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

  1. Sep 12, 2014 #1
    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.
     
  2. jcsd
  3. Sep 12, 2014 #2
    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.
     
  4. Sep 12, 2014 #3

    jtbell

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    Staff: Mentor

    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".
     
  5. Sep 12, 2014 #4
    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.
     
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