魅影直播

Write the expression for the potential at the surface of a sphere.

${\mathbf{V}}_{\mathbf{0}}\mathbf{=}\mathbf{kcos}\mathbf{鈦�}\mathbf{3}\mathbf{胃}$ 鈥︹�� (1)

Here, $\mathrm{k}$is constant.

Write the trigonometry formula for $\cos 鈦�3\mathrm{胃}$.

$\cos 鈦�3\mathrm{胃}=4{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}$ 鈥︹�� (2)

Now, substitute $4{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\text{for}\cos 鈦�3\mathrm{胃}$for in equation (1).

${\mathrm{V}}_0=\mathrm{k}4{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}$鈥︹�� (3)

Write the general expression for the Legendre polynomial.

${\mathbf{P}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{n}}}{{\mathbf{dx}}^{\mathbf{n}}}{\left(x^2鈭�1\right)}^{\mathbf{n}\mathbf{/}\mathbf{2}}$ 鈥︹�� (4)

Thus, the value of ${\mathrm{P}}_1(\cos 鈦�\mathrm{胃})$and ${\mathrm{P}}_3(\cos 鈦�\mathrm{胃})$is,

${\mathrm{P}}_1(\cos 鈦�\mathrm{胃})=\frac{{\mathrm{d}}^1}{{\mathrm{dx}}^1}{\left({\cos }^2鈦�\mathrm{胃}鈭�1\right)}^{\frac{1}{2}}$

$=\frac{{\mathrm{d}}^1}{{\mathrm{dx}}^1}(\sin 鈦�\mathrm{胃})$

role="math" localid="1657465852226" $=\mathrm{肠辞蝉胃}$

${\mathrm{P}}_3(\cos 鈦�\mathrm{胃})=\frac{1}{2}\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)$

Write the expression potential of the sphere of the sphere of the radius $\mathrm{R}$in terms of Legendre polynomials.

${\mathrm{V}}_0\left(\mathrm{胃}\right)=\mathrm{k}\left[4{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right]$

$=\mathrm{k}\left[{\mathrm{伪笔}}_3(\cos 鈦�\mathrm{胃})+{\mathrm{尾笔}}_1(\cos 鈦�\mathrm{胃})\right]$鈥︹�� (5)

Here, $\mathrm{伪}$and $\mathrm{尾}$are the Legendre polynomial constants.

Substitute $\mathrm{肠辞蝉胃}$for ${\mathrm{P}}_1\left(\mathrm{肠辞蝉胃}\right)$and $\frac{1}{2}\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)$for ${\mathrm{P}}_3\left(\mathrm{肠辞蝉胃}\right)$in equation (5).

$\mathrm{k}\left[4{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right]=\mathrm{k}\left[{\mathrm{伪笔}}_3(\cos 鈦�\mathrm{胃})+{\mathrm{尾笔}}_1(\cos 鈦�\mathrm{胃})\right]$

$4{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}=\mathrm{k}\left(\mathrm{伪}\left[\frac{1}{2}\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)\right]+\mathrm{尾肠辞蝉}鈦�\mathrm{胃}\right)$

$4{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}=\frac{5\mathrm{伪}}{2}{\cos }^3鈦�\mathrm{胃}+\left(\mathrm{尾}鈭�\frac{3\mathrm{a}}{2}\cos 鈦�\mathrm{胃}\right)$

Compare the above equation on both sides, then the value of constant $\mathrm{伪}$is

$\frac{5\mathrm{伪}}{2}=4$

$\mathrm{伪}=\frac{8}{5}$

Calculate the value of $\mathrm{尾}$.

$\mathrm{尾}鈭�\frac{3\mathrm{伪}}{2}=鈭�3$

$\mathrm{尾}鈭�\frac{3}{2}\left(\frac{8}{5}\right)=鈭�3$

$\mathrm{尾}鈭�\frac{24}{10}=鈭�3$

$\mathrm{尾}=鈭�3+\frac{24}{10}$

Simplifying further,

$\mathrm{尾}=\frac{鈭�30+24}{10}$

localid="1657467164052" $\mathrm{尾}=\frac{鈭�6}{10}$

Substitute for $\frac{8}{5}$and $\frac{-3}{5}$for in equation (5).

${\mathrm{V}}_0\left(\mathrm{胃}\right)=\mathrm{k}\left[\frac{8}{5}{\mathrm{P}}_3(\cos 鈦�\mathrm{胃})鈭�\frac{3}{5}{\mathrm{P}}_1(\cos 鈦�\mathrm{胃})\right]$

$=\frac{\mathrm{k}}{5}\left[8{\mathrm{P}}_3(\cos 鈦�\mathrm{胃})鈭�3{\mathrm{P}}_1(\cos 鈦�\mathrm{胃})\right]$鈥︹�� (6)

Write the expression the potential inside the sphere.

