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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?

Amsterdam6483
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Neel Shejwalkar
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  • 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. – G. Smith Apr 04 '20 at 02:01
  • Gauss's Law is a law, you don't use Columb's law to derive Gauss's Law – SK Dash Apr 04 '20 at 02:14
  • 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. – Charlie Apr 04 '20 at 03:12
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    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. – Charles Francis Apr 04 '20 at 19:54

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Yes, your surface integral would change depending on the shape of the surface you're talking about, if you take a highly complex surface you're going to have to solve a highly complex surface integral. Gauss' law applies in all cases though. If you have a closed surface, the flux of the electric field through it is always proportional to the charge contained within it, regardless of how simple or complex the surface is, that is what Guass' law says, it makes no comment on the shape of the surface.

Charlie
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