Question

A parallel plate capacitor is to be designed with a voltage rating1 kV, using a material of dielectric constant 3 and dielectric strengthabout 107 Vm–1. (Dielectric strength is the maximum electric field amaterial can tolerate without breakdown, i.e., without starting toconduct electricity through partial ionisation.) For safety, we shouldlike the field never to exceed, say 10% of the dielectric strength.What minimum area of the plates is required to have a capacitanceof 50 pF?

Answer 1

Given: The potential rating of the parallel plate capacitor is 1000 V, the dielectric constant of a material is 3, the dielectric strength is 10 7  V/m and the electric field intensity is 10% of 10 7 or 10 6  V/m and the capacitance of the parallel plate capacitor is 50 pF.

The distance between the plates is given as,

d= V E

Where, the potential rating is V and the electric intensity is E.

By substituting the given values in the above formula, we get

d= 1000 10 6 = 10 −3  m

The capacitance is given as,

C= ε 0 ε r A d

Where, the area of each plate is A the dielectric constant of the material is ε r .

By substituting the given values in above formula, we get

C= ε 0 ε r A d 50× 10 −12 = 8.85×3× 10 −12 ×A 10 −3 A=1.89× 10 −3   m 2 =18.9  cm 2

Thus, the area of each plate is about 18.9  cm 2 or 19  cm 2 .