Griffiths, when he's talking about the method of images, shows how to calculate the potential distribution when given a point charge that is outside of a spherical conductor which is held at a fixed potential.

If you are given a similar scenario, where a point is outside a spherical conductor but the conductor is not held at a fixed potential, how would you calculate the potential of the conductor?

It's basically solving laplace's equation when the boundary conditions aren't known. But intuitively, it seems like a solution would exist, so I'm guessing the boundary condition can be extracted somehow.

  • $\begingroup$ What is known about the potential of the conductor? For example, did it start out at a known potential before the point charge was moved into place? $\endgroup$
    – David Z
    Feb 5, 2012 at 6:25
  • $\begingroup$ This is my own problem, add whatever constraints you like. :) $\endgroup$ Feb 5, 2012 at 15:18

1 Answer 1


If the potential of the conductor is not fixed, the total charge must be fixed or there will be no unique solution. Let's assume that the total charge is fixed as zero.

First suppose the conductor is grounded and has thus zero potential, then you can calculate the electric field outside the sphere with method of images. With Gauss' law you can easily prove that the charge the sphere carries is exactly the same as that of the imaginary image. Denote that as Q.

Now remove the grounding line and add -Q to the sphere to satisfy our boundary condition. Since the point charge and the induced charge result in a net zero potential for the sphere, the only contributor will be the -Q charge we introduce. Thus it is clear the potential of the conductor sphere is $$\varphi=-\frac{Q}{4\pi\varepsilon_0 R}$$

  • $\begingroup$ Ok, I see what you're saying, but we are justified in using the method of images because of the uniqueness of solutions to laplace's equation when boundary values are specified. In your post we are deriving the boundary condition from the method of images, which I'm not sure is legitimate. $\endgroup$ Feb 5, 2012 at 15:16
  • 1
    $\begingroup$ @user19192: The boundary conditions are not derived, but specified. They are that the total charge of the sphere is fixed as zero, that there is a point charge at a specific location, and that infinity has a zero potential. $\endgroup$
    – Siyuan Ren
    Feb 6, 2012 at 3:34

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