Electric Flux out of Closed Surface surrounding Point Charge/Lemma

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

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:

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