Consider an isolated capacitor (no battery). Since the electrostatic force is conservative, the infinitesimal work done on the capacitor plates can be written as $$\mathrm{d} W=-\mathrm{d} U_e $$

Where $U_e$ is the electrostatic potential energy contained in the capacitor.

Suppose that now the capacitor is linked to a battery. On textbook I found that in this case, if I want to find the work done on the plates I must write$$\mathrm{d} W=-\mathrm{d} U_{tot}=-(\mathrm{d} U_e+\mathrm{d} U_{battery}) \tag{*}$$ Where $U_{battery}$ is called "internal energy of the battery".

My question is the following. Since $U_{battery}$ is probably some sort of energy linked to chemical reactions inside the battery, I do not think that it can be seen as a potential energy of some force field (which should be conservative). But then how is it possible to write $(*)$?

Only the infinitesimal work of a conservative force can be seen as the (exact) differential of a scalar field which represent a potential energy, but this is not true for non conservative processes (like the ones happening in a battery).

So how to justify $(*)$ from the point of view of the non reversibility of processes involved here?

  • $\begingroup$ The process inside the battery is of no concern here. You can extract work from many irreversible process. The only assumption is that the battery produces an electric field, which is conservative. If what happens inside the battery is irreversible, all it means is that you cannot use the energy created by the voltage difference to recharge the battery to the same level. $\endgroup$
    – user126422
    Dec 23, 2016 at 20:30

1 Answer 1


You always have conservation of energy, so the only difference with "conservative" potentials is that they are reversible and path-independent. But if you burn gasoline and it releases energy, and you put that energy into a heat engine and generate a known amount of work, you can keep track of that work on some system and you will still have conservation of energy, even if neither the burning nor the heat engine are themselves conservative potentials.

  • $\begingroup$ Thanks for the answer! I agree with you, but saying $(*)$ is like stating that the equilibrium "state" for the system (capacitor+battery) is the one where the total energy is minimum. Now, this is clear in the case we are talking about the potential energy of a conservative field (which, by definition, tries to go to a minimum), but if I include also the "internal energy of the battery" I don't know how to explain that minimum energy still implies equilibrium. How to explain this fact? $\endgroup$
    – Sørën
    Dec 24, 2016 at 1:18
  • $\begingroup$ The * statement is not about a minimum, or even an equilibrium, it is just about energy conservation. It just says that if the system does work, that energy must come from somewhere, and it either comes from the capacitor (conservative potential) or the battery (not necessarily conservative, since it's not necessarily rechargeable). $\endgroup$
    – Ken G
    Dec 24, 2016 at 2:33
  • $\begingroup$ Thanks for the answer, but in the case of a conservative force $F$ associated to a potential energy $U$ the relation $d W=-dU$ imply that the equilibrium positions are the ones where $dU$ is minimum (or in other words, that the force "tries" to make $U$ as small as possible). I understand that $*$ is about energy conservation but it can also be interpreted as the fact that (whatever) force does work on capacitor plates "tries" to reduce $U_{battery}$, therefore the equilibrium should be when $U_{battery}$ is minimum, and I do not explain this only with conservation of energy.. $\endgroup$
    – Sørën
    Dec 27, 2016 at 0:34
  • $\begingroup$ Should I totally forget about this? Or is there something like this behind the force on a capacitor? $\endgroup$
    – Sørën
    Dec 27, 2016 at 0:35
  • $\begingroup$ I'm not seeing why dW = -dU ever implies anything about a U minimum. The rules that determine dU are quite separate from the rule dW = -dU, the latter is nothing other than conservation of energy. If a capacitor plate does work by dumping its U, then whatever caused it to dump its U is something quite different from dW. It's like if you write a check, your bank account drops by the amount of your check, but that's not the reason you wrote the check. $\endgroup$
    – Ken G
    Dec 27, 2016 at 5:52

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