In a parallel plate capacitor with air between the plates,

Solution:

Area of each plate of the parallel plate capacitor, A = 6 × 10−3 m2

Distance between the plates, d = 3 mm = 3 × 10−3 m

Supply voltage, V = 100 V

Capacitance C of a parallel plate capacitor is given by,

$C=\frac{\in_{0} A}{d}$

Where,

$\epsilon_{0}=$ Permittivity of free space

$=8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \mathrm{C}^{-2}$

$\therefore C=\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}$

$=17.71 \times 10^{-12} \mathrm{~F}$

$=17.71 \mathrm{pF}$

Potential $V$ is related with the charge $q$ and capacitance $C$ as

$V=\frac{q}{C}$

$\therefore q=V C$

$=100 \times 17.71 \times 10^{-12}$

$=1.771 \times 10^{-9} \mathrm{C}$

Therefore, capacitance of the capacitor is $17.71 \mathrm{pF}$ and charge on each plate is $1.771 \times 10^{-9} \mathrm{C}$.