Q. If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to
A) $E$
B) $E^{2}$
C) $1/E^{2}$
D) $E^{3}$
Explanation
(b) Electrostatic energy density $\frac{dU}{dV} = \frac{1}{2}K\varepsilon_{0}E^{2}$
$\therefore\frac{dU}{dV} \propto E^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.