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

(N/A) Given:
Area of each plate,$A = 6 \times 10^{-3} \, m^{2}$
Distance between the plates,$d = 3 \, mm = 3 \times 10^{-3} \, m$
Supply voltage,$V = 100 \, V$
Permittivity of free space,$\varepsilon_{0} = 8.854 \times 10^{-12} \, F/m$
$1$. Calculation of Capacitance $(C)$:
The formula for the capacitance of a parallel plate capacitor is $C = \frac{\varepsilon_{0} A}{d}$.
Substituting the values:
$C = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}$
$C = 8.854 \times 10^{-12} \times 2 = 17.708 \times 10^{-12} \, F \approx 17.71 \, pF$.
$2$. Calculation of Charge $(q)$:
The charge on each plate is given by $q = CV$.
$q = 17.71 \times 10^{-12} \, F \times 100 \, V$
$q = 1.771 \times 10^{-9} \, C$.
Thus,the capacitance is $17.71 \, pF$ and the charge on each plate is $1.771 \times 10^{-9} \, C$.