# Gauss' Law Generalization to all closed surfaces [duplicate]

I'm currently in an introductory physics class, and we've learned about Gauss Law defining the flux as $$\int d\phi = \oint EdA = \frac{q}{\epsilon_0}$$ and from what I understand the way to arrive at this equation is to assume a source at the center of a sphere, so the electric field at every point is the same. $$\oint EdA = E\oint dA = EA = (\frac{1}{4\pi\epsilon_0})(\frac{q}{r^2})(4\pi r^2) = \frac{q}{\epsilon_0}$$ So I understand how to derive Gauss' Law, and I see how it can be applied to spheres, but I can't see mathematically how this law can be generalized for any enclosed surface. Wouldn't there be a change in derivation for any other shape? Or am I missing something?

• the way to arrive at this equation is to assume a source at the center of a sphere No, this approach doesn’t prove Gauss’ Law for arbitrary closed surfaces. Apr 4, 2020 at 2:01
• Gauss's Law is a law, you don't use Columb's law to derive Gauss's Law Apr 4, 2020 at 2:14
• The way you've written your integrals is confusing, $d\phi$ isn't really standard notation and you've done something strange in your second equation, $(1/4\pi\epsilon_0)(Q/4^2)(4\pi r^2)$ is definitely not right, you should have $EA=4\pi r^2 E$, from which Coulomb's law follows. Apr 4, 2020 at 3:12
• All you are missing is that this is an introductory treatment! Coulomb's law is really just a consequence of Gauss's law. When you do the full Maxwell equations, you will understand better. Apr 4, 2020 at 19:54