2Electrostatics
IB Electromagnetism
2.3 Electrostatic energy
We want to calculate how much energy is stored in the electric field. Recall from
IA Dynamics and Relativity that a particle of charge
q
in a field
E
=
−∇φ
has
potential energy U(r) = qφ(r).
U
(
r
) can be thought of as the work done in bringing the particle from infinity,
as illustrated below:
work done = −
Z
r
∞
F · dr
= −
Z
r
∞
E · dr
= q
Z
r
∞
∇φ · dr
= q[φ(r) − φ(∞)]
= U(r)
where we set φ(∞) = 0.
Now consider
N
charges
q
i
at positions
r
i
. The total potential energy stored
is the work done to assemble these particles. Let’s put them in one by one.
(i) The first charge is free. The work done is W
1
= 0.
(ii) To place the second charge at position r
2
takes work. The work is
W
2
=
q
1
q
2
4πε
0
1
r
1
− r
2

.
(iii) To place the third charge at position r
3
, we do
W
3
=
q
3
4πε
0
q
1
r
1
− r
3

+
q
2
r
2
− r
3

(iv) etc.
The total work done is
U =
N
X
i=1
W
i
=
1
4πε
0
X
i<j
q
i
q
j
r
i
− r
j

.
Equivalently,
U =
1
4πε
0
1
2
X
i6=j
q
i
q
j
r
i
− r
j

.
We can write this in an alternative form. The potential at point
r
i
due to all
other particles is
φ(r
i
) =
1
4πε
0
X
j6=i
q
j
r
i
− r
j

.
So we can write
U =
1
2
N
X
i=1
q
i
φ(r
i
).
There is an obvious generalization to continuous charge distributions:
U =
1
2
Z
ρ(r)φ(r) d
3
r.
Hence we obtain
U =
ε
0
2
Z
(∇ · E)φ d
3
r
=
ε
0
2
Z
[∇ · (Eφ) − E · ∇φ] d
3
r.
The first term is a total derivative and vanishes. In the second term, we use the
definition E = −∇φ and obtain
Proposition.
U =
ε
0
2
Z
E · E d
3
r.
This derivation of potential energy is not satisfactory. The final result shows
that the potential energy depends only on the field itself, and not the charges.
However, the result was derived using charges and electric potentials — there
should be a way to derive this result directly with the field, and indeed there is.
However, this derivation belongs to a different course.
Also, we have waved our hands a lot when generalizing to continuous dis
tributions, which was not entirely correct. If we have a single point particle,
the original discrete formula implies that there is no potential energy. However,
since the associated field is nonzero, our continuous formula gives a nonzero
potential.
This does not mean that the final result is wrong. It is correct, but it
describes a more sophisticated (and preferred) conception of “potential energy”.
Again, we shall not go into the details in this course.