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) =
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:
+
-