State and prove Gauss theorem
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Gauss's Theorem and its Proof. The surface integral of electrostatic field E produce by any source over any closed surface S enclosing a volume V in vacuum i.e. total electric flux over the closed surface S in vacuum is 1/epsilon times the total charge Q contained inside S.
Gauss's Theorem and its Proof. The surface integral of electrostatic field E produce by any source over any closed surface S enclosing a volume V in vacuum i.e. total electric flux over the closed surface S in vacuum is 1/epsilon times the total charge Q contained inside S.
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Explanation:
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
∮E⃗ .d⃗ s=1∈0q .
According to Gauss Law,
Φ = → E.d → A
Φ = Φcurved + Φtop + Φbottom
Φ = → E . d → A = ∫E . dA cos 0 + ∫E . dA cos 90° + ∫E . dA cos 90°
Φ = ∫E . dA × 1
Due to radial symmetry, the curved surface is equidistant from the line of charge and the electric field in the surface has a constant magnitude throughout.
Φ = ∫E . dA = E ∫dA = E . 2πrl
The net charge enclosed by the surface is:
qnet = λ.l
Using Gauss theorem,
Φ = E × 2πrl = qnet/ε0 = λl/ε0
E × 2πrl = λl/ε0
E = λ/2πrε0
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