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:

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

$\blacksquare$

Sources

• 1990: I.S. Grant and W.R. Phillips: Electromagnetism (2nd ed.) ... (previous) ... (next): Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law: $1.4.2$ The flux of the electric field out of a closed surface: $(1.14 \, \text b)$