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Is there a finite object whose charge distribution is a perfect dipole (only has a dipole moment) yet doesn't involve any infinities or discontinuities in the fields? And if there happens to be a fairly "standard" object in textbooks or literature, that would be even better.

One example of a perfect dipole is a sphere of material with a uniform polarization. Outside of the sphere, in a multipole expansion the potential only has a dipole moment, so this configuration looks like a perfect dipole externally.

Another example of a perfect dipole is a pair of opposite charges ($\pm q$) in the limit that the distance between them ($d$) goes to zero but their charge increases such that $p = qd$ remains constant.

Neither of these objects is easy to analyze rigorously. The uniformly polarized sphere at least doesn't involve a limit procedure with charges going to infinity, but it does have discontinuities in the electric field at the surface (although the potential is still continuous; I was initially incorrect on that).


In case any such object is very complicated, and so the motivation behind the question would help frame an answer, I was thinking about this after reading Electromagnetic field energy "paradox" and wondering if the answer is hidden in the discontinuities of the field.

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  • $\begingroup$ Related: Is there a finite-sized charge distribution whose electric field is exactly that of a point dipole?. Note that OP's request to keep the electric field continuous makes them different enough to keep this one open. $\endgroup$ – Emilio Pisanty Oct 6 '17 at 21:08
  • $\begingroup$ Does it need to be finite, or are exponential tails in the charge distribution OK? $\endgroup$ – Emilio Pisanty Oct 6 '17 at 21:09
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    $\begingroup$ @EmilioPisanty Actually, expanding on the central statement of that other answer I think gives an answer to this too. Choose $\rho(\mathbf{r}) = f(r)g(\theta,\phi)$ where $g$ is the spherical harmonic we want. If $\rho$ is finite, the electric field should be continuous (or at least I believe so). $\endgroup$ – JJMalone Oct 6 '17 at 21:14
  • $\begingroup$ Yeah, that is correct. Maybe you can answer your own question, then? $\endgroup$ – Emilio Pisanty Oct 6 '17 at 21:33

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