A sphere of radius R carries a charge density $\mathit{\rho }\left(r\right)\mathbf{=}\mathit{k}\mathit{r}$(where $\mathit{k}$ is a constant). Find the energy of the configuration. Check your answer by calculating it in at least two different ways.

Let’s consider that, R is the radius of uniformly charged sphere, the charge density of the sphere is,

$\mathit{\rho }\left(r\right)\mathbf{=}\mathit{k}\mathit{r}$ ……. (1)

Here, k is constant.

Assume a point $\mathit{r}\mathbf{<}\mathit{R}$. Consider dris the one elementary part of thickness, the volume of its elementary part is $\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{dr}$. Therefore the charge of this elementary part is

$\mathit{\rho }\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{\rho }\mathbf{\right)}\mathbf{\left(}\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{dr}\mathbf{\right)}$ …… (2)

Substitute $\mathbf{\rho }\mathbf{=}\mathbf{kr}$in equation (2)

$$\begin{gathered} \mathbf{\rho }\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{kr}\mathbf{\right)}\mathbf{\left(}\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{dr}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{k}\mathbf{4}{\mathbf{\pi r}}^{\mathbf{3}}\mathbf{dr} \end{gathered}$$

Write the expression for total charge enclosed within sphere.

$$\begin{gathered} {\mathit{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}\mathbf{l}\mathbf{o}\mathbf{s}\mathbf{e}\mathbf{d}}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{r}}{\mathbf{\int }}}\mathit{\rho }\mathit{d}\mathit{\zeta } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{r}}{\mathbf{\int }}}\mathit{k}\mathit{r}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathit{\pi }\mathit{k}\underset{\mathbf{0}}{\overset{\mathbf{r}}{\mathbf{\int }}}{\mathit{r}}^{\mathbf{3}}\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathit{\pi }\mathit{k}{\left(\frac{r^4}{4}\right)}_{\mathbf{0}}^{\mathbf{r}} \end{gathered}$$

Therefore, total charge enclosed within the sphere is,

${\mathit{q}}_{\mathbf{e}\mathbf{n}\mathbf{c}\mathbf{l}\mathbf{o}\mathbf{s}\mathbf{e}\mathbf{d}}\mathbf{=}{\mathbf{\pi kr}}^{\mathbf{4}}$.

According to statement of Gauss’s law, the electric field is directly proportional to the ${\mathit{q}}_{\mathbf{enclosed}}$in to the Gaussian sphere.

Write the expression for electric flux of the sphere.

$$\begin{gathered} {\mathit{\Phi }}_{\mathbf{E}}\mathbf{=}\mathbf{\int }{\mathit{E}}_{\mathbf{1}}\mathit{d}\mathit{A} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{i}\mathbf{n}\mathbf{s}\mathbf{i}\mathbf{d}\mathbf{e}}}{{\mathbf{\epsilon }}_{\mathbf{0}}} \end{gathered}$$ …… (3)

Write the formula for the area with the Gaussian surface of radius r.

$\mathit{A}\mathbf{=}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}$

Substitute $\mathit{A}\mathbf{=}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}$and ${\mathit{q}}_{\mathbf{encloses}}\mathbf{=}\mathit{\pi }\mathit{\pi }\mathit{k}{\mathit{r}}^{\mathbf{4}}$in equation (3),

$$\begin{gathered} \mathit{E}\mathbf{\left(}\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{\pi kr}}^{\mathbf{4}}}{{\mathbf{\epsilon }}_{\mathbf{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{E}\mathbf{=}\frac{\mathbf{k}{\mathbf{r}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}} \end{gathered}$$

Assume a point $\mathit{r}\mathbf{<}\mathit{R}$.

