# Taking into consideration the non-0 value of an equipotential in a method of images problem

If the infinite conducting plane in the diagram above is grounded, then $$V=0$$ on the plane and the image problem is easy to solve - just put the dotted $$-q$$ charge there and the $$V=0$$ equipotential between the two charges satisfies the grounded plane boundary condition.

However, what if the infinite conducting plane is not grounded and so $$V$$ is not necessarily $$0$$?

Duffin, in Electricity and Magnetism 4th Edition, simply sets the conducting plane (which, as a conductor, must be an equipotential) to be the point of 0 potential. He then points out how the equipotential between the $$+q$$ charge and the image $$-q$$ charge is valued at $$V=0$$, so the boundary condition is satisfied. However, he set the plane as the potential reference point first when considering the initial configuration of plane and $$+q$$ charge and then used infinity as the reference point when he added the mirror $$-q$$ charge, so I don't see how his argument holds up.

How can we argue that the above image set up, with the dotted $$-q$$ charge, is valid even if the the potential on the conducting surface is not $$0$$?

• Never mind - I think it just clicked. We need the boundary to be an equipotential for the method of images to work and so it doesn't necessarily need to be 0. Any constant offset will be lost in Poisson's equation and also if the boundary is an equipotential it can be set to 0 but if it is not then it cannot be (because it varies across the boundary). If anyone else is struggling with the same thing and reading this question later I suggest you draw out a example in the diagram but with different charges replacing the $-q$ and this will make you realise the importance of the equipotential. – Pancake_Senpai Jun 1 at 12:12