Question

Energy stored per unit volume of a parallel plate capacitor having plate area $A$ and plate separation d when charged to a potential of $V$ volts is (air space in between the plates)

• A. $\frac{1}{2}{{C}^{2}}{{V}^{2}}$
• B. $\frac{{{q}^{2}}}{4\,C}$
• C. $\frac{1}{2}{{\epsilon}_{0}}\left( \frac{V}{d} \right)$
• D. $\frac{1}{2}{{\epsilon}_{0}}\left( \frac{{{V}^{2}}}{{{d}^{2}}} \right)$

Answer 1

Answer: (D)

$\frac{1}{2}{{\epsilon}_{0}}\left( \frac{{{V}^{2}}}{{{d}^{2}}} \right)$

Answer 2

Energy stored in capacitor of capacitance $C =\frac{1}{2} CV ^{2}$
Capacitance of a parallel plate capacitor $= C =\frac{ A \epsilon_{0}}{ d }$
Volume of parallel plate capacitor $=A d$
Thus energy stored per unit volume $=\frac{\frac{1}{2} CV ^{2}}{ Ad }$
$=\frac{1}{2} \epsilon_{0} \frac{ V ^{2}}{ d ^{2}}$