Write the expression for total charge enclosed within the sphere.

$$\begin{gathered} {\mathit{q}}_{\mathbf{enclosed}}\mathbf{=}\stackrel{\mathbf{R}}{\mathbf{\int }}\mathit{\rho }\mathit{d}\mathit{\zeta } \\ \mathbf{=}\underset{\mathbf{r}\mathbf{=}\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}\mathit{k}\mathit{r}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathit{\pi }\mathit{k}\underset{\mathbf{r}\mathbf{=}\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}{\mathit{r}}^{\mathbf{3}}\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathit{\pi }\mathit{k}{\mathbf{\left(}\frac{{\mathbf{r}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{\right)}}_{\mathbf{0}}^{\mathbf{R}} \end{gathered}$$

Solve further as,

${\mathit{q}}_{\mathbf{enclosed}}\mathbf{=}\mathit{\pi }\mathit{k}{\mathit{R}}^{\mathbf{4}}$

Substitute the value $\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}$for $\mathit{A}$and $\mathit{\pi }\mathit{k}{\mathit{r}}^{\mathbf{4}}$for ${\mathit{q}}_{\mathbf{enclosed}}$in equation (3),

$$\begin{gathered} \mathbf{E}\mathbf{\left(}\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{\pi kR}}^{\mathbf{4}}}{{\mathbf{\epsilon }}_{\mathbf{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{E}\mathbf{=}\frac{{\mathbf{kR}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}} \end{gathered}$$

Hence the electric filed is,

$\mathbf{E}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{ll}\frac{{\mathbf{kr}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}}& \mathbf{r}\mathbf{<}\mathbf{R}\\ \frac{{\mathbf{kR}}^{\mathbf{4}}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}& \mathbf{r}\mathbf{>}\mathbf{R}\end{array}$

Method 1:

Write the expression for the energy configuration.

$\mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\int }{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{E}}^{\mathbf{2}}\mathit{d}\mathit{\zeta }$ …… (4)

Now, write the expression for the total energy of the uniformly charged sphere is obtained by using equation (4) and substituting the limits of electric field$\mathit{E}\left(r\right)$.

$$\begin{gathered} \mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}{\mathbf{\left(}\frac{{\mathbf{kr}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{\epsilon }}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\mathbf{\left(}\frac{{\mathbf{kR}}^{\mathbf{4}}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{\right)}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}\frac{{\mathbf{k}}^{\mathbf{2}}{\mathbf{r}}^{\mathbf{4}}}{\mathbf{16}{\mathbf{\epsilon }}_{\mathbf{0}}^{\mathbf{2}}}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\frac{{\mathbf{k}}^{\mathbf{2}}{\mathbf{R}}^{\mathbf{8}}}{\mathbf{16}{{\mathbf{\epsilon }}_{\mathbf{0}}}^{\mathbf{2}}{\mathbf{r}}^{\mathbf{4}}}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{\pi k}}^{\mathbf{2}}}{\mathbf{8}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{[}\frac{{\mathbf{R}}^{\mathbf{7}}}{\mathbf{7}}\mathbf{+}{\mathbf{R}}^{\mathbf{8}}{\left(-\frac{1}{r}\right)}_{\mathbf{R}}^{\mathbf{\infty }}\mathbf{]} \end{gathered}$$

Solve further as,

$$\begin{gathered} \mathit{W}\mathbf{=}\frac{{\mathbf{\pi k}}^{\mathbf{2}}}{\mathbf{8}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{(}\frac{{\mathbf{R}}^{\mathbf{7}}}{\mathbf{7}}\mathbf{+}{\mathbf{R}}^{\mathbf{7}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{\pi k}}^{\mathbf{2}}{\mathbf{R}}^{\mathbf{7}}}{\mathbf{7}{\mathbf{\epsilon }}_{\mathbf{0}}} \end{gathered}$$

Hence, the total energy is $\frac{{\mathbf{\pi k}}^{\mathbf{2}}{\mathbf{R}}^{\mathbf{7}}}{\mathbf{7}{\mathbf{\epsilon }}_{\mathbf{0}}}$.

Method 2:

Write the expression for the energy configuration.

