State and prove Gauss Theorem in electrostatics

Gauss’s Theorem Statement:

According to Gauss’s theorem the net-outward normal electric flux through any closed surface of any shape is equivalent to 1/ε0 times the total amount of charge contained within that surface.

Proof of Gauss’s Theorem Statement:

• Let the charge be = q
• Let us construct the Gaussian sphere of radius = r

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

Normal having the flux at ds:

Flux at ds:

d e = E (vector) d s (vector) cos θ

But , θ = 0

Therefore, Total flux:

C = f d Φ

E 4 π r2

Therefore,

σ = 1 / 4πɛo q / r2 × 4π r2

σ = q / ɛo

Gauss law

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

$$\oint{\vec{E}.\vec{d}s=\frac{1}{{{\in }_{0}}}q}$$

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