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Answer in Brief

State Gauss's law in electrostatics. Show, with the help of a suitable example along with the figure, that the outward flux due to a point charge 'q'. in vacuum within a closed surface, is independent of its size or shape and is given by `q/ε_0`

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#### Solution

**Statement: **The electric flux linked with a closed surface is equal to `(1)/ε_0` times the net charge enclosed by a closed surface.

**Mathematical expression :**

`Ø_"E" = oint vec"E".dvec"s" = (1)/(ε_0) (q_"net")`

Consider two spherical surfaces of radius r and 2r respectively and a charge 1 is enclosed in it. According to gauss theorem, the total electric flux linked with a closed surface depends on the charge enclosed in it so

(a)

(b)

`Ø_E = q/ε_0 "and for fig"("b")`

`Ø_E = q/ε_0`.

Concept: Gauss’s Law

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