Any two conductors separated by an insulator (or a vacuum) form a capacitor. A capacitor is a device that stores electric charge and electric potential energy.

The electric field at any point in the region between the conductors

$$E \propto Q$$

Hence, the potential difference between the conductors:

$$V_{ab} \propto Q$$

The **capacitance** of the capacitor:

$$C = \frac{Q}{V_{ab}}$$

C is the proportionality constant between Q and V.

In general, the value of the capacitance depends only on the shapes and sizes of the conductors, and on the nature of the insulating material between them – independent of Q and $V_{ab}$.

The greater the capacitance C of a capacitor, the greater the magnitude Q of charge on either conductor for a given potential difference $V_{ab}$ and hence the greater the amount of stored energy.

Capacitance is a measure of the ability of a capacitor to store energy.

The **SI unit** of capacitance is farad (F).

$$1 \, F \equiv 1 \, C \, V^{-1} \equiv 1 \, C^{2} \, J^{-1} \equiv 1 \, C^{2} \, N^{-1} \, m^{-1}$$

Note that $\epsilon_{0} \approx 8.854 \times 10^{-12} \, C^{2} \, N^{-1} \, m^{-2} = 8.854 \times 10^{-12} \, F \, m^{-1}$

### Parallel-plate capacitor in vacuum

This consists of two parallel conducting plates, each with area A, separated by a distance d that is small in comparison with their dimension. From previous analysis of parallel conducting plates, we know that:

$$\begin{aligned} E &= \frac{\sigma}{\epsilon_{0}} \\ &= \frac{Q}{\epsilon_{0} A} \end{aligned}$$

$$\begin{aligned} V_{ab} &= E d \\ &= \frac{Qd}{\epsilon_{0} A} \end{aligned}$$

Hence,

$$\begin{aligned} C &= \frac{Q}{V_{ab}} \\ &= \frac{\epsilon_{0} A}{d} \end{aligned}$$

C for a parallel plate capacitor depends only on the geometry of the capacitor.

**Small note:** You might have realised that the electric field is slightly wrong as that expression is derived for the case of **infinite** parallel conducting plates. Hence, there is the condition – distance d that is small in comparison with their dimension, such that the electric field approximates that for the infinite case. Notice that for finite parallel conducting plates, there will be fringing of electric field at the sides of the plates.

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