Question

The energy stored per unit volume in an electric field of strength $E$ V/m in a medium of dielectric constant $K\left( {inJ{m^{ - 3}}} \right)$ is:
A) $\dfrac{1}{2}{\varepsilon _0}{E^2}$
B) $\dfrac{1}{2}K{\varepsilon _0}{E^2}$
C) $\dfrac{1}{2} \cdot \dfrac{{{\varepsilon _0}{E^2}}}{K}$
D) $\dfrac{1}{2}{K^2}{\varepsilon _0}E$

Answer 1

We know that, when a material is stored in an electric field, it stores electric energy. When that material is placed in a medium other than vacuum, then the dielectric constant of that medium changes the electric energy stored in it. For any material stored in vacuum carries energy which is given by $U = \dfrac{1}{2}{\varepsilon _0}{E^2}V$ where electric field of strength is given $E$ and the volume is given by $V$. Now, in the given question we are provided with the per unit volume that we can easily calculate by dividing the energy by the volume. Then, we can simply find the energy in the given medium by the dielectric constant.

Complete step by step solution:
According to the question, we are given:
$E$ = electric field of strength
$K$ = dielectric constant of the medium
Now, to calculate the energy stored we can use the formula which is:
$U = \dfrac{1}{2}{\varepsilon _0}{E^2}V$
Where, ${\varepsilon _0}$ is the permittivity of free space or vacuum
$E$ is the electric field of strength in volt per meter
$V$ is the volume of field

Now, we need to find the energy stored per unit volume in the given electric field.
So, we will divide the above energy stored by the volume of the field.
We get,
$U = \dfrac{1}{2}{\varepsilon _0}{E^2}$
The above energy stored is for vacuum and we need to find the energy stored in a medium whose dielectric constant is given $K$ $J{m^{ - 3}}$.
We have to replace vacuum with the given medium. So the permittivity of the medium will be .
Therefore, the energy stored per unit volume in an electric field of strength $E$ volt per meter in a medium of dielectric constant $K$ $J{m^{ - 3}}$ is:
$U = \dfrac{1}{2}K{\varepsilon _0}{E^2}$

Hence, option (B) is correct.

Note: These types of questions can be solved by using the derivation of energy stored in the capacitor. We should always be cautious doing such calculations because a small calculation mistake can lead us to an incorrect answer. Also, we need to be precise with the details given in the question.