Question

State and prove Gauss's theorem in electrostatics.

Answer 1

Gauss Theorem in electrostatics

1. Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.
2. According to Gauss’s theorem, the net-outward normal electric flux through any closed surface of any shape is equivalent to $\frac{1}{{\epsilon }_0}$ times the total amount of charge contained within that surface.
3. The equation of Gauss's law is given by $\varphi =\frac{q}{{\epsilon }_0}$ where $\varphi$ is the electric flux, $q$ is the charge enclosed and ${\epsilon }_0$ is the permittivity of free space$\left({\epsilon }_0=8.85\times {10}^{-12}\right)$

Proof of Gauss’s Theorem Statement:

Let a charge $q$be situated at a point $O$ inside a closed surface $S$ as shown.

A point $P$ is situated on the closed surface $S$ at a distance $r$ from $O$.

Then the intensity of electric field at point $P$ will be
$E=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}-------------------\left(1\right)$
Electric flux passing through area $ds$ enclosing the point $P$,
$$\begin{gathered} d\varphi =\overrightarrow{E}·\overrightarrow{ds} \\ \Rightarrow d\varphi =Eds\cos \theta \end{gathered}$$

where $\theta$ is the angle between $\overrightarrow{E}$ and $\overrightarrow{ds}$
Flux passing through the surface $S$,
${\iint }_{S}d\varphi ={\iint }_{S}Eds\cos \theta --------------\left(2\right)$
Substituting the value of $E$ from $\left(1\right)$ in $\left(2\right)$, we get
$$\begin{gathered} \varphi ={\iint }_S\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}ds\cos \theta \left(Here,{\iint }_{S}d\varphi =\varphi \right) \\ \Rightarrow \varphi =\frac{1}{4{\mathrm{\pi \epsilon }}_0}q{\iint }_S\frac{ds\cos \theta }{r^2} \\ \Rightarrow \varphi =\frac{1}{4{\mathrm{\pi \epsilon }}_0}q\omega \left(\because {\iint }_S\frac{ds\cos \theta }{r^2}=\omega \right) \end{gathered}$$
Where $\omega$ is the solid angle.
But here the solid angle subtended by the closed surface $S$ at $O$ is $4\mathrm{\pi }$, thus
$$\begin{gathered} \varphi =\frac{1}{4{\mathrm{\pi \epsilon }}_0}q\times 4\mathrm{\pi } \\ \Rightarrow \mathrm{\varphi }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0} \end{gathered}$$

Hence the Gauss's theorem is proved.