Use Eq. 2.29 to calculate the potential inside a uniformly charged

solid sphere of radiusRand total charge q.Compare your answer to Pro b. 2.21.

Consider the sphere for the given condition.

Write the formula for volume charge density of the sphere,

$\mathit{p}=\frac{q}{V}$

Here, q Charge on the solid sphere, Vis the volume of the sphere.

Write the potential to the infinitesimal element at the point P.

$\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{p}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{d\tau }}{\mathbf{r}}$

Here, r is the distance between point P and element and qis the constant charge density.

$$\begin{gathered} V=\frac{1}{2{\epsilon }_0}\left(\frac{q}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}\right)\left(R^2-\frac{z^2}{3}\right) \\ =\frac{3q}{\left(8{\mathrm{\pi R}}^3\right){\mathrm{\epsilon }}_0}\left(R^2-\frac{z^2}{3}\right) \\ =\frac{3q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{3R^2}\right) \\ =\frac{q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{R^2}\right) \\ \mathrm{Therefore},\mathrm{the}\mathrm{potential}\mathrm{inside}\mathrm{a}\mathrm{uniformly}\mathrm{charged}\mathrm{solid}\mathrm{sphere}\mathrm{is}\frac{q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{R^2}\right). \end{gathered}$$Consider the spherical co-ordinates; an infinitesimal volume element is described as,

$\mathit{d}\mathit{\tau }\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{r}\mathit{d}\mathit{\theta }\mathit{d}\mathit{\phi }$

At point P the potential to the infinitesimal element is,

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{p}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{d}\mathbf{\tau }}{\mathbf{r}}$

Substitute $r^2\sin \theta drd\theta d\phi \mathrm{for}d\tau \mathrm{and}\sqrt{r^2+z^2-2rz\cos \mathit{\theta }}\mathrm{for}r$for and for in above equation.

$V=\frac{p}{4{\mathrm{\pi \epsilon }}_0}{\int }_{\varphi -0}^{2\mathrm{\pi }}d\varphi {\int }_{r=0}^R{r}^{2}dr{\int }_{r=0}^{\mathrm{\pi }}\frac{\sin \theta d\theta }{\sqrt{r^2+z^2-2rz\cos \theta }}$

Consider the expression.

$r^2+z^2-2rz\cos \mathit{\theta }=t$

Differentiate the above equation,

$$\begin{gathered} 2rz\sin \mathbf{}\mathit{\theta }d\mathit{\theta }=dt \\ \sin \mathit{\theta }d\mathit{\theta }=\frac{dt}{2rz} \end{gathered}$$

If $\mathit{\theta }=0$, then solve as,

$$\begin{gathered} t=r^2+z^2-2rz\cos \left(0\right) \\ =r^2+z^2-2rz \\ ={(r-z)}^2 \end{gathered}$$

If $\theta =\mathrm{\pi }$, then solve as,

$$\begin{gathered} t=r^2+z^2-2rz\cos \left(\mathrm{\pi }\right) \\ ={\mathrm{r}}^2+{\mathrm{z}}^2+2\mathrm{rz} \\ ={(\mathrm{r}+\mathrm{z})}^2 \end{gathered}$$

Substitute $r^2+z^2-2rz\cos \mathit{\theta }\mathbf{}\mathit{f}\mathit{o}\mathit{r}t\mathrm{and}\sin \mathit{\theta }\mathit{d}\mathit{\theta }\mathbf{}\mathrm{for}\frac{dt}{2rz}$values for and for into the equation.

$$\begin{gathered} {\int }_0^x\frac{\sin \theta }{\sqrt{r^2+z^2-2rz\cos \theta }}d\theta =\frac{1}{2rz}{\int }_{{(r-z)}^2}^{{(r+z)}^2}\frac{1}{\sqrt{t}}dt \\ =\frac{2}{2rz}{\left(\sqrt{t}\right)}_{{(r-z)}^2}^{{(r+z)}^2} \\ =\frac{1}{rz}(r+z-\left|r-z\right|) \end{gathered}$$

