Gauss' Law applicability on any closed surface

[torito_verdejo]

TL;DR Summary

How does universal applicability of Gauss Law follow from divergence theorem and the conservative nature of the electric field?

I have read multiple threads on Physics Forums, Stackexchange and Quora, as well as the explanation of Gauss Law, but still don't understand the most fundamental aspect of it: its applicability for any kind of surface. More precisely, I don't get how this follows from the fact that

$$\iiint_V(\nabla\cdot\textbf{F})d\tau = \oiint_S\textbf{F}\cdot d\textbf{s}$$

What I have guessed is that, since electric field is conservative

$$\oiint_S\textbf{E}\cdot d\textbf{s}=\iiint_V(0)d\tau = \frac{Q}{\epsilon_0}=\iiint_{V'}(0)d\tau'=\oiint_{S'}\textbf{E}\cdot d\textbf{s'}$$

Which would imply, if my reasoning is right, that if something is true for the integral of a given spherical surface $S$ enclosing a volume $V$, on which we integrate a null function ($\nabla\cdot\textbf{E}=0$), it stays true for some other integral over any other surface $S'$ enclosing a volume $V'$, for their triple integral would reduce to the same: integrating zero.

But this reasoning is far from mathematical, and I wonder if its correct at all. Am I right? If I'm not, I would like to know why and I would really appreciate a formal explanation.

Thank you very much.

 

[PeroK]

You ought to fix your latex.

The divergence theorem is true for non-spherical surfaces, if that's what is concerning you.

What don't you understand on this page, for example:

https://physics.stackexchange.com/q...m-arrived-at-from-coulombs-law-and-how-is-the

 

[torito_verdejo]

You ought to fix your latex.

The divergence theorem is true for non-spherical surfaces, if that's what is concerning you.

What don't you understand on this page, for example:

https://physics.stackexchange.com/q...m-arrived-at-from-coulombs-law-and-how-is-the

I get that derivation, but I think it doesn't explicitly answer what I'm asking. I'm asking why what is obtained from the integral of a closed sphere (i.e. $\Phi_E=\frac{Q_{enclosed}}{\epsilon_0}$) holds for any other closed surface. Where does the shape independence come from?

 

[PeroK]

I get that derivation, but I think it doesn't explicitly answer what I'm asking. I'm asking why what is obtained from the integral of a closed sphere (i.e. $\Phi_E=\frac{Q_{enclosed}}{\epsilon_0}$) holds for any other closed surface. Where does the shape independence come from?

That page does not assume a spherical surface. It takes an arbitrary "weird" surface, and uses an argument to show that $1/r^2$ integrated over any surface comes to $4\pi$.

This is analagous to the 2D argument that $1/r$ integrated round a closed curve equals $2\pi$. This one is easier to see/prove, so you might start with that.

 

[torito_verdejo]

That page does not assume a spherical surface. It takes an arbitrary "weird" surface, and uses an argument to show that $1/r^2$ integrated over any surface comes to $4\pi$.

This is analagous to the 2D argument that $1/r$ integrated round a closed curve equals $2\pi$. This one is easier to see/prove, so you might start with that.

You're right, the second answer does so! (hard palm face on me)

https://physics.stackexchange.com/a/339364/230605
Thank you very much. :)

 

[vanhees71]

I get that derivation, but I think it doesn't explicitly answer what I'm asking. I'm asking why what is obtained from the integral of a closed sphere (i.e. $\Phi_E=\frac{Q_{enclosed}}{\epsilon_0}$) holds for any other closed surface. Where does the shape independence come from?

Gauß's Law in integral form follows from Gauß's Integral theorem,
$$\int_V \mathrm{d}^3 x \mathrm{div} \vec{E}(\vec{x})=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}$$
and one of Maxwell's equations (Gauß's Law in differential form),
$$\mathrm{div} \vec{E}=\frac{1}{\epsilon_0} \rho.$$
Now for the Coulomb field it's easy to show that for a spherical shell of radius $a$
$$\int_{\partial K_a} \mathrm{d}^2 \vec{f} \cdot \frac{Q \vec{r}}{4 \pi \epsilon_0 r^3}=\frac{Q}{\epsilon_0}.$$
This holds for any (sic!) radius $a>0$.

Now consider a volume $V$ containing the origin of the coordinate system in its interior (if the charge sits on the surface the issue is less trivial). Then you can always find a radius $a$ such that the sphere $K_a$ is completely in the interior of $V$. Then just use Gauß's integral theorem for $V \setminus K_a$, i.e., for the volume with the sphere taken out. Since at this volume $\rho(\vec{x})=0$, you get
$$\int_{\partial_V} \mathrm{d}^2 \vec{f} \cdot \vec{E}-\int_{\partial K_a} \mathrm{d}^2 \vec{f} \cdot \vec{E}=0.$$
The - sign comes from the fact that by definition in Gauß's theorem you have to direct the surface normal outside of the volume you integrate over. That means for the surface of the sphere for the here considered integral the surface normal has to point towards the origin, but in my notation the integral over the surface sphere is taken with the normal pointing out of the sphere, i.e., in opposite direction.

Finally you get
$$\int_{\partial_V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\int_{\partial K_a} \mathrm{d}^2 \vec{f} \cdot \vec{E} = \frac{Q}{\epsilon_0}.$$