Question

A parallel plate capacitor has plates of area $200\,cm^{2}$ and separation $0.05\, cm$. It has been charged to a potential difference of $300\, V$, the energy of the capacitor is :

• A. $ 2.0\times 10^{-6}J$
• B. $ 2.4\times 10^{-5} J$
• C. $ 0.8\times 10^{-5}J$
• D. $ 1.6\times 10^{-5}J$

Answer 1

Answer: (D)

The capacitance of a parallel plate capacitor of plate area $A$ and plate separated by distance d is given by
$C=\frac{\varepsilon _{0}A}{d}$
Given, $A=200\,cm^{2}=200\times 10^{-4}m^{2}$,
$d=0.05\,cm=0.05\times 10^{-2}m$,
$\varepsilon _{0}=8.86\times 10^{-12}C^{2}/N-m^{2}$
$\therefore C=\frac{8.86\times 10^{-12}\times 200\times 10^{-4}}{0.05\times 10^{-2}}$
$ =3.54\times 10^{-10}\mu F$
Energy stored in capacitor is
$U=\frac{1}{2}CV^{2}=\frac{1}{2}\times 3.54\times 10^{-10}\times 9\times 10^{4}$
$=1.6\times 10^{-5}J$
Note: This energy resides in the electric field created between the plates of the charged capacitor.