# Capacitance

### What is a capacitor?

A capacitor is an electrical component that stores charge. It is usually made by having two parallel plates thinly separated by an insulating material. This material is also known as a dielectric. Current flows through an insulator if the voltage across it is sufficient. In a capacitor, the conducting plates are close enough for the current to flow, even through it is separated by the insulating dielectric. However, because of the dielectric, some charges are stored on the plates as the current flows, resulting in higher potential difference as time passes. When the potential difference across the capacitor equals to the e.m.f of the circuit, current stops flowing.

### Physics of capacitance

Capacitance is defined as

Capacitance is the ratio of the charge stored to potential difference across a capacitor.

Mathematically,

\begin{aligned}C=\frac{Q}{V} \end{aligned}

The SI  unit of capacitance is farad (F).

A capacitor of 1 farad has a stored charge of 1 C when the potential difference across it is 1 V.

### Combined Capacitance

Capacitors can be connected either in series or parallel.

In a series connection, the combined capacitance can be found from the relationship

\begin{aligned} V_\text{total} &= V_1 + V_2 + ...\\\frac{Q}{C_\text{total}} &= \frac{Q}{C_1}+\frac{Q}{C_2}+...\end{aligned}

Since the charge $Q$ stored in each capacitance is the same,

\begin{aligned}\frac{1}{C_\text{total}} &= \frac{1}{C_1}+\frac{1}{C_2}+...\end{aligned}

In a parallel connection, the combined resistance is derived as follow:

\begin{aligned} Q_\text{total}&=Q_1+Q_2+...\\C_\text{total}V&=C_1V+C_2V+...\\\end{aligned}

Since the potential difference $V$ is the same in a parallel circuit,

\begin{aligned} C_\text{total}=C_1+C_2+... \end{aligned}

## Summary

1. Capacitance is the ratio of the charge stored to the potential difference across the component.
2. SI unit of capacitance is farad (F).
3. The total capacitance of series connected capacitors is \begin{aligned} \frac{1}{C_\text{total}}=\frac{1}{C_1}+\frac{1}{C_2}+... \end{aligned}
4. The total capacitance of parallel connected capacitors is \begin{aligned} C_\text{total} = C_1+C_2+... \end{aligned}

## Review

### Question 1

The figure shows three capacitors connected in series with a cell of e.m.f. 3.0 V.

Calculate the p.d. across each capacitor.

### Solution

Since the charge stored in each capacitor is the same as the charge stored across all the capacitors, we shall go ahead and find the charge stored in the combined capacitors and then use it to calculate the potential difference across each capacitor.

\begin{aligned} \frac{1}{C_\text{total}} &= \frac{1}{100 \times 10^{-3}}+\frac{1}{200 \times10^{-3}} + \frac{1}{400 \times 10^{-3}}\\C_\text{total}&= 0.0571\text{ F}\\ Q_\text{total}&=3.0 \times 0.0571\\&=0.1713\text{ C}\end{aligned}

Hence,

\begin{aligned} V_1&=\frac{0.1713}{100 \times 10^{-3}}\\&=1.713\text{ V}\\V_2&=\frac{0.1713}{200 \times 10^{-3}}\\&=0.857\text{ V} \\V_3&=\frac{0.1713}{400 \times 10^{-3}}\\&=0.428\text{ V}\end{aligned}

You can observe that the sum of all the three p.d. is the e.m.f. of the cell.