Capacitance of a Capacitor

Divide the stored charge in coulombs by the applied voltage in volts to get the capacitance in farads. Example: Q = 6×10⁻⁴ C at U = 12 V gives C = 5×10⁻⁵ F = 50 µF. Usually the capacitance is known and you seek the charge (Q = C·U) or the voltage (U = Q/C). Since one farad is enormous, real components carry values in micro-, nano- or picofarads; convert the prefixes consistently: µ = 10⁻⁶, n = 10⁻⁹, p = 10⁻¹². A 100 µF capacitor at 9 V stores Q = 10⁻⁴ × 9 = 9×10⁻⁴ C. The unit check coulomb per volt = farad secures every rearrangement.

Capacitance is the holding capacity for charge per volt: a capacitor of 1 F takes exactly 1 C of charge at 1 V. A helpful picture is a water tank: the capacitance corresponds to the base area of the tank, the voltage to the water level, the charge to the amount of water. A wide tank (large capacitance) stores more water at the same level; a narrow tank reaches a high level with little water, just as little charge on a small capacitance already creates a high voltage. The distinction matters: C describes the storage capability of the component (depending only on geometry), Q the actually stored charge (depending on the applied voltage).

On three construction quantities: C = ε₀·ε_r·A/d. A larger plate area A offers more room for charge and raises C linearly. A smaller plate separation d strengthens the attraction between the opposite charges and also raises C; C is inversely proportional to d. A dielectric between the plates (plastic, ceramic) polarises in the field, weakens it and multiplies the capacitance by the factor ε_r (ceramics up to over 1,000). The natural constant ε₀ = 8.854×10⁻¹² F/m sets the scale and explains why centimetre-sized capacitors reach only pico- to microfarads. Worked example: A = 0.01 m², d = 1 mm, air: C = 8.854×10⁻¹² × 0.01/0.001 ≈ 88.5 pF.

This is the most important case distinction in capacitor problems. If the capacitor stays connected to the voltage source, U is pinned; if you then change the capacitance (say by inserting a dielectric), charge flows in: Q = C·U grows along. If you disconnect it first, the charge Q is trapped, since it has nowhere to flow, and now the voltage adapts: U = Q/C. Inserting a dielectric with ε_r = 4 into a disconnected capacitor quadruples C and the voltage drops to a quarter. If you instead pull the plates apart, C falls and the voltage rises; the work needed for this is done by your hand against the attraction of the plates.

Exactly the other way round from resistors. Capacitors in parallel add directly: C_total = C₁ + C₂; both sit at the same voltage and their plate areas act like one large one. Two 100 µF capacitors in parallel give 200 µF. In series the reciprocals add: 1/C_total = 1/C₁ + 1/C₂; the same charge sits on all capacitors, the voltages add, and the total capacitance is smaller than the smallest individual one. Two 100 µF capacitors in series give 50 µF. The rule "parallel adds, series uses reciprocals" is thus swapped compared with resistors, the most common error in mixed exam problems. Series connection is also used to increase the voltage rating.