Capacitance of a Parallel Plate Capacitor

Finds the capacitance as ε₀ × relative permittivity × area ÷ plate separation. Wider plates or a smaller gap both increase it.

Explanation

A capacitor stores charge on two facing metal plates. Its capacitance says how much it can hold for a given voltage.

C = \varepsilon_0 \varepsilon_r \dfrac{S}{d}

$\varepsilon_0$ — the permittivity of free space, $8.854 \times 10^{-12}$ F/m
$\varepsilon_r$ — the relative permittivity of whatever sits between the plates
$S$ — the plate area, $d$ — the separation

Wider plates and a smaller gap both increase it. The opposite charges on the two plates attract each other, and the closer they are, the more of them the plates will hold.

Example

Take two plates of 10 cm × 10 cm (100 cm²), 1 mm apart, with air between them (relative permittivity essentially 1).

C = 8.854 \times 10^{-12} \times \dfrac{0.01}{0.001} = 8.85 \times 10^{-11}\ \text{F} = 88.5\ \text{pF}

Plates the size of your hand give a mere 88 picofarads. A pico is a trillionth. Real capacitors reach a microfarad and beyond — ten thousand times more — by rolling metres of ultra-thin dielectric film into a cylinder, squeezing the gap down while winning back the area.

Dielectrics multiply it

Slipping an insulator between the plates multiplies the capacitance by its relative permittivity.

Air — about 1
Paper — about 3
Mica — about 6
Water — about 80

Watch out

A smaller gap buys capacitance, but it also makes the insulation easier to break down. That is why every capacitor carries a voltage rating. Capacitance and working voltage always trade against one another.