Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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: $\Phi=\frac {Q}{ \epsilon_0}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.