Q. 334.3( 6 Votes )

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

Dielectric constant of a material(ϵr) = 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

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