$\mathrm{V}\left({\mathrm{r}}_1\mathrm{胃}\right)=\underset{\mathrm{l}=0}{\overset{\mathrm{e}}{鈭�}}鈥�{\mathrm{A}}_{\mathrm{l}}{\mathrm{r}}^{\mathrm{鈥�}}{\mathrm{P}}_{\mathrm{i}}(\cos 鈦�\mathrm{胃})$ 鈥︹��. (7)

Multiply the above equation with ${\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})$on both sides.

${\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{V}(\mathrm{r},\mathrm{胃})=\underset{\mathrm{t}=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�{\mathrm{A}}_{\mathrm{l}}{\mathrm{r}}^{\mathrm{i}}{\mathrm{P}}_1(\cos 鈦�\mathrm{胃}){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})$$鈭�\mathrm{V}(\mathrm{r},\mathrm{胃}){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}={\mathrm{A}}_{\mathrm{l}}{\mathrm{r}}^{\mathrm{鈥�}}鈭�\underset{\mathrm{l}=0}{\overset{\mathrm{胃}}{鈭�}}鈥�{\mathrm{P}}_{\mathrm{l}}(\cos 鈦�\mathrm{胃}){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})$${\mathrm{A}}_{\mathrm{i}}{\mathrm{r}}^{\mathrm{鈥�}}\left(\frac{2}{2\mathrm{l}+1}\right)={鈭�}_0^{\mathrm{a}}鈥�{\mathrm{V}}_0\left(\mathrm{胃}\right){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}$ 鈥︹�� (8

Use the condition,

${鈭�}_0^{\mathrm{a}}鈥�{\mathrm{V}}_0\left(\mathrm{胃}\right){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}=\left\{\begin{array}{l}\frac{2}{2\mathrm{l}+1}鈥呪赌呪赌呪赌�\text{if}\mathrm{I}=1\\ 0鈥呪赌呪赌呪赌�\text{if}\mathrm{I}=0\end{array}\right.$ in equation (8).

${\mathrm{A}}_1=\frac{2\mathrm{l}+1}{2{\mathrm{r}}^{\mathrm{鈥�}}}{鈭�}_0^{\mathrm{蟺}}鈥�{\mathrm{V}}_0\left(\mathrm{胃}\right){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}$

Find the value of ${\mathrm{A}}_1$.

${\mathrm{A}}_1=\frac{2\mathrm{l}+1}{2{\mathrm{R}}^{\mathrm{鈥�}}}{鈭�}_0^{\mathrm{蟺}}鈥�{\mathrm{V}}_0\left(\mathrm{胃}\right){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}$$=\frac{2\mathrm{l}+1}{2{\mathrm{R}}^{\mathrm{鈥�}}}\left[{鈭�}_0^{\mathrm{蟺}}鈥�\frac{\mathrm{k}}{5}\left\{8{\mathrm{P}}_3(\cos 鈦�\mathrm{胃})鈭�3{\mathrm{P}}_1(\cos 鈦�\mathrm{胃})\right\}{\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}\right]$

$=\frac{2\mathrm{l}+1}{2{\mathrm{R}}^{\mathrm{鈥�}}}\frac{\mathrm{k}}{5}\left[8{鈭�}_0^{\mathrm{蟺}}鈥�{\mathrm{P}}_3(\cos 鈦�\mathrm{胃}){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}鈭�3{鈭�}_0^{\mathrm{蟺}}鈥�{\mathrm{P}}_1(\cos 鈦�\mathrm{胃}){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}\right]$鈥︹�� (9)

Use the Legendre polynomial and the integral equation (9).

${鈭�}_{鈭�1}^{+1}鈥�{\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{m}}\left(\mathrm{x}\right)\mathrm{dx}={\mathrm{未}}_{\mathrm{mn}}\frac{2}{2\mathrm{n}+1}$

$=0(\text{for}\mathrm{m}鈮�\mathrm{n})$

$=\frac{2}{2\mathrm{n}+1}(\text{for}\mathrm{m}鈮�\mathrm{n})$

Use the above equation.

