Q. Consider a parallel plate capacitor with plates $20\ cm$ by $20\ cm$ and separated by 2 mm. The dielectric constant of the material between the plates is 5. The plates are connected to a voltage sources of 500 $V.$ The energy density of the field between the plates will be close to
A) $2.65\ J/m^{3}$
B) $1.95\ J/m^{3}$
C) $1.38\ J/m^{3}$
D) $0.69\ J/m^{3}$
Explanation
(c) $\frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9} \Rightarrow \varepsilon_{0} = \frac{1}{4\pi \times 9 \times 10^{9}}$
Electric field density or energy per unit volume is
$u = \frac{k\varepsilon_{0}}{2}\left( \frac{V}{d} \right)^{2}$
$k = 5,\ V = 500\ volts,\ d = 2 \times 10^{- 3}m$
$\therefore u = \frac{5}{2} \times \frac{1}{4\pi \times 9 \times 10^{9}} \times \frac{(500)^{2}}{\left( 2 \times 10^{- 3} \right)^{2}} = 1.38\ Jm^{- 3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.