Conventionally one define electric energy as $$ U = \frac{1}{2} \int \vec{E}(r') \cdot \vec{E}(r') d^3 x' $$ where $\vec{E}$ is a Electric field.

And from textbook like Griffith, we know that electric field generated by dipole is given as $$ \vec{E}_{dipole}(r) = \frac{1}{4\pi} \frac{1}{r^3}(3 (\vec{p} \cdot \hat{r}) \hat{r} - \vec{p} ) $$ I am trying to plug this and obtain explicit formula for electric energy $U$.

I think the result should be on some textbook or papers but I couldn't find.

Do you know the formula?

The purpose of this question is actually to compute the $U$ with $\vec{E}$.

My trial was

$$ U = \frac{1}{2} \frac{1}{16 \pi^2} \int \frac{1}{r^6} \left( 3 (\vec{p}\cdot \hat{r})^2 - \vec{p}\cdot\vec{p} \right) r^2 dr\sin(\theta) d\theta d\varphi$$

$$ = - \frac{1}{32\pi r^3} \int_{0}^{\pi} \left( 3 (\vec{p} \cdot \hat{r})^2 - (\vec{p}\cdot \vec{p}) \right) \sin(\theta) d\theta $$

  • $\begingroup$ So the electrostatic potential of this setup is simply that of two point charges separated by some distance $d$. I imagine one can find the electrostatic energy of this setup by noting that $U = qV$? $\endgroup$
    – talrefae
    Commented Dec 24, 2019 at 5:04
  • $\begingroup$ How do you wish to cope with infinite E-field when $r=0$? $\endgroup$
    – ProfRob
    Commented Dec 24, 2019 at 11:02
  • $\begingroup$ How did you integrate over $r$ and end up with $1/r^3$? $\endgroup$
    – G. Smith
    Commented Dec 24, 2019 at 11:21
  • $\begingroup$ There is a sign error : it should be $+p^2 $. $\endgroup$
    – my2cts
    Commented Dec 24, 2019 at 12:04

1 Answer 1


The electric field of a dipole can be written in spherical coordinates (using your unit system) as $$\vec{E} =\frac{1}{4\pi r^3}\left( 2p\cos \theta\ \hat{r} + p\sin \theta\ \hat{\theta}\right)$$ $$ E^2 =\frac{p^2}{16\pi^2 r^6}\left(1 + 3\cos^2 \theta \right)$$

Integrating over a spherical volume (as you propose) from an inner radius $r_1$ to an outer radius $r_2$, yields $$U = \frac{p^2}{4\pi}\left[\frac{1}{r^3}\right]^{r=r_1}_{r=r_2}.$$

As you can see there is no problem to integrate out to $r_2 =\infty$, but you get an infinite energy if you allow $r_1 =0$ (because the electric field is infinite when $r=0$).

The expression for $U$ in terms of $E^2$ cannot be used at the position of the point charges, or in this case at the position of an (assumed) point dipole.


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