Question

(a) Suppose the potential is a constant ${\mathit{V}}_{\mathbf{0}}$over the surface of the sphere. Use the results of Ex. 3.6 and Ex. 3.7 to find the potential inside and outside the sphere. (Of course, you know the answers in advance-this is just a consistency check on the method.)

(b) Find the potential inside and outside a spherical shell that carries a uniform surface charge ${\mathit{\sigma }}_{\mathbf{0}}$, using the results of Ex. 3.9.

Short answer

Answer

1. The potential outside the sphere is $\frac{\mathbf{R}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{r}}$.

2. The expression for the potential inside and outside the potential is

$\mathit{V}\left(r,\theta \right)\mathbf{=}\{\begin{array}{ll}\frac{{\sigma }_{0}R}{{\epsilon }_0}& ifr\le R\\ \frac{{\sigma }_0{R}^2}{{\epsilon }_{0}r}& ifr\ge R\end{array}$

Step-by-step solution

Define function

Write the expression potential inside the sphere.

$\mathit{V}\left(r,\theta \right)\mathbf{=}\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\left(cos\theta \right)$ ….. (1)

Here, ${\mathit{P}}_{\mathbf{I}}$is the Legendre polynomial, ${\mathit{A}}_{\mathbf{I}}$is a constant which changes its value according to $\mathit{I}$, and $\mathit{r}$ is the radius.

Determine potential inside and outside the sphere

Write the expression for the potential is constant over the surface of the sphere.

$$\begin{gathered} \mathit{A}\mathit{t}\mathbf{}\mathit{r}\mathbf{=}\mathit{R} \\ \mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{A}}_{\mathbf{i}}{\mathit{r}}^{\mathbf{i}}{\mathit{P}}_{\mathbf{i}}\mathbf{(}\mathbf{cos}\mathbf{}\mathbf{\theta }\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathit{V}}_{\mathbf{0}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)} \end{gathered}$$ …… (2)

The ${\mathit{A}}_{\mathbf{i}}$is expressed as,

role="math" localid="1657278374527" $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{A}}_{\mathbf{i}}\mathbf{=}{\frac{(2I+1)}{\mathbf{2}\mathbf{R}}}^{\mathbf{I}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{V}}_{\mathbf{0}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta } \\ {\mathit{V}}_{\mathbf{0}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}{\mathit{V}}_{\mathbf{0}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{A}}_{\mathbf{I}}\mathbf{=}\frac{(2I+1)}{\mathbf{2}{\mathbf{R}}^{\mathbf{I}}}{\mathbf{V}}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{i}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta } \end{gathered}$$ …… (3)

Hence, it is called as Legendre polynomials for $\mathit{I}\mathbf{=}\mathbf{0}$is 1. Thus,

${\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{=}\mathbf{1}$

Now, write the Legendre polynomial for orthogonal functions.

$\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta }\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{0}\mathbf{,}& \mathbf{if}\mathbf{}\mathbf{I}\mathbf{\text{'}}\mathbf{\ne }\mathbf{I}\mathbf{,}\\ \frac{\mathbf{2}}{\mathbf{2}\mathbf{I}\mathbf{+}\mathbf{1}}\mathbf{,}& \mathbf{if}\mathbf{}\mathbf{I}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\end{array}$

For $\mathit{I}\mathbf{=}\mathbf{0}$,

$\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta }\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{0}\mathbf{,}& \mathbf{if}\mathbf{}\mathbf{I}\mathbf{\text{'}}\mathbf{\ne }\mathbf{0}\mathbf{,}\\ \mathbf{0}\mathbf{,}& \mathbf{if}\mathbf{}\mathbf{I}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\end{array}$

Now,

${\mathit{A}}_{\mathbf{I}}\mathbf{=}\frac{\left(2I+1\right)}{\mathbf{2}{\mathbf{R}}^{\mathbf{I}}}{\mathit{V}}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta }$is written as,

${\mathit{A}}_{\mathbf{I}}\mathbf{=}\frac{(2I+1)}{\mathbf{2}{\mathbf{R}}^{\mathbf{I}}}{\mathbf{V}}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta }$ …… (4)

Substitute ${\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}$for 1 in equation (4).

