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 ?

Answer

`19 cm^(2)`

Explanation

Here, `V = 1 kV = 10^(3)` volt, `K = 3`,
dielctric strength `= 10^(7) Vm^(-1)`
Electric field, `E = 10% xx10^(7) V//m = 10^(6) V//m`
`A = ?, C = 50 pF = 50xx10^(-12) F`
As `C = (K in_(0) A)/(d)` and `d = (V)/(E)`
`:. C = (K in_(0) AE)/(V)`
`A = (CV)/(K in_(0) E)`
`A = (50xx10^(-12)xx10^(3))/(3xx8.85xx10^(-12)xx10^(6)) = 19xx10^(-4) m^(2)`
`= 19 cm^(2)`