## Flux Out of a Surface Contaioning Point Charges

Theorem
The flux out of any closed surface surrounding a system of charges
$\{ q_i \}_i$
is
$\int \int S \mathbf{E} \cdot \mathbf{n} dS = \sum_i \frac{q_i}{\epsilon_0}$

Proof
The potential at the point
$(x,y,z)$
on the surface due to the charge
$q_i$
is
$\phi_i = \frac{q_i}{4 \pi \epsilon_0 r_i}$
where
$r_i$
is the position vector of
$(x,y,z)$
relative to the charge.
The electric field at
$on the surface due to the charge \[q_i$
is
$\mathbf{E_i} = - \mathbf{\nabla} \phi_i = - \frac{q_i}{4 \pi \epsilon_0} \mathbf{\nabla} (\frac{1}{r_i}) = \frac{q_i}{4 \pi \epsilon_0} \frac{\mathbf{r_i}}{r^3_i}$

The electric field due to the collection of charges is
$\mathbf{E} = \sum_i \mathbf{E_i} = \sum_i \frac{q_i}{4 \pi \epsilon_0} \frac{\mathbf{r_i}}{r^3_i}$

The flux out of the surface
$S$
is
$\int \int_S \mathbf{E} \cdot \mathbf{n} dS = \frac{1}{4 \pi \epsilon_0} \int \int_S \sum_i \frac{q_i \mathbf{r_i} \cdot \mathbf{n}}{r^3_i} dS$

Now use Gauss's Theorem
to give
$\frac{}{4 \pi \epsilon_0} \int \int_S \sum_i \frac{q_i \mathbf{r_i} \cdot \mathbf{n}}{r^3_i} dS=\frac{}{4 \pi \epsilon_0} \sum_i \frac{4 \pi q_i} = \sum_i \frac{q_i}{\epsilon_0}$