Electric Flux out of Closed Surface/Examples/Sphere

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Example of Use of Electric Flux out of Closed Surface

Let $B$ be a spherical body in space with radius $R$.

Let $Q$ be the total electric charge within $B$


Then the total electric flux out of $B$ is given by:

$F = \dfrac Q {\varepsilon_0}$

where $\varepsilon_0$ denotes the vacuum permittivity.


Proof

Let the center of $B$ be located at the origin of a spherical polar coordinate system.

Let $\mathbf r$ denote the position vector of an arbitrary point $P$ on the surface of $B$.

Let the spherical polar coordinates of $P$ be denoted $\polar {r, \theta, \phi}$.

The electric charge density and electric field caused by $B$ are expressible in terms of $\polar {r, \theta, \phi}$ as:

$\map \rho {r, \theta, \phi}$

and:

$\map {\mathbf E} {r, \theta, \phi}$

Let us define $\d \mathbf S$ as being the infinitesimal area element of the surface of $B$ demarcated by the arcs on $B$ subtending:

the polar angles $\theta$ and $\d \theta$ and

and:

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

where $\d \theta$ and $\d \phi$ are likewise infinitesimal.

From Area Element in Spherical Polar Coordinate System:

$\d S = R^2 \sin \theta \rd \theta \rd \phi$

Integrating over all space, we have:

\(\ds F\) \(=\) \(\ds \int_{\phi \mathop = 0}^{2 \pi} \int_{\sigma \mathop = 0}^\pi \map E {R, \theta, \phi} R^2 \sin \theta \rd \theta \rd \phi\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\varepsilon_0} \int_{\phi \mathop = 0}^{2 \pi} \int_{\sigma \mathop = 0}^\pi \int_{r \mathop = 0}^R \map \rho {r, \theta, \phi} r^2 \sin \theta \rd r \rd \theta \rd \phi\)

from which the result follows.

$\blacksquare$


Sources