Result

Parallel plates

C = \frac{ε _{0} A}{d}

Statement

Bigger plates or a smaller gap store more charge per volt.

Symbols and units

When it applies

Gap small compared with plate size, so edge effects can be ignored.

Don't use when

Ignores fringing field bulging at the plate edges — valid only while the gap is small compared with the plate size.

Derivation

Put ±Q on the plates. The surface density is

σ = \frac{Q}{A}

Between two oppositely charged sheets the fields add (see the infinite-sheet result):

E = \frac{σ}{ε _{0}} = \frac{Q}{ε _{0} A}

The field is uniform, so the voltage across the gap is just E times d:

Δ V = E d = \frac{Q d}{ε _{0} A}

Divide, and Q cancels — as it must:

C = \frac{Q}{Δ V} = \frac{ε _{0} A}{d}

Worked examples

Realistic plate capacitor

A = 0.010 m², d = 1.0 mm:

C = \frac{( 8.85 \times 1 0 ^{- 12} ) ( 0.010 )}{0.0010} \approx 89 pF

Area required for one farad

A = \frac{C d}{ε _{0}} = \frac{( 1 ) ( 0.001 )}{8.85 \times 1 0 ^{- 12}} \approx 1.1 \times 1 0^{8} m^{2}

About 100 km². Supercapacitors reach farads through porous electrodes of very large internal area.

Watch out for

Ignores the field bulging at the edges, which is fine while the gap is small compared with the plate size.

Connects to