Energy changes due increasing plate separation of unconnected charged capacitor

Question

If a capacitor is charged by a source with constant voltage, $V$, the energy, $W$, stored by a capacitor of capacitance $C$ is given by:

$$ W = \frac{1}{2} QV = \frac{1}{2} V^{2}C = \frac{Q^{2}}{2C} $$

where $Q$ is the charge stored on the plates.

If the capacitor is now disconnected, and its capacitance is changed by increasing the distance between the plates, its energy must increase since work is being done by moving charge under the influence of an electrical field. Furthermore, since the plates are not connected to anything, it can be assumed that charge is conserved.

The method for calculating the change in energy that I have seen in textbooks and on the internet is to calculate $Q$ for the pre-separated case, calculate $C$ for both cases, then use:

$$ W = \frac{Q^{2}}{2C} $$

to give $$ \delta W = \frac{Q^{2}}{2} \left( \frac{1}{C_{2}} - \frac{1}{C_{1}} \right) $$

where $C_{2}$ is capacitance post-separation, $C_{1}$ is capacitance pre-separation and $\delta W$ is the change in energy.

This seems naive to me. That formula was for the process of charging a capacitor from a constant voltage with the $\frac{1}{2}$ arising from the potential difference that charge is raised by decreasing as more charge is deposited.

With constant $Q$, we can say that:

$$ dW = QdV $$

and integrate to give:

$$ \delta W = Q(V_{2} - V_{1}) $$

where $V_{2}$ is the post-separation voltage and $V_{1}$ is the pre-separation voltage. Substituting $Q = CV$ then gives:

$$ \delta W = Q^{2} \left( \frac{1}{C_{2}} - \frac{1}{C_{1}} \right) $$

i.e. double the textbook answer.

Am I missing something?

Answer 1

With constant Q, we can say that:

dW=QdV

This is 'double counting'.

For simplicity, assume that the plate with charge $-Q$ is fixed and that the plate with charge $Q$ is movable.

Making the usual simplifying assumptions for parallel plate capacitors, note that the electric field, due to the fixed plate, at the location of the movable plate is

$$E_\mathrm{fixed} = -\frac{Q}{2\epsilon A}$$

where $A$ is the area of a plate. It follows that the work done slowly moving the plate a small distance is

$$\Delta W = F\delta d = (- Q E_\mathrm{fixed}) \Delta d = \frac{Q^2}{2\epsilon A} \Delta d = \frac{Q^2}{2}\left(\frac{1}{C_f} - \frac{1}{C_i}\right)$$

Thus, the work done equals the change in energy stored in the capacitor.

Relevant reading: Force between the plates of a capacitor

Answer 2

You need to do work when you want to seperate two plate. Assuming the area of plate is S and distance between the plates is d(also assume $S\gg d^2$). Then $C=\frac {S}{\epsilon_0 d}$ in vacuum. The area charge density is $Q \over S$ and it makes electric field. $$E=\frac {Q}{2\epsilon_0 S}$$ So there is attractive force between the plates and you need to do work. Difference between two calculation equals to the amount of work.