Energy of two charges in front of a plane

In summary, the electrostatic energy of the system is zero because the system is shielded by the charges on the opposite side of the plane.
  • #1
mrteo
2
0
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):

Utot=U1,2+U1,1{img}+U2,2{img}+U2,1{img}+U1,2{img}+U1{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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  • #2
mrteo said:
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):

Utot=U1,2+U1,1{img}+U2,2{img}+U2,1{img}+U1,2{img}+U1{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.
 
  • #3
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.
 

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1. What is the energy of two charges in front of a plane?

The energy of two charges in front of a plane refers to the electrostatic potential energy that exists between two point charges placed in front of an infinite conducting plane.

2. How is the energy of two charges in front of a plane calculated?

The energy of two charges in front of a plane is calculated using the formula: E = q1q2/(4πε0d), where q1 and q2 are the magnitudes of the two charges, ε0 is the permittivity of free space, and d is the distance between the charges.

3. What is the significance of a conducting plane in the energy calculation?

A conducting plane acts as a boundary condition for the electric field created by the two charges. It ensures that the electric field lines terminate at the plane, resulting in a finite energy calculation.

4. How does the distance between the charges affect the energy?

The energy between two charges in front of a plane is inversely proportional to the distance between the charges. As the distance increases, the energy decreases, and vice versa.

5. Can the energy of two charges in front of a plane be negative?

Yes, the energy of two charges in front of a plane can be negative if the charges have opposite signs. This indicates that the two charges are attracted to each other and the system has a lower potential energy compared to when the charges are infinitely far apart.

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