Which value is closest to the capacitance of a parallel-plate capacitor with a dielectric layer of relative permittivity 3000 and thickness 2.0 µm between electrodes of area 1.0 mm²? Take the permittivity of free space as 8.85×10⁻¹² F/m.

Answer: B. About 13 nF

Substitute into C = ε₀εᵣA/d in SI units. A = 1.0 mm² = 1.0×10⁻⁶ m² and d = 2.0 µm = 2.0×10⁻⁶ m, so C = 8.85×10⁻¹² × 3000 × 1.0×10⁻⁶ / (2.0×10⁻⁶) ≈ 1.3×10⁻⁸ F, about 13 nF. About 13 pF and about 13 µF are off by a factor of 1000 from a slip in converting mm² or µm. BaTiO₃-based dielectrics have relative permittivities in the thousands, and multilayer ceramic capacitors (MLCCs) stack hundreds of such thin layers connected in parallel to obtain a large capacitance in a small volume.