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Flux integral over a closed surface

  1. Apr 17, 2012 #1
    So we recently began electrostatics and here you encounter Gauss' law saying that the flux integral of an electric field E over a closed surface is only dependent on the charge confined within the surface.

    Now for a sphere that's pretty obvious why. Because since the field gets weaker proportional to 1/r2 but the area gets bigger proportional to r2 evidently those two things should cancel.

    However! It is common knowledge that Gauss' law works for all kinds of surfaces, as long as they are closed. How can I realize that must be true? Because somehow it all hinges on the fact that the area gets bigger proportional to r2, and it is definately not intuitive for me, that that should be true.
     
  2. jcsd
  3. Apr 17, 2012 #2
    I don't think that you can make that kind of sense of gauss law, but you always choose the surface exhibiting the most symmetry in your problem i.e. you make up a surface sourrounding the charge and find the field. And not all fields goes like 1/r^2, how about a line charge exhibiting cylindrical symmetry?
     
  4. Apr 17, 2012 #3
    hmm.. I don't see what you mean. Everything hinges the symmetry of the 1/r2 and you can always break your field up into a superposition of fields that go like 1/r2.

    Indeed Gauss' law holds for all closed surfaces.
     
  5. Apr 17, 2012 #4
    Gauss's law is nothing but the divergence theorem which says the flux of a vector field over any closed surface is equal to the divergence of the vector field integrated over the volume. For electric field, the divergence is the charge density, which integrates to total charge enclosed. If you deform your surface from one to another, as long as you don't cross any charges, the total flux over the two surfaces are the same.
     
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