# Capacitance of parallel plate capacitor – formula derivation

At a point between two plates of the parallel plate capacitor, the Electric field generated by the combined effect of the 2 plates of the capacitor can be expressed as: E = σ /2 ε_{0} + σ /2 ε_{0} = σ / ε_{0} [ note that the electric field produced by a plane sheet of charge = σ /2 ε_{0} ]

Air is there between the 2 plates. Hence ε_{0} is used in the equation.

σ = surface charge density in each plate = total amount of charge in each plate / area of each plate = Q/A

Hence, E = σ / ε_{0} = σ A/(A ε_{0}) = Q/(A ε_{0})

E = Q/(A ε_{0}) …………… (1)

The potential difference between two plates of the parallel plate capacitor V = Ed

=> V = Qd/(A ε_{0}) ……………. (2)

As, Q = CV

=> C = Q/V = Q / [ Qd/(A ε_{0})] = A ε_{0} / d

**C = A ε _{0} / d ……………. (3)** [for air capacitor]

If any dielectric with dielectric constant K is used instead of air as the medium between the 2 plates, then the equation of capacitance (equation 3) can be rewritten as:

C = A ε / d = A Kε_{0} / d = K A ε_{0} / d

**C = K A ε _{0} / d …………. (4)**

For air K = 1. Hence for air capacitor we get equation (3) from equation (4) by putting the value of K as 1 in equation (4).

C = **A ε _{0} / d** (for air capacitor)