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
\[\frac{}{4 \pi \epsilon_0} \int \int_S \sum_i \frac{q_i \mathbf{r_i} \cdot \mathbf{n}}{r^3_i} dS=\frac{}{4 \pi \epsilon_0} \sum_i \frac{4 \pi q_i} = \sum_i \frac{q_i}{\epsilon_0} \]

You have no rights to post comments