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?
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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