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

Hint
Electrostatic energy density (u) is given as-
$u = \dfrac{1}{2}{\varepsilon _\circ }{E^2}$ -- (1)
Where, ${\varepsilon_\circ }$ = permittivity of free space
$E$ = electric field intensity.
From the equation we can find out the dependence of electrostatic energy density in electrostatic fields.

Step By Step Solution
To derive equation (1) we have to consider the case of parallel plate capacitor.
+Q and –Q charge is distributed into two plates of capacitor respectively. A is the area of plates and d is the distance between them. So electric field due either of the plate is given by
$\vec E = \dfrac{Q}{{2A{\varepsilon _ \circ }}}$ -- (2)
So force due to either of the plate on the other plate is given as
$\vec F = Q\vec E$ -- (3)
Put the value of equation (3) in equation (2) we get
$\vec F = \dfrac{{{Q^2}}}{{2A{\varepsilon _ \circ }}}$ -- (4)
Since these plates are of opposite charges so they attract each other. So to make the given arrangement, external agents must do work against the attractive force between them. The force due to an external agent must be in the opposite direction to the attractive force between them, such that work done by an external agent is equal to potential energy stored in the system (body cannot accelerate or it is in quasi static equilibrium).
Work done by the external agent is given as
$W = Fd$ -- (5)
Where d = distance between the plate
Put the value of equation (4) in equation (5) we get
$W = \dfrac{{{Q^2}d}}{{2A{\varepsilon _ \circ }}}$
This work done is stored as potential energy of the system
$U = W = \dfrac{{{Q^2}d}}{{2A{\varepsilon _ \circ }}}$
Energy density is defined as energy per unit volume and volume of the system is product of area of either of the plate and distance between the plates.
Energy density (u) is given as
$u = \dfrac{U}{{Ad}} = \dfrac{{{Q^2}}}{{2{A^2}{\varepsilon _ \circ }}}$
Multiply by ${\varepsilon _\circ }$ in denominator and numerator we get
$u = \dfrac{1}{2}{\varepsilon _ \circ }\dfrac{{{Q^2}}}{{{A^2}{\varepsilon _ \circ }^2}}$
On further simplification we get
$u = \dfrac{1}{2}{\varepsilon _ \circ }{(\dfrac{Q}{{A{\varepsilon _ \circ }}})^2}$
$\dfrac{Q}{{A{\varepsilon _ \circ }}}= E$ (net electric field between the plates of capacitor.)
$\Rightarrow u = \dfrac{1}{2}{\varepsilon _ \circ }{(E)^2}$.
From equation (1) it is clear that electrostatic energy density is directly proportional to ${E^2}$ .
Correct option: (B) ${E^2}$

Note
Electrostatic Energy Density is the amount of energy stored in a given system or field of charge per unit mass. It depends on field intensity.