Cambridge Syllabus Notes

Capacitors and Capacitance – A-Level Physics 9702

Capacitors and Capacitance

Recall and use the energy stored in a capacitor: $$W = \frac{1}{2} QV = \frac{1}{2} CV^{2}$$

The capacitance $C$ of a device is the ability to store charge per unit potential difference:

$$C = \frac{Q}{V}$$ where $Q$ is the charge on one plate and $V$ is the potential difference between the plates.

Derivation:

Work required to move a small charge $dq$ onto a plate when the existing charge is $q$: $dW = V\,dq = \frac{q}{C}\,dq$.
Integrate from $0$ to $Q$: $$W = \int_{0}^{Q} \frac{q}{C}\,dq = \frac{1}{2}\frac{Q^{2}}{C} = \frac{1}{2} QV = \frac{1}{2} C V^{2}.$$

Quantity	Symbol	SI Unit	Typical Range (A‑Level)
Capacitance	$C$	farad (F)	pF – μF (parallel‑plate), mF – F (electrolytic)
Charge	$Q$	coulomb (C)	10⁻⁹ – 10⁻³ C
Voltage	$V$	volt (V)	1 – 500 V
Energy	$W$	joule (J)	10⁻⁹ – 10⁻¹ J

Problem: A 47 µF capacitor is charged to 12 V. Calculate the energy stored.

Identify the given values: $C = 47 \times 10^{-6}\,\text{F}$, $V = 12\,\text{V}$.
Use the formula $W = \frac{1}{2} C V^{2}$.
Calculate: $$W = \frac{1}{2} (47 \times 10^{-6}) (12)^{2} \approx 3.4 \times 10^{-3}\,\text{J}.$$
Interpretation: The capacitor stores about 3.4 mJ of energy.

Confusing $W = \frac{1}{2} QV$ with $W = QV$ – the factor $\frac{1}{2}$ arises from the integration.
Using the voltage of the source instead of the voltage across the capacitor when it is partially charged.
Mixing units – always convert µF to F and m \cdot to \cdot before substitution.

The energy formula is useful for:

Estimating discharge energy in flash lamps.
Designing timing circuits (RC circuits) where energy considerations affect component choice.
Understanding energy density in capacitors versus batteries.

Suggested diagram: Parallel‑plate capacitor showing plate area $A$, separation $d$, electric field $E$, and labelled $V$, $Q$, $C$.