Suppose we have a conducting plate and a positive charge adjacent to it (say to the left of it). From what I understand, we can introduce a 'virtual' negative 'mirror' charge on the right and work out the electric field.
I don't understand how this works. Why are we introducing this virtual charge? Surely my calculation now is the electric field due to both charges, which is not what I want. Why could I just not use the positive charge alone?
The Laplacian equations for electrostatics are such that if you have to find out the electric field (or potential) at any point in a volume, such that you are given the charge distribution in the volume and the boundary conditions on the surface of the volume (the potential at the surface), you are guaranteed to get a unique solution.
Thus, no matter what the situation is outside the volume if the boundary conditions are the same your answer for inside the volume would be same.
This is the principle due to which the method of images works:
For your example, the volume that we are interested in is the volume to that side of the plane sheet which contains the "real charge". You have been asked to calculate the potential at some point on this volume. Now, you know that the electric field at the boundary is perpendicular and the potential is zero. Can you come up with a different setup for the other side of the volume for which these boundary conditions remain unchanged?
Yes! Replace the conductor with a charge(of equal magnitude but opposite nature) placed just opposite the "real" charge. If you draw the electric field diagrams you will see that the field is perpendicular all over the plane where the conductor was lying. Also, the potential is zero there. So, the solution you get for this configuration is the same unique solution you would have got for the conductor situation.
Remember, you have to replace the conductor with the virtual charge not just add the virtual charge to the initial settings.
