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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 $0$?


Remember that absolute values of the electrical potential have no physical significance. Only potential differences have a physical meaning. Mathematically, you should note that this method is derived from Poisson's equation, and adding any constant value to the potential (i.e., shifting its reference value) will not change the equation, because the derivative will cancel it.

  • $\begingroup$ Yes, but the potentials from the two charges will have to be calculated with respect to the same reference point as the potential on the plane. So if we set the plane's potential to be 0 then we've changed the reference point and so the potentials from the two charges will no longer add to give 0 as they did before. I don't see how the method of images configuration can work unless the potential at the boundary is 0 with respect to a reference point at infinity. $\endgroup$ – Pancake_Senpai May 31 '19 at 3:57
  • $\begingroup$ I am sorry, but I still can't see the problem. The conductor plane must only have a finite constant potential (i.e., be an equipotential plane region). Its value does not need to be necessarily zero. $\endgroup$ – Woe Jun 1 '19 at 0:20
  • $\begingroup$ For the method of images to work the boundary condition must be met with our image system for it to be equivalent in the region above the plane. If the potential across the conductor is 0 then the two image charges add to give a 0 potential at the boundary and so everything is fine. If the potential across the conductor is not 0 then the method of images should not work because the potential sum from both charges is still 0 at the boundary, so the boundary condition is not met. $\endgroup$ – Pancake_Senpai Jun 1 '19 at 12:05
  • $\begingroup$ 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. $\endgroup$ – Pancake_Senpai Jun 1 '19 at 12:12

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