A common textbook problem in electromagnetism is to place a point charge $q$ outside a dielectric material with permittivity $\epsilon$ .

In the case of a semi-infinite dielectric material, the induced image charge $q'$ is $$q' = -q \frac{\epsilon-1}{\epsilon+1}$$

My question is, what is the right way to deal with the image charge when the permittivity $\epsilon$ is complex (i.e. $\epsilon = \epsilon_1 + i \epsilon_2$)?

Is the following expression the right solution? If so, what would be the interpretation of the imaginary part?

$$q' = -q \,\,\mathrm{Re}[\frac{\epsilon-1}{\epsilon+1}]$$

By taking the permittivity to be complex, I am presumably making the implicit assumption that the external charge is "oscillating" at a fixed frequency $\omega$ such that $\epsilon(\omega) = \epsilon_1 + i \epsilon_2$. At some point the magnetic field caused by the oscillating charge may be important, but I am interested mainly in the regime where the magnetic field can be neglected.

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1 Answer 1


As you say, with a complex permittivity, you are assuming that the charge is oscillating. This, of course, means that if you start with a charge $q$, it will eventually change to a charge $-q$. That is, $-2q$ charge had to flow in from somewhere. So an image charge is not well defined for this case. Instead, the problem that is usually studied is image charges for an oscillating dipole either electric or magnetic. You can then put these together to get any current conserving charge density.

There are no simple solutions. Generally, even in the quasistatic case, the space has to be extended to complex values and the image sources are placed in this extended space. An old example with references to previous work is given in I.V. Lindell and E. Alanen, "Exact Image Theory for the Sommerfeld Half-Space Problem, Part III: General Formulation", IEEE Transactions on Antennas and Propagation, Volume AP-32, (1984), DOI:10.1109/TAP.1984.1143204.

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    $\begingroup$ @KF Gauss I've deleted the answer I posted earlier. It was incorrect and misleading. The complex permittivity results in a finite skin depth which you can think of as a distributed mirror surface. So there is no sharp image. $\endgroup$
    – Roger Wood
    Jan 1, 2022 at 8:51
  • $\begingroup$ @Roger Wood, I am not sure the skin depth is the right explanation either, since even in the idealized static case of a metal, the induced charge is all localized at the surface as a Delta function. $\endgroup$
    – KF Gauss
    Jan 1, 2022 at 17:07
  • $\begingroup$ @user200143 thanks for your answer! It seems the paper you linked is solving a more difficult problem. Is there no limiting case (simple semi-infinite system and zero B field) where the oscillating dipole gives an analytic expression? $\endgroup$
    – KF Gauss
    Jan 1, 2022 at 17:09

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