Question

Assume that a stationary electron is a point of charge.

(a)What is the energy density of its electric field at radial distances $\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathit{m}\mathit{m}$ ?

(b)What is the energy density of its electric field at radial distances$\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathit{\mu }\mathit{m}\mathbf{?}$

(c)What is the energy density of its electric field at radial distances$\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathit{n}\mathit{m}\mathbf{?}$

(d)What is the energy density of its electric field at radial distances $\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathit{p}\mathit{m}\mathbf{?}$

(e)What is uin the limit as $\mathit{r}\mathbf{\to }\mathbf{0}$?

Short answer

1. Energy density at radial distance $r=1.00mm$is $9.16x{10}^{-18}J/m^3$.
2. Energy density at radial distance$r=1.00\mu m$is$9.16x{10}^{-6}J/m^3$.
3. Energy density at radial distance$r=1.00nm$is$9.16x{10}^{6}J/m^3$.
4. Energy density at radial distance $r=1.00pm$is $9.16x{10}^{18}J/m^3$.
5. Energy density in the limit as $r\to 0$is $\infty$.

Step-by-step solution

The given data

1. A stationary point charge of electron is assumed.
2. Distance,$r=1.00mm$
3. Distance,$r=1.00\mu m$
4. Distance,$r=1.00nm$
5. Distance,$r=1.00pm$

Understanding the concept of the energy density

The quantity of energy that is contained in a system or area of space per unit volume is known as the energy density. We use the equation of energy density stored in the electrostatic field. Substituting the value of the magnitude of the electric field in density, we can calculate the energy density for the different radial distances.

Formulae:

The energy stored per unit volume that is energy density ( u ) between the capacitors,

$u=\frac{1}{2}{\epsilon }_0{E}^2$ …(i)

The electric field at a point due to stored charge, $E=\frac{q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$ …(ii)

Where,

q is charge of electron $q=1.6\times {10}^{-19}C$

${\epsilon }_0=8.85\times {10}^{-12}\frac{C^2}{N.m^2}$ is permittivity of free space

(a) Calculation of the energy density at  r = 1 mm

For the formula of energy density due to electric field, we substitute the value of equation (ii) in equation (i) as follows:

$$\begin{gathered} u=\frac{1}{2}{\epsilon }_0{\left(\frac{q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\right)}^2 \\ =\frac{q}{32{\pi }^2{\epsilon }_0\mathrm{r}4} \end{gathered}$$

…(iii)

Substituting all the given values in equation (iii), we get the energy density at r = 1.0 mm as follows:

$$\begin{gathered} u=\frac{{\left(1.6\times {10}^{-19}C\right)}^2}{32{\mathrm{\pi }}^2\times 8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}}\times {\left({10}^{-3}\mathrm{m}\right)}^4} \\ =\frac{2.56\times {10}^{-38}{C}^2}{2.795\times {10}^{-21}{\displaystyle \frac{C^2.m^2}{N}}} \\ =9.16\times {10}^{-18}J/m^3 \end{gathered}$$

Hence, the value of the energy stored per volume is $9.16\times {10}^{-18}J/m^3$.

(b) Calculation of the energy density at r=1 μm

Substituting all the given values in equation (a), we get the energy density at $r=1.00\mu m$as follows:

$$\begin{gathered} u=\frac{{\left(1.6\times {10}^{-19}C\right)}^2}{32{\mathrm{\pi }}^2\times 8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}}\times {\left({10}^{-6}\right)}^4} \\ =\frac{2.56\times {10}^{-38}{C}^2}{2.795\times {10}^{-33}{\displaystyle \frac{C^2.m^2}{N}}} \\ =9.16\times {10}^{-6}J/m^3 \end{gathered}$$

Hence, the value of the energy stored per volume is $9.16\times {10}^{-6}J/m^3$.

(c) Calculation of the energy density at r = 1 nm

Substituting all the given values in equation (a), we get the energy density at r = 1.00 nm as follows:

$$\begin{gathered} u=\frac{{\left(1.6\times {10}^{-19}C\right)}^2}{32{\mathrm{\pi }}^2\times 8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}}\times {\left({10}^{-9}\right)}^4} \\ =\frac{2.56\times {10}^{-38}{C}^2}{2.795\times {10}^{-45}{\displaystyle \frac{C^2.m^2}{N}}} \\ =9.16\times {10}^{6}J/m^3 \end{gathered}$$

Hence, the value of the energy stored per volume is $9.16\times {10}^{-6}J/m^3$.

(d) Calculation of the energy density at r = 1 pm

Substituting all the given values in equation (a), we get the energy density at r = 1 pm as follows:

$$\begin{gathered} u=\frac{{\left(1.6\times {10}^{-19}C\right)}^2}{32{\mathrm{\pi }}^2\times 8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}}\times {\left({10}^{-12}\right)}^4} \\ =\frac{2.56\times {10}^{-38}}{2.795\times {10}^{-57}} \\ =9.16\times {10}^{18}J/m^3 \end{gathered}$$

Hence, the value of the energy stored per volume is $9.16\times {10}^{18}J/m^3$.

(e) Calculation of the energy density

From equation (a), we can get the relation of energy density to distance as follows:

$u\propto r^{-4}$

As the value of r approaches $r\to 0$, the value of the energy density is $\infty$.