What is meant by the capacitance of a capacitor? Derive an expression for the capacitance of a parallel plate capacitor, and show how a dielectric changes it.

1. 1 · Define. C = Q/V: the charge on either plate divided by the potential difference between the plates. Unit: farad.
2. 2 · Draw the figure. [Two parallel plates of area A and gap d; + signs on one, − signs on the other; equally spaced arrows for E between them]
3. 3 · Field between the plates. σ = q/A, so E = σ/ε₀ = q/(ε₀A), uniform when d is small (edge effects ignored).
4. 4 · Potential difference. V = Ed = qd/(ε₀A).
5. 5 · Capacitance in air. C₀ = q/V = ε₀A/d.
6. 6 · With a dielectric. The dielectric is polarised and weakens the field K times, so V falls and C = Kε₀A/d = KC₀.
7. 7 · Real life. Variable air capacitors tune radios, and electrolytic capacitors smooth the supply in phone chargers.
8. 8 · Close in one line. So C rises with plate area and dielectric constant, and falls as the gap widens.

Scoring tip: State the assumption that d is small compared with the plates, so the field is uniform. It is a quick mark many answers miss.