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?

  • $\begingroup$ In your integration you've assumed $Q$ is constant, but you know it's a function of the potential difference, $Q=CV$. So $dW=CV\,dV$. Now integrate... $\endgroup$
    – lemon
    Jun 28, 2016 at 11:15
  • $\begingroup$ @lemon, if the plates are disconnected, how can $Q$ not be constant? $\endgroup$
    – BowlOfRed
    Jun 29, 2016 at 17:10

2 Answers 2


With constant Q, we can say that:


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

  • $\begingroup$ Since this demonstrates that all the work done in moving the plate ends up in the capacitor, my premise that it was possible to consider the charge on the plate as having moved from one potential to another and hence calculate an energy change was overly simplistic. I assume that this is because $W=QV$ is based on a test charge changing position in an electrical field rather than the charge causing the electrical field changing position. I think I need to do some more reading. Thanks for your help. $\endgroup$
    – Jon Hurst
    Jun 29, 2016 at 17:27

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.

  • $\begingroup$ As I understand it, then, moving the charge, Q, up a potential gradient through the mechanism of decreasing capacitance from C1 to C2 requires an amount of work according to my last equation. Half of this ends up in the electrical field, as would be calculated for a black box capacitor of capacitance C2 subject to a PD of Q/C2. The other half ends up in the physical position of Q relative to its own electrical field. $\endgroup$
    – Jon Hurst
    Jun 28, 2016 at 13:38
  • $\begingroup$ On further rumination, I'm not sure that this answers the question. The work done in separating the plates is the source of the extra energy manifested in the increased voltage, it is not a sink for that energy. A capacitor with charge Q and capacitance C2 stores the same energy regardless of physical configuration or the path it took to that situation. $\endgroup$
    – Jon Hurst
    Jun 29, 2016 at 7:43

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