Solution :
Potential rating of a parallel plate capacitor, $V = 1\; kV = 1000\; V $
Dielectric constant of a material,$\in =3$
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 $
Capacitance of the parallel plate capacitor,
$C = 50 \;pF = 50 \times 10^{−12} F $
Distance between the plates is given by,
$d= \large\frac{V}{E}$
$\quad= \large\frac{1000}{10^6}$
$\quad=10^{-3}\;m$
Capacitane is given by the relation
$c= \large\frac{\in_0 \in _rA}{d}$
Where, A = Area of each plate
$\in_0 $ = Permittivity of free space = $8.85 \times 10^{-12}N^{-1}C^{2} m^{-2}$
$A=\large\frac{Cd}{\in_0 \in_r}$
$\qquad= \large\frac{50 \times 10^{-12} \times 10^{-3}}{8.85 \times 10^{-12} \times 3}$
Hence, the area of each plate is about $19\; cm^2. $