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Deducing potential from symmetry (from Griffiths EM)

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data

    It's example 3.8 in the Griffiths book in case someone has it. Basically the problem involves a uncharged metal sphere in a uniform field in the +z direction. Naturally, there will be induced positive charges on top and negative charges at the bottom. The question asks for potential outside the sphere.


    2. Relevant equations


    3. The attempt at a solution

    What I don't get is how he deduces that the xy plane is at 0 potential using symmetry, after setting the metal sphere to be 0 potential.

    What I'm thinking is maybe it's because the E field (even with distortion from induced charges) is perpendicular to the xy plane at all points so moving out in xy-plane causes no change in potential. This satisfy the symmetry. But there could also be an E field with a component pointing inward towards the z-axis. Or one with a component outward from the z-axis. And these latter possibilities would also work but would not give 0 potential from the xy-plane. I need help finding a convincing argument to eliminate these two options. Intuitively, I can see that they violates symmetry of the Coulomb force between positive and negative charges (ie E field acting on a positive charge is same as for negative charge but only in opposite direction) but they are symmetric geometrically.
     
  2. jcsd
  3. Apr 3, 2012 #2

    tiny-tim

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    welcome to pf!

    hi louielouie welcome to pf! :smile:
    no

    try drawing the E field lines …

    they must cross z = 0 at 90° …

    if they cross at some other angle, why should it be inward rather than outward? :wink:
     
  4. Apr 3, 2012 #3
    Thank you, tiny-tim.

    I think it could be inward because the situation is not top-down symmetric. Since we have a positive charge on top the field tend to diverge outward in the +z direction. But the by "the symmetry of the Coulomb force law" (I dont know how to say it better), the field will have a converging component inward (as much the top has outward diverging component) because of the same amount of charge distributed at the bottom. Is this what you mean by saying why it should prefer the inward direction to the outward direction? So I can equally argue both ways. The only independent way is to have it be perpendicular.


    Or I think I found a better explanation: It's clear that z is the axis of symmetry. So if there are any components directed out or in, I can make a loop in the xy plane center around the sphere and get work out of the field. Therefore the electric field is not conservative. But the net electric field comes from the electrostatic field of the induced charges and the externally applied electric field by superposition. {Since the external field is conservative, the electrostatic field (from induced charges must be nonconservative)}. Contradiction.

    I have doubts about the claim in { }.
     
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