Home NCERT Solutions NCERT Exemplar CBSE Sample Papers NCERT Books Class 12 Class 10
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?


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,



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.

Popular Questions of Class 12 Physics

Recently Viewed Questions of Class 12 Physics

1 Comment(s) on this Question

Write a Comment: