# Electric Flux out of Closed Surface surrounding Point Charge

## Theorem

Let $q$ be a point charge.

Let $S$ be a closed surface surrounding $q$.

The total electric flux through $S$ is given by:

$F = \dfrac q {\varepsilon_0}$

## Proof

### Lemma

Let $q$ be a point charge located at the origin of a spherical polar coordinate system.

Let $S$ be a closed surface surrounding $q$.

Let $\delta \mathbf S$ be an area element of $S$.

Let $\delta \Omega$ be the solid angle subtended by the projection of $\delta \mathbf S$ from $q$.

The total electric flux through $\delta \Omega$ is given by:

$F = \dfrac q {4 \pi \varepsilon_0} \delta \Omega$

$\Box$

From this lemma, we can transform the surface integral for the flux through the entire surface $S$ into a double integral between angles:

 $\ds \int_S \mathbf E \cdot \rd \mathbf S$ $=$ $\ds \dfrac q {4 \pi \varepsilon_0} \int_0^\pi \sin \theta \rd \theta \int_0^{2 \pi} \rd \phi$ $\ds$ $=$ $\ds \dfrac q {4 \pi \varepsilon_0} \int_0^\pi \sin \theta \rd \theta \bigintlimits \phi 0 {2 \pi}$ $\ds$ $=$ $\ds \dfrac q {4 \pi \varepsilon_0} \int_0^\pi \sin \theta \rd \theta \times 2 \pi$ $\ds$ $=$ $\ds \dfrac q {2 \varepsilon_0} \int_0^\pi \sin \theta \rd \theta$ $\ds$ $=$ $\ds \dfrac q {2 \varepsilon_0} \bigintlimits {-\cos \theta} 0 \pi$ $\ds$ $=$ $\ds \dfrac q {2 \varepsilon_0} \paren {-\paren {\cos \pi - \cos 0} }$ $\ds$ $=$ $\ds \dfrac q {2 \varepsilon_0} \paren {-\paren {-1 - 1} }$ $\ds$ $=$ $\ds \dfrac q {\varepsilon_0}$

$\blacksquare$