Question

State and prove Gauss's theorem in electrostatics

Answer 1

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Answer 2

Suppose the charge qq is located in the origin. Then the flux of E⃗ E→ through anyclosed Gaussian surface ΣΣ can be represented by the surface integral:

Φ=q4πε0∮Σr⃗ r3.dσ⃗ Φ=q4πε0∮Σr→r3.dσ→

To prove that this integral is independent of the surface ΣΣ, you need the following theorem, which you can easily verify by a simple calculation or you can find it online.

Let SS be any surface and suppose S′S′ is the projection of SS onto the sphere x2+y2+z2=R2x2+y2+z2=R2, then:

∬Sr⃗ r3.dσ⃗ =area(S′)R2∬Sr→r3.dσ→=area(S′)R2

If you apply this theorem with R=1R=1 to equation one, then the area of S′S′ is that of the entire sphere with R=1R=1 and is equal to 4π4π. Therefor:

Φ=(q4πϵ0)4π=qε0Φ=(q4πϵ0)4π=qε0

Note: area(S′)R2area(S′)R2 is the solid angle of area S