Question

State and prove Gauss's Theorem.

Answer 1

Gauss's Theorem: The net electric flux passing through any closed surface is $\frac{1}{{\epsilon }_o}$ times, the total charge q present inside it.
Mathematically, $\Phi =\frac{1}{{\epsilon }_o}\cdot q$
Proof: Let a charge q be situated at a point O within a closed surface S as shown. Point P is situated on the closed surface at a distance r from O. The intensity of electric field at point P will be
$\overline{E}=\frac{1}{4\pi {\epsilon }_o}\cdot \frac{q}{r^2}$ .......(1)
Electric flux passing through area ds enclosing point P,
$d\Phi =\overrightarrow{E}\cdot \overrightarrow{ds}$
or $d\Phi =E\cdot dscos\theta$
[where $\theta$ is the angle between E and ds]
Flux passing through the whole surface S,
$\int {\int }_{s}d\Phi =\int {\int }_{s}E\cdot dscos\theta$ ........(2)

Substituting the value of E from eqn. (1) in eqn. (2),
$\Phi =\int {\int }_s\frac{1}{4\pi {\epsilon }_o}\frac{q}{r^2}dscos\theta$ $[Hereint{\int }_{s}d\Phi =\Phi ]$
$\Phi =\frac{1}{4\pi {\epsilon }_o}q\int {\int }_s\frac{dscos\theta }{r^2}$
$\Rightarrow \Phi =\frac{1}{4\pi {\epsilon }_o}q\cdot \omega$ $[\because \int {\int }_s\frac{dscos\theta }{r^2}=\omega ]$
Here $\omega =$ solid angle.

But here the solid angle subtended by the closed surface S at O is $4\pi$, thus
$\Phi =\frac{1}{4\pi {\epsilon }_o}\times q\times 4\pi$
or $\Phi =\frac{1}{{\epsilon }_o}\cdot q$.