Derive an expression for the capacitance of a parallel plate capacitor. How does the capacitance change when a dielectric of constant K completely fills the space between the plates? Derive the new value.

1. 1 · Define capacitance. C = Q/V, the charge stored per unit potential difference.
2. 2 · Draw the capacitor. [Two parallel plates, area A, separation d; +Q and −Q; uniform field arrows between them; a dielectric slab with induced charges shown]
3. 3 · Field between the plates. Each plate gives σ/(2ε₀); between them the fields add: E = σ/ε₀, with σ = Q/A. Edge effects are ignored.
4. 4 · Capacitance in vacuum. V = Ed = Qd/(ε₀A), so C₀ = ε₀A/d.
5. 5 · With the dielectric. Polarisation charges reduce the field: E = σ/(Kε₀). Then V = Qd/(Kε₀A).
6. 6 · New capacitance. C = Q/V = Kε₀A/d = KC₀: it rises K times.
7. 7 · Real life. A phone touchscreen senses the change in capacitance caused by your finger.
8. 8 · Close in one line. C depends only on A, d and K: area up, distance down, dielectric in.

Scoring tip: Say 'ignoring edge effects' when you take the field as uniform. It shows you know the assumption.