Solving Laplace's Eqn: Setting Zero of Potential

  1. Nov 20, 2010 #1
    1. The problem statement, all variables and given/known data
    I have two quick related questions that I think will help clear something up for me.

    (1) An uncharged metal sphere of radius R is placed in an otherwise uniform
    electric field E = E0 zhat. The field induces charge. Find the potential in the region outside the sphere.

    (2) Find the potential outside an infitely long metal pipe of radius R placed at right angle to an otherwise uniform electric field E0.


    2. Relevant equations



    3. The attempt at a solution

    Okay, so I am mostly comfortable with the solutions to these problems, with the exception of one key concept.

    Problem 1: If E=E0 zhat (for r>>R, then V for r>>R is -E0z + C)
    The one hang-up I have is that for problem (1), you set V=0 on the xy-plane "by symmetry." Thus we can say C=0. I'm not sure I understand the reasoning behind setting V=0 for all z=0.


    Similarly, from problem (2), if we suppose that the electric field is pointing along z direction, so again, the form of the potential is -E0z + C. Again, one should set V=0 on the xy-plane, so C=0. I don't have a clear picture in my head of why you set V=0 on the xy-plane.


    Any explanations of this concept would be greatly appreciated
     
  2. jcsd
  3. Nov 20, 2010 #2

    fzero

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    Since the electric field is the gradient of the potential, [tex]\vec{E} = - \nabla V[/tex] (for electrostatics), a constant shift in the potential [tex]V' = V + c[/tex] doesn't change the electric field. Therefore we are usually free to define the location of [tex]V=0[/tex] by adding a constant term to the potential. In your examples, choosing [tex]V(z=0) = 0[/tex] is just the most convenient choice, but it was not the only one.
     
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