魅影直播

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

$\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{胃}\mathbf{)}\mathbf{=}\underset{\mathbf{l}\mathbf{=}\mathbf{0}}{\overset{\mathbf{鈭�}}{\mathbf{鈭�}}}\mathbf{鈥�}\frac{{\mathbf{B}}_{\mathbf{i}}}{{\mathbf{r}}^{\mathbf{t}\mathbf{鈭�}\mathbf{1}}}{\mathbf{P}}_{\mathbf{i}}\mathbf{(}\mathbf{cos}\mathbf{鈦�}\mathbf{胃}\mathbf{)}${for $\text{r<R}$] 鈥︹�� (1)

Given that, the potential on the axis is,

$\mathbf{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\frac{\mathbf{蟽}}{\mathbf{2}{\mathbf{蔚}}_{\mathbf{0}}}\left(\sqrt{r^2+R^2}鈭�r\right)$ 鈥︹�� (2)

a)

From the equation (1)

$\mathrm{V}(\mathrm{r},\mathrm{胃})=\underset{\mathrm{i}=0}{\overset{\mathrm{n}=}{鈭�}}鈥�\frac{{\mathrm{B}}_{\mathrm{i}}}{{\mathrm{r}}^{\mathrm{l}=1}}{\mathrm{P}}_1(\cos 鈦�\mathrm{胃})$

$=\underset{\mathrm{i}=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�\frac{{\mathrm{B}}_{\mathrm{i}}}{{\mathrm{r}}^{\mathrm{l}=1}}{\mathrm{P}}_{\mathrm{i}}\left(1\right)$

$\mathrm{V}(\mathrm{r},\mathrm{胃})=\underset{\mathrm{i}=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�\frac{{\mathrm{B}}_{\mathrm{i}}}{{\mathrm{r}}^{\mathrm{i}=1}}$

Then,$\underset{\mathrm{j}=0}{\overset{\mathrm{ir}}{鈭�}}鈥�\frac{{\mathrm{B}}_{\mathrm{j}}}{{\mathrm{r}}^{\mathrm{j}鈭�1}}=\frac{\mathrm{蟽}}{2{\mathrm{蔚}}_0}\left[\sqrt{{\mathrm{r}}^7+{\mathrm{R}}^{\mathrm{鈥�}}}鈭�\mathrm{r}\right]$

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

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

$=\mathrm{r}\left[1+\frac{1}{2}\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}鈭�\frac{1}{8}\frac{{\mathrm{R}}^4}{{\mathrm{r}}^4}+鈥�鈥�\right]$

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

$\sqrt{{\mathrm{r}}^2+{\mathrm{R}}^2}=1+\frac{1}{2}\mathrm{x}鈭�\frac{1}{2^3}{\mathrm{x}}^2+鈥�鈥�$

$\underset{\mathrm{l}=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�\frac{{\mathrm{B}}_{\mathrm{j}}}{{\mathrm{r}}^{\mathrm{l}鈭�1}}=\frac{\mathrm{蟽}}{2{\mathrm{蔚}}_0}\left[1+\frac{1}{2}\frac{{\mathrm{R}}^2}{{\mathrm{r}}^2}鈭�\frac{1}{8}\frac{{\mathrm{R}}^4}{{\mathrm{r}}^4}+鈥�鈥�鈭�1\right]$

$=\frac{\mathrm{蟽}}{2{\mathrm{蔚}}_0}\left[\frac{{\mathrm{R}}^2}{2\mathrm{r}}鈭�\frac{1}{8}\frac{{\mathrm{R}}^4}{{\mathrm{r}}^3}+鈥�鈥�\right]$鈥︹��. (3)

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

${\mathrm{B}}_0=\frac{{\mathrm{蟽搁}}^2}{4{\mathrm{蔚}}_0}$

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

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

Now,

$\mathrm{V}(\mathrm{r},0)=\frac{{\mathrm{蟽搁}}^2}{4{\mathrm{蔚}}_0\mathrm{r}}鈭�\frac{{\mathrm{蟽搁}}^4}{16{\mathrm{蔚}}_0{\mathrm{r}}^3}鈭�{\mathrm{P}}_2(\cos 鈦�0)+鈥�鈥�$

$=\frac{{\mathrm{蟽搁}}^2}{4{\mathrm{蔚}}_0\mathrm{r}}\left[\frac{1}{\mathrm{r}}鈭�\frac{{\mathrm{R}}^2}{4{\mathrm{r}}^3}{\mathrm{P}}_2(\cos 鈦�\mathrm{胃})+鈥�鈥�\right]$

$=\frac{{\mathrm{蟽搁}}^2}{4{\mathrm{蔚}}_0\mathrm{r}}\left[1鈭�\frac{{\mathrm{R}}^2}{8{\mathrm{r}}^2}\left(3{\cos }^2鈦�\mathrm{胃}鈭�1\right)+鈥�鈥�\right]$

Hence, the first three terms in the expansion for the potential$\mathrm{r}>\mathrm{R}$ is $\frac{{\mathrm{蟽搁}}^2}{4{\mathrm{蔚}}_0\mathrm{r}}\left[1鈭�\frac{{\mathrm{R}}^2}{8{\mathrm{r}}^2}\left(3{\cos }^2鈦�\mathrm{胃}鈭�1\right)+鈥�鈥�\right]$