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In A-level physics, one is told that:

The energy stored in a capacitor with charge $Q$, capacitance $C$ and voltage across the plates $V$ is: $$\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV=\frac{1}{2}CV^2$$

With no further elaboration on why, or what that actually means. I've tried to reason about it like this (which I think is pretty standard):

Take a capacitor with charge $Q_0$, capacitance $C$. To say how much energy is stored in it is surely to describe how many joules of energy this capacitor can give off to another electrical circuit. So let's discharge our capacitor by using it as a power source for an external circuit with overall resistance $R$.

Let $t$ be the time in seconds since the capacitor was first attached to the circuit. Assuming the circuit is Ohmic, we know that: $$\frac{\mathrm{d}Q}{\mathrm{d}t}=-I=-\frac{V(t)}{R}=-\frac{1}{RC}Q(t)$$Holds, with $Q(t)$ being the charge stored on the capacitor at time $t$. It is easy to solve this differential equation: $$Q(t)=Q_0\cdot\exp\left(-\frac{t}{RC}\right)$$For all $t\ge0$.

The total energy supplied will be the integral from $0\to\infty$ of the instantaneous power developed in the external circuit, which will be $\frac{1}{R}V^2$ if the circuit is Ohmic. Also, $V=Q/C$, so: $$E=\int_0^\infty\frac{Q_0^2}{RC^2}\exp\left(-\frac{2t}{RC}\right)\,\mathrm{d}t=\frac{1}{2}\frac{Q_0^2}{C}$$

Which is the expected answer.

The following is a question that has appeared in past papers:

A capacitor is connected to a battery and is fully charged: it stores an energy of $E$. If we double the distance between the plates, without disconnecting the battery, how much energy does the capacitor store?

The answer is:

Because $C\propto\frac{1}{d}$ and $E\propto\frac{1}{C}$, $E\propto d$ and the energy stored is doubled.

I'm very puzzled why we need to assume the battery is not being disconnected. The line of reasoning used to calculate $E=\frac{1}{2}\frac{Q^2}{C}$ only assumes three things (as far as I can tell):

  • The circuit the capacitor is being discharged through is Ohmic with constant resistance
  • The capacitance equation always holds: $Q=CV$
  • The capacitor has a constant capacitance $C$ and no internal resistance to discharging

When we double the distance between the plates, I accept we have halved the effective capacitance. We have not changed the charge stored on the capacitor (right?), and the above three bullet points all hold (for the same total charge $Q$, but with a halved value of $C$) so it follows that $E$ is doubled, irrespective of whether or not the capacitor is currently being charged.

But that reasoning doesn't make sense either, because it would imply that energy has been created from nowhere. The question presumably assumes the battery is connected so that there is a source that this extra energy could come from.

Where is the flaw in the reasoning in the above, greyed-out paragraph?

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  • $\begingroup$ One is told that … with no further elaboration on why. I’ve never seen a textbook that doesn’t explain why this is the energy stored in the capacitor. Which textbook are you using? $\endgroup$
    – Ghoster
    Jan 8 at 17:32
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    $\begingroup$ @Ghoster The standard AQA physics textbook. The quality of A level physics is up for debate, let's say $\endgroup$
    – FShrike
    Jan 8 at 17:34
  • $\begingroup$ No need to assume an ohmic discharge device. Connect your C to a dissipative component X. Suppose that C loses charge $-dQ$ when the pd across it (and across X) is $V$. The accompanying energy loss by C is therefore $-VdQ=-\frac 1C QdQ$. So the energy lost by C as it discharges from $Q=Q_0$ to $Q=0$ is $$U=-\int_{Q_0}^0 \frac 1C Q \ dQ= \frac 1{2C} {Q_0}^2$$ It's slightly nicer to consider the capacitor charging rather than discharging. $\endgroup$
    – Philip Wood
    Jan 8 at 17:58

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If you keep the battery connected and double the spacing between the plates then $V$ is unchanged but $Q$ changes $Q\to Q/2$. If you disconnect the battery then $Q$ stays the same but $V$ doubles. In this case the energy increases because you have to do work to pull the plates apart.

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    $\begingroup$ Why should the charge halve? We haven't 'touched' the accumulated electrons on the -ve plate, or the accumulated +ve charge on the +ve plate. I'd appreciate more detail on what the connection of the battery has to do with anything: it's not at all obvious to me that $V$ should be constant in one scenario, but double in the other $\endgroup$
    – FShrike
    Jan 8 at 17:09
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    $\begingroup$ The battery has voltage $V$, so if it is connected and no current is flowing the voltage between the plates must be $V$. Double the spacing at fixed $Q$ increases the voltage between the plates, so the extra $V$ across the plates pushes charge back into the battery and current will flow until the plates end up at the same $V$ as the battery. In other words we are " touching" the electrons. $\endgroup$
    – mike stone
    Jan 8 at 17:27
  • $\begingroup$ That makes more sense, thank you. The voltage doubles because... the electric field strength between the plates must be constant (why?), and it is $V/d$, so a doubled $d$ makes for a doubled $V$? $\endgroup$
    – FShrike
    Jan 8 at 17:56
  • $\begingroup$ By Gauss' the electric field $E$ is determined by the charge density on the plates and $V=Ed$. Sof if $Q$ does not change then $V$ doubles. $\endgroup$
    – mike stone
    Jan 8 at 18:37
  • $\begingroup$ You have to do work to pull the plates apart in both cases. In the open circuit case, the energy of the field increases. With the battery connected, the work charges the battery. $\endgroup$
    – John Doty
    Jan 8 at 18:42

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