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 `10^(7) 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?

Explanation

Potential rating of a parallel plate capacitor with dielectric = 1 kV = `10^(3)` V
For safety the field intensity never exceeds 10 % of the dielectric strength. Hence electric field intensity = 10 % of `10^(7)`
`= 10(7) xx(10)/(100)`
`=10^(6) Vm^(-1)`
The minimum necessary distance between two plates
`E = (V)/(d) implies d =(V)/(E) = (10^(3))/(10^(6))= 10^(-3)` m
Capacitance of parallel plate capacitor
`C = (kin_(0)A)/(d)`
`:. A = (Cd)/(KE_(0))=(50xx10^(-12)xx10^(-3))/(3xx8.85xx10^(-12))`
`:. A = 18.8 xx10^(-4) m^(2)`
`:. A = 19 cm^(2)`