91Ó°ÊÓ

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.

$\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}}\mathbf{∫}\frac{\mathbf{p}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{»åÏ„}}{\mathbf{r}}$

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

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

$\mathit{d}\mathit{τ}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathit{d}\mathit{r}\mathit{d}\mathit{θ}\mathit{d}\mathit{φ}$

At point P the potential to the infinitesimal element is,

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathbf{∫}\frac{\mathbf{p}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{d}\mathbf{Ï„}}{\mathbf{r}}$

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

$V=\frac{p}{4{\mathrm{Ï€Î\mu }}_0}{∫}_{\mathrm{Ï•}-0}^{2\mathrm{Ï€}}d\mathrm{Ï•}{∫}_{r=0}^R{r}^{2}dr{∫}_{r=0}^{\mathrm{Ï€}}\frac{\sin \mathrm{θ}d\mathrm{θ}}{\sqrt{r^2+z^2-2rz\cos \mathrm{θ}}}$

Consider the expression.

$r^2+z^2-2rz\cos \mathit{θ}=t$

Differentiate the above equation,

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

If $\mathit{θ}=0$, 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 $\mathrm{θ}=\mathrm{π}$, then solve as,

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

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

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

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

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

Thus, potential at point P inside solid sphere is,

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

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