${\mathit{A}}_{\mathbf{I}}\mathbf{=}\frac{(2I+1)}{\mathbf{2}{\mathbf{R}}^{\mathbf{I}}}{\mathbf{V}}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta }$

For calculating ${\mathit{A}}_{\mathbf{I}}$using the Legendre polynomial for orthogonal function,

${\mathit{A}}_{\mathbf{I}}\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{0}& \mathbf{if}\mathbf{}\mathbf{I}\mathbf{\ne }\mathbf{0}\\ {\mathbf{V}}_{\mathbf{0}}& \mathbf{if}\mathbf{=}\mathbf{0}\end{array}$

Substitute the ${\mathit{V}}_{\mathbf{0}}$for ${\mathit{A}}_{\mathbf{0}}$in the equation $\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}{\mathit{A}}_{\mathbf{0}}{\mathit{r}}^{\mathbf{0}}{\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}$

$$\begin{gathered} \mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}{\mathit{V}}_{\mathbf{0}}{\mathit{r}}^{\mathbf{0}}{\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathit{V}}_{\mathbf{0}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathit{V}}_{\mathbf{0}} \end{gathered}$$

Therefore, the potential inside the sphere is ${\mathit{V}}_{\mathbf{0}}$.

Now, Calculate potential outside the sphere,

Write the expression for the potential outside the sphere.

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{B}}_{\mathbf{I}}}{{\mathbf{r}}^{\mathbf{i}\mathbf{+}\mathbf{I}}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}$ …… (5)

The value of ${\mathit{B}}_{\mathbf{I}}$is given as,

${\mathit{B}}_{\mathbf{I}}\mathbf{=}\frac{(2I+1)}{\mathbf{2}}{\mathit{R}}^{\mathbf{i}\mathbf{+}\mathbf{I}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{V}}_{\mathbf{0}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{sin}\mathit{\theta }\mathit{d}\mathit{\theta }$

Here, ${\mathit{V}}_{\mathbf{0}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{V}}_{\mathbf{0}}$, so

${\mathit{B}}_{\mathbf{I}}\mathbf{=}\frac{(2I+1)}{\mathbf{2}}{\mathit{R}}^{\mathbf{i}\mathbf{+}\mathbf{I}}{\mathit{V}}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{sin}\mathit{\theta }\mathit{d}\mathit{\theta }$

Substitute ${\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}$for 1 in above equation.

${\mathit{B}}_{\mathbf{I}}\mathbf{=}\frac{(2I+1)}{\mathbf{2}}{\mathit{R}}^{\mathbf{i}\mathbf{+}\mathbf{I}}{\mathit{V}}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{sin}\mathit{\theta }\mathit{d}\mathit{\theta }$

For calculating ${\mathit{B}}_{\mathbf{I}}$using the Legendre polynomial for orthogonal function,

For $\mathit{I}\mathbf{=}\mathbf{0}$

$$\begin{gathered} {\mathit{B}}_{\mathbf{I}}\mathbf{=}\frac{(2I+1)}{\mathbf{2}}{\mathit{R}}^{\mathbf{i}\mathbf{+}\mathbf{I}}{\mathit{V}}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{sin}\mathit{\theta }\mathit{d}\mathit{\theta } \\ {\mathit{B}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{RV}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ \mathbf{=}\mathit{R}{\mathit{V}}_{\mathbf{0}} \end{gathered}$$

Substitute the $\mathit{R}{\mathit{V}}_{\mathbf{0}}$for ${\mathit{B}}_{\mathbf{0}}$in the equation $\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\frac{{\mathbf{B}}_{\mathbf{0}}}{{\mathbf{r}}^{\mathbf{0}\mathbf{+}\mathbf{1}}}{\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}$

$$\begin{gathered} \mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\frac{{\mathbf{RV}}_{\mathbf{0}}}{\mathbf{r}}{\mathit{P}}_{\mathbf{0}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{RV}}_{\mathbf{0}}}{\mathbf{r}}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{RV}}_{\mathbf{0}}}{\mathbf{r}} \end{gathered}$$

Therefore, the potential outside the sphere is $\frac{{\mathbf{RV}}_{\mathbf{0}}}{\mathbf{r}}$.

Determine the potential inside and outside a spherical shell

b)

Write the expression for the interior region of the sphere.

