State Gauss's Theorem and prove it. - Vidyadip

Question

State Gauss's Theorem and prove it.

✓

Answer

Gauss's Law : It states that 'The total electric flux $\Phi_{E}$ through any closed surface is $(\frac{1}{\in_{0}})$ times the total charge $q$ enclosed inside the closed surface, where ɛ₀ is the permittivity of free space."
That is $\Phi_{E}=\oint_{S}\vec{E}.\vec{dS}=\frac{q}{\in_{0}}$
This is integral form of Gauss's law.
Proof : Consider a closed surface S, surrounding a point charge $+q$ and also consider a small area $d S$ on the closed surface S . (adjacent fig.) Let $d \omega$ be the solid angle subtended by area $d S$ at $O$. Then, electric flux over the area $d S$ is
$d \Phi_{E}=\overrightarrow{E} \cdot \hat{n} d S$
where $\hat{n}$ is the unit vector normal to $d S$.
The electric flux or the total electric normal induction through the whole of closed surface $S$ is
$\Phi_{E}=\oint_{S} \overrightarrow{E} \cdot \hat{n} d S$ ...(1)
The electric intensity at a point on $d S$ due to charge $q$ is given by :
$\overrightarrow{E}=\frac{q \hat{r}}{4 \pi \varepsilon_0 r^2}$ ...(2)
where $\hat{r}$ is a unit vector in the direction of $\vec{E}$.

Substituting relation (2) in (1),
$\Phi_{E}=\oint_{S}\left(\frac{q \hat{r} \cdot \hat{n}}{4 \pi \varepsilon_0 r^2}\right) d S $
$\Rightarrow \Phi_{E}=\oint_{S}\left(\frac{q d S \cos \theta}{4 \pi \varepsilon_0 r^2}\right) $ ...(3)
$\text { But } d \omega=\frac{d S \cos \theta}{r^2}$
∴ Relation (3) takes the following form:
$\Phi_{E}=\frac{q}{4 \pi \varepsilon_0} \oint d \omega .$ ...(4)
The solid angle subtended by any closed surface at any point inside the surface $=4 \pi$ staradian i.e., $\oint d \omega= 4 \pi$
Substituting it in relation (4),
$\Phi_{E}=\frac{q}{4 \pi \varepsilon_0}(4 \pi) $
$\Rightarrow \Phi_{ E }=\frac{q}{\varepsilon_0}$ ...(5)

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