${\mathrm{A}}_1=\frac{2\mathrm{I}+1}{2{\mathrm{R}}^{\mathrm{鈥�}}}\frac{\mathrm{k}}{5}\left[8{鈭�}_0^{\mathrm{n}}鈥�{\mathrm{P}}_3(\cos 鈦�\mathrm{胃}){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}鈭�3{鈭�}_0^{\mathrm{n}}鈥�{\mathrm{P}}_1(\cos 鈦�\mathrm{胃}){\mathrm{P}}_1(\cos 鈦�\mathrm{胃})(\sin 鈦�\mathrm{胃})\mathrm{诲胃}\right]$${\mathrm{A}}_{\mathrm{l}}=\frac{\mathrm{k}}{5}\frac{2\mathrm{l}+1}{2{\mathrm{R}}^{\mathrm{鈥�}}}\left\{8\left(\frac{2\mathrm{l}+1}{2{\mathrm{R}}^{\mathrm{鈥�}}}\right){\mathrm{未}}_{13}鈭�3\left(\frac{2\mathrm{l}+1}{2{\mathrm{R}}^{\mathrm{l}}}\right){\mathrm{未}}_{11}\right\}$
$=\frac{\mathrm{k}}{5{\mathrm{R}}^{\mathrm{鈥�}}}\left(8{\mathrm{未}}_{13}鈭�3{\mathrm{未}}_{11}\right)$
$=\frac{鈭�3\mathrm{k}}{5\mathrm{R}}\text{if}\mathrm{I}=1$
$=\frac{8\mathrm{k}}{5{\mathrm{R}}^3}\text{if}\mathrm{I}=3$

Thus, write the potential inside the sphere.

$\mathrm{V}(\mathrm{r},\mathrm{胃})=\underset{\mathrm{i}=0}{\overset{\mathrm{a}}{鈭�}}鈥�{\mathrm{A}}_{\mathrm{i}}{\mathrm{r}}^{\mathrm{l}}{\mathrm{P}}_{\mathrm{t}}\cos 鈦�\mathrm{胃}$

localid="1657520350567" $=\frac{鈭�3\mathrm{k}}{5\mathrm{R}}{\mathrm{rP}}_1(\cos 鈦�\mathrm{胃})+\frac{8\mathrm{k}}{5{\mathrm{R}}^3}{\mathrm{r}}^3{\mathrm{P}}_3(\cos 鈦�\mathrm{胃})$鈥︹�� (10)

Substitute $\mathrm{肠辞蝉胃}$for ${\mathrm{P}}_1\left(\mathrm{肠辞蝉胃}\right)$and $\frac{1}{2}\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)$for ${\mathrm{P}}_3$in equation (10).

$\mathrm{V}(\mathrm{r},\mathrm{胃})=\frac{\mathrm{k}}{5}\left[鈭�3\left(\frac{\mathrm{r}}{\mathrm{R}}\right)\cos 鈦�\mathrm{胃}+8{\left(\frac{\mathrm{r}}{\mathrm{R}}\right)}^3\frac{\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)}{2}\right]$$=\frac{\mathrm{k}}{5}\left(\frac{\mathrm{r}}{\mathrm{R}}\right)\cos 鈦�0\left[4{\left(\frac{\mathrm{r}}{\mathrm{R}}\right)}^2\left(5{\cos }^2鈦�0鈭�3\right)鈭�3\right]$

Thus, the electric potential inside the sphere is $\frac{\mathrm{k}}{5}\left(\frac{\mathrm{r}}{\mathrm{R}}\right)\cos 鈦�0\left[4{\left(\frac{\mathrm{r}}{\mathrm{R}}\right)}^2\left(5{\cos }^2鈦�0鈭�3\right)鈭�3\right]$.

Write the potential outside the sphere $(\mathrm{r}鈮�\mathrm{R})$.

$\mathrm{V}(\mathrm{r},\mathrm{胃})=\underset{\mathrm{i}=0}{\overset{\text{a}}{鈭�}}鈥�\frac{{\mathrm{B}}_{\mathrm{i}}}{{\mathrm{r}}^{\mathrm{j}/1}}{\mathrm{P}}_1(\cos 鈦�\mathrm{胃})$ 鈥︹�� (11)

The value of role="math" localid="1657520636990" ${\mathrm{B}}_{\mathrm{i}}$is given as ,

