Energy stored in an electrostatic field exam question
An isolated sphere has radius R and relative permittivity r. The sphere carries charge Q, which is
uniformly distributed throughout the sphere’s volume.
a) Show that the radial component of electric field intensity at a radius r from the
sphere’s centre is:
E
Qr
4 r  0 R 2
(r  R)
b) Derive an expression for the energy stored in the electrostaic field: i) within the
sphere, and ii) external to the sphere.
c) Show that the total energy stored is greater than would be stored by a sphere of the
same radius carrying the same charge uniformly distributed over its surface by a factor of 1
5 r
Solution
a)
For the uniformly-charged sphere, the charge contained within a spherical volume
of radius r is:
Qr 
Qr 3
R3
(r  R)
the electric flux density Dr is uniform over the surface area (4r2) of the spherical volume and
from Gauss’s Law:
Qr 
Qr 3
Qr 3
Qr
2

4

r
D

D


r
r
3
2 3
R
4r R
4R 3
E
D
 Er 
 r 0
Qr
4r  0 R 3
b)
Energy stored per unit volume = ½DE
i)
Now consider a spherical shell of radius r and thickness r. Volume = 4r2.r and the
energy stored in the shell is:
energy = ½DE. 4r2.r  1
Qr
Qr
Q2r 4
2
.
.
4

r

r

.r
2 4R 3 4  R 3
8r  0 R 6
r 0
so the total energy stored within the sphere of radius R is:
R
R
Q2r 4
Q2
1 r6 
internal energy  
.
dr

6
6
5 0
8  R
8  R

0
ii)
r
0
r

0
Q2
40r  0 R
External to the sphere:
Q
4r 2
Q
 Er 
40 r 2
Q  4r 2 Dr  Dr 
E
D
0
Now for a spherical shell of radius r (r>R) and thickness r. Volume = 4r2.r and the
energy stored in the shell is:
energy = ½DE. 4r2.r  1
Q
Q
Q2
2
.
.
4

r

r

.r
2 4r 2 4 r 2
80 r 2
0
so the total energy stored outside the sphere of radius R is:


Q2
Q2
Q2
1
.
dr



external energy  
2
r R
80
80 R
R 8 0 r
 
c)
Total energy stored by the uniformly-charged sphere = internal energy + external energy
=
Q2
40r  0 R

Q2
80 R

Q2
40 r  0 R
1  5 r 
For the surface-charged sphere, the internal field is zero, so there is zero internal energy
stored. The analysis for the external energy stored is the same as above, so energy stored
=
Q2
80 R
Q2
1  5 r 
40 r  0 R
1
 1
Ratio of energy stored =
2
5 r
Q
80 R