Gauss' Law -- is there a general proof for all geometries?

[davidbenari]

I know how to derive Gauss's law considering only one point charge and a sphere.

I've seen other derivations for other geometrical shapes and I would say this is way too tedious as a method to prove that Gauss's law always holds true.

I was wondering if there is a general proof that says this has to be the case for all charge distributions and all geometrical shapes? Namely,

Θ=Q/εo holds true always.

Also, I'm not looking for proofs that refer to the fact that the "irregular shape is equivalent or reducible to the spherical case". I'm considering cylinders, cubes, and other polygons which as far as I know are not reducible to the spherical case.

Thanks.

 

[elegysix]

If you draw your gaussian surface big enough, not matter what object/shape is inside, it will be comparatively small enough to be assumed a point source.

I don't think there is proof that anything always holds true.

 

[jtbell]

http://farside.ph.utexas.edu/teaching/em/lectures/node30.html

This covers both the generalization from a spherical surface of integration to an arbitrary surface, and from point charges to arbitrary charge distributions. Warning: requires vector calculus, and you'll have to learn about the Dirac delta "function".

 

[davidbenari]

Elegysix:

I liked what you said however:

suppose you're calculating flux out of a charged long rod and create a gaussian surface (sphere in this case) so big such that the rod can be consider a point charge. This makes the calculations not simple at all because the electric field is not always parallel to the differential of area on your spherical surface. This is because the electric field on a charged long rod is mostly radial and outwards.

It's important to consider just the sphere in this case because this is would be the basis for a general proof because of its simplicity. So I wouldn't be convinced by "consider a gaussian cylinder then".

Thanks by the way, I didn't think of what you said.

 

[pmacias]

I can say that there is indeed a general proof for Gauss' Law that holds true for all charge distributions and geometries. This proof is known as the divergence theorem, which is a fundamental theorem in vector calculus. It states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of that field over the enclosed volume. In simpler terms, this means that the total amount of a vector field coming out of a closed surface is equal to the amount of that field flowing into the enclosed volume.

In the case of Gauss' Law, the vector field is the electric field, and the closed surface is a Gaussian surface that encloses the charge distribution. Using the divergence theorem, we can show that the flux of the electric field through the Gaussian surface is equal to the charge enclosed by that surface divided by the permittivity of free space (εo). This is the same equation as Θ=Q/εo, which is Gauss' Law.

Therefore, we can say with confidence that Gauss' Law holds true for all charge distributions and geometries, as it is a consequence of the more general divergence theorem. This eliminates the need for tedious derivations for different geometries, as the proof is applicable to all cases.

Furthermore, the divergence theorem does not rely on the shape of the Gaussian surface, so it is not necessary to reduce irregular shapes to the spherical case. This makes it a more robust and general proof for Gauss' Law.

In conclusion, there is indeed a general proof for Gauss' Law that holds true for all charge distributions and geometries, and it is based on the fundamental divergence theorem.