A parallel plate capacitor is to be designed with a voltage rating 1kV, using a material of dielectric constant 3 and dielectric strength about `10^(7)Vm^(-1)`. 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

Given V=1`xx10^(3)V,epsi_(r)=K=3,E=10^(7)Vm^(-1)`
dielectric break down at 10% of `E=(10)/(100)xx10^(7)Vm^(-1)=1xx10^(6)Vm^(-1)and C=50xx10^(-12)F`.
By definition, `E=(V)/(d), i.e., d=(V)/(E)=(10^(3))/(10^(6))m`
i.e., the distance between the plates 'd'=`10^(-3)m`
we also know that `C=(epsi_(0)epsi_(r)A)/(d)`
`therefore`Area of the plates`A=(Cd)/(epsi_(0)epsi_(r))`
i.e., `A=(50xx10^(-12)xx10^(-13))/(8.854xx10^(-12)xx3)m^(2)=1.8xx10^(-3)m^(2) or 18.8cm^(2)`