bjnartowt
Aug26-10, 12:31 PM
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:
\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}}
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:
\Delta W = \frac{{ - {Q^2}}}{{8\pi {\varepsilon _0}d}}
...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.
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:
\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}}
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:
\Delta W = \frac{{ - {Q^2}}}{{8\pi {\varepsilon _0}d}}
...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.