In Prob. 2.25, you found the potential on the axis of a uniformly charged disk:

$\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\left(\sqrt{r^2+R^2}-r\right)$

(a) Use this, together with the fact that ${\mathbf{P}}_{\mathbf{i}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$to evaluate the first three terms

in the expansion (Eq. 3.72) for the potential of the disk at points off the axis, assuming $\mathbf{r}\mathbf{>}\mathbf{R}$.

(b) Find the potential for $\mathbf{r}\mathbf{<}\mathbf{R}$by the same method, using Eq. 3.66. [Note: You

must break the interior region up into two hemispheres, above and below the

disk. Do not assume the coefficients ${\mathbf{A}}_{\mathbf{1}}$are the same in both hemispheres.]

Write the expression for the potential outside the disk in spherical polar co-ordinates.

$\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{l}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{ }\frac{{\mathbf{B}}_{\mathbf{i}}}{{\mathbf{r}}^{\mathbf{t}\mathbf{-}\mathbf{1}}}{\mathbf{P}}_{\mathbf{i}}\mathbf{(}\mathbf{cos}\mathbf{}\mathbf{\theta }\mathbf{)}${for $\text{r<R}$] …… (1)

Given that, the potential on the axis is,

$\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\left(\sqrt{r^2+R^2}-r\right)$ …… (2)

a)

From the equation (1)

$\mathrm{V}(\mathrm{r},\mathrm{\theta })=\sum _{\mathrm{i}=0}^{\mathrm{n}=} \frac{{\mathrm{B}}_{\mathrm{i}}}{{\mathrm{r}}^{\mathrm{l}=1}}{\mathrm{P}}_1(\cos \mathrm{\theta })$

$=\sum _{\mathrm{i}=0}^{\mathrm{\infty }} \frac{{\mathrm{B}}_{\mathrm{i}}}{{\mathrm{r}}^{\mathrm{l}=1}}{\mathrm{P}}_{\mathrm{i}}\left(1\right)$

$\mathrm{V}(\mathrm{r},\mathrm{\theta })=\sum _{\mathrm{i}=0}^{\mathrm{\infty }} \frac{{\mathrm{B}}_{\mathrm{i}}}{{\mathrm{r}}^{\mathrm{i}=1}}$

Then,$\sum _{\mathrm{j}=0}^{\mathrm{ir}} \frac{{\mathrm{B}}_{\mathrm{j}}}{{\mathrm{r}}^{\mathrm{j}-1}}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left[\sqrt{{\mathrm{r}}^7+{\mathrm{R}}^{\mathrm{\prime }}}-\mathrm{r}\right]$

As, $\mathrm{r}>\mathrm{R}$for this region

$\sqrt{{\mathrm{r}}^2+{\mathrm{R}}^2}=\mathrm{r}{\left(1+\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}\right)}^2$

$=\mathrm{r}\left[1+\frac{1}{2}\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}-\frac{1}{8}\frac{{\mathrm{R}}^4}{{\mathrm{r}}^4}+\dots \dots \right]$

This is done by $(\mathrm{x}+1{)}^{\frac{1}{2}}$.

$\sqrt{{\mathrm{r}}^2+{\mathrm{R}}^2}=1+\frac{1}{2}\mathrm{x}-\frac{1}{2^3}{\mathrm{x}}^2+\dots \dots$

$\sum _{\mathrm{l}=0}^{\mathrm{\infty }} \frac{{\mathrm{B}}_{\mathrm{j}}}{{\mathrm{r}}^{\mathrm{l}-1}}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left[1+\frac{1}{2}\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}-\frac{1}{8}\frac{{\mathrm{R}}^4}{{\mathrm{r}}^4}+\dots \dots -1\right]$

$=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}\left[\frac{{\mathrm{R}}^2}{2\mathrm{r}}-\frac{1}{8}\frac{{\mathrm{R}}^4}{{\mathrm{r}}^3}+\dots \dots \right]$……. (3)

Substitute $\mathrm{I}=0$in equation (3), then

${\mathrm{B}}_0=\frac{{\mathrm{\sigma R}}^2}{4{\mathrm{\epsilon }}_0}$

${\mathrm{B}}_1=0$

${\mathrm{B}}_2=\frac{{\mathrm{\sigma R}}^4}{16{\mathrm{s}}_0}$

Now,

$\mathrm{V}(\mathrm{r},0)=\frac{{\mathrm{\sigma R}}^2}{4{\mathrm{\epsilon }}_0\mathrm{r}}-\frac{{\mathrm{\sigma R}}^4}{16{\mathrm{\epsilon }}_0{\mathrm{r}}^3}-{\mathrm{P}}_2(\cos 0)+\dots \dots$

$=\frac{{\mathrm{\sigma R}}^2}{4{\mathrm{\epsilon }}_0\mathrm{r}}\left[\frac{1}{\mathrm{r}}-\frac{{\mathrm{R}}^2}{4{\mathrm{r}}^3}{\mathrm{P}}_2(\cos \mathrm{\theta })+\dots \dots \right]$

$=\frac{{\mathrm{\sigma R}}^2}{4{\mathrm{\epsilon }}_0\mathrm{r}}\left[1-\frac{{\mathrm{R}}^2}{8{\mathrm{r}}^2}\left(3{\cos }^2\mathrm{\theta }-1\right)+\dots \dots \right]$

Hence, the first three terms in the expansion for the potential$\mathrm{r}>\mathrm{R}$ is $\frac{{\mathrm{\sigma R}}^2}{4{\mathrm{\epsilon }}_0\mathrm{r}}\left[1-\frac{{\mathrm{R}}^2}{8{\mathrm{r}}^2}\left(3{\cos }^2\mathrm{\theta }-1\right)+\dots \dots \right]$