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Question

# State and prove Gauss's Theorem.

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Solution

## Gauss's Theorem: The net electric flux passing through any closed surface is 1εo times, the total charge q present inside it.Mathematically, Φ=1εo⋅qProof: Let a charge q be situated at a point O within a closed surface S as shown. Point P is situated on the closed surface at a distance r from O. The intensity of electric field at point P will be¯E=14πεo⋅qr2 .......(1)Electric flux passing through area ds enclosing point P,dΦ=→E⋅→dsor dΦ=E⋅dscosθ[where θ is the angle between E and ds]Flux passing through the whole surface S,∫∫sdΦ=∫∫sE⋅dscosθ ........(2)Substituting the value of E from eqn. (1) in eqn. (2),Φ=∫∫s14πεoqr2dscosθ [Hereint∫sdΦ=Φ]Φ=14πεoq∫∫sdscosθr2⇒Φ=14πεoq⋅ω [∵∫∫sdscosθr2=ω]Here ω= solid angle.But here the solid angle subtended by the closed surface S at O is 4π, thusΦ=14πεo×q×4πor Φ=1εo⋅q.

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