2Electrostatics
IB Electromagnetism
2.2 Electrostatic potential
In the most general case, we will have to solve both
∇·E
=
ρ/ε
0
and
∇×E
=
0
.
However, we already know that the general form of the solution to the second
equation is E = −∇φ for some scalar field φ.
Definition
(Electrostatic potential)
.
If
E
=
−∇φ
, then
φ
is the electrostatic
potential.
Substituting this into the first equation, we obtain
∇
2
φ =
ρ
ε
0
.
This is the Poisson equation, which we have studied in other courses. If we are
in the middle of nowhere and ρ = 0, then we get the Laplace equation.
There are a few interesting things to note about our result:
– φ
is only defined up to a constant. We usually fix this by insisting
φ
(
r
)
→
0
as
r → ∞
. This statement seems trivial, but this property of
φ
is actually
very important and gives rise to a lot of interesting properties. However,
we will not have the opportunity to explore this in this course.
–
The Poisson equation is linear. So if we have two charges
ρ
1
and
ρ
2
, then
the potential is simply
φ
1
+
φ
2
and the field is
E
1
+
E
2
. This is the
principle of superposition. Among the four fundamental forces of nature,
electromagnetism is the only force with this property.
2.2.1 Point charge
Consider a point particle with charge Q at the origin. Then
ρ(r) = Qδ
3
(r).
Here δ
3
is the generalization of the usual delta function for (3D) vectors.
The equation we have to solve is
∇
2
φ = −
Q
ε
0
δ
3
(r).
Away from the origin
r
=
0
,
δ
3
(
r
) = 0, and we have the Laplace equation. From
the IA Vector Calculus course, the general solution is
φ =
α
r
for some constant α.
The constant
α
is determined by the delta function. We integrate the equation
over a sphere of radius r centered at the origin. The left hand side gives
Z
V
∇
2
φ dV =
Z
S
∇φ · dS =
Z
S
−
α
r
2
ˆ
r · dS = −4πα
The right hand side gives
−
Q
ε
0
Z
V
δ
3
(r) dV = −
Q
ε
0
.
So
α =
Q
4πε
0
and
E = −∇φ =
Q
4πε
0
r
2
ˆ
r.
This is just what we get from Coulomb’s law.
2.2.2 Dipole
Definition
(Dipole)
.
A dipole consists of two point charges, +
Q
and
−Q
at
r = 0 and r = −d respectively.
To find the potential of a dipole, we simply apply the principle of superposition
and obtain
φ =
1
4πε
0
Q
r
−
Q
r + d
.
This is not a very helpful result, but we can consider the case when we are far,
far away, i.e.
r d
. To do so, we Taylor expand the second term. For a general
f(r), we have
f(r + d) = f(r) + d · ∇f(r) +
1
2
(d · ∇)
2
f(r) + ··· .
Applying to the term we are interested in gives
1
r + d
=
1
r
− d · ∇
1
r
+
1
2
(d · ∇)
2
1
r
+ ···
=
1
r
−
d · r
r
3
−
1
2
d · d
r
3
−
3(d · r)
2
r
5
+ ··· .
Plugging this into our equation gives
φ =
Q
4πε
0
1
r
−
1
r
+
d · r
r
3
+ ···
∼
Q
4πε
0
d · r
r
3
.
Definition
(Electric dipole moment)
.
We define the electric dipole moment to
be
p = Qd.
By convention, it points from ve to +ve.
Then
φ =
p ·
ˆ
r
4πε
0
r
2
,
and
E = −∇φ =
1
4πε
0
3(p ·
ˆ
r)
ˆ
r − p
r
3
.
2.2.3 General charge distribution
To find
φ
for a general charge distribution
ρ
, we use the Green’s function for the
Laplacian. The Green’s function is defined to be the solution to
∇
2
G(r, r
0
) = δ
3
(r − r
0
),
In the section about point charges, We have shown that
G(r, r
0
) = −
1
4π
1
r − r
0

.
We assume all charge is contained in some compact region V . Then
φ(r) = −
1
ε
0
Z
V
ρ(r
0
)G(r, r
0
) d
3
r
0
=
1
4πε
0
Z
V
ρ(r
0
)
r − r
0

d
3
r
0
Then
E(r) = −∇φ(r)
= −
1
4πε
0
Z
V
ρ(r
0
)∇
1
r − r
0

d
3
r
0
=
1
4πε
0
Z
V
ρ(r
0
)
(r − r
0
)
r − r
0

3
d
3
r
0
So if we plug in a very complicated ρ, we get a very complicated E!
However, we can ask what φ and E look like very far from V , i.e. r r
0
.
We again use the Taylor expansion.
1
r − r
0

=
1
r
+ r
0
· ∇
1
r
+ ···
=
1
r
+
r · r
0
r
3
+ ··· .
Then we get
φ(r) =
1
4πε
0
Z
V
ρ(r
0
)
1
r
+
r · r
0
r
3
+ ···
d
3
r
0
=
1
4πε
0
Q
r
+
p ·
ˆ
r
r
2
+ ···
,
where
Q =
Z
V
ρ(r
0
) dV
0
p =
Z
V
r
0
ρ(r
0
) dV
0
ˆ
r =
r
krk
.
So if we have a huge lump of charge, we can consider it to be a point charge
Q
,
plus some dipole correction terms.
2.2.4 Field lines and equipotentials
Vectors are usually visualized using arrows, where longer arrows represent larger
vectors. However, this is not a practical approach when it comes to visualizing
fields, since a field assigns a vector to every single point in space, and we don’t
want to draw infinitely many arrows. Instead, we use field lines.
Definition
(Field line)
.
A field line is a continuous line tangent to the electric
field E. The density of lines is proportional to E.
They begin and end only at charges (and infinity), and never cross.
We can also draw the equipotentials.
Definition
(Equipotentials)
.
Equipotentials are surfaces of constant
φ
. Because
E = −∇φ, they are always perpendicular to field lines.
Example.
The field lines for a positive and a negative charge are, respectively,
+

We can also draw field lines for dipoles:
+
