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The distribution of charges in a conductor

  1. Oct 5, 2009 #1
    I was taught in my Eletricmagnetics Lessons that the density of electric charge is higher near the surface with big curvature and lower near the flat surface.
    Why does the density of electric charge has something to do with the surface curvature?
    Is this because such a system has the minimal energy?

  2. jcsd
  3. Oct 7, 2009 #2


    Staff: Mentor

    Consider that the charges are constrained to move on the surface. Then you can see that their motion is determined by the tangential component of the Coulomb force. In a region with high curvature the tangential component is less so it requires more charge to make the same force.

    That is a rather "hand-waving" argument, but I find it rather intuitive.
  4. Oct 7, 2009 #3
    The charges in a conductor repell and go apart as far as possible. If a ball has a niddle on its surface, some charges will go to the niddle. But if a ball has a niddle hole inwards, they will not accumulate there.
  5. Oct 7, 2009 #4


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    Science Advisor
    Gold Member

    I think the question has been pretty well answered but since I love to hear myself talk I'm going to add a bit.

    Since charges can move in a conductor any electric field inside the conductor will not last as it means a force on the charges which then move in reaction to the force. Eventually the charges must move until they cancel out any internal E-fields.

    Now we come to the surface where the charges cannot move outward. There thus can be an E-field at the surface but it must be normal to this surface since any lateral component again means a component of force on charges which can move in this lateral direction. So picture a blob shaped conductor as having 0 E field inside and at the surface the E-field points straight out.

    Finally recall that the divergence of the E field (the degree to which it radiates outward rather than lining up parallel) is proportional to the charge density. Given the E field is normal to the surface the more curved the surface the higher the divergence of the E field and thus the more concentrated the charge density. This isn't so much a causal explanation but a way to see that this is how it must be.

    By the same equation we see that since the divergence of the zero E field inside the conductor is also zero we know that none of the charge is distributed through the interior of the conductor and all must reside at the surface.
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