Now recall that flux is the scalar product of a vector field and a bit of
surface
F
l
u
x
=
F
⃗
⋅
a
⃗
F
l
u
x
=
F
⃗
⋅
a
⃗
where
F
⃗
F
⃗
is some vector field and
a
⃗
a
⃗
is a surface with the direction defined by the normal to the surface. For a
series of connected surfaces
a
⃗
j
a
⃗
j
the total flux through the combined surface would be the sum of the individual
elements. For a vector field
E
⃗
E
⃗
passing through the surface this leads to
Φ
=
∑
E
⃗
j
⋅
a
⃗
j
Φ
=
∑
E
⃗
j
⋅
a
⃗
j
or when we go to infinitesimal areas
Φ
=
∫
s
u
r
f
a
c
e
E
⃗
⋅
d
a
⃗
Φ
=
∫
s
u
r
f
a
c
e
E
⃗
⋅
d
a
⃗
Now lets consider a charge
q
q
in the middle of a sphere

Φ
=
∮
E
⃗
⋅
d
a
⃗
=
∮
E
⋅
ⅆ
a
=
E
∮
ⅆ
a
=
E
(
4
π
r
2
)
Φ
=
∮
E
⃗
⋅
d
a
⃗
=
∮
E
⋅
ⅆ
a
=
E
∮
ⅆ
a
=
E
(
4
π
r
2
)
but
E
=
k
q
r
2
E
=
k
q
r
2
then
Φ
=
4
π
k
q
Φ
=
4
π
k
q
ε
0
≡
1
4
π
k
ε
0
≡
1
4
π
k
So for this case we get
∮
E
⃗
⋅
d
a
⃗
=
q
ε
0
∮
E
⃗
⋅
d
a
⃗
=
q
ε
0
We can generalize this to any closed surface. It is clear that for an
arbitrary closed source, we can draw a sphere around the source within the
arbitrary surface.. Think of bullets being fired from a gun, it is clear that
the bullets originating in the inner sphere all pass through the outer surface
and so one would expect that the flux would be the same. For example consider
a
⃗
a
⃗
to be a patch on the inner sphere and
A
⃗
A
⃗
to be its projection onto the outer arbitrary surface (with its normal making
an angle
θ
θ
with
respect to the normal to
a
⃗
.
a
⃗
.

On the inner patch
Φ
r
=
E
⃗
r
⋅
a
⃗
=
E
r
a
Φ
r
=
E
⃗
r
⋅
a
⃗
=
E
r
a
and at the outer patch
Φ
R
=
E
⃗
R
⋅
A
⃗
=
E
R
A
cos
θ
=
[
E
r
(
r
R
)
2
]
[
a
(
R
r
)
2
1
cos
θ
]
cos
θ
=
E
r
a
=
Φ
r
Φ
R
=
E
⃗
R
⋅
A
⃗
=
E
R
A
cos
θ
=
[
E
r
(
r
R
)
2
]
[
a
(
R
r
)
2
1
cos
θ
]
cos
θ
=
E
r
a
=
Φ
r
So the two have equivalent fluxes.

Any electric field is the sum of fields of its individual sources so we can
write
Φ
=
∮
E
⃗
⋅
d
a
⃗
=
∮
∑
i
E
⃗
i
d
a
⃗
=
1
ε
0
∑
i
q
i
Φ
=
∮
E
⃗
⋅
d
a
⃗
=
∮
∑
i
E
⃗
i
d
a
⃗
=
1
ε
0
∑
i
q
i
or for charge distributed throughout the volume
∮
E
⃗
⋅
d
a
⃗
=
1
ε
0
∫
ρ
ⅆ
V
∮
E
⃗
⋅
d
a
⃗
=
1
ε
0
∫
ρ
ⅆ
V

Now we can apply Gauss' Theorem
∮
E
⃗
⋅
d
a
⃗
=
∫
∇
⃗
⋅
E
⃗
ⅆ
V
=
1
ε
0
∫
ρ
ⅆ
V
∮
E
⃗
⋅
d
a
⃗
=
∫
∇
⃗
⋅
E
⃗
ⅆ
V
=
1
ε
0
∫
ρ
ⅆ
V

The equation
∫
∇
⃗
⋅
E
⃗
ⅆ
V
=
1
ε
0
∫
ρ
ⅆ
V
∫
∇
⃗
⋅
E
⃗
ⅆ
V
=
1
ε
0
∫
ρ
ⅆ
V
must be true for any volume of any size, shape or location.
The only way that can be true is if:

∇
⃗
⋅
E
⃗
=
ρ
ε
0
∇
⃗
⋅
E
⃗
=
ρ
ε
0
Initially one may think that this is a much less clear way of posing Gauss'
Law. In practice it is much more useful than the integral form. Given an
arbitrary distribution of charge we can calculate the electric field anywhere
in space.

Comments:"This book covers second year Physics at Rice University."