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathbf{(}\mathbf{r}\mathbf{\le }\mathbf{R}\mathbf{)}$

The coefficient ${\mathit{A}}_{\mathbf{I}}$is given as,

${\mathit{A}}_{\mathbf{I}}\mathbf{=}\frac{(2I+1)}{\mathbf{2}{\mathbf{R}}^{\mathbf{I}}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{V}}_{\mathbf{0}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta }$

Write the expression for the potential at the exterior region of the sphere.

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{B}}_{\mathbf{I}}}{\mathbf{i}\mathbf{+}\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathbf{(}\mathbf{r}\mathbf{\ge }\mathbf{R}\mathbf{)}$

At the boundary to justify the continuity of the potential,
$\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{B}}_{\mathbf{I}}}{\mathbf{i}\mathbf{+}\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{}\mathbf{(}\mathbf{r}\mathbf{\ge }\mathbf{R}\mathbf{)}$

$$\begin{gathered} \mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{B}}_{\mathbf{I}}}{\mathbf{i}\mathbf{+}\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{A}}_{\mathbf{I}}{\mathit{R}}^{\mathbf{I}}\mathbf{=}\frac{{\mathbf{B}}_{\mathbf{j}}}{{\mathbf{R}}^{\mathbf{I}\mathbf{+}\mathbf{1}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{B}}_{\mathbf{I}}\mathbf{=}{\mathit{A}}_{\mathbf{I}}{\mathit{R}}^{\mathbf{2}\mathbf{I}\mathbf{+}\mathbf{1}} \end{gathered}$$

Write the expression for the radial derivative of V is discontinuous at the surface.

${\overline{)\left(\frac{\partial V_{out}}{\partial r}-\frac{\partial V_{in}}{\partial r}\right)}}_{\mathbf{r}\mathbf{=}\mathbf{R}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathit{\sigma }}_{\mathbf{0}}\left(\theta \right)$ …… (6)

$$\begin{gathered} \frac{\mathbf{\partial }{\mathbf{V}}_{\mathbf{out}}}{\mathbf{\partial }\mathbf{r}}\mathbf{=}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{B}}_{\mathbf{I}}}{{\mathbf{r}}^{\mathbf{i}\mathbf{+}\mathbf{1}}}{\mathit{p}}_{\mathbf{i}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{(}\mathbf{I}\mathbf{+}\mathbf{1}\mathbf{)}\frac{{\mathbf{B}}_{\mathbf{I}}}{{\mathbf{r}}^{\mathbf{i}\mathbf{+}\mathbf{1}}}{\mathit{p}}_{\mathbf{i}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)} \end{gathered}$$

For inner potential,

$$\begin{gathered} \frac{\mathbf{\partial }{\mathbf{V}}_{\mathbf{i}\mathbf{n}}}{\mathbf{\partial }\mathbf{r}}\mathbf{=}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\left(cos\theta \right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\sum }_{\mathbf{I}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathit{I}{\mathit{R}}^{\mathbf{I}\mathbf{+}\mathbf{1}}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\left(cos\theta \right) \end{gathered}$$

Now, substitute $\mathbf{-}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{(}\mathbf{I}\mathbf{+}\mathbf{1}\mathbf{)}\frac{{\mathbf{B}}_{\mathbf{I}}}{{\mathbf{r}}^{\mathbf{i}\mathbf{+}\mathbf{1}}}{\mathit{p}}_{\mathbf{i}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{r}}$,$\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathit{I}{\mathit{R}}^{\mathbf{I}\mathbf{+}\mathbf{1}}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}$for $\frac{\mathbf{\partial }{\mathbf{V}}_{\mathbf{in}}}{\mathbf{\partial }\mathbf{r}}$in equation${\overline{)(\frac{\partial V_{\mathrm{out}}}{\partial r}-\frac{\partial V_{\mathrm{in}}}{\partial r})}}_{\mathbf{r}\mathbf{=}\mathbf{R}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathit{\sigma }}_{\mathbf{0}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}$role="math" localid="1657282694438" $$\begin{gathered} \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{}\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{}{\overline{)(\frac{\partial V_{\mathrm{out}}}{\partial r}-\frac{\partial V_{\mathrm{in}}}{\partial r})}}_{\mathbf{r}\mathbf{=}\mathbf{R}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathit{\sigma }}_{\mathbf{0}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)} \\ \mathbf{(}\mathbf{-}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{(}\mathbf{I}\mathbf{+}\mathbf{1}\mathbf{)}\frac{{\mathbf{B}}_{\mathbf{I}}}{{\mathbf{r}}^{\mathbf{i}\mathbf{+}\mathbf{1}}}{\mathbf{p}}_{\mathbf{i}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathbf{-}\mathbf{-}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{IR}}^{\mathbf{I}\mathbf{+}\mathbf{1}}{\mathbf{A}}_{\mathbf{I}}{\mathbf{r}}^{\mathbf{I}}{\mathbf{p}}_{\mathbf{i}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathbf{)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathit{\sigma }}_{\mathbf{0}}\mathbf{\left(}\mathbf{\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{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{(}\mathbf{2}\mathbf{I}\mathbf{+}\mathbf{1}\mathbf{)}{\mathit{R}}^{\mathbf{I}\mathbf{+}\mathbf{1}}{\mathit{A}}_{\mathbf{I}}{\mathit{r}}^{\mathbf{I}}{\mathit{P}}_{\mathbf{I}}\mathbf{\left(}\mathbf{cos\theta }\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathit{\sigma }}_{\mathbf{0}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)} \end{gathered}$$

