A charge $Q$ is located at a distance $r>R$ from the center of a non-grounded conducting spherical shell with radius $R$ and total charge $q_s$. The field external to the shell can be mimicked by the combination of the image charge plus a second image charge. What is this second charge, and where is it located?

It's problem $3.16$ in Electricity and Magnetism by Purcell.

Purcell argued Since the image charges produce the same external field as the shell, Gauss's law implies that the charge on the actual shell is $-QR/r$, whereas we are told that the charge is $q_s$. We can remedy this by placing another image charge of $q_s+QR/r$ at the center. The total charge on the actual shell is now $q_s$, as desired. Furthermore, the boundary condition of constant potential on the shell is still satisfied, by symmetry, because the second image charge is located at the center. So by the uniqueness theorem, the field from our two image charges mimic the (external) field from the shell.

The external field is the same as produced by the charge $Q$ and $q=-QR/r$ for the grounded shell. I don't understand How gauss law implies the charge on the shell to be $-QR/r$? There wasn't any net charge on the shell initially. I also don't get how the boundary conditions are satisfied in the two problem?


1 Answer 1


This configuration is just the superposition of two known configurations:

  1. The grounded shell and the real charge $Q$ at $r$

  2. A second image charge

By the method of images, the first one can be changed into a real charge $Q$ at $r$ and image charge $-QR/r$ inside the shell. Remember that a conductor is an equipotential surface. The first configuration (the grounded case) already produces a uniform potential of zero at all points on the shell. Therefore, the only point the second image charge can be placed without making the shell potential non-uniform is at the center of the shell.

The next question is to find the magnitude of this second image charge. Consider a spherical Gaussian surface enclosing the shell but not the real charge $Q$. This surface encloses a total real charge of $q_s$. Since the two image charges are enclosed inside, their sum must equal this real charge. As such, the second image charge must be the difference between $q_s$ and the first image charge $-QR/r$, so we get that it is $$q_s - (-\frac{QR}{r}) = q_s + \frac{QR}{r}$$

There wasn't any net charge on the shell initially.

The net charge on the shell is defined to be $q_s$. No matter how it is distributed across the shell, it cannot change as the shell is not connected to anything else (isolated). Compare this to the grounded case, where the amount of charge that flows between the ground and shell is exactly correct such that it makes the potential on the shell zero.

  • $\begingroup$ What's the potential on the shell in this case? $\endgroup$ Jun 1, 2021 at 17:18
  • $\begingroup$ @YoungKindaichi Potential obeys superposition. The real charge $Q$ and the first image charge together produce zero potential. The second image charge is simply at the center of the shell. Therefore the potential on the shell is just the potential due to the second image charge. $\endgroup$ Jun 2, 2021 at 2:59
  • $\begingroup$ I mean what's the potential on a shell due to if there is $q_f$ charge on it and there is a real charge $Q$. $\endgroup$ Jun 2, 2021 at 3:22
  • $\begingroup$ @YoungKindaichi I already addressed this in my previous comment. The potential on the shell is just the potential due to the second image charge. So imagine a point charge $q_s + QR/r$. What would be the potential at a distance $R$? $\endgroup$ Jun 2, 2021 at 3:37
  • $\begingroup$ Let's not first get into image charges, So first I need to see the boundary conditions. So I need to find out what's the potential of the shell is. Like in the grounding shell problem, I look for boundary conditions then find out the problem that fits these boundary conditions that's what I'm asking. How do you know the potential of the shell? $\endgroup$ Jun 2, 2021 at 3:45

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