91Ó°ÊÓ

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

$\mathbf{ÒÏ}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{kr}$ ……. (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 . $4\mathrm{Ï€}{r}^{2}dr$Therefore the charge of this elementary part is

$\mathbf{ÒÏ}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{ÒÏ}\mathbf{\left(}\mathbf{4}{\mathbf{Ï„Ï„°ù}}^{\mathbf{2}}\mathbf{dr}\mathbf{\right)}$ …… (2)

Substitute p = kr in equation (2)

$$\begin{gathered} \mathrm{ÒÏ}\left(r\right)=\left(kr\right)\left(\mathrm{Ï€}{r}^{2}dr\right) \\ =k4\mathrm{Ï€}{r}^{3}dr \end{gathered}$$

Write the expression for total charge enclosed within sphere.

$$\begin{gathered} q_{enclosed}={∫}_0^r\mathrm{ÒÏ}d\mathrm{Ï„} \\ ={∫}_{r=0}^{r}kr4\mathrm{Ï€}{r}^{2}dr \\ =4\mathrm{Ï€}k{∫}_{r=0}^r{r}^{3}dr \\ =4\mathrm{Ï€}k{\left(\frac{r^4}{4}\right)}_0^r \end{gathered}$$

Therefore, total charge enclosed within the sphere is,

$q_{enclosed}=\mathrm{Ï€}k{r}^4$

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.

$$\begin{gathered} {\mathrm{Φ}}_E=∫E_{1}dA \\ =\frac{q_{inside}}{{\mathrm{Î\mu }}_0}.....\left(3\right) \end{gathered}$$

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

$A=4\mathrm{Ï€}{r}^2$

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

$$\begin{gathered} E\left(4\mathrm{Ï€}{r}^2\right)=\frac{\mathrm{Ï€}k{r}^4}{{\mathrm{Î\mu }}_0} \\ E=\frac{k{r}^2}{4{\mathrm{Î\mu }}_0} \end{gathered}$$

$Assumeapointr<R.$

Write the expression for total charge enclosed within the sphere.

$$\begin{gathered} q_{enclosed}={∫}_0^r\mathrm{ÒÏ}d\mathrm{Ï„} \\ ={∫}_{r=0}^{r}kr4\mathrm{Ï€}{r}^{2}dr \\ =4\mathrm{Ï€}k{∫}_{r=0}^r{r}^{3}dr \\ =4\mathrm{Ï€}k{\left(\frac{r^4}{4}\right)}_0^r \end{gathered}$$

Solve further as,

$q_{enclosed}=\mathrm{Ï€}k{r}^4$

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

$$\begin{gathered} E\left(4\mathrm{Ï€}{r}^2\right)=\frac{\mathrm{Ï€}k{r}^4}{{\mathrm{Î\mu }}_0} \\ E=\frac{k{r}^2}{4{\mathrm{Î\mu }}_0{r}^2} \end{gathered}$$

Hence the electric filed is,

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

Method 1:

Write the expression for the energy configuration.

