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State and prove Gauss's theorem in electrostatics.


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Solution

Gauss Theorem in electrostatics

  1. Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.
  2. According to Gauss’s theorem, the net-outward normal electric flux through any closed surface of any shape is equivalent to 1ε0 times the total amount of charge contained within that surface.
  3. The equation of Gauss's law is given by ϕ=qε0 where ϕ is the electric flux, q is the charge enclosed and ε0 is the permittivity of free spaceε0=8.85×10-12

Proof of Gauss’s Theorem Statement:

Let a charge qbe situated at a point O inside a closed surface S as shown.

A point P is situated on the closed surface S at a distance r from O.

Then the intensity of electric field at point P will be
E=14πε0qr2-------------------(1)
Electric flux passing through area ds enclosing the point P,
dϕ=E·dsdϕ=Edscosθ

where θ is the angle between E and ds
Flux passing through the surface S,
Sdϕ=SEdscosθ--------------(2)
Substituting the value of E from (1) in (2), we get
ϕ=S14πε0qr2dscosθHere,Sdϕ=ϕϕ=14πε0qSdscosθr2ϕ=14πε0qωSdscosθr2=ω
Where ω is the solid angle.
But here the solid angle subtended by the closed surface S at O is 4π, thus
ϕ=14πε0q×4πϕ=qε0

Hence the Gauss's theorem is proved.


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