# 2.8) 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?

Given,

Area of the capacitor plate $A$ = $6 \times 10^{-3}m^{2}$

Distance between the plates $d=3mm$

Now,

Capacitance of the parallel plate capacitor is

$C=\frac{\epsilon_0 A}{d}$

$\epsilon _0$= permittivity of  free space =  $8.854*10^{-12}N^{-1}m^{-2}C^{_2}$

putting all know value we get,

$C=\frac{8.854*10^{-12}*6*10^{-3}}{3*10^{-3}}=17.71*10^{-12}F=17.71pF$

Hence capacitance of the capacitor is 17.71pF.

Now,

Charge on the plate of the capacitor :

$Q=CV=17.71*10^{-12}*100=1.771*10^{-9}C$

Hence charge on each plate of the capacitor is $1.771*10^{-9}C$.

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