Capacitance C=\frac{Q}{V} Farads (F). Charge stored per volt. Energy stored E=\tfrac12 CV^2=\tfrac12 QV=\frac{Q^2}{2C} Three equivalent forms. Capacitors in parallel C<sub>T</sub> = C₁ + C₂ + … ⚠ Opposite to resistors: parallel capacitors ADD. Capacitors in series \frac{1}{C_T}=\frac{1}{C_1}+\frac{1}{C_2}+\cdots ⚠ Series capacitors combine like parallel resistors.
Capacitance from the plates
Parallel plates C=\frac{\varepsilon_0\varepsilon_r A}{d} A is the area of one plate (m²), d the gap between them (m), εr the dielectric’s relative permittivity, and ε0 = 8.85 × 10⁻¹² F m⁻¹. Bigger plates or a better dielectric raise C; a wider gap lowers it. ✕
The combination trap
Capacitor rules are the reverse of resistor rules. Parallel capacitors add; series capacitors give a smaller total. Mixing these up is the most common exam error here. Worked example Combined capacitor network A 3.0 µF and 6.0 µF capacitor in series, then that combination in parallel with a 2.0 µF. Find total C. Series pair: 1/C = 1/3 + 1/6 = 1/2 ⟹ C = 2.0 µF. Now in parallel with 2.0 µF: C_T = 2.0 + 2.0 = 4.0 µF. ✓ C_total = 4.0 µF.