Worked example: A parallel-plate capacitor has plates of area A = 0.02 m² separated by d = 1 mm of air. (a) Find C. (b) What charge sits on it at V = 12 V? (c) If the gap is halved, what happens to C? Step 1: C = ε0·A/d = (8.854 × 10⁻¹²)(0.02)/0.001 = 1.77 × 10⁻¹⁰ F = 177 pF. Step 2: Q = CV = (1.77 × 10⁻¹⁰)(12) = 2.12 × 10⁻⁹ C ≈ 2.1 nC. That is about 1.3 × 10¹⁰ electrons — a huge number of particles, yet a tiny charge by everyday standards. Step 3: C ∝ 1/d, so halving d doubles C to 354 pF; at the same 12 V the plates would hold twice the charge. This is why capacitor design pushes plates as close as the dielectric strength allows.