Question 33: 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?
Answer:
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, ∈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 × 10-12 F
Distance between the plates is given by,

Capacitance is given by the relation,

Where,
A = Area of each plate
∈0 = Permittivity of free space = 8.85 x10-12 N-1C2m-2

Hence, the area of each plate is about 19 cm2.
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Thank u saral study