Question

State and prove Gauss's theorem in electrostatics.

Answer 1

Statement : The net-outward normal electric flux through any closed surface of any shape is equal to $1/{ϵ}_0$ times the total charge contained within that surface, $1/{ϵ}_0$ i.e.,
$\oint S\overrightarrow{E}\cdot d\overrightarrow{S}=\frac{1}{{ϵ}_0}\sum q$
Where ${\oint }_S$ indicates the surface integral over the whole of the closed surface, $\sum q$
Is the algebraic sum of all the charges (i.e., net charge in coulombs) enclosed by surface $S$ and remain unchanged with the size and shape of the surface.
Proof : Let a point charge $+q$ be placed at centre $O$ of a sphere $S$. Then $S$ is a Gaussian surface.
Electric field at any point on $S$ is given by
$E=\frac{1}{4\pi {ϵ}_0}\cdot \frac{q}{r^2}$
The electric field and area element points radially outwards, so $\theta =0^{\circ }$
Flux through area $\overrightarrow{dS}$ is
$d\phi =\overrightarrow{E}\cdot dS=EdS\cos 0^{\circ }=EdS$
Total flux through surface $S$ is
$\phi ={\oint }_{S}d\phi ={\oint }_{S}EdS=E{\oint }_{S}dS=E\times AreaofSphere$
$\frac{1}{4\pi {ϵ}_0}\cdot \frac{q}{x^2}\cdot 4\pi r^2$
or, $\phi =\frac{q}{{ϵ}_0}$ which proves Gauss's theorem.