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State and Prove Gauss theorem in electrostatics.

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Statement: The net-outward normal electric flux through any closed surface of any shape is equal to 1/ε0 times the total charge contained within that surface, 1/ε0 i.e.,

Where \(\oint S\) indicates the surface integral over the whole of the closed surface, \(\Sigma\,q\)

Is the algebraic sum of all the charges (i.e., net charge in coulombs) enclosed by surface S and remain unchanged with the size and shape of the surface.

Proof: Let a point charge +q be placed at centre O of a sphere S. Then S is a Gaussian surface.

Electric field at any point on S is given by

The electric field and area element points radially outwards, so θ = 0°.

Flux through area \(\vec {dS}\) is

or, \(\varphi=\cfrac{q}{\varepsilon_0}\) which proves Guass's theorem.

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