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Laplace Equation with Charged Conductor

  1. Feb 27, 2015 #1
    I know the following works, but I cannot explain why.

    Say you have an infinite conducting cylinder (radius r=a) with charge Q/(unit length) in an electric field E.

    To find the potential, you just match the boundary conditions V=0 at a, and V→ -Er cosφ far away. Then you add in the potential of a *uniformly charged cylinder*.

    The same goes for a charged anything. You just add in the potential from the *uniformly charged version* of the shape, after.

    Why does this work? I am imagining that it is something like: it works because the uncharged solution works and gets the induced charge right to cancel the field, and that cancels the field inside and on the surface of the conductor, so there is no reason for the charges Q to move, but I wish I could better formulate/explain/understand this.
  2. jcsd
  3. Mar 4, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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