Electric Flux out of Closed Surface surrounding Body with Continuous Charge Distribution

Theorem
Let $B$ be a body of matter which has a continuous macroscopic charge density $\map \rho {\mathbf r}$.

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

The total electric flux through $S$ generated by the electric charge on $B$ is given by:
 * $\ds F = \dfrac 1 {\varepsilon_0} \int_V \map \rho {\mathbf r} \rd \tau$

where:
 * $V$ is the total volume enclosed by $S$
 * $\d \tau$ is an infinitesimal volume element
 * $\mathbf r$ is the position vector of $\d \tau$
 * $\map \rho {\mathbf r}$ is the macroscopic charge density of the macroscopic electric field at $\mathbf r$
 * $\varepsilon_0$ denotes the vacuum permittivity.

Proof
Let $\d \tau$ be an arbitrary infinitesimal volume element.

$\d \tau$ can be considered as an infinitesimal point charge $\d q$.

From Electric Flux out of Closed Surface surrounding Point Charge:
 * $\d F_\tau = \dfrac {\d q} {\varepsilon_0}$

where $\d F_\tau$ is the infinitesimal part of $F$ brought about by $\d q$.

By definition of macroscopic charge density:


 * $\map \rho {\mathbf r} = \dfrac {\d q} {\d \tau}$

Hence:
 * $\d F_\tau = \dfrac {\map \rho {\mathbf r} \rd \tau} {\varepsilon_0}$

Integrating over all space:


 * $\ds F = \dfrac 1 {\varepsilon_0} \int_V \map \rho {\mathbf r} \rd \tau$