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I'm trying to understand what happens, when the capacity of a capacitor connected to a battery changes.

I have this circuit formed by a battery that provides a potential difference of $\Delta V$ and a capacitor with capacity $C$, and I want to compute the work necessary to increase the distance between the parallel plates from $d$ to $2d$.

In this situation $\Delta V$ between the plates remains the same as that provided by the battery, so $C$ becomes $\frac{C}{2}$. Therefore, the potential energy stored in the capacitor changes:

$$\Delta U = \frac{1}{2}(\Delta V)^2(C_f-C)=-\frac{1}{4}(\Delta V)^2C \ .$$

Since, $C_f$ became half of $C$, the capacitor can now accumulate only half of the initial charge $Q=C \Delta V$, so half of that charge needs to be moved.

What I don't understand is what happens now: I seem to understand that the battery does an additional work:

$$W=\frac{Q}{2}\Delta V \ ,$$

but I don't understand why since I don't really get what physically happens. Before all of this happens we have a charge $Q$ on the plate of the capacitor that is connected to the positive pole of the battery, and an induced charge $-Q$ on the other plate. Does the positive charge $Q$ go back to the positive pole of the battery?

Does it need to be brought back to the negative pole?

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Some minus charges move back to the battery, charges move back to the battery, and some negative charges move to the positive side, since the positive charges in the wire can not move.

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but I don't understand why since I don't really get what physically happens.

Two things are happening when the plates are separated while connected to a constant voltage source.

First, external mechanical work is being done on the capacitor to move the plates apart against the electrostatic attraction force between the plates. That work adds energy to the capacitor. Note that if the capacitor were not attached to the battery its voltage would increase.

Second, as you are already aware, separating the plates halves the capacitor and from the relationship

$$C=\frac{Q}{V}$$

That means, for the voltage to remain constant, the charge is halved. For that to happen current has to flow to the battery. In effect, separating the plates adds energy to, or "charges" the battery. That takes energy away from the capacitor. As noted above, if the battery were not present separating the plates would increase the voltage. Charging of the constant voltage battery prevents that from happening.

The difference between the energy transferred to the capacitor by the work done separating the plates, and the energy transferred out of the capacitor to the battery in charging the battery equals the loss of stored electrical potential energy of

$$E=\frac{1}{4}CV^2$$

I leave it to you to do the detailed calculations as then this becomes a homework and exercise problem.

Hope this helps.

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  • $\begingroup$ I don't understand though why the external mechanical work adds energy to the capacitor: since its energy is $U=\frac{1}{2}CV^2$ and $C=\epsilon_0 \frac{A}{d}$, if the external force doubles d and therefore halves the capacity, isn't it decreasing the potential energy? $\endgroup$
    – Fede
    Commented Aug 4, 2023 at 19:51
  • $\begingroup$ Do you agree that an external force is needed to separate the plates? $\endgroup$
    – Bob D
    Commented Aug 4, 2023 at 19:55
  • $\begingroup$ I understand that. So, if we compute the $\Delta U$ due to the distance increase between the plates, we obtain: $$\Delta U = \frac{1}{2} \Delta C \left( \Delta V \right) ^2 = -\frac{1}{4} C \left( \Delta V \right) ^2$$ It is negative, which actually means it is not a spontaneous process: an external work equal to $W=\frac{1}{4} C \left( \Delta V \right) ^2$ is needed. Therefore the charge halves and we have $\Delta Q = \frac{C \Delta V}{2}$ that spontaneously goes back to the positive pole of the battery, and from there is brought to the negative pole spontaneously by $\Delta V$. $\endgroup$
    – Fede
    Commented Sep 18, 2023 at 8:37
  • $\begingroup$ Therefore the battery does a work $W(battery)=\frac{C \Delta V}{2}*\Delta V$: this means the external work is this case is $-W(battery)$. However, doing so we obtain an overall external work that is negative, so I guess something's wrong. $\endgroup$
    – Fede
    Commented Sep 18, 2023 at 8:37
  • $\begingroup$ @Fede I'm having a hard time following you. But the bottom line is the amount of positive mechanical work done by an external agent pulling the plates apart, plus an equal amount of stored electrical potential energy taken from the capacitor, has gone into recharging the battery. $\endgroup$
    – Bob D
    Commented Sep 18, 2023 at 17:26

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