Energy of two charges in front of a plane

[mrteo]

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.

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!

 

[vkash]

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.

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!

Both charges on the opposite of the plane. so they are electrostaticaly shielded.
As you know the potential of a system is negative of work done by internal forces in setting up that system.Here first of all consider there is a +vs charged particle standalone, there was nothing in surrounding. so U=0. Now take a large plate towards it. Due to induction some -ve charge will induce on closer area are +ve in farther area. After all i can't calculate this work since it is too complicated for me. Let me say it is W(assume). Now second turn, take the second +ve charge from infinity to it(on opposite side) The effect of first charge on this will zero(electrostatic shielding ). So all the things will due to plate charge. Here seen is just opposite. In first due due to induction plate might have been attracted towards the charge but here it will repel. so work done in this case should -W. Total work done=w-w=0 so change in potential energy is zero.

Hope i am correct. And you understand my answer.

 

[mrteo]

Seems to me that you haven't exactly gotten what the situation is: the charges are not on different sides of the plane. They're both on the same side and on the other side I only put the image charges. Here is a drawing to explain what I mean.