Use your result in Prob. 2.7 to find the field inside and outside a solidsphere of radius $\mathit{R}$that carries a uniform volume charge density$\mathit{p}$.Express your answers in terms of the total charge of the sphere,$\mathit{q}$.Draw a graph of lEIas a function of the distance from the center.

The radius of sphere is $R$.

The sphere carries a uniform volume charge$p$.

Electric field due to charge $\mathbf{q}$at a distance $\mathbf{r}$is proportional to the charge $\mathbf{q}$and inversely proportional to the square of the distance $\mathbf{r}$ as,

$\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}$

The sphere of radius $R$is divided into small spherical shell of thickness $dx$at a radius $x$from the center, as shown below:

The differential charge $dq$tis the product of charge density $p$and the differential volume $dv$written as $\mathrm{dq}=pdV$. Thus the differential electric field is obtained as

$$\begin{gathered} dE=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi x^{2}pdx}{r^2} \\ =\frac{p{x}^2}{{\epsilon }_0{r}^2}dx \end{gathered}$$

Integrate above differential integral as,

$$\begin{gathered} E=\int dE \\ =\int \frac{p{x}^2}{{\epsilon }_0{r}^2}dx \\ =\frac{p{R}^3}{3{\epsilon }_0{r}^2} \end{gathered}$$

The volume charge density $p$is the ratio of total charge to the volume of the sphere, that is, $p=\frac{q}{\frac{4}{3}\pi R^3}$.

Substitute $\frac{q}{{\displaystyle \frac{4}{3}}\pi R^3}$for $p$into $E=\frac{p{R}^3}{3\epsilon r^2}$

$$\begin{gathered} \begin{array}{l}E=\frac{\left({\displaystyle \frac{q}{{\displaystyle \frac{4}{3}}\pi R^3}}\right)R^3}{3{\epsilon }_0{r}^2}\end{array} \\ =\frac{q}{4\pi {\epsilon }_0{r}^2}\hat{r} \end{gathered}$$

Thus, the electric field outside the sphere is obtained as$\frac{q}{4\pi {\epsilon }_0{r}^2}\hat{r}$

The differential charge $dq$tis the product of charge density $p$and the differential volume $dV$written as$dq=pdV$. Thus the differential electric field is obtained as

$$\begin{gathered} dE=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi x^{2}pdx}{r^2} \\ =\frac{p{x}^2}{{\epsilon }_0{r}^2}dx \end{gathered}$$

Integrate above differential integral as,

$$\begin{gathered} E=\int dE \\ =\int \frac{p{x}^2}{{\epsilon }_0{r}^2}dx \\ =\frac{p}{{\epsilon }_0{r}^2}{\int }_0^R{x}^{2}dx \\ =\frac{p}{{\epsilon }_0{r}^2}\left(\frac{r^3}{3}\right) \end{gathered}$$

Simplify further as

$$\begin{gathered} E=\frac{pr}{3{\epsilon }_0} \\ =\frac{qr}{4{\pi }_0{R}^3} \end{gathered}$$

Thus, the electric field inside the sphere is obtained as$E=\frac{qr}{4{\pi }_0{R}^3}$.

The plot of $\left|E\right|$versus $r$is plotting using the equation of obtained electric field is shown below: