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⁻. (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 rating of a parallel plate capacitor, V = 1 kV = 1000 V

Dielectric constant of a material,  ε_r =3

Dielectric strength = 10⁷ V/m

For safety, the field intensity never exceeds 10% of the dielectric strength.

Hence, electric field intensity, E = 10% of 10⁷ = 10⁶ V/m

Capacitance of the parallel plate capacitor, C = 50 pF = 50 x 10^-12 F Distance between the plates is given by,

d =V/E

= 1000/10⁶

=10^–³m

Capacitance is given by the relation,

C=ε₀ ε_r A/d

Where,

A = Area of each plate

ε₀ = Permittivity of free space = 8.85×10^-12 N⁻¹C²m⁻²

Therefore A =Cd/ε₀ ε_r

= (50 x 10^-12x 10^–³)/( 8.85×10^-12 x 3)

≈ 19 cm²

Hence, the area of each plate is about 19 cm².