The electric potential energy U stored in a charged capacitor is equal to the amount of work W required to charge it.

Recall that $V = \frac{q}{C}$,

$$\begin{aligned} dw &= V \, dq \\ dw &= \frac{q}{C} \, dq \\ \int\limits_{0}^{W} dw &= \int\limits_{0}^{Q} \frac{q}{C} \, dq \\ W &= \frac{Q^{2}}{2C} \end{aligned}$$

You can manipulate the final result into other forms:

$$\begin{aligned} U &= \frac{Q^{2}}{2C} \\ &= \frac{1}{2} C V^{2} \\ &= \frac{1}{2} QV \end{aligned}$$

Notice that 50% of the energy is wasted as the work done by the battery is QV over the whole charging process. The main culprit to the energy loss is the heating of wire. You might say “OK, what if superconducting wires are used in place of conventional wires? There would be no resistance – no energy loss due to heating of wire.”. But then, in this case, radiation would be the contributor to the energy loss. Since the resistance is zero, the current will be infinite. Large amount of energy will be radiated away as electromagnetic energy – exactly 50%.

**What?! If it is so inefficient, why are capacitors still in use today?**

There is a workaround to this problem. This 50% energy loss does not occur if you add an inductor into the circuit.

**You might ask “Where is the energy of a capacitor stored?”**

We can think of he electric potential energy U as being stored in the electric field in the region between the plates.

Recalling that $C = \epsilon_{0} \frac{A}{d}$ for parallel plate capacitors:

$$\begin{aligned} U&= \frac{1}{2} CV^{2} \\ &= \frac{1}{2}\epsilon_{0}\frac{A}{d}V^{2}\\&=\frac{1}{2}\epsilon_{0}\left(\frac{V}{d}\right)^{2}\times Ad\\&=\frac{1}{2}\epsilon_{0}E^{2}\times Ad \end{aligned}$$

Notice that $Ad$ is just the volume of the space between the plates. Hence, the energy per unit volume or energy density, u in the space between the plates:

$$u = \frac{1}{2} \epsilon_{0} E^{2}$$

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