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 break down, i.e, without starting to conduct electrically through partial ionization. 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 , V=1kv =1000 V
dielectric constant of a meterial `in_(r)=3`
dielectric strength `=10^(7)V//m`
For safety, the field intensity never exceeds 10 % of the dielectric strength. Hence, electric field intensity, `E=10 % ` of `10^(7) =10^(6) V//m`
Capactiance of the parallel plate capacitor , C=50 pF =`50xx10^(-12)F`
distance between the plates in given by ,
`d=V/E`
`=1000/(10^(6))=10^(-3) m`
Capacitance is given by the relation,
`C=(in_(o)in_(r)A)/d`
where, A= area of each plate
`in_(0)`=permitivity of free space =`8.85xx10^(-12) N^(-1)C^(2)m^(-2)`
`:. A=(Cd)/(in_(0)in_(r))`
`=(50xx10^(-12)xx10^(-3))/(8.85xx10^(-17)xx3) ~~19cm^(2)`
Hence, the area of each plate is about `19 cm^(2)`.