Electric Flux out of Closed Surface surrounding Point Charge

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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.


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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:

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

$\blacksquare$


Sources

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