Q. A parallel plate capacitor having a plate separation of $2\ mm$ is charged by connecting it to a $300\ V$ supply. The energy density is
A) $0.01\ J/m^{3}$
B) $0.1\ J/m^{3}$
C) $1.0\ J/m^{3}$
D) $10\ J/m^{3}$
Explanation
(b) The energy density of parallel plate capacitor is given by
$U = \frac{1}{2}\varepsilon_{0}E^{2} = \frac{1}{2}\varepsilon_{0}\left( \frac{V}{d} \right)^{2}$
$= \frac{1}{2} \times 8.85 \times 10^{- 12}C^{2}/Nm^{2} \times \left( \frac{300\ volt}{2 \times 10^{- 3}m} \right)^{2} = 0.1\ J/m^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.