A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm^-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength.

What minimum area of the plates is required to have a capacitance of 50 pF?

Potential difference of a parallel plate capacitor(V) = 1kV = 1000V

Dielectric constant of a material(ϵ_r) = 3

Dielectric strength = 10⁷V/m

Electric field intensity(E) = 10%of 10⁷

⇒ E = 10⁶V/m

(since, the field intensity never exceeds 10% of the dielectric strength)

Capacitance of the parallel plate capacitor(C) = 50pF = 50 × 10^-12F

Distance between the plates (d),

⇒

⇒ d = 10^-3m

∵ Capacitance =

where, ϵ₀ = Permittivity of free space = 8.85 × 10^-12N^-1 m^-2 C²

A = area of plates

∴

⇒ A = 50 × 10^-12F × 10^-3m/8.85 × 10^-12N^-1 m^-2 C² × 3

⇒ A = 1.88 × 10^-3m²

∴ Minimum area of the plates required = 1.88 × 10^-3m²