$\mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\int }\mathit{\rho }\mathit{V}\left(r\right)\mathit{d}\mathit{\zeta }$ ……. (5)

For$\mathit{r}\mathbf{<}\mathit{R}$,

The relation between the electric potential and intensity is,

$$\begin{gathered} \mathit{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\underset{\mathbf{\infty }}{\overset{\mathbf{r}}{\mathbf{\int }}}\mathit{E}\mathbf{·}\mathit{d}\mathit{l} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\underset{\mathbf{\infty }}{\overset{\mathbf{R}}{\mathbf{\int }}}\mathit{E}\mathbf{·}\mathit{d}\mathit{l}\mathbf{-}\underset{\mathbf{\infty }}{\overset{\mathbf{r}}{\mathbf{\int }}}\mathit{E}\mathbf{·}\mathit{d}\mathit{l} \end{gathered}$$

Now, write the expression for the total energy of the uniformly charged sphere is obtained by using equation (5) and substituting the limits of electric field$\mathit{E}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$.

$$\begin{gathered} \mathit{V}\left(r\right)\mathbf{=}\mathbf{-}\underset{\mathbf{\infty }}{\overset{\mathbf{R}}{\mathbf{\int }}}\left(\frac{k{R}^4}{4{\epsilon }_0{r}^2}\right)\mathit{d}\mathit{r}\mathbf{-}\underset{\mathbf{\infty }}{\overset{\mathbf{r}}{\mathbf{\int }}}\left(\frac{k{r}^2}{4{\epsilon }_0}\right)\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\frac{\mathbf{k}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}}\left[{\overline{)R^4\left(-\frac{1}{r}\right)}}_{\infty }^R+\overline{){\frac{r^3}{3}}_R^r}\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\frac{\mathbf{k}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}}\left(-R^3+\left(\frac{r^3}{3}-\frac{R^3}{3}\right)\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\frac{\mathbf{k}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}}\left(-\frac{4}{3}{R}^3+\frac{r^3}{3}\right) \end{gathered}$$

Solve further as,

$\mathit{V}\left(r\right)\mathbf{=}\frac{\mathbf{k}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}\left(R^3-\frac{r^3}{4}\right)$

Substitute $\mathit{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{k}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{(}{\mathbf{R}}^{\mathbf{3}}\mathbf{-}\frac{{\mathbf{r}}^{\mathbf{3}}}{\mathbf{4}}\mathbf{)}$and $\mathit{d}\mathit{\zeta }\mathbf{=}\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r}$in equation (5)

$$\begin{gathered} \mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\underset{\mathbf{\infty }}{\overset{\mathbf{R}}{\mathbf{\int }}}\left(kr\right)\left[\frac{k}{3{\epsilon }_0}\left(R^3-\frac{r^3}{4}\right)\right]\mathbf{4}\mathit{\pi }{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }{\mathbf{k}}^{\mathbf{2}}}{\mathbf{3}{\mathbf{\epsilon }}_{\mathbf{0}}}\underset{\mathbf{\infty }}{\overset{\mathbf{R}}{\mathbf{\int }}}\left(R^3{r}^3-\frac{1}{4}{r}^6\right)\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}{\mathbf{\pi k}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{\epsilon }}_{\mathbf{0}}}\left[R^3\left(\frac{R^3}{4}\right)-\frac{1}{4}\left(\frac{R^7}{7}\right)\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }{\mathbf{k}}^{\mathbf{2}}{\mathbf{R}}^{\mathbf{7}}}{\mathbf{2}\left(3{\epsilon }_0\right)}\mathbf{\left(}\frac{\mathbf{6}}{\mathbf{7}}\mathbf{\right)} \end{gathered}$$

Solve as further,

$\mathit{W}\mathbf{=}\frac{{\mathbf{\pi k}}^{\mathbf{2}}{\mathbf{R}}^{\mathbf{7}}}{\mathbf{7}{\mathbf{\epsilon }}_{\mathbf{0}}}$

Hence, the total energy is $\mathit{W}\mathbf{=}\frac{{\mathbf{\pi k}}^{\mathbf{2}}{\mathbf{R}}^{\mathbf{7}}}{\mathbf{7}{\mathbf{\epsilon }}_{\mathbf{0}}}$.