss’s theorem to be
Q
r/R3 (r &lt; R)
E(r) =
&times;
(2)
1/r2 (r &gt; R)
4πǫ0
The two integrals we need are
Z R 2
4π
r
4πr2 dr =
,
3
R
5R
0
Z
∞
R
1
r2
2
4πr2 dr =
4π
,
R
(3)
3k0 Q2
.
5 R
(4)
so that we obtain
U0 =
Z
0
∞
ǫ0 2
ǫ0
E (r)4πr2 dr =
2
2
Q
4πǫ0
2 4π
4π
+
5R
R
=
c) The electrostatic potential is given by
Q
(−r2 /2R3 + 3/2R) (r &lt; R)
V (r) =
&times;
1/r
(r &gt; R)
4πǫ0
(5)
Since ρ = 0 for r &gt; R, we only need to compute the integral of ρV /2 from 0 to R:
Z
Q
3
r2
Q
R2
1 R Q
2
2
+R
ρ − 3+
4πr dr =
−
U0 =
2 0 4πǫ0
2R
2R
4ǫ0 4πR3 /3
5
3 Q2 4R2
3k0 Q2
=
=
.
(6)
4 4πǫ0 R3 5
5 R
It is noticeable that, while methods b) and c) both give the correct result (as they should), they
remind us that it is incorrect to give to the “energy density” of the electric field a physical interpretation as “energy contained in a certain region of space”. In fact, if we identify the energy density
with ǫ0 E2 /2 as in b), we would conclude that the energy spreads out over the whole space, up to
infinity; if, on the other hand, we take ρV as the energy density as in c), it is “contained” only in the
volume of the sphere, i.e. “where the charge is”. Thus, the concept of energy density is ambiguous,
while the total electrostatic energy of the system is well defined.
2
```