The textbook I am reading claims that Gauss's law is a fundamental law of nature, however is that really true? For example, would it hold up if inside the closed surface there was a negative charge and outside a positive charge?
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4$\begingroup$ gauss law is just an application of the "divergence theorem" en.wikipedia.org/wiki/Divergence_theorem so it is not so much a physical law but a mathematical one. So asking if there are exceptions is a bit like asking if there exception to 1+1=2 $\endgroup$– mondCommented May 7 at 11:23
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$\begingroup$ Related questions concerning charges outside of surfaces and Gauss's Law: Why do outside charges do not contribute to net flux of a Gaussian Surface?; Gauss' Law and an external charge; Does Gauss's law not hold when there are charges outside of the Gaussian surface?. You can find several more related questions by clicking through on these links and looking at the "Linked" and "Related" questions in the sidebar. $\endgroup$– Michael SeifertCommented May 7 at 11:35
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1$\begingroup$ @mond I don't agree. It is not just a mathematical tautology. For example, if the photon had a mass, then you would have an additional term linear to the electric potential. You therefore need to confirm it with experiments. $\endgroup$– LPZCommented May 7 at 11:43
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$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotCommented May 7 at 11:46
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$\begingroup$ @LPZ On the other hand if Gauss's Law was invalid that it would mean one or the other of the integral and differential forms of Maxwell's equations was wrong. And that would imply something fundamentally wrong with Maxwell's equations. $\endgroup$– The PhotonCommented May 7 at 16:33
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2 Answers
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The trick is that Gauss's law tells you about the flux of the Electric field through your surface, it doesn't actually tell you the electric field directly.
So regardless of your charge distribution and any weird closed surface you may choose, Gauss's law will correctly give you $$\oint \vec{E} \cdot d\vec{A}=\frac{Q}{\varepsilon_0}$$
But $\oint \vec{E} \cdot d\vec{A}$ is actually sort of useless to know unless we know the problem has a symmetry such that $\vec{E}\cdot d\vec{A}$ must be the same everywhere on the surface (if we choose the right surface), and we want to know what the magnitude of E is at that surface. That is how Guass's law is typically applied. But without that symmetry it just gives us the flux, and I'm not sure if there are any cases where the flux of the field through some arbitrary surface is actually useful for anything.
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$\begingroup$ There are a few oddball problems where you can use Gauss's Law over a surface where $\vec{E}$ varies to say something meaningful about the electric field. See this problem, for example. But 99% of the problems you'll see in an intro physics class involve the assumption that $\vec{E} \cdot d \vec{A}$ is (piecewise) constant over some surface. $\endgroup$ Commented Jun 7 at 14:09
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I wouldn’t use the word “fundamental” but yes, Gauss’ law appears to be a law of nature. It does hold even in the case you described, with a negative charge inside and a positive charge outside.
Any field lines from the positive charge will go in one side and out another side. The field lines to the negative charge will go in any side but terminate on the charge instead of leaving.
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$\begingroup$ How would they cancel out though? Since the radius would be different going in and out, so the field magnitude would thus also differ. Also ty. $\endgroup$– WerCommented May 7 at 18:55