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This kind of a configuration struck me when I read this question.

The original question asked how the force on the two charges would change if a metal plate was inserted between them. This question addresses what would happen if it was an infinite plane (those are so much more convenient).

I gave it some thought and the result is rather interesting.

enter image description here EDIT: This question is very different from the original in that:

  • The infinite plane is grounded, a plate with definite dimensions does not have to be
  • One can't use image charges with the plate
  • If the plate is uncharged, positive charges will accumulate on the edges and create their own field, in case of an infinite plane these positive charges are at infinity and do not affect the configuration.
  • A plate does not provide electrostatic shielding or a simple boundary region to apply the uniqueness theorem in.
  • The charge distribution on a finite plate is much harder to model.
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marked as duplicate by sammy gerbil, Yashas, ZeroTheHero, Jon Custer, John Rennie Apr 4 '17 at 7:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What is the difference between a metal plate and a metal plane? You appear to be duplicating the question you cite. If you wish to provide an answer to that question you should post it in response to that question, not open a new question. $\endgroup$ – sammy gerbil Apr 3 '17 at 15:26
  • $\begingroup$ @sammygerbil@yashas samaga I've added edits to explain why this question is unique. I did not want to post it as an answer to the other question because I thought I'd be using too many assumptions. $\endgroup$ – GeeJay Apr 3 '17 at 16:28
  • $\begingroup$ I don't see the difference. Your "plane" must have 2 parallel surfaces and finite thickness, otherwise it could not lie "between" two points. There was no suggestion in the original question that this plate had finite dimensions (other than thickness), hence the issue of whether it is grounded is also irrelevant. The method of images can be used for charged, ungrounded and finite conductors (eg spherical shell). Your other objections are dealt with by not assuming the plate is finite (which was not assumed in the other question). $\endgroup$ – sammy gerbil Apr 3 '17 at 17:05
  • $\begingroup$ @sammy gerbil The other question does not explicitly state that the plate is infinite. Maybe I misinterpreted it but when I read "plate" unaccompanied by "infinite" I usually imagine it as finite. The difference is that a finite plate does not provide complete electrostatic shielding. The field from one positive charge on the other is not wholly cancelled by the plate so this result does not hold good for a finite plate. And also, importantly, the other question asks how the force on one charge due to the other changes. This one asks how the net force on either charge changes. $\endgroup$ – GeeJay Apr 4 '17 at 3:39
  • $\begingroup$ "not assuming the plate is finite" is the same as assuming it's infinite "(which was not assumed in the other question)" $\endgroup$ – GeeJay Apr 4 '17 at 3:40
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Let's say the conducting plane is the x-y plane. One charge is in the $+z$ half and the other in the $-z$ half.

Let's focus on a configuration that consists of just the charge in the top half and the conducting plane. For a point charge $q$ above a conducting plane, the charge distribution on the plane can be found from the image charge method (this is very well illustrated in chapter 3 of Purcell's Electricity and Magnetism).

The sum of the fields due to the plate and $+q$ is zero for $z<0$. For $z>0$, the electric field is like that of a dipole.

The image charge method will also tell you that the force on the charge $+q$ is that of an image charge $-q$ at the same distance from the xy plane but in the bottom half.

enter image description here

enter image description here

enter image description here

Now let us consider just the charge in the bottom half and the conducting plane. The field is going to be just the opposite of what was described above. Zero for $z>0$ and that of a dipole for $z<0$. Now using the superposition theorem, we can put these two configurations together.

What is most fascinating here is that the field due to one charge does not penetrate into the region with the other charge. This is electrostatic shielding. So the force on each positive charge is as if there was a negative charge of the same magnitude where the other charge is sitting.

When you sum this up it's a truly astounding result: if there are two positive charges and you insert an infinite conducting plane between them, the force on either charge changes sign!

enter image description here

*All images from Purcell's Electricity and Magnetism

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  • $\begingroup$ As drawn the conducting plate has a negative charge on it. Where has that charge come from? $\endgroup$ – Farcher Apr 3 '17 at 10:51
  • $\begingroup$ There's an explanation on p 138-139 of Purcell. That charge was always there. There is also a positive charge of the same magnitude on the plane, so the net charge is zero. $\endgroup$ – GeeJay Apr 3 '17 at 10:59
  • $\begingroup$ Where is the equivalent positive charge to balance the negative charges on the surface situated? $\endgroup$ – Lelouch Apr 3 '17 at 16:02
  • $\begingroup$ "imagine that the conducting plane is actually a metal disk, not infinite but finite with a radius R>>h. If a charge +Q were to be spread uniformly over this disk, on both sides (so Q/2 is on each side), the resulting surface density on each side would be Q/2πR^2, which would cause an electric field of strength $Q/2π\epsilon_0R^2$ normal to the plane of the disk. Since our disk is a conductor, on which charge can move, the charge density and the resulting field strength will be even less than $Q/2π\epsilon_0R^2$ near the center of the disk because of the tendency of the charge to $\endgroup$ – GeeJay Apr 3 '17 at 16:37
  • $\begingroup$ spread out toward the rim. In any case the field of this distribution smaller in order of magnitude by a factor h^2/R^2 than the field described because the latter field behaves like 1/h^2 in the vicinity of r = 0. As long as R >>h we were justified in ignoring the former field, and of course it vanishes completely for an unbounded conducting plane with R = ∞." $\endgroup$ – GeeJay Apr 3 '17 at 16:37

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