A parallel plate capacitor is to be desigined with a voltage rating 1kV, 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 partical 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

Here V = 1kv = 1000 volt, `K = epsilon_(r ) = 3`
Dielectric strength `= 10^(7)` V/m
As electric field at the most should be 10% of dielectric strength due to reasons of saftely.
`E = 10% "of" 10^(7) V//m = 10^(6) V//m A= ?`
`C = 50 pF = 50 xx 10^(-12) F`
As `E = (V)/(d)`
`:. d = (V)/(E) = (10^(3))/(10^(6)) = 10^(-3) m`
Now `C = (epsilon_(0) epsilon_(1) A)/(d)`
`A = (Cd)/(epsilon_(0) epsilon_(r )) = (50 xx 10^(-12) xx 10^(-3))/(8.85 xx 10^(-12) xx 3)`
`1.9 xx 10^(-3) m^(2)`