Question

Charge density

$\mathit{\sigma }\left(\varphi \right)\mathbf{=}\mathit{a}\mathit{s}\mathit{i}\mathit{n}\left(5\varphi \right)$

(whereais a constant) is glued over the surface of an infinite cylinder of radiusR

(Fig. 3.25). Find the potential inside and outside the cylinder. [Use your result from Prob. 3.24.]

Short answer

Answer

The potential inside the infinite cylinder of radius R having surface charge density $\mathit{\sigma }\mathbf{\left(}\mathbf{\varphi }\mathbf{\right)}\mathbf{=}\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathbf{5}\mathbf{\varphi }\mathbf{\right)}\mathbf{10}{\mathit{\epsilon }}_{\mathbf{0}}{\mathit{R}}^{\mathbf{4}}$is $\frac{\mathbf{asin}\mathbf{5}\mathbf{\varphi }\mathbf{}}{\mathbf{10}{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{{\mathbf{s}}^{\mathbf{5}}}{{\mathbf{R}}^{\mathbf{4}}}$. The potential outside the cylinder is $\frac{\mathbf{asin}\mathbf{5}\mathbf{\varphi }\mathbf{}}{\mathbf{10}{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{{\mathbf{R}}^{\mathbf{6}}}{{\mathbf{s}}^{\mathbf{5}}}$.

Step-by-step solution

Given data

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

Potential inside and outside a cylinder

In cylindrical coordinates, the potential inside a cylinder is

${\mathit{V}}_{\mathbf{i}\mathbf{n}}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{s}}^{\mathbf{k}}\left(a_{k}cosk\varphi +b_{k}sink\varphi \right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here,,andare constants.

In cylindrical coordinates, the potential outside a cylinder is

${\mathit{V}}_{\mathbf{out}}\mathbf{=}{\overline{\mathbf{a}}}_{\mathbf{0}}\mathbf{+}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{s}}^{\mathbf{-}\mathbf{k}}\mathbf{(}{\mathbf{c}}_{\mathbf{k}}\mathbf{cosk\varphi }\mathbf{+}{\mathbf{d}}_{\mathbf{k}}\mathbf{sink\varphi }\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

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

Potential inside and outside cylinder for σ(ϕ)=asin(5ϕ) on the surface

The surface charge on the cylinder is

$\mathit{\sigma }\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}{\left(\frac{\partial V_{out}}{\partial s}-\frac{\partial V_{in}}{\partial s}\right)}_{\mathbf{s}\mathbf{r}}$

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

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

$$\begin{gathered} \mathit{a}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(5\varphi \right)\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}{\left[\begin{array}{c}\frac{\partial }{\partial s}\left(a_0+\sum _{k=1}^{\infty }{s}^k\left(a_{k}cosk\varphi +b_{k}sink\varphi \right)\right)\\ -\frac{\partial }{\partial s}\left({\overline{a}}_0+\sum _{k=1}^{\infty }{s}^{-k}\left(c_{k}cosk\varphi +d_{k}sink\varphi \right)\right)\end{array}\right]}_{\mathbf{s}\mathbf{=}\mathbf{R}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\left[-k{R}^{-k-1}\left(c_{k}cosk{\varphi }_{k}sink\varphi \right)-k{R}^{k-1}\left(a_{k}cosk\varphi +b_{k}sink\varphi \right)\right] \end{gathered}$$

Compare both sides to get

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

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

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

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

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

Compare the two sides and get

${\mathit{a}}_{\mathbf{0}}\mathbf{=}{\overline{\mathbf{a}}}_{\mathbf{0}}$

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),

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

Thus, the potential is

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