I know that, in vacuum, the electrostic energy is:

$$ U_E = \frac12 \int \rho(\mathbf r)\varphi(\mathbf r) d^3\mathbf{r} $$

But I don't know how to pass to the matter version? The formula would be the same, but what is $\rho(\mathbf r)$ now?, Is it free charge or total charge (free plus bound charge) density?

P.D.: I am studying classical electrodynamics


1 Answer 1


It depends on the context (which you may want to add to your question). In principle the formula is correct everywhere, provided that the potential takes account for all the charges.

In the context of classical electrodynamics this means accounting for the induced local polarization which is often reduced to dividing the field and the potential due to free charges by the dielectric constant.

In quantum solid state theory the equation is used as is, taking account for all the charges present (i.e. the electrons and the ions composing the matter.)

  • $\begingroup$ I understand your point, but i'm not understand how to apply it. I'm studying classical electrodynamics. Thanks for your answer. $\endgroup$ Jun 6, 2020 at 22:00
  • $\begingroup$ @Elborito it is hard to be more precise, since it depends on what is $\rho$ and $\varphi$ in your equation. Yours is just general equation for electrostatic energy. $\endgroup$
    – Roger V.
    Jun 7, 2020 at 6:27
  • $\begingroup$ And if you have a polarization density $\mathbf P$ no dependent of the electric field $\mathbf E$, what do you suggest? $\endgroup$ Jun 7, 2020 at 15:06
  • $\begingroup$ @BorisValderrama in this case, you have to evaluate the work (and if the contribution is conservative, the energy) of that polarization field. As an example, let's assume you have a prescribed "a priori" uniform polarization $\mathbf{P}^0$. You evaluate that contribution to work when you build the configuration of the system. The electric field due to this polarization is $\mathbf{E}^0 = -\frac{\mathbf{P^0}}{\epsilon_0}$, the elementary work reads $dW = -q \frac{\mathbf{P^0}}{\epsilon_0} \cdot d\mathbf{r} = - q d \left(\frac{\mathbf{P^0}}{\epsilon_0} \cdot \mathbf{r} \right) = q d \Phi^0$ $\endgroup$
    – basics
    Aug 25, 2022 at 13:05
  • 1
    $\begingroup$ @RogerVadim maybe I was drunk, because now the concept seems to be extremely easy. Thank you all for your replies $\endgroup$ Aug 25, 2022 at 20:20

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