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2D Gauss' Law applied to an infinite cylinder

  1. Oct 30, 2011 #1

    Geofleur

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    Hello,

    I've been working my way through Mathematics for Physicists by Dennery and Krzywicki and, on page 65, they assert that Gauss' law applied to a 2D cross-section along an infinite charged cylinder is:

    ∫E.n dl = 4πσ

    where E is the electric field on the Gauss surface (a circle around the cylinder), n is the unit normal to this surface, dl is an element of length along the circumference of the Gauss surface, and σ is the charge per unit length along the cylinder.

    The right side of the 2D Gauss' law should be the charge enclosed times some constant, and the charge enclosed is the circumference of the cylinder times the charge density, no? But then I get

    ∫E.n dl = 2πrσ (times some constant)

    where r is the radius of the cylinder enclosed. Why is there no radius factor on the right hand side of the equation given in the text? Perhaps I'm misinterpreting the problem setup?
     
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  3. Oct 30, 2011 #2

    Doc Al

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    Note that σ is the charge per unit length along the axis of the cylinder, so the total charge is σ*L. (But they divide out the L on both sides.)
     
  4. Oct 30, 2011 #3

    Geofleur

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    Yes, I see it now. Thanks!

    I did fiddle with the idea that σ was along axis but, because the cylinder was said to be infinite I thought that the charge contained in a 2D slice would have to be zero. It seemed to me like calculating the probability that a particle would be at an exact point when it's position is governed by a (non-delta function) probability density - you'd get zero, right? I'll have to think about why the analogy doesn't carry over...
     
  5. Oct 30, 2011 #4
    Even though the answer is numerically right, there's a severe flaw. Gauss law is about the flux of the electric field through a closed surface. It stems from the divergence theorem. In a very long charged cylinder the field lines point radially and we choose, as a gaussian closed surface another cylinder, coaxial with the first one. The cylinder's ends contribute with zero to the flux because the normal vector is perpendicular to the field lines and so you get the result posted in the OP. To the best of my knowledge there's no 2D Gauss law; at best it would only be a simplified version of the right one
     
  6. Oct 31, 2011 #5

    Doc Al

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    That's correct. I have no idea why the problem statement says "2D Gauss' Law". It's the same old 3D Gauss' law applied to an infinite cylinder of charge.
     
  7. Oct 31, 2011 #6

    Doc Al

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    Of course, a real 2D slice would have zero charge, but that's not what's going on. They are taking just taking a Gaussian surface--a cylinder of some unspecified length--and applying the usual 3D Gauss' law. Nothing really 2D about it.
     
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