The confident ${\mathit{A}}_{\mathbf{I}}$is determined as,

$$\begin{gathered} {\mathit{A}}_{\mathbf{I}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{R}}^{\mathbf{I}\mathbf{+}\mathbf{1}}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{\sigma }}_{\mathbf{0}}\left(\theta \right){\mathit{P}}_{\mathbf{I}}\left(cos\theta \right)\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{R}}^{\mathbf{I}\mathbf{+}\mathbf{1}}}{\mathit{\sigma }}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{I}}\left(cos\theta \right)\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta } \end{gathered}$$

For $\mathit{I}\mathbf{=}\mathbf{0}$,

$$\begin{gathered} {\mathit{A}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{R}}^{\mathbf{I}\mathbf{+}\mathbf{1}}}{\mathit{\sigma }}_{\mathbf{0}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}{\mathit{P}}_{\mathbf{I}}\left(cos\theta \right)\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{R}{\mathbf{\sigma }}_{\mathbf{0}}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\underset{\mathbf{0}}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{d}\mathit{\theta } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{R}{\mathbf{\sigma }}_{\mathbf{0}}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}{\left[-cos\theta \right]}_{\mathbf{0}}^{\mathbf{\pi }} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{R}{\mathbf{\sigma }}_{\mathbf{0}}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}{\left[cos\theta \right]}_{\mathbf{\pi }}^{\mathbf{0}} \end{gathered}$$

Solve further simplification,

$$\begin{gathered} {\mathit{A}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{R\sigma }}_{\mathbf{0}}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\cos \left(\theta \right)-\cos \left(\pi \right)\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{R\sigma }}_{\mathbf{0}}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\left[1-\left(-1\right)\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{R\sigma }}_{\mathbf{0}}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}} \end{gathered}$$

Therefore, role="math" localid="1657276189854" ${\mathit{A}}_{\mathbf{1}}\mathbf{=}\mathbf{\{}\begin{array}{ll}\frac{{\mathbf{R\sigma }}_{\mathbf{0}}}{{\mathbf{\epsilon }}_{\mathbf{0}}}& \mathbf{if}\mathbf{}\mathbf{I}\mathbf{=}\mathbf{0}\\ \mathbf{0}& \mathbf{if}\mathbf{}\mathbf{I}\mathbf{\ne }\mathbf{0}\end{array}$

Thus, the expression for the potential inside and outside the potential is

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{\{}\begin{array}{ll}\frac{{\mathbf{\sigma }}_{\mathbf{0}}\mathbf{R}}{{\mathbf{\epsilon }}_{\mathbf{0}}}& \mathbf{if}\mathbf{}\mathbf{r}\mathbf{}\mathbf{\le }\mathbf{}\mathbf{R}\\ \frac{{\mathbf{\sigma }}_{\mathbf{0}}{\mathbf{R}}^{\mathbf{2}}}{{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{r}}& \mathbf{if}\mathbf{}\mathbf{r}\mathbf{}\mathbf{\ge }\mathbf{}\mathbf{R}\end{array}$