91Ó°ÊÓ

A Charge density$\mathit{σ}\mathbf{\left(}\mathbf{ϕ}\mathbf{\right)}\mathbf{=}\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{5}\mathbf{ϕ}\mathbf{\right)}$ (where a is a constant) is glued over the surface of an infinite cylinder of radius R.

In cylindrical coordinates, the potential inside a cylinder is

${\mathit{V}}_{\mathbf{i}\mathbf{n}}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}\underset{\mathbf{k}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathit{s}}^{\mathbf{k}}\left(a_{k}cosk\mathrm{ϕ}+b_{k}sink\mathrm{ϕ}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here,,andare constants.

In cylindrical coordinates, the potential outside a cylinder is

${\mathit{V}}_{\mathbf{out}}\mathbf{=}{\stackrel{\mathbf{¯}}{\mathbf{a}}}_{\mathbf{0}}\mathbf{+}\underset{\mathbf{k}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathit{s}}^{\mathbf{-}\mathbf{k}}\mathbf{(}{\mathbf{c}}_{\mathbf{k}}\mathbf{³¦´Ç²õ°ìÏ•}\mathbf{+}{\mathbf{d}}_{\mathbf{k}}\mathbf{²õ¾\pm ²Ô°ìÏ•}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here, ${\stackrel{\mathbf{¯}}{\mathbf{a}}}_{\mathbf{0}}$, ${\mathit{c}}_{\mathbf{k}}$and ${\mathit{d}}_{\mathbf{k}}$are constants.

The surface charge on the cylinder is

$\mathit{σ}\mathbf{=}\mathbf{-}{\mathit{Î\mu }}_{\mathbf{0}}{\left(\frac{∂V_{out}}{∂s}-\frac{∂V_{in}}{∂s}\right)}_{\mathbf{s}\mathbf{r}}$

Here, ${\mathit{Î\mu }}_{\mathbf{0}}$is the permittivity of free space.

Substitute in the above equation from equations (1) and (2)

$$\begin{gathered} \mathit{a}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(5\mathrm{Ï•}\right)\mathbf{=}\mathbf{-}{\mathit{Î\mu }}_{\mathbf{0}}{\left[\begin{array}{c}\frac{∂}{∂s}\left(a_0+\underset{k=1}{\overset{∞}{∑}}{s}^k\left(a_{k}cosk\mathrm{Ï•}+b_{k}sink\mathrm{Ï•}\right)\right)\\ -\frac{∂}{∂s}\left({\stackrel{¯}{a}}_0+\underset{k=1}{\overset{∞}{∑}}{s}^{-k}\left(c_{k}cosk\mathrm{Ï•}+d_{k}sink\mathrm{Ï•}\right)\right)\end{array}\right]}_{\mathbf{s}\mathbf{=}\mathbf{R}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}{\mathit{Î\mu }}_{\mathbf{0}}\underset{\mathbf{k}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}\left[-k{R}^{-k-1}\left(c_{k}cosk{\mathrm{Ï•}}_{k}sink\mathrm{Ï•}\right)-k{R}^{k-1}\left(a_{k}cosk\mathrm{Ï•}+b_{k}sink\mathrm{Ï•}\right)\right] \end{gathered}$$

Compare both sides to get

$$\begin{gathered} {\mathit{a}}_{\mathbf{k}}\mathbf{=}\mathbf{0} \\ {\mathit{c}}_{\mathbf{k}}\mathbf{=}\mathbf{0} \\ {\mathit{b}}_{\mathbf{k}}\mathbf{=}\mathbf{0}\left(k≠5\right) \\ {\mathit{d}}_{\mathbf{k}}\mathbf{=}\mathbf{0}\left(k≠5\right) \\ \mathit{a}\mathbf{=}\mathbf{5}{\mathit{Î\mu }}_{\mathbf{0}}\left(\frac{d_5}{R^6}+R^4{b}_5\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(3\right) \end{gathered}$$

Put these back in equations (1) and (2) to get

$$\begin{gathered} {\mathit{V}}_{\mathbf{i}\mathbf{n}}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{s}}^{\mathbf{5}}{\mathit{b}}_{\mathbf{5}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{5}\mathit{ϕ} \\ {\mathit{V}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\mathbf{=}{\stackrel{\mathbf{¯}}{\mathbf{a}}}_{\mathbf{0}}\mathbf{+}{\mathit{s}}^{\mathbf{-}\mathbf{5}}{\mathit{d}}_{\mathbf{5}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{5}\mathit{ϕ} \end{gathered}$$

Also, the potential has to be continuous at the surface, that is,

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\overline{){\mathbf{V}}_{\mathbf{i}\mathbf{n}}}}_{\mathbf{s}\mathbf{=}\mathbf{R}}\mathbf{=}{\overline{){\mathbf{V}}_{\mathbf{o}\mathbf{u}\mathbf{t}}}}_{\mathbf{s}\mathbf{=}\mathbf{R}} \\ {\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{R}}^{\mathbf{5}}{\mathit{b}}_{\mathbf{5}}\mathit{s}\mathit{i}\mathit{n}\mathbf{5}\mathit{ϕ}\mathbf{=}{\stackrel{\mathbf{¯}}{\mathbf{a}}}_{\mathbf{0}}\mathbf{+}{\mathit{R}}^{\mathbf{-}\mathbf{5}}{\mathit{d}}_{\mathbf{5}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{5}\mathit{ϕ} \end{gathered}$$

Compare the two sides and get

${\mathit{a}}_{\mathbf{0}}\mathbf{=}{\stackrel{\mathbf{¯}}{\mathbf{a}}}_{\mathbf{0}}$

Both of these can be chosen to be zero

${\mathit{R}}^{\mathbf{5}}{\mathit{b}}_{\mathbf{5}}\mathbf{=}{\mathit{R}}^{\mathbf{-}\mathbf{5}}{\mathit{d}}_{\mathbf{5}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(4\right)$

From equations (3) and (4),

$$\begin{gathered} {\mathit{b}}_{\mathbf{5}}\mathbf{=}\frac{\mathbf{a}}{\mathbf{10}{\mathbf{Î\mu }}_{\mathbf{0}}{\mathbf{R}}^{\mathbf{4}}} \\ {\mathit{d}}_{\mathbf{5}}\mathbf{=}\frac{\mathbf{a}\mathbf{R}}{\mathbf{10}{\mathbf{Î\mu }}_{\mathbf{0}}} \end{gathered}$$

Thus, the potential is

$$\begin{gathered} {\mathit{V}}_{\mathbf{i}\mathbf{n}}\mathbf{=}\frac{\mathbf{asin}\mathbf{5}\mathbf{Ï•}\mathbf{}}{\mathbf{10}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{{\mathbf{s}}^{\mathbf{5}}}{{\mathbf{R}}^{\mathbf{4}}} \\ {\mathit{V}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\mathbf{=}\frac{\mathbf{asin}\mathbf{5}\mathbf{Ï•}\mathbf{}}{\mathbf{10}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{{\mathbf{R}}^{\mathbf{6}}}{{\mathbf{s}}^{\mathbf{5}}} \end{gathered}$$.