Find the charge density $\mathrm{\sigma }\left(\mathrm{\theta }\right)$ on the surface of a sphere (radius R ) that

produces the same electric field, for points exterior to the sphere, as a charge qat the point $\mathrm{a}<\mathrm{R}$ onthe zaxis.

The radius of the sphere is $R.$

The charge $q$produces a field similar to the surface charge.

The potential of a charge $q$is

${\mathbf{V}}_{\mathbf{q}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{r}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\left(\frac{a}{r}\right)}^{\mathbf{n}}{\mathbf{p}}_{\mathbf{n}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, $P_n$is the Legendre polynomial of order n .

The potential for a charge distribution $\sigma$is

${\mathbf{V}}_{\mathbf{\sigma }}\mathbf{=}\mathbf{\sum }_{\mathbf{l}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{B}}_{\mathbf{l}}}{{\mathbf{r}}^{\mathbf{l}\mathbf{+}\mathbf{l}}}{\mathbf{p}}_{\mathbf{l}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, ${\mathbf{B}}_{\mathbf{l}}$ is the coefficient of a Legendre polynomial.

The relation between coefficients of Legendre polynomial are

role="math" localid="1657698194599" ${\mathbf{B}}_{\mathbf{l}}\mathbf{=}{\mathbf{A}}_{\mathbf{l}}{\mathbf{R}}^{\mathbf{2}\mathbf{l}\mathbf{+}\mathbf{1}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

The expression for charge density is

$\mathbf{\sigma }\mathbf{=}{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{\sum }_{\mathbf{l}\mathbf{=}\mathbf{0}}^{\mathbf{\yen }}\mathbf{(}\mathbf{2}\mathbf{l}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{A}}_{\mathbf{l}}{\mathbf{R}}^{\mathbf{l}\mathbf{-}\mathbf{1}}{\mathbf{P}}_{\mathbf{l}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{4}\mathbf{\right)}$

Compare equations (1) and (2) and use (3) to get

$$\begin{gathered} {\mathrm{B}}_{\mathrm{l}}=\frac{{\mathrm{qa}}_{\mathrm{l}}}{4{\mathrm{\pi \epsilon }}_0} \\ {\mathrm{A}}_{\mathrm{l}}=\frac{{\mathrm{qa}}_{\mathrm{l}}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^{2\mathrm{l}+1}} \end{gathered}$$

Substitute expression of $A_l$in equation (4) and get

$\sigma =\frac{q}{4{\mathrm{\pi R}}^2}\sum _{l-0}^{\infty }(2l+1){\left(\frac{a}{R}\right)}^l{p}_l\left(\cos \theta \right).......\left(5\right)$

But

$\frac{1}{\sqrt{{\mathrm{a}}^2+{\mathrm{R}}^2-2\mathrm{aRcos\theta }}}=\frac{1}{\mathrm{R}}\sum _{\mathrm{l}=0}^{\infty }{\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}^{\mathrm{l}}{\mathrm{p}}_{\mathrm{l}}\left(\mathrm{cos\theta }\right)$

Differentiate the above equation with respect to a

$-\frac{(\mathrm{a}-\mathrm{Rcos\theta })}{{({\mathrm{a}}^2+{\mathrm{R}}^2-2\mathrm{aRcos\theta })}^{3/2}}=\frac{1}{\mathrm{aR}}\sum _{\mathrm{l}=0}^{\infty }\mathrm{l}{\left(\frac{\mathrm{a}}{\mathrm{R}}\right)}^{\mathrm{l}}{\mathrm{p}}_{\mathrm{l}}\left(\mathrm{cos\theta }\right)$

Substitute this expression in equation (5) and get

$$\begin{gathered} \sigma =\frac{q}{4{\mathrm{\pi R}}^2}\left[-2aR\frac{(a-R\cos \theta )}{{(a^2+R^2-2aR\cos \theta )}^{3/2}}+\frac{R}{{(a^2+R^2-2aR\cos \theta )}^{1/2}}\right] \\ =\frac{q}{4\mathrm{\pi R}}\frac{(R^2-a^2)}{{(a^2+R^2-2aR\cos \theta )}^{3/2}} \end{gathered}$$

Thus, the surface charge density is role="math" localid="1657699312244" $\frac{q}{4\mathrm{\pi R}}\frac{(R^2-a^2)}{{(a^2+R^2-2aR\cos \theta )}^{3/2}}.$