Question

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. $E^3$

Answer 1

The correct option is B $E^2$

Electrostatic potential energy $U=\frac{C{V}^2}{2}$
As $C=\frac{{ϵ}_{0}A}{d},V=Ed$
$\Rightarrow U=\frac{{ϵ}_0{E}^2}{2}\left(Ad\right)$
$\Rightarrow \frac{U}{\left(Ad\right)}=\frac{{ϵ}_0{E}^2}{2}$

Electrostatic energy per unit volume $\frac{1}{2}{ϵ}_0{E}^2$
Electrostatic energy density is $\propto E^2$