Let be the potential. Now, is invariant under addition of constants, so WLOG, we can assume that (the potential of a metal plate is constant, and we are setting it to be 0). Now, we are solving the PDE
in the region with boundary condition .

Now, consider the entirely different problem where we have a charge at a height , and a charge (this is the infamous "image charge") below it at . One can check that in the region , we are solving the same PDE as the above with the same boundary conditions. Thus, in that region must be the same as in this new problem.

Thus, the field is the same, so it is the field of the situation descried in the new problem. To find the force on the charge, we have to subtract the field of that charge alone, multiplied by . This is just the force between and the imaginary image charge, which is .

Note that the energy of this setup is
This is half of what one might naively guess, which is just the energy formula for the image charge and the charge, which is just
The reason this is wrong is because this counts the work it takes to move the image charge to , but by simply moving the regular charge, we are automatically moving the image charge, so there is no need to account extra work for it.

Another reason why the answer is wrong is by considering the energy in terms of the field, i.e.
where is the infinitesimal volume element. For the case of the actual setup, we are only integrating over the region , since the field is for , so we get only half the answer.

of Apollonius circles