Question

Derive an expression for the energy stored per unit volume (energy density) in an electric field. Or obtain an expression for energy density of a medium.

Answer 1

Hint:Energy density of a medium is defined as the potential energy within the medium per unit volume of the medium. For a constant electric field or constant potential energy through the volume of the medium, energy density of the medium is equal to the total energy divided by the volume of the medium.

Formula used:
$dU=\dfrac{1}{2}C{{V}^{2}}$
dU is potential energy, C is capacitance, V is potential difference.
$V=Edl$ E is electric field, dl is width
$C=\dfrac{dAk{{\varepsilon }_{0}}}{dl}$
 dA is the area of face and k is dielectric constant.

Complete step by step answer:
Now, consider a medium with a dielectric constant k placed in an electric field E (not necessarily constant). The medium can be considered of many small cubes of length dl.
Consider one small cube of the medium. Since the volume of the cube is very small, the electric field inside it can be considered as constant. Let the capacitance of this cube be C and the potential difference across two opposite faces be V. Then the potential energy stored within the cube is given as $dU=\dfrac{1}{2}C{{V}^{2}}$ …. (i)
The potential difference for constant electric field across length dl is given as $V=Edl$.
Substitute V in (i).
$dU=\dfrac{1}{2}C{{\left( Edl \right)}^{2}}\\
\Rightarrow dU =\dfrac{1}{2}C{{E}^{2}}{{(dl)}^{2}}$ …. (ii).
The capacitance of the cube is given as $C=\dfrac{dAk{{\varepsilon }_{0}}}{dl}$, where dA is the area of the opposite faces.
In this case, $dA={{\left( dl \right)}^{2}}$.
$\Rightarrow C=\dfrac{{{\left( dl \right)}^{2}}k{{\varepsilon }_{0}}}{dl}=dlk{{\varepsilon }_{0}}$.
Substitute the value of C in (ii).
$dU=\dfrac{1}{2}\left( dlk{{\varepsilon }_{0}} \right){{E}^{2}}{{(dl)}^{2}}\\
\Rightarrow dU =\dfrac{1}{2}{{\varepsilon }_{0}}k{{E}^{2}}{{(dl)}^{3}}$.
Now, the volume of the cube is $dV={{(dl)}^{3}}$.
The electric energy density for the cube is,
$u=\dfrac{dU}{dV}=\dfrac{\dfrac{1}{2}{{\varepsilon }_{0}}k{{E}^{2}}{{(dl)}^{3}}}{dV}\\
\Rightarrow u=\dfrac{\dfrac{1}{2}{{\varepsilon }_{0}}k{{E}^{2}}{{(dl)}^{3}}}{{{(dl)}^{3}}}$
$\therefore u=\dfrac{1}{2}{{\varepsilon }_{0}}k{{E}^{2}}$.

Hence, we found an expression for the energy charge density for a medium with an electric field inside it.

Note: The term is ${{\varepsilon }_{0}}k$ is written as ${{\varepsilon }_{0}}k=\varepsilon $, where is $\varepsilon $ is called permittivity of the medium.
Therefore, the energy density of a medium can be written as $u=\dfrac{1}{2}\varepsilon {{E}^{2}}$.
This expression of energy density is helpful in finding the potential energy in the case of constant electric fields.