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Infinite Plane with Point Charge Above - Method of Images - Uniqueness Dogma

  1. Aug 26, 2010 #1
    1. The problem statement, all variables and given/known data

    You have point charge a distance "d" above infinite conducting plane held at V = 0. What is the potential when you remove charge to infinity?

    2. Relevant equations

    3. The attempt at a solution

    I think I incorrectly used Coulomb's law between the charge (+q, distance "+d" away from infinite-plane) and image charge (-q, distance "-d" away from infinite plane), and got:

    [tex]\Delta W = - \frac{{{Q^2}}}{{16\pi {\varepsilon _0}}}\int_d^\infty {\frac{1}{{{z^2}}}dz} = - \frac{{{Q^2}}}{{16\pi {\varepsilon _0}}}\left( {\frac{{ - 1}}{\infty } - \frac{{ - 1}}{d}} \right) = \frac{{ - {Q^2}}}{{16\pi {\varepsilon _0}d}}[/tex]

    But: I am told that this is wrong by the discussion on p. 124 of Griffiths: with two point charges and no conductor, I am told:

    [tex]\Delta W = \frac{{ - {Q^2}}}{{8\pi {\varepsilon _0}d}}[/tex]

    ...while with single charge and conducting plane, the energy is half of this: which is what I calculated above.

    I read Griffiths discussion about physical justification for factor of (1/2). But how does that not thwart the First and Second uniqueness theorems that we love so much? They guarantee the uniqueness of the field, so why do I need to consult another field to get the work to remove charge to infinity from "d"?

    I mean, I know I got "the right answer" by looking at Griffiths, but the availability of the other "wrong answer" that looks like the "right answer" if we were to naively apply it without knowing about the uniqueness theorem worries me. Can you see the disagreeing vertices of the triangle of things I'm thinking about? Can you help me resolve this?

    This may be an ill-posed question, but offer what thoughts and critiques may come to mind anyway in spite of the lack of specificness of my question.
  2. jcsd
  3. Aug 26, 2010 #2
    Oh, wait...I think I know why. Is it because the electric field inside a conductor (where the image-charge "is") is zero, thus preserving the uniqueness of the and potential on the real-charge side of the plane? Hence, the factor of 1/2, which does not thwart uniqueness?
  4. Aug 26, 2010 #3


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    Let me see if I get this correctly. The final picture has a grounded plane and no charges anywhere near it. Is that correct? If so, how can the potential be anything other than zero?
  5. Aug 26, 2010 #4

    Hi again, Kuruman! Anyway, I have metal/conducting plate held at V = 0, so strictly equipotential. I have charge nearby: namely, a distance "d" above the plane. There's a unique field, E, and unique potential, V, in the space above this plane where the charge is, which I calculated and it matches Griffiths results.
  6. Aug 26, 2010 #5


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    Fine. However the question you posted is
    The charge is now longer at distance "d" above the plane, it is at infinity. There are no charges anywhere in finite space and the plane is grounded, so what should the potential be in this case?
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