91影视

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

$\mathit{V}\left(r,胃\right)\mathbf{=}\underset{\mathbf{I}\mathbf{=}\mathbf{0}}{\overset{\mathbf{鈭�}}{\mathbf{鈭�}}}\frac{{\mathbf{B}}_{\mathbf{I}}}{{\mathbf{r}}^{\mathbf{I}\mathbf{=}\mathbf{1}}}{\mathit{P}}_{\mathbf{I}}\left(cos胃\right)$ $\left[forr>R\right]$ 鈥︹�� (1)

Given that, the potential on the axis is,

role="math" localid="1655811849704" $V(\mathrm{r},0)=\frac{蟽}{2{蔚}_0}(\sqrt{{\mathrm{r}}^2+{\mathrm{R}}^2}-\mathrm{r})$ 鈥︹�� (2)

a)

From the equation (1)

$$\begin{gathered} V\left(r,胃\right)=\underset{I=0}{\overset{鈭�}{鈭�}}\frac{B_I}{r^{I=1}}{P}_I\left(\cos 胃\right) \\ =\underset{I=0}{\overset{鈭�}{鈭�}}\frac{B_I}{r^{I=1}}{P}_I\left(1\right) \\ V\left(r,胃\right)=\underset{I=0}{\overset{鈭�}{鈭�}}\frac{B_I}{r^{I=1}} \end{gathered}$$

Then, $\underset{I=0}{\overset{鈭�}{鈭�}}\frac{B_I}{r^{I=1}}=\frac{蟽}{2{蔚}_0}\left[\sqrt{r^2+R^2}-r\right]$

As, $r>R$for this region

$$\begin{gathered} \sqrt{r^2+R^2}=r{\left(1+\frac{R^2}{r^2}\right)}^{\frac{1}{2}} \\ =r\left[1+\frac{1}{2}\frac{R^2}{r^2}-\frac{1}{8}\frac{R^4}{r^4}+......\right] \end{gathered}$$

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

$$\begin{gathered} \sqrt{r^2+R^2}=1+\frac{1}{2}x-\frac{1}{2^3}{x}^2+...... \\ \underset{I=0}{\overset{鈭�}{鈭�}}\frac{B_I}{I=1}=\frac{蟽}{2{蔚}_0}\left[1+\frac{1}{2}\frac{R^2}{r^2}-\frac{1}{8}\frac{R^4}{r^4}+......-1\right] \\ =\frac{蟽}{2{蔚}_0}\left[\frac{R^2}{2r}-\frac{1}{8}\frac{R^4}{r^3}+.......\right] \end{gathered}$$

鈥︹��. (3)

Substitute $I=0$in equation (3), then

$$\begin{gathered} B_0=\frac{蟽R^2}{4{蔚}_0} \\ {B}_I=0 \\ {B}_2=\frac{蟽R^4}{16{蔚}_0} \end{gathered}$$

Now,

$$\begin{gathered} V\left(r,胃\right)=\frac{蟽R^2}{4{蔚}_{0}r}-\frac{蟽R^4}{16{蔚}_0{r}^3}-P_2\left(\cos 胃\right)+...... \\ =\frac{蟽R^2}{4{蔚}_{0}r}\left[\frac{1}{r}-\frac{R^2}{4r^3}{P}_2\left(\cos 胃\right)+......\right] \\ =\frac{蟽R^2}{4{蔚}_{0}r}\left[1-\frac{R^2}{8r^2}\left(3{\cos }^2胃-1\right)+......\right] \end{gathered}$$

Hence, the first three terms in the expansion for the potential $r>R$is $=\frac{蟽R^2}{4{蔚}_{0}r}\left[1-\frac{R^2}{8r^2}\left(3{\cos }^2胃-1\right)+......\right]$.

b)

