Solution :
Area of each plate of the parallel plate capacitor, $A = 6 \times 10^{−3} m^2 $
Distance between the plates, $d = 3\; mm = 3 \times 10^{−3 }m$
Supply voltage, V = 100 V
Capacitance C of a parallel plate capacitor is given by, $C= \large\frac{E_0 A}{d}$
where
$\in_0$= Permittivity of free space $= 8.854 \times 10^{-12} N^{-1}m^{-2}C^{-2}$
$C= \large\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}$
$\qquad= 17.71 \times 10^{-12}F$
$\qquad =17.71$
So, charge on each plate of the capacitor
$q=VC= 100 \times 17.71 \times 10^{-12}C= 1.771\times 10^{-9}\;C$
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is $1.771 \times 10^{−9} C.$