Am I correct to say that since the divergence of the following E field

$$ \mathbf{E} = \frac{1}{4\pi\epsilon_0}\left[ \frac{3(\mathbf{p_0}\cdot\hat{r})\hat{r}-\mathbf{p_0}}{r^3} \right] - \frac{1}{3\epsilon_0}\mathbf{p_0} \delta^3(\mathbf{r}) $$

satisfies Gauss’ law for electricity where


for an electric point dipole, the usual electric field associated with the point dipole

$$ \mathbf{E} = \frac{1}{4\pi\epsilon_0}\left[ \frac{3(\mathbf{p_0}\cdot\hat{r})\hat{r}-\mathbf{p_0}}{r^3} \right] $$

is thus incomplete?

Incidentally, what is the divergence of the usual electric field of point dipole?

  • $\begingroup$ If I'm understanding your question, the usual electric field you've written is the correct one everywhere except for the origin and usually that's where we're interested to look. The divergence will then be, as you said, given by Gauss' law, just for $\boldsymbol{r}\neq0$ $\endgroup$ Jan 19, 2021 at 12:11
  • $\begingroup$ Does that mean that if I want to include $\mathbf{r}=0$, I will need the Dirac term? Also, what is the divergence of the usual electric field? $\endgroup$
    – user161157
    Jan 19, 2021 at 12:16
  • 2
    $\begingroup$ The delta term fixes spatial integrals of the electric field to physically correct value (so they are the limit of what we get with physical non-point dipoles when the distance is shrinked to 0). You only ever need this term when you calculate spatial integrals of electric field vector (for example, for spatial averaging). $\endgroup$ Jan 19, 2021 at 12:49
  • $\begingroup$ @JánLalinský That sounds like an answer. $\endgroup$ Jan 19, 2021 at 13:57

1 Answer 1


As @Ján Lalinský mentions, the addition of the $\delta$ contribution comes out when discussing a volume integral of the field. A finer point that I think is worth mentioning is that the addition of the $\delta$ term comes as relevant once we set the volume integral of the singular non-$\delta$ term to $0$ (you can see this in the discussion at the end of section $4.1$ in Jackson's Classical Electrodynamics book). As mentioned there, this makes sense when looking at the angular integration, but it is less obvious when one looks at the radial integration.

When asking about the divergence of the field, it is somehow more straightforward to talk about the Laplacian of the point dipole potential, which you can find in this answer. When you compute the electric field via $-\nabla \Phi$ from that potential, you find only the non-$\delta$ term, as long as you ignore all subtleties with the singular point at the location of the dipole. Interestingly enough, when computing $-\nabla \Phi$, it is possible to somehow motivate the $\delta$ term, by splitting up double partial derivatives $\partial_i \partial_j$ into a trace and a traceless terms, in which the trace brings out a Laplacian of a $\frac{1}{r}$. In the end, this is an attempt to fully capture the singular behavior of the expression.

You can also address the issue of the divergence with a tangential computation: by definition of the dipole moment, $ \mathbf{p} = \int \mathbf{x'} \rho(\mathbf{x'}) d^3x'$. Demanding $ \mathbf{p} = \epsilon_0 \int \mathbf{x'} \,\nabla'\cdot \mathbf{E}(\mathbf{x'}) d^3x'$ does give you another angle to tackle the problem and test your expressions for the field and charge density. This is another integral that is ultimately related to the volume integral of the field. It also touches on the importance of preserving the contributions from the boundary terms. By imposing this constraint and using the prescription set for the volume integral, you can find some consistency in the end.


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