# Why is electric flux through any closed surface $q/\epsilon_0$?

Why is electric flux through any closed surface $q/\epsilon_0$? In schools we are only taught of its simplest case, i.e. flux through a sphere with charge centered at origin. And then it is generalised to all closed surfaces. Is there really any proof of flux through all closed surfaces.

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Gauss's theorem: en.wikipedia.org/wiki/Divergence_theorem – Dave Nov 1 '13 at 16:41
Possible duplicate: physics.stackexchange.com/q/38404/2451 – Qmechanic Nov 1 '13 at 21:42

A simple derivation would be this one. Suppose you have a single charge $q_i$ inside a closed surface.

The flux across a closed surface is: $\Phi=\oint\vec{E} d\vec{A}$

Coulomb's law says: $\vec E=\frac{q}{4\pi\epsilon_0r^3}\vec r$

In this case, it means that the flux is: $\Phi_i =\oint \frac{q_i}{4\pi\epsilon_0r^3}\vec r d\vec{A}$

If you solve the integral, it yields: $\Phi_i=\frac {q_i}{ \epsilon_0}$

By the superpositon principle: $\Phi=\frac {Q}{ \epsilon_0}$

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