Question

How much energy is stored in $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$of air due to the “fair weather” electric field of magnitude 150 V/m?

Short answer

The energy stored in $1.00{\mathrm{m}}^3$ of air due to the “fair weather” electric field of magnitude 150 V/m is $9.96\times {10}^{-8}J$ .

Step-by-step solution

The given data

a) Volume of air, $V=1.00{\mathrm{m}}^3$

b) Magnitude of electric field,E =150 V/m

Understanding the concept of the stored energy

We need to use the equation of the energy density and the energy stored per unit volume in the magnitude E of the electric field. Using the given formulae, we can get the required value of the stored energy.

Formulae:

The energy density of an electric field,$u=\frac{1}{2}{\epsilon }_0{E}^2$ ...(i)

The energy stored per unit volume in the electric field,$u=\frac{U}{V}$ ...(ii)

Step 3: Calculation of the stored energy

Using equations (i) and (ii), we can get the energy stored in the electric field as follows: (${\mathrm{\epsilon }}_0=8.85\times {10}^{-12}\frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}$is permittivity of free space)

$$\begin{gathered} U=\frac{1}{2}{\epsilon }_0{E}^{2}V \\ =\frac{1}{2}\left(8.85\times {10}^{-12}\frac{C^2}{N.m^2}\right){\left(150\frac{V}{m}\right)}^2\left(1.00m^3\right) \\ =9.96\times {10}^{-8}J \end{gathered}$$

Hence, the value of the stored energy is $9.96\times {10}^{-8}\mathrm{J}$.