Suppose you have the classic method of images problem. You have a point charge $q$ sitting a distance $d$ above grounded metal plate. You can then find the field as though there were 2 point charges with + and - q, respectively, at a distance $2d$ apart. Great.

My question is, how does the energy of these two systems compare?

In principle, if there is no dielectric, the energy should be something like the integral of $E^2$ over all space. I have seen it stated that the energy in the metal plate case is half that of the energy in the 2 point charge case, but it isn't at all clear to me why. Secondary question, does any of this change if the plate isn't grounded?

  • $\begingroup$ Please check your space key, probably it is glued into your keyboard. :-) $\endgroup$ – peterh - Reinstate Monica Aug 13 '16 at 15:10

Yes, the equation is indeed "something like the integral of E2 "

To be precise, it is

W=(ε0/2) ∫all space E2 dV

The electric field described by the images only holds wherever the boundary conditions are satisfied by both the original configuration AND your simpler image charges scenario. In the case of a grounded conductor that region is above the plane, on the side of the point charge q. Assuming the the plane is at z=0, the field at z<0 is:

  • Zero in the case of the grounded plane and charge
  • Perfectly symmetrical (dipole field) in the case of the image charges The z<0 region falls outside the boundary condition-satisfying region. The field E is not the same in both cases here

So, just intuitively, without solving the integral, since the E-field exists only in half of all space with the conducting plane, and in all space with the image charge, it should be fairly obvious to you why the energy "in the metal plate case is half that of the energy in the 2 point charge case".

In answer to your secondary question, this kind of problem usually involves an infinite plate. If it isn't infinite, then you have edge conditions and all the standard problems that realistic geometries bring with them to deal with. So we assume it is infinite. That implies that it is grounded, even if not specifically mentioned. The plate goes to infinity=> potential=0 at infinity=> plate is an equipotential surface=> plate is grounded

Hope that helped!

  • $\begingroup$ So if the plane was not infinite the field on the z<0 would not be zero, but also not the dipole case but something more complicated. I hadn't fully considered whether or not the plane was infinite. I felt like there had to be a connection between the case of the plane and the case of a charge in the cavity of a metal conductor where the charge is "visible" on the surface because you could conceivably wrap a planar conductor around a charge. $\endgroup$ – Cogitator Aug 13 '16 at 7:40
  • $\begingroup$ In the intermediary case of "a finite bit of metal in the neighborhood of a charge" we can say that the free charge will induce some charge on the metal and the field all over will be complicated. Would you agree? Does this change if we ground the finite bit of metal ? $\endgroup$ – Cogitator Aug 13 '16 at 7:47
  • 2
    $\begingroup$ Yes, I would agree. If you ground it, that will change the charge distribution on the metal. $\endgroup$ – GeeJay Aug 14 '16 at 5:39

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