Hello everybody and merry Christmas! Here is the text of the problem which has been puzzling me for the last few days: there are two charges distant d from one another and at distance a from a conducting plane. Find the electrostatic energy of the system.(adsbygoogle = window.adsbygoogle || []).push({});

The first thing I find pretty strange is the fact that they don't tell me whether the plane is grounded or not. Actually up to now I haven't found any problem which involves an insulated plane when using image charges, so I tried to solve it supposing it's at V=0. Doing so I rewrite the charge distribution (two original charges plus the induced charge -2q on the plane) as the two charges plus the two image charges symmetrical with them and now to find out the energy I simply have to consider all the terms of mutual energy (6 of them):

U_{tot}=U_{1,2}+U_{1,1{img}}+U_{2,2{img}}+U_{2,1{img}}+U_{1,2{img}}+U_{1{img},2{img}}

Just to attempt an interpretation of the solution I think that the last term (energy interaction between the two image charges) tells us how much energy we need to induce another -q charge on the -q charged plane. Now, what I'd like to know from you is, first of all, if this solution is correct, second what would have happened if the plane would have been insulated instead of grounded (total charge on the plane =0). I'd say that we would have found a -2q charge on the side of the two charges and a symmetrical positive distribution on the other one, but what would have been the energy of the system in that case?

Thank you very much for your help!

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# Energy of two charges in front of a plane

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