The energy stored in a capacitor


In the prerequisite reading, the capacitor, we've described a capacitor as "a way to store electrical energy as a separation of charge." Let's work out the details.

The work you have to do to charge a capacitor by moving charges from one side of the plate to the other is equal to the amount of potential energy stored in it (by the work-energy theorem). This energy can be extracted to move charges around, light bulbs, or send signals down an axon.

Getting the form by dimensional analysis

Since electric potential, $V$, is defined as electrical energy $U^E$ divided by the test charge, 

$$\Delta V = \frac{\Delta U}{q}$$

so it's clear that the electric potential has the dimensions of energy/charge.

Since we have a quantity with the dimensions of $\mathrm{Q}$, the magnitude of charge that has been separated (into a $Q$ and a $-Q$), and a quantity with the dimensions of energy/charge, we expect that our energy stored will be proportional to ($\propto$) the product

$$U^E_{stored} \propto Q\Delta V.$$

We only have to figure out the proportionality constant.

If we think about moving charges from one side of the pair of plates to another, we start with a pair of uncharged plates as shown in the figure below left, and we end with a pair of oppositely charge plates as shown in the figure below right.


When we start charging the plates, it looks like the figure at the left. If we take a small bit of positive charge, $dq$ from the gray (uncharged) plate on the right side and move it to the gray (uncharged) plate on the left, the left slide will be slightly positively charged (by an amount $dq$) while the rich side will now slightly negatively charged (by an amount $-dq$). The potential difference between the two plates was 0, so we didn't have to do any work.

When we are just about finished, the plates will look like the figure on the right. The left plate will be positively charged (blue) and the right plate will be negatively charged (red). There will be an electric field between the plates pointing to the right as shown. If we take our last small bit of positive charge, $dq$ from the right plate and move it to the left, we will have to push against the E field, moving our charge up the potential hill, which now has a value of almost $\Delta V$. The amount of work we do will be $dq \Delta V$.

So through the process, the bits of charge we are moving sees a potential difference that starts at 0 and ends at $\Delta V$. We might expect the correct answer to be the average of the two, $\frac{1}{2}\Delta V$. This turns out to be correct giving the energy stored in a capacitor as

$$\Delta U^E_{stored} = \frac{1}{2} Q \Delta V$$

This is the easiest form to remember, since we know that in general, a quantity that looks like $QV$ is an energy.

But sometimes, we know the capacitance and either the charge or the potential difference rather than both. In that case, it might be useful to rewrite our energy equation using $Q = C \Delta V$. Eliminating one or the other of the variables in favor of the parameter ($C$), we get

$$\Delta U^E_{stored} = \frac{Q^2}{2C}$$


$$\Delta U^E_{stored} = \frac{1}{2} C \Delta V^2$$.

You are guided to work through finding the factor by explicit calculation in the associated problem linked at the bottom of the page.

Joe Redish & Wolfgang Losert 2/20/12, 2/22/13, 4/30/19

Associated Problem

Article 647
Last Modified: May 16, 2019