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 `10^7V m^(-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

`:.` Dielectric strength `= 10^7 Vm^(-1)` and field up to 10% of the dielectric strength is to be applied.
Hence, electric field employed E = 10% of `10^7 =10^6 V m^(-1)`
As voltage rating V = 1 kV = 1000V and for a capacitor `E = V/d`
`:. d = V/E = 1000/10^6 =10^(-3)m`
As `C = (K epsi_0A)/d` and here`C = 50 pF = 50 xx 10^(-12) F and K = 3`
`:.` Area of the plates `A = (C.d)/(K epsi_0) = (50 xx 10^(-12) xx 10^(-3))/(3 xx 8.85 xx 10^(-12)) = 1.9 xx 10^(-3) m^2 = 19 cm^2`.