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 107 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?
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, 
=3
Dielectric strength = 107 V/m
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, E = 10% of 107 = 106 V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 × 10−12 F
Distance between the plates is given by,
$d=\frac{V}{E}$
$=\frac{1000}{10^{6}}=10^{-3} \mathrm{~m}$
Capacitance is given by the relation,
$C=\frac{\in_{0} \in, A}{d}$
Where,
A = Area of each plate
$€_{0}=$ Permittivity of free space $=8.85 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$
$\therefore A=\frac{C d}{\epsilon_{0} \in_{r}}$
$=\frac{50 \times 10^{-12} \times 10^{-3}}{8.85 \times 10^{-12} \times 3} \approx 19 \mathrm{~cm}^{2}$
Hence, the area of each plate is about $19 \mathrm{~cm}^{2}$.