A sphere of radius Rcarries a charge density $\mathit{\rho }\left(r\right)\mathbf{=}\mathit{k}\mathit{r}$(where kis 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,

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

Here, kis constant.

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

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

Substitute p = kr in equation (2)

$$\begin{gathered} \rho \left(r\right)=\left(kr\right)\left(\pi r^{2}dr\right) \\ =k4\pi r^{3}dr \end{gathered}$$

Write the expression for total charge enclosed within sphere.

$$\begin{gathered} q_{enclosed}={\int }_0^r\rho d\tau \\ ={\int }_{r=0}^{r}kr4\pi r^{2}dr \\ =4\pi k{\int }_{r=0}^r{r}^{3}dr \\ =4\pi k{\left(\frac{r^4}{4}\right)}_0^r \end{gathered}$$

Therefore, total charge enclosed within the sphere is,

$q_{enclosed}=\pi k{r}^4$

According to statement of Gauss’s law, the electric field is directly proportional to the $q_{enclosed}$in to the Gaussian sphere.

Write the expression for electric flux of the sphere.

$$\begin{gathered} {\Phi }_E=\int E_{1}dA \\ =\frac{q_{inside}}{{\epsilon }_0}.....\left(3\right) \end{gathered}$$

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

$A=4\pi r^2$

Substitute $A=4\pi r^2$and $q_{enclosed}=\pi k{r}^4$in equation (3),

$$\begin{gathered} E\left(4\pi r^2\right)=\frac{\pi k{r}^4}{{\epsilon }_0} \\ E=\frac{k{r}^2}{4{\epsilon }_0} \end{gathered}$$

$Assumeapointr<R.$

Write the expression for total charge enclosed within the sphere.

$$\begin{gathered} q_{enclosed}={\int }_0^r\rho d\tau \\ ={\int }_{r=0}^{r}kr4\pi r^{2}dr \\ =4\pi k{\int }_{r=0}^r{r}^{3}dr \\ =4\pi k{\left(\frac{r^4}{4}\right)}_0^r \end{gathered}$$

Solve further as,

$q_{enclosed}=\pi k{r}^4$

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

$$\begin{gathered} E\left(4\pi r^2\right)=\frac{\pi k{r}^4}{{\epsilon }_0} \\ E=\frac{k{r}^2}{4{\epsilon }_0{r}^2} \end{gathered}$$

Hence the electric filed is,

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

Method 1:

Write the expression for the energy configuration.

$W=\frac{1}{2}\int {\epsilon }_0{E}^{2}d\tau .....\left(4\right)$

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 E (r).

localid="1657610076562" $$\begin{gathered} W=\frac{1}{2}{\epsilon }_0{\int }_0^R\left(\frac{k{r}^2}{4{\epsilon }_0}\right)4\pi r^{2}dr+\frac{1}{2}{\epsilon }_0{\int }_R^{\infty }{\left(\frac{k{r}^2}{4{\epsilon }_0{r}^2}\right)}^{2}4\pi r^{2}dr \\ =\frac{1}{2}{\epsilon }_0{\int }_0^R\frac{k^2{r}^4}{16{\epsilon }_0^2}4\pi r^{2}dr+\frac{1}{2}{\epsilon }_0{\int }_R^{\infty }\frac{k{r}^2}{16{{\epsilon }_0}^2{r}^4}4\pi r^{2}dr \\ =\frac{4\pi {\epsilon }_0}{2}{\left(\frac{k}{4{\epsilon }_0}\right)}^2\left[{\int }_0^R{r}^{6}dr+R^8{\int }_R^{\infty }\frac{1}{r^2}dr\right] \\ =\frac{\pi k}{8{\epsilon }_0}\left[\frac{R^7}{7}+R^8{\left(-\frac{1}{r}\right)}_R^{\infty }\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} W=\frac{\pi k}{8{\epsilon }_0}\left(\frac{R^7}{7}+R^7\right) \\ =\frac{\pi k^2{R}^7}{7{\epsilon }_0} \end{gathered}$$

Hence, the total energy is $\frac{\pi k^2{R}^7}{7{\epsilon }_0}$.

Method 2:

Write the expression for the energy configuration.

$$\begin{gathered} W=\frac{1}{2}\int \rho V\left(r\right)d\tau .......\left(5\right) \\ Forr<R, \end{gathered}$$

The relation between the electric potential and intensity is,

$$\begin{gathered} V\left(r\right)=-{\int }_{\infty }^{r}E.dl \\ =-{\int }_{\infty }^{r}E.dl-{\int }_R^{r}E.dl \end{gathered}$$

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 E(r).

$$\begin{gathered} V\left(r\right)=-{\int }_{\infty }^R\left(\frac{k{R}^4}{4{\epsilon }_0{r}^2}\right)dr-{\int }_R^r\left(\frac{k{R}^2}{4{\epsilon }_0}\right)dr \\ =-\frac{k}{4{\epsilon }_0}\left[R^4\left(-\frac{1}{r}\right){\left|{}_{\infty }^R+\frac{r^3}{3}\right|}_R^r\right] \\ =-\frac{k}{4{\epsilon }_0}\left(-R^3+\left(\frac{r^3}{3}-\frac{R^3}{3}\right)\right) \\ =-\frac{k}{4{\epsilon }_0}\left(-\frac{4}{3}{R}^3+\frac{r^3}{3}\right) \end{gathered}$$

Solve further as,

$V\left(r\right)=\frac{k}{3{\epsilon }_0}\left(R^3-\frac{r^3}{4}\right)$

Substitute V (r) $-\frac{k}{4{\epsilon }_0}\left(-\frac{4}{3}{R}^3+\frac{r^3}{3}\right)$and $d\tau =4\pi r^{2}dr$in equation (5)

$$\begin{gathered} W=\frac{1}{2}{\int }_0^R\left(kr\right)\left[\frac{k}{3{\epsilon }_0}\left(R^3-\frac{r^3}{4}\right)\right]4\pi r^{2}dr \\ =\frac{2\pi k^2}{3\epsilon }{\int }_0^R\left(R^3{r}^3-\frac{1}{4}{r}^6\right)dr \\ =\frac{2\pi k^2}{3\epsilon }\left[R^3\left(\frac{R^3}{4}\right)-\frac{1}{4}\left(\frac{R^7}{7}\right)\right] \\ =\frac{\pi k^2{R}^7}{2\left(3{\epsilon }_0\right)}\left(\frac{6}{7}\right) \end{gathered}$$

Solve as further,

$W=\frac{\pi k^2{R}^7}{7{\epsilon }_0}$

Hence, the total energy is $W=\frac{\pi k^2{R}^7}{7{\epsilon }_0}$.