For $r<R$

$$\begin{gathered} V\left(r,胃\right)=\underset{I=0}{\overset{鈭�}{鈭�}}{A}_I{r}^I{P}_I\left(\cos 胃\right) \\ V\left(r,胃\right)=\underset{I=0}{\overset{鈭�}{鈭�}}{A}_I{r}^I \end{gathered}$$ 鈥︹�� (4)

Then, $\underset{I=0}{\overset{鈭�}{鈭�}}{A}_I{r}^I=\frac{蟽}{2{蔚}_0}\left[\sqrt{r^2+R^2}-r\right]$

As, $r<R$for this region

$$\begin{gathered} \sqrt{r^2+R^2}=R\sqrt{1+\frac{r^2}{R^2}} \\ =R{\left(1+\frac{R^2}{r^2}\right)}^{\frac{1}{2}} \\ =R\left[1+\frac{1}{2}\frac{r^2}{R^2}-\frac{1}{8}\frac{r^4}{R^4}+......\right] \\ =R+\frac{1}{2}\frac{r^2}{R}-\frac{1}{8}\frac{r^4}{R^3}+...... \\ \underset{I=0}{\overset{鈭�}{鈭�}}{A}_I{r}^I=\frac{蟽}{2{蔚}_0}\left[R+\frac{1}{2}\frac{r^2}{R}-\frac{1}{8}\frac{r^4}{R^3}+......-r\right] \end{gathered}$$

Comparing powers of r,

$$\begin{gathered} A_0=\frac{蟽R}{2{蔚}_0} \\ {A}_I=\frac{-蟽}{2I_0} \\ {A}_2=\frac{蟽}{4{蔚}_{0}R} \end{gathered}$$

Then,

$$\begin{gathered} V\left(r,胃\right)=\frac{蟽}{2{蔚}_0}\left[R-r{P}_I\left(\cos 胃\right)+\frac{r^2}{2R}{P}_2\left(\cos 胃\right)+......\right] \\ =\frac{蟽}{2{蔚}_0}\left[R-r\cos 胃+\frac{r^2}{8R}\left(3{\cos }^2胃-1\right)+......\right] \end{gathered}$$

For southern hemisphere,

$$\begin{gathered} P_I\left(-1\right)={\left(-1\right)}^I \\ V\left(r,\mathrm{蟺}\right)=\underset{I=0}{\overset{鈭�}{鈭�}}{\left(-1\right)}^I{A}_I{r}^I \\ =\frac{蟽}{2{蔚}_0}\left[\sqrt{r^2+R^2}-r\right] \end{gathered}$$

Then,

$\underset{I=0}{\overset{鈭�}{鈭�}}{\left(-1\right)}^I{A_I}^{\text{'}}{r}^I=\frac{蟽}{2{蔚}_0}\left[R+\frac{1}{2}\frac{r^2}{R}-\frac{1}{8}\frac{r^4}{R^3}+......-r\right]$

Comparing powers of,

$$\begin{gathered} A_0\text{'}=\frac{蟽R}{2{蔚}_0} \\ {A}_1\text{'}=\frac{蟽}{2{蔚}_0} \\ {A}_1\text{'}=\frac{蟽}{2{蔚}_{0}R} \end{gathered}$$

Then, Write the expression for $V\left(r,胃\right)$for $r<R$for southern hemisphere.

$$\begin{gathered} V\left(r,胃\right)=\frac{蟽}{2{蔚}_0}\left[R+r{P}_I\left(\cos 胃\right)+\frac{r^2}{2R}{P}_2\left(\cos 胃\right)+......\right] \\ =\frac{蟽}{2{蔚}_0}\left[R+r\cos 胃+\frac{r^2}{8R}\left(3{\cos }^2胃-1\right)+......\right] \end{gathered}$$

Thus, the first three terms in the expansion for the potential $r<R$is

$=\frac{蟽}{2{蔚}_0}\left[R+r\cos 胃+\frac{r^2}{8R}\left(3{\cos }^2胃-1\right)+......\right]$.