Namely that the Flux of the electric field through any surface equals the charge enclosed/vacuum permittivity.
Without any coordinates or reference to axis.
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2$\begingroup$ What are you trying to prove it from? What equations are on the table? $\endgroup$– Cort AmmonFeb 26, 2020 at 14:56
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$\begingroup$ So we're starting from Maxwell's equations as a given? $\endgroup$– Cort AmmonFeb 26, 2020 at 15:10
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$\begingroup$ No I don't think so. First one is basically gauss law. $\endgroup$– Kugutsu-oFeb 26, 2020 at 15:20
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1$\begingroup$ Then what is your starting point? What do you want to prove it from? $\endgroup$– fqqFeb 26, 2020 at 16:06
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$\begingroup$ Coulombs law... $\endgroup$– Kugutsu-oFeb 26, 2020 at 16:17
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3 Answers
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If you start with Maxwell's equation you can apply the divergence theorem.
$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$ $$\int_V(\nabla \cdot \vec{E})dV = \frac{1}{\epsilon_0}\int_V \rho dV$$ $$\oint_A \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$$
Where $\oint_A \vec{E} \cdot d\vec{A}$ is the electric flux.
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$\begingroup$ No that's trivial.thats not a proof of the law it's just equating different forms of stating it. $\endgroup$ Feb 26, 2020 at 15:54
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$\begingroup$ This is a proof if you start with Maxwell's equations and the divergence theorem being given. What is your starting point then? Also, are you just downvoting answers you don't like, even if they aren't wrong? $\endgroup$– roshokaFeb 26, 2020 at 15:55
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$\begingroup$ Check my inswer how I would prove this. Yours is not the proof as I said. Using my answer, ( I found it in a textbook) you can prove divergence theorem which is equivalent to gauss law. Starting with divergence theorem to prove gauss law is circular. $\endgroup$ Feb 26, 2020 at 16:05
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$\begingroup$ I guess I don't understand. How is the divergence theorem equivalent to Gauss' law? It seems like Gauss' integral law is a consequence of the divergence theorem, which is something you prove in Real Analysis class $\endgroup$– roshokaFeb 26, 2020 at 16:08
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$\begingroup$ To calculate divergence you need to calculate the Flux though the boundary of small volume first than divide by the volume. $\endgroup$ Feb 26, 2020 at 16:15
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physic is proved by experiments, so enclose a charge in some surface and measure the flux, it is easier taking a conducting surface and measure the charge of the surface. this and of measurements finally came to create the Maxwell equations. and sometimes laws are explained by them. So you should say from what standpoint you want to proof the law.
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This I think answers it. The point is just to show that the Flux through any surface is equal to the Flux through a sphere because any surface patch will always be (R/r) ^2 /cos¶ bigger that the corresponding sphere patch but it will be tilted by an angle ¶ reducing the Flux by cos¶. Since the electric field also drops as distance squared everything cancels out and you just get the electric field times the area of the sphere.
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1$\begingroup$ ie $\oint \hat r\cdot d\vec A / r^2 = 4\pi$ $\endgroup$ Feb 26, 2020 at 17:04