Electric Flux out of Closed Surface surrounding Point Charge/Lemma

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Theorem

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:

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


Proof

Consider the area element $\delta \mathbf S$ at the position $P$ with position vector $\mathbf r$ whose spherical coordinates are $\polar {r, \theta, \phi}$.

Let $\delta S_p$ be the projected area of $\delta \mathbf S$ to the plane perpendicular to $\mathbf r$.


Area-Element-Spherical.png


Let us define $\delta \mathbf S$ as being the subset of the surface of $S$ demarcated by the arcs on $S$ subtending:

the polar angles $\theta$ and $\delta \theta$ and

and:

the azimuthal angles $\phi$ and $\delta \phi$.

From the above diagram:

$\delta S_p = r^2 \sin \theta \rdelta \theta \rdelta \phi$

Let $\mathbf E$ be the electric field generated by $q$ at $P$.

According to Coulomb's Law of Electrostatics:

the magnitude of $\mathbf E$ is:
$E = \dfrac q {4 \pi \varepsilon_0 r^2}$
the direction of $\mathbf E$ is along $\mathbf r$ away from the origin.

Hence the electric flux in the direction away from the origin through $\delta \mathbf S$ is:

\(\ds \mathbf E \cdot \delta \mathbf S\) \(=\) \(\ds E \rdelta S_p\)
\(\ds \) \(=\) \(\ds \dfrac q {4 \pi \varepsilon_0 r^2} r^2 \sin \theta \rdelta \theta \rdelta \phi\)
\(\ds \) \(=\) \(\ds \dfrac q {4 \pi \varepsilon_0} \sin \theta \rdelta \theta \rdelta \phi\)
\(\ds \) \(=\) \(\ds \dfrac q {4 \pi \varepsilon_0} \rdelta \Omega\)

where $\Omega$ is that solid angle so subtended.

$\blacksquare$


Sources

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