Question

The energy density in a parallel plate capacitor is given as $2.1 \times 10^{-9} J/m^3$. The value of the electric field in the region between the plates is
(A) $2.1 NC^{-1}$
(B) $21.6 NC^{-1}$
(C) $72 NC^{-1}$
(D) $8.4 NC^{-1}$

Answer 1

Hint
As here energy density is given and we have to find the electric field between the plates so we can use the direct formula of energy density and electric field i.e. $\eta = \dfrac{1}{2}{ \in _0}{E^2}$, where $E$ is the electric field and $η$ is the energy density and $ϵ_0$ is the permittivity of the free space.

Complete step by step answer
It is given that that the energy density between the parallel plate capacitor$\eta = 2.1 \times {10^{ - 9}}J/{m^3}$ and we have to find the electric field between the plates then we know that-
$\Rightarrow \eta = \dfrac{1}{2}{ \in _0}{E^2}$……..(1) where, ϵ0 is the permittivity of free space ie. ${ \in _0} = 8.85 \times {10^{ - 12}}$faraday/m
E is the electric field between the plates of capacitor
Put the values in the equation (1), we get
$\Rightarrow 2.1 \times {10^{ - 9}} = \dfrac{1}{2} \times 8.85 \times {10^{ - 12}} \times {E^2}$
$\Rightarrow {E^2} = \dfrac{{4.2 \times {{10}^{ - 9}}}}{{8.85 \times {{10}^{ - 12}}}} = 0.47 \times {10^3}$
$\Rightarrow {E^2} = 470$
$\Rightarrow E = 21.6N/C$.
Hence, the electric field between the plates of the capacitor is $21.6 N/C$.
Option (B) is the correct option.

Note
Energy density is the amount of energy stored in a given system or region of space per unit volume. It's S.I. unit is $J/m^3$. Permittivity is defined as the ability of the substance to store electrical energy in an electrical field. Here, just remember that the value of permittivity for free space is used in the formula of energy density and also notice the units of each term.