A parallel plate capacitor is to be designed with a voltage rating 1 KV using a material of dielectrical constant 3 and dielectric strength about `10^(7) Vm^(-1)`. [Dielectric strength is the maximum electric field a material can tolerate without break down, i.e, without starting to conduct electrically 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 ?

Explanation

As electric field should not be more than `10%` of the dielectric strength, the required electric field can be calculated as `E=10%` of `10^(7)Vm^(-1)=10^(6)Vm^(-1)`.
Plate separation for the capacitor will be:
`d=(V)/(E )=(10^(3))/(10^(6))=10^(-3)`
We know, `C=epsilon_(0)epsilon_(r ) (A)/(d)`
`rArr A=(dC)/(epsilon_(0)epsilon_(r ))=(10^(-3)xx50xx10^(-12))/(3xx8.85xx10^(-12))=18.87xx10^(-4)m^(2)`