In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{-3} \mathrm{~m}^{2}$ and the distance between the plates is $3 \mathrm{~mm}$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100 \mathrm{~V}$ supply, what is the charge on each plate of the capacitor?
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 $1.771 \times 10^{-9} \mathrm{C}$.