$W=\frac{1}{2}∫{\mathrm{Î\mu }}_0{E}^{2}d\mathrm{Ï„}.....\left(4\right)$

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" $$\begin{gathered} W=\frac{1}{2}{\mathrm{Î\mu }}_0{∫}_0^R\left(\frac{k{r}^2}{4{\mathrm{Î\mu }}_0}\right)4\mathrm{Ï€}{r}^{2}dr+\frac{1}{2}{\mathrm{Î\mu }}_0{∫}_R^{∞}{\left(\frac{k{r}^2}{4{\mathrm{Î\mu }}_0{r}^2}\right)}^{2}4\mathrm{Ï€}{r}^{2}dr \\ =\frac{1}{2}{\mathrm{Î\mu }}_0{∫}_0^R\frac{k^2{r}^4}{16{\mathrm{Î\mu }}_0^2}4\mathrm{Ï€}{r}^{2}dr+\frac{1}{2}{\mathrm{Î\mu }}_0{∫}_R^{∞}\frac{k{r}^2}{16{{\mathrm{Î\mu }}_0}^2{r}^4}4\mathrm{Ï€}{r}^{2}dr \\ =\frac{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{2}{\left(\frac{k}{4{\mathrm{Î\mu }}_0}\right)}^2\left[{∫}_0^R{r}^{6}dr+R^8{∫}_R^{∞}\frac{1}{r^2}dr\right] \\ =\frac{\mathrm{Ï€}k}{8{\mathrm{Î\mu }}_0}\left[\frac{R^7}{7}+R^8{\left(-\frac{1}{r}\right)}_R^{∞}\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} W=\frac{\mathrm{Ï€}k}{8{\mathrm{Î\mu }}_0}\left(\frac{R^7}{7}+R^7\right) \\ =\frac{\mathrm{Ï€}{k}^2{R}^7}{7{\mathrm{Î\mu }}_0} \end{gathered}$$

Hence, the total energy is $\frac{\mathrm{Ï€}{k}^2{R}^7}{7{\mathrm{Î\mu }}_0}$.

Method 2:

Write the expression for the energy configuration.

$$\begin{gathered} W=\frac{1}{2}∫\mathrm{ÒÏ}V\left(r\right)d\mathrm{Ï„}.......\left(5\right) \\ Forr<R, \end{gathered}$$

The relation between the electric potential and intensity is,

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

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).

$$\begin{gathered} V\left(r\right)=-{∫}_{∞}^R\left(\frac{k{R}^4}{4{\mathrm{Î\mu }}_0{r}^2}\right)dr-{∫}_R^r\left(\frac{k{R}^2}{4{\mathrm{Î\mu }}_0}\right)dr \\ =-\frac{k}{4{\mathrm{Î\mu }}_0}\left[R^4\left(-\frac{1}{r}\right){\left|{}_{∞}^R+\frac{r^3}{3}\right|}_R^r\right] \\ =-\frac{k}{4{\mathrm{Î\mu }}_0}\left(-R^3+\left(\frac{r^3}{3}-\frac{R^3}{3}\right)\right) \\ =-\frac{k}{4{\mathrm{Î\mu }}_0}\left(-\frac{4}{3}{R}^3+\frac{r^3}{3}\right) \end{gathered}$$

Solve further as,

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

Substitute V (r) $-\frac{k}{4{\mathrm{Î\mu }}_0}\left(-\frac{4}{3}{R}^3+\frac{r^3}{3}\right)$and $d\mathrm{Ï„}=4\mathrm{Ï€}{r}^{2}dr$in equation (5)

$$\begin{gathered} W=\frac{1}{2}{∫}_0^R\left(kr\right)\left[\frac{k}{3{\mathrm{Î\mu }}_0}\left(R^3-\frac{r^3}{4}\right)\right]4\mathrm{Ï€}{r}^{2}dr \\ =\frac{2\mathrm{Ï€}{k}^2}{3\mathrm{Î\mu }}{∫}_0^R\left(R^3{r}^3-\frac{1}{4}{r}^6\right)dr \\ =\frac{2\mathrm{Ï€}{k}^2}{3\mathrm{Î\mu }}\left[R^3\left(\frac{R^3}{4}\right)-\frac{1}{4}\left(\frac{R^7}{7}\right)\right] \\ =\frac{\mathrm{Ï€}{k}^2{R}^7}{2\left(3{\mathrm{Î\mu }}_0\right)}\left(\frac{6}{7}\right) \end{gathered}$$

Solve as further,

$W=\frac{\mathrm{Ï€}{k}^2{R}^7}{7{\mathrm{Î\mu }}_0}$

Hence, the total energy is $W=\frac{\mathrm{Ï€}{k}^2{R}^7}{7{\mathrm{Î\mu }}_0}$.