${\mathrm{B}}_{\mathrm{j}}={\mathrm{A}}_1{\mathrm{R}}^{2/11}$

If ,$\mathrm{j}=1$then value of ${\mathrm{B}}_{\mathrm{j}}$,

${\mathrm{B}}_{\mathrm{j}}={\mathrm{A}}_{\mathrm{i}}{\mathrm{R}}^3$

${\mathrm{B}}_{\mathrm{f}}=\frac{鈭�3\mathrm{k}}{5\mathrm{R}}{\mathrm{R}}^3$

$=\frac{鈭�3{\mathrm{kR}}^2}{5}$

If $\mathrm{I}=3$then the value of ${\mathrm{B}}_2$

${\mathrm{B}}_3={\mathrm{A}}_3{\mathrm{R}}^7$

Substitute $\frac{8\mathrm{k}}{5{\mathrm{R}}^3}$for ${\mathrm{A}}_3$

${\mathrm{B}}_3=\frac{8\mathrm{k}}{5{\mathrm{R}}^3}{\mathrm{R}}^7$

$=\frac{8\mathrm{k}}{5{\mathrm{R}}^4}$

Write the expression for potential outside the sphere.

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

$=鈭�\left(\frac{3{\mathrm{kR}}^2}{5}\right)\frac{1}{{\mathrm{r}}^2}{\mathrm{P}}_1(\cos 鈦�\mathrm{胃})+\left(\frac{8{\mathrm{kR}}^4}{5}\right)\frac{1}{{\mathrm{r}}^4}{\mathrm{P}}_3(\cos 鈦�\mathrm{胃})$鈥︹�� (12

Substitute $\mathrm{肠辞蝉胃}$for ${\mathrm{P}}_1\left(\mathrm{肠辞蝉胃}\right)$and $\frac{1}{2}\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)$for ${\mathrm{P}}_3$in equation (12).

$\mathrm{V}(\mathrm{r},\mathrm{胃})=\frac{\mathrm{k}}{5}\left[鈭�3{\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2\cos 鈦�\mathrm{胃}+8{\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^4\left(\frac{5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}}{2}\right)\right]$ $=\frac{\mathrm{k}}{5}{\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2\cos 鈦�\mathrm{胃}\left[4{\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\right)鈭�3\right]$

Hence, the potential outside the sphere $\frac{\mathrm{k}}{5}{\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2\cos 鈦�\mathrm{胃}\left[4{\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\right)鈭�3\right]$.

Using the equation 3.83, write the expression surface charge density on the sphere,

$\mathrm{蟽}\left(\mathrm{胃}\right)={\mathrm{蔚}}_0\underset{\mathrm{i}=0}{\overset{\mathrm{g}}{鈭�}}鈥�(2\mathrm{l}+1){\mathrm{A}}_1{\mathrm{R}}^{\mathrm{l}鈭�1}{\mathrm{P}}_{\mathrm{i}}(\cos 鈦�\mathrm{胃})$ 鈥︹�� (13)

$={\mathrm{蔚}}_0\left(3{\mathrm{A}}_1{\mathrm{P}}_1(\cos 鈦�\mathrm{胃})+7{\mathrm{A}}_3{\mathrm{R}}^2{\mathrm{P}}_3\right)$

Substitute $\frac{鈭�3\mathrm{k}}{5\mathrm{R}}$for ${\mathrm{A}}_1$, $\cos 鈦�\mathrm{胃}$for ${\mathrm{P}}_1(\cos 鈦�\mathrm{胃})$, $\frac{8\mathrm{k}}{5{\mathrm{R}}^3}$for ${\mathrm{A}}_3$and $\frac{1}{2}\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)$ for ${\mathrm{P}}_3$in equation (13).

$\mathrm{蟽}\left(\mathrm{胃}\right)={\mathrm{蔚}}_0\left[3\left(\frac{鈭�3\mathrm{k}}{5\mathrm{R}}\right)\cos 鈦�\mathrm{胃}+7\left(\frac{8\mathrm{k}}{5{\mathrm{R}}^3}\right){\mathrm{R}}^2\frac{\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)}{2}\right]$

$=\frac{{\mathrm{蔚}}_0\mathrm{k}}{5\mathrm{R}}\left[鈭�9\cos 鈦�\mathrm{胃}+\frac{56}{2}\left(5{\cos }^3鈦�\mathrm{胃}鈭�3\cos 鈦�\mathrm{胃}\right)\right]$

$=\frac{{\mathrm{蔚}}_0\mathrm{k}}{5\mathrm{R}}\cos 鈦�\mathrm{胃}\left[140{\cos }^2鈦�\mathrm{胃}鈭�93\right]$

Hence, the surface charge density on the sphere is $\frac{{\mathrm{蔚}}_0\mathrm{k}}{5\mathrm{R}}\cos 鈦�\mathrm{胃}\left[140{\cos }^2鈦�\mathrm{胃}鈭�93\right]$.