three surfaces are
(
)(
)
φ
ρ
ρ
d
d
ˆ
d
1
z
=
a
r
,
(
)
φ
ρ
d
d
ˆ
d
2
z
ρ
=
a
r
and
(
)(
)
φ
ρ
ρ
d
d
ˆ
d
3
z

=
a
r
. Then
Figure 2.10
:
A Gaussian disc straddling a uniform planar distribution of electric scalar charges
E
E
2
0
0
2
0
0
3
0
2
0
1
0
e
2
d
d
2
d
d
d
3
2
1
ρ
πε
φ
ρ
ρ
ε
ε
ε
ε
ρ
ρ
π
φ
=
=
⋅
+
⋅
+
⋅
=
Φ
∫ ∫
∫
∫
∫
=
=
a
a
a
a
a
a
r
r
r
r
r
r
E
E
E
The portion of the charged plane enclosed by the Gaussian cylinder is a circular surface
of area
a
′
. Hence the net charge within this cylinder is
2
0
2
0
face
circular
enc
d
d
d
πσρ
φ
ρ
ρ
σ
σ
ρ
ρ
π
φ
=
′
′
′
=
′
=
∫ ∫
∫
=
′
=
a
q
(2.41)
Substituting these results into Gauss’ law gives
.
.
.
.
.
.
.
.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
1
d
a
r
2
d
a
r
3
d
a
r
E
r
E
r
E
r
σ
l
2
d
a
r
E
r
17
<

=

>
+
=
+
=
0
z
,
4
ˆ
2
ˆ
0
z
,
4
ˆ
2
ˆ
2
0
cps
0
2
0
cps
0
z
q
z
q
πε
ε
σ
πε
ε
σ
z
z
z
z
E
r
where, for any field point at a distance
z
from the charged plane,
a
σ
=
cps
q
arises from a
circular area
2
2
z
π
=
a
that extends to a variable radius of
z
2
1
2
within the plane.
Exercise 5
:
A thin spherical shell of radius
a
r
has a uniform surface charge density of
σ
.
Find the electric field
E
r
and potential
V
due to this charge distribution.
Exercise 6
:
Two infinitely wide planar distributions of electric scalar charges cut the
z
–axis
at positions
a
z
2
1
±
=
and have uniform surface charges densities of
σ
±
. Find their
combined electric field
E
r
everywhere. Sketch the field
E
r
against position
z
.
2.6.4.
Differential form of Gauss’ law and Poisson’s equation
In equations (2.39) and (2.40) the integrals are over the source scalar charges to remind
ourselves that our interest is to determine the
E
r
due to these sources only. Using the divergence
theorem to the LHS of (2.32), and expressing the RHS as a volume integral over the source
distribution, Gauss’ law becomes
(
)
∫
∫
∫
≡
′
=
=
⋅
∇
=
Φ
′
v
v
v
v
v
v
d
d
d
enc
0
e
ρ
ρ
ε
q
E
r
(2.42)
The equivalence is valid for
v
v
≤
′
if there are no charges in the rest of the Gaussian volume
v
.
Then the integrands over the same volume on the two sides of the equation must be equal, that is:
ρ
ε
=
⋅
∇
E
r
0
(2.43)
This is the differential form of Gauss’ law.
Exercise 7
:
An electric scalar charge
q
is deposited on a conducting sphere of radius
0
r
. The
potential at a radial distance
s
r
from the centre of this conducting sphere is
s
V
. Express the
charge density everywhere in terms of
s
V
,
s
r
and
0
r
. Find the electric field
E
r
everywhere.
2.7
Work and energy in electrostatics
If we have a collection of electric scalar point charges
n
i
q
i
,
.
.
.
,
3
,
2
,
1
,
=
, the
minimum work done to assemble them is
(
)
∑
=
=
n
i
i
i
V
q
W
1
2
1
r
r
where
(
)
i
V
r
r
is the potential at the position of the charge
i
q
due to all the other
1

n
charges in
the system. For a continuous electric scalar charge distribution the work done is
∫
=
q
V
W
d
2
1
[ J ]
(2.44)
18
where
V
is the potential at the position of the elemental charge
q
d
.
The work represents the energy stored in the assembly of charges. Thus the electric
potential energy is
∫
=
q
V
U
d
2
1
[
J ]
(2.45)
We can also rewrite this potential energy in terms of the electric field at the position of the
charge distribution as
(
)
∫
∫
=
=
space
All
2
0
2
1
space
All
0
2
0
d
d
2
v
v
E
ε
ε
ε
E
U
(2.46)
where the integration is over the entire space. It follows that the electric potential energy density