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Charge density at pointed ends

  1. Nov 13, 2016 #1
    1. The problem statement, all variables and given/known data:
    IMG_0868.JPG

    2. Relevant equations: A conductor is an equipotential surface. The charge density near a conductor is proportional to the electric field. Electric field is the negative gradient of potential and thus electric field is in a direction normal to the surface.


    3. The attempt at a solution: Since electric field is the gradient of potential, if the surface of a conductor has sharp turns, the gradient will be undefined (cannot draw a tangent plane at such points) and thus the charge density is undefined.
    So I think it is option (a). Am I right?

    Thanks.
     
  2. jcsd
  3. Nov 13, 2016 #2

    kuruman

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    If the electrostatic potential is defined everywhere in space, so is the charge density. Why don't you sketch a couple of equipotentials very near the surface of the conductor? The charge density on the surface is higher where the electric field is stronger and the electric field is stronger where the equipotentials come closer together.
     
  4. Nov 13, 2016 #3

    mfb

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    I agree with your analysis of point A, but I think for point B the charge density should have a well-defined value.
     
  5. Nov 13, 2016 #4
    @kuruman : Thanks for the help. Maybe I should not be too rigorous about the pointed cones.
    If I round off the cones a little, then A will be a concave surface and B will be a convex surface. It looks like the charges would be crowded at B compared to A. Therefore B has a lesser charge density than A.

    How do I draw an equipotential for this surface that wouldn't look symmetric at both A and B?

    @mfb But there is a sharp edge at B as well. Why doesn't the reasoning at A work at B as well?
     
  6. Nov 13, 2016 #5

    mfb

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    I would expect the field to vanish around B because it is an inner corner. Zero field gives zero charge density and zero divergence problems.

    If we round off the corners a bit, then both A and B have defined charge densities and B has a smaller one than A, sure.
     
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