Suppose the potential ${\mathit{V}}_{\mathbf{0}}\left(0\right)$at the surface of a sphere is specified,

and there is no charge inside or outside the sphere. Show that the charge density on the sphere is given by

$\mathit{\sigma }\left(\theta \right)\mathbf{=}\frac{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{2}\mathbf{R}}\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\left(2I+1\right)}^{\mathbf{2}}{\mathit{C}}_{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\left(cos\theta \right)$

Where,

${\mathit{C}}_{\mathbf{I}}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{V}}_{\mathbf{0}}\left(\theta \right){\mathit{P}}_{\mathbf{I}}\left(cos\theta \right)\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta }$

Let’s consider that, the potential difference inside the conducting sphere at a distance r from the origin is,

$$\begin{gathered} V\left(r,\theta \right)=\sum _{I=0}^{\infty }\left(A_I{r}^I\right)P_I\cos \theta \\ =V\left(0\right) \end{gathered}$$ ……. (1)

Multiplying both sides by $P_I\left(\cos \theta \right)\sin \theta$in equation (1) and integrate with respect to $\theta$.

Take the limit $0to\pi$. Thus we get the value of coefficient of $A_I$.

For the region $r<R$

$$\begin{gathered} \underset{0}{\overset{\pi }{\int }}{V}_0\left(\theta \right)P_I\left(\cos \theta \right)\sin \theta d\theta =\underset{0}{\overset{\pi }{\int }}\sum _{I=0}^{\pi }{P}_I\left(\cos \theta \right)\sin \theta d\theta \\ =\frac{2}{2I+1}\left(A_I{r}^I\right) \end{gathered}$$

If $I=I^I$, then

$A_I=\frac{2I+1}{2}\underset{0}{\overset{\pi }{\int }}{V}_0\left(\theta \right)P_I\left(\cos \theta \right)\sin \theta d\theta$ …… (2)

Now, write the expression for the exterior portion.

For region $r>R$of the sphere, then the expression is,

$V_{\left(r,\theta \right)}=\sum _{I=0}^{\infty }\frac{B_I}{I+1}{P}_I\left(\cos \theta \right)$ …… (3)

Write the expression for the radial derivative of potential difference of surface.

${\left(\frac{\partial V_{out}}{\partial r}-\frac{\partial V_{in}}{\partial r}\right)}_{r=R}=\frac{\sigma \left(\theta \right)}{{\epsilon }_0}$ …… (4)

Or

${\left[-\sum _{}\left(I+1\right)\frac{B_I}{r^{I+2}}{P}_I\left(\cos \theta \right)-\sum _{}I{A}_I{R}^{I=1}{P}_I\left(\cos \theta \right)\right]}_{r=R}=-\frac{\sigma \theta }{{\epsilon }_0}$ ……. (5)

In equation (5) the value of $B_I$is $A_I{R}^{2I+1}$.

Thus,

$-\frac{\sigma \theta }{{\epsilon }_0}={\left[-\sum _{}\left(I+1\right)\frac{A_I{R}^{2I+1}}{R^{I+2}}{P}_I\left(\cos \theta \right)-\sum _{}I{A}_I{R}^{I=1}{P}_I\left(\cos \theta \right)\right]}_{r=R}$

Therefore the value of $\sigma \left(\theta \right)={\epsilon }_0{\displaystyle \sum _{I=0}^{\infty }}\left(2I+1\right)A_I{R}^{I=1}{P}_I\left(\cos \theta \right)$ …… (6)

As we calculate,

$A_I=\frac{2I+1}{2}\underset{0}{\overset{\pi }{\int }}{V}_0\left(\theta \right)P_I\left(\cos \theta \right)\sin \theta d\theta$

Substitute in equation (6)

$\sigma \left(\theta \right)={\epsilon }_0\sum _{I=0}^{\infty }{\left(2I+1\right)}^2{C}_I{P}_I\left(\cos \theta \right)$

Where, $C_I=\underset{0}{\overset{\pi }{\int }}{V}_0\left(\theta \right)P_I\left(\cos \theta \right)\sin \theta d\theta$

Thus, the charge density is $\sigma \left(\theta \right)={\epsilon }_0\sum _{I=0}^{\infty }{\left(2I+1\right)}^2{C}_I{P}_I\left(\cos \theta \right)$.