If r < z, then $\left|r-z\right|=-(r-z)$then, solve as,

$$\begin{gathered} {\int }_0^x\frac{\sin \mathbf{}\mathit{\theta }}{\sqrt{r^2+z^2-2rz\cos \mathit{\theta }}}d\mathit{\theta }\mathbf{}\mathbf{=}\frac{1}{rz}(r+z+(r-z\left)\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=\frac{2}{z} \\ \mathrm{If}\mathrm{r}>\mathrm{z},\mathrm{then}\left|\mathrm{r}-\mathrm{z}\right|=(\mathrm{r}-\mathrm{z})\mathrm{solve}\mathrm{as}, \\ {\int }_0^{\mathrm{x}}\frac{\sin \mathbf{}\mathbf{\theta }}{\sqrt{{\mathrm{r}}^2+{\mathrm{z}}^2-2\mathrm{rz}\cos \mathbf{\theta }}}\mathrm{d}\mathbf{\theta }\mathbf{}\mathbf{=}\frac{1}{\mathrm{rz}}(\mathrm{r}+\mathrm{z}-(\mathrm{r}-\mathrm{z}\left)\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=\frac{2}{\mathrm{z}} \end{gathered}$$

Thus, potential at point P inside solid sphere is,

$$\begin{gathered} V=\frac{p}{4{\mathrm{\pi \epsilon }}_0}{\int }_{\varphi -0}^{2x}d\varphi {\int }_{r-0}^{r-R}{r}^{2}dr{\int }_{\theta -0}^x\frac{\sin \theta d\theta }{\sqrt{r^2+z^2-2rzcos\theta }} \\ =\frac{p}{4{\mathrm{\pi \epsilon }}_0}{\left(\varphi \right)}_0^{2x}\left({\int }_{r-0}^{r-r}{r}^{2}dr\left(\frac{2}{z}\right)+{\int }_{r-z}^{r-R}{r}^{2}dr\left(\frac{2}{z}\right)\right) \\ =\frac{p}{4{\mathrm{\pi \epsilon }}_0}(2\mathrm{\pi }-0)\left(\left(\frac{2}{z}\right){\left(\frac{{\mathrm{r}}^3}{3}\right)}_0^{\mathrm{z}}+2{\left(\frac{r^2}{2}\right)}_z^R\right) \\ =\frac{\mathrm{p}}{4{\mathrm{\pi \epsilon }}_0}\left(\left(\frac{2}{\mathrm{z}}\right)\left(\frac{{\mathrm{z}}^3}{3}-0\right)+2\left(\frac{{\mathrm{R}}^2}{2}-\frac{{\mathrm{r}}^2}{2}\right)\right) \\ \mathrm{Solving}\mathrm{further}, \\ \mathrm{V}=\frac{\mathrm{p}}{2{\mathrm{\epsilon }}_0}\left(\frac{2{\mathrm{z}}^2}{3}+({\mathrm{R}}^2-{\mathrm{z}}^2)\right) \\ =\frac{\mathrm{p}}{2{\mathrm{\epsilon }}_0}\left({\mathrm{R}}^2-\frac{{\mathrm{z}}^2}{3}\right) \\ \mathrm{Substitute}\frac{\mathrm{q}}{{\displaystyle \frac{4}{3}}\mathbf{\pi }{\mathrm{R}}^3}\mathrm{for}\mathit{p}\mathrm{in}\mathrm{the}\mathrm{equation}. \end{gathered}$$

$$\begin{gathered} V=\frac{1}{2{\epsilon }_0}\left(\frac{q}{{\displaystyle \frac{4}{3}{\mathrm{\pi R}}^3}}\right)\left(R^2-\frac{z^2}{3}\right) \\ =\frac{3q}{\left(8{\mathrm{\pi R}}^3\right){\mathrm{\epsilon }}_0}\left(R^2-\frac{z^2}{3}\right) \\ =\frac{3q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{3R^2}\right) \\ =\frac{q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{R^2}\right) \\ \mathrm{Therefore},\mathrm{the}\mathrm{potential}\mathrm{inside}\mathrm{a}\mathrm{uniformly}\mathrm{charged}\mathrm{solid}\mathrm{sphere}\mathrm{is}\frac{q}{8\pi {\epsilon }_{0}R}\left(1-\frac{z^2}{R^2}\right). \end{gathered}$$