Why is energy absorbed by the battery when the plates of a parallel plate capacitor are pulled apart?

Question

It is said that when the plates of a parallel plate capacitor connected to a battery are pulled apart to increase the separation, energy is absorbed by the battery and no heat is produced during this process.

For example, let us consider a parallel plate capacitor of capacitance $C$ with plates having area $A$ and separated by a distance $d$. Suppose the plates are maintained at a potential difference $V$ by a battery and the plates are pulled apart to increase the separation to $2d$, the capacitance reduces by a factor of two in accordance to $C=A\epsilon_0/d$. The energy stored in the electric field also reduces to half its original value in accordance to $U=CV^2/2$. I determined the work done by the external force which increases the separation to be half of the energy stored in the field initially i.e., $U/2$. The work done by the external agent increases the energy of the system.

Initially, an energy $U$ is stored in the field and after separation $U/2$ is stored in the field. The work done by the external agent increases the energy of the system by $U/2$. So, on the whole, the energy of the system must increase by $U/2+U/2=U$.

Why is this energy $U$ absorbed by the battery? Why shouldn't it be liberated as thermal energy (heat) as it happens during charging of a capacitor?

I read the answers to the following questions, but still my doubt on why should the energy be absorbed by the battery instead be liberated as heat exists:

• What happens to half of the energy in a circuit with a capacitor?

• Where does all the voltage go?

• How is Capacitance affected by a seperation distance and with or without a battery?

• Energy developed during change of capacitance

Answer 1

Why is this energy $U$ absorbed by the battery? Why shouldn't it be liberated as thermal energy (heat) as it happens during charging of a capacitor?

The answer is that your analysis has no resistance in the circuit.

Assume that the initial charge on the capacitor is $Q$, the final charge on the capacitor is $\dfrac Q2$ and the emf of the battery is $V$.
You have worked out the work done in sending charge through the battery during the separation process is $\dfrac Q2 V = U$ and there is a balanced energy equation.

If there is a resistive component in the circuit then suppose that during the separation of the plates phase the potential difference across the resistive part is $v$.
This means that the total potential difference across the battery and the resistive part of the circuit is $V+v$.

As the plates of the capacitor are being separated $V+v$ must also be the potential difference across the plates of the capacitor.
This means that the force between the plates of the capacitor, which depends on the potential difference across the plates, is increased which in turns means more external work need to be done in separating the plates.

That extra work done by the external force in separating the plates is the source of the heat dissipated by the resistive part of the circuit.

Answer 2

When the plates of a parallel plate capacitor are pulled apart while battery remains connected, the potential difference between the plates remains equal to that of the EMF of the battery as the plates are connected to the terminals of the battery through conducting wires. The battery gets charged during this process. Heat is not liberated when there is no resistance in the circuit, but in the presence of resistance some amount of heat is liberated as discussed in the answer above.

Now lets focus on the question "Why does the battery absorb energy in the first place?". When the plates of the capacitor are pulled apart, its capacitance or the ability to store charge decreases and so the magnitude of charge on either plates reduces. So as we gradually increase the separation between the plates, excess charge on one plate slowly moves to the other plate through the battery as given below:

It can be seen that electrons go from the negative terminal of the battery to its positive terminal. Usually, battery gets discharged when electrons move from the positive terminal to the negative terminal. And here, the case is opposite, and since electrons go from the negative to the positive terminal, the battery gets charged. Further, when the plates are pulled slowly, the energy stored in the battery increases as a cause of decrease in the energy stored in the electric field and the work done by the external agent. Also the electrons move very slowly. In the case of charging a capacitor, the electrons gain velocity in the process and energy is lost as heat as a result of dampening of electron flow oscillations as described in this answer.

On the whole, the following equation gives a good idea of how is energy stored and how much of it is stored (in parenthesis) in the process described in the question:

$$\text{Energy stored in the electric field before separation} \ (U) + \text{Work done by the external agent in pulling the plates} \ (U/2) = \text{Energy stored in the electric field after separation} \ (U/2)+ \text{Energy absorbed by the battery} \ (U)$$