A sphere of radiusR,centered at the origin, carries charge density

$\mathit{\rho }\left(r,\theta \right)\mathbf{=}\mathit{k}\frac{\mathbf{R}}{{\mathbf{r}}^{\mathbf{2}}}\left(R-2r\right)\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

where k is a constant, and r, $\mathit{\theta }$are the usual spherical coordinates. Find the approximate potential for points on the z axis, far from the sphere.

There is a sphere of radiusR,centered at the origin carrying a charge density

$\mathit{\rho }\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathit{k}\frac{\mathbf{R}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{(}\mathbf{R}\mathbf{-}\mathbf{2}\mathbf{r}\mathbf{)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$R .

where k is a constant.

The volume element in spherical polar coordinates is

$\mathit{d}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{r}\mathit{d}\mathit{\theta }\mathit{d}\mathit{\varphi }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

The monopole term is

${\mathit{V}}_{\mathbf{m}\mathbf{o}\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\int }\mathit{\rho }\mathit{d}\mathit{\zeta }$

Here, ${\mathit{\epsilon }}_{\mathbf{0}}$is the permittivity of free space.

Substitute form of charge density and use equation (1),

$$\begin{gathered} {\mathit{V}}_{\mathbf{m}\mathbf{o}\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{z}}\mathit{k}\mathit{R}\underset{\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}\frac{\left(R-2r\right)}{{\mathbf{r}}^{\mathbf{2}}}{\mathit{r}}^{\mathbf{2}}\mathit{d}\mathit{r}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathit{d}\mathit{\theta }\underset{\mathbf{0}}{\overset{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}}\mathit{d}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{z}}\mathit{k}\mathit{R}{\left[Rr-r^2\right]}_{\mathbf{0}}^{\mathbf{R}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathit{d}\mathit{\theta }\underset{\mathbf{0}}{\overset{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}}\mathit{d}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

The dipole term is

${\mathit{V}}_{\mathbf{d}\mathbf{i}\mathbf{p}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{2}}}\mathbf{\int }\mathit{r}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathit{\rho }\mathit{d}\mathit{\zeta }$

Substitute form of charge density and use equation (1),

$$\begin{gathered} {\mathit{V}}_{\mathbf{d}\mathbf{i}\mathbf{p}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{2}}}\mathit{k}\mathit{R}\underset{\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}\frac{\left(R-2r\right)}{{\mathbf{r}}^{\mathbf{2}}}{\mathit{r}}^{\mathbf{3}}\mathit{d}\mathit{r}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathit{d}\mathit{\theta }\underset{\mathbf{0}}{\overset{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}}\mathit{d}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{2}}}\mathit{k}\mathit{R}\underset{\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}\left(R-2r\right)\mathit{r}\mathit{d}\mathit{r}\left[\frac{si{n}^3\theta }{3}\right]\underset{\mathbf{0}}{\overset{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}}\mathit{d}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \end{gathered}$$

The quadrupole term is

${\mathit{V}}_{\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{d}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{3}}}\mathbf{\int }{\mathit{r}}^{\mathbf{2}}\left(\frac{3}{2}co{s}^2\theta -\frac{1}{2}\right)\mathit{\rho }\mathit{d}\mathit{\zeta }$

Substitute form of charge density and use equation (1),

$$\begin{gathered} {\mathit{V}}_{\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{d}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{3}}}\mathit{k}\mathit{R}\underset{\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}\frac{\left(R-2r\right)}{{\mathbf{r}}^{\mathbf{2}}}{\mathit{r}}^{\mathbf{4}}\mathit{d}\mathit{r}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\left(\frac{3}{2}co{s}^2\theta -\frac{1}{2}\right)\mathit{d}\mathit{\theta }\underset{\mathbf{0}}{\overset{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}}\mathit{d}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{3}}}\mathit{k}\mathit{R}\underset{\mathbf{0}}{\overset{\mathbf{R}}{\mathbf{\int }}}\frac{\left(R-2r\right)}{{\mathbf{r}}^{\mathbf{2}}}{\mathit{r}}^{\mathbf{4}}\mathit{d}\mathit{r}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\left(\frac{3}{2}co{s}^2\theta -\frac{1}{2}\right)\mathit{d}\mathit{\theta }\underset{\mathbf{0}}{\overset{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}}\mathit{d}\mathit{\varphi } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi }}_{\mathbf{0}}{\mathbf{z}}^{\mathbf{3}}}\mathit{k}\mathit{R}\mathbf{\times }\left(\frac{R^4}{6}\right)\mathbf{\times }\left(\frac{3}{8}\times \frac{\pi }{2}-\frac{1}{}\times \frac{\pi }{2}\right)\mathbf{\times }\mathbf{2}\mathit{\pi } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi }}_{\mathbf{0}}}\frac{\mathbf{k}{\mathbf{R}}^{\mathbf{5}}{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{48}{\mathbf{z}}^{\mathbf{3}}} \end{gathered}$$

Thus, the approximate potential is

$$\begin{gathered} \mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{m}\mathbf{o}\mathbf{n}}\mathbf{+}{\mathit{V}}_{\mathbf{d}\mathbf{i}\mathbf{p}}\mathbf{+}{\mathit{V}}_{\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{d}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{k}{\mathbf{R}}^{\mathbf{5}}{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{48}{\mathbf{z}}^{\mathbf{3}}} \end{gathered}$$

Thus, the potential is $\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{{\mathbf{kR}}^{\mathbf{5}}{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{48}{\mathbf{z}}^{\mathbf{3}}}$.