Question

State and prove Gauss' theorem in electrostatics.

Answer 1

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Gauss theorem states that the electric flux ΦE through any closed surface is equal to 1 / ɛo times the 'net' charge q is enclosed by the surface .

Proof :-

Let q be the charge .

Let us construct the Gaussian sphere of radius r .

Consider , A surface or area ds having having ds(vector) .

Normal having the flux at ds .

Flux at ds .

d e = E (vector) d s (vector) cos θ

But , θ = 0 .

Total flux

C = f d Φ

E 4 π r^2

σ = 1 / 4πɛo q / r^2 × 4π r^2

σ = q / ɛo (proved)

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Answer 2

Gauss Theorem :- according to Gauss theorem, electric flux through Gaussian surface is the ratio of total charged enclosed the Gaussian surface to permittivity of medium.
mathematically,
$\bf{\Phi=\frac{Q_{in}}{\epsilon_0}}$

proof :- Let's take a charge q in vaccum space. we want to find electric field due to charge q , r distance from it .
from Coulomb's law,
E = Kq/r²

where K is kappa constant and its value is
$\kappa=\frac{1}{4\pi\epsilon_0}$

so, $E=\frac{q}{4\pi\epsilon_0r^2}$

$E=\frac{q}{(4\pi r^2)\epsilon_0}$

$E=\frac{q}{A\epsilon_0}$ [ here A denotes spherical area made by electric flux at r distance from position of charge]

or, $EA=\frac{q}{\epsilon_0}$

we know, EA = electric flux

so, $\Phi=\frac{Q_{in}}{\epsilon_0}$
hence proved//