91Ó°ÊÓ

Write the expression for electric filed,

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{2}\mathbf{Ï€}\mathbf{s}\mathbf{I}\mathbf{}{\mathbf{Î\mu }}_{\mathbf{0}}}\hat{\mathbf{s}}for\mathit{a}\mathbf{}\mathbf{<}\mathit{s}\mathbf{}\mathbf{<}\mathit{b}$

Here, ${}^q$is the charge, ${}^E$is the electric filed.

Consider the given figure.

Now consider${}^I$is the length of the tube.

Let’s give the chargelocalid="1654330497372" ${}^{+q}$to the inner cylinder and${}^{-q}$to the outer cylinder.

Now finding the potential difference between inner and outer faces of cylinder and it is given by,

$$\begin{gathered} \mathit{V}\left(a\right)\mathbf{-}\mathit{V}\left(b\right)\mathbf{=}\mathbf{-}{\mathbf{∫}}_{\mathbf{b}}^{\mathbf{a}}\left(\frac{q}{2\mathrm{Ï€}sI{\mathrm{Î\mu }}_0}\right)\mathit{d}\mathit{s} \\ \mathit{V}\left(a\right)\mathbf{-}\mathit{V}\left(b\right)\mathbf{=}\mathit{V} \end{gathered}$$

Solve further as,

$$\begin{gathered} V=-{∫}_b^a\left(\frac{q}{2\mathrm{Ï€}sI{\mathrm{Î\mu }}_0}\right)ds \\ =-\frac{q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}I}\left(\log a-\log b\right) \\ =\frac{q}{2\mathrm{Ï€}{\mathrm{Î\mu }}_{0}I}In\left(\frac{b}{a}\right) \end{gathered}$$

$\frac{C}{I}=\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}{In\left({\displaystyle \frac{b}{a}}\right)}$Consider the expression for the potential.

Now comparing the above equation with localid="1654328482139" $V\frac{q}{C}$, here ${}^C$is the capacitance. Then,

localid="1654328603077" $\frac{C}{I}=\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}{In\left({\displaystyle \frac{b}{a}}\right)}$

${}^{\frac{C}{I}=\frac{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}{In\left({\displaystyle \frac{b}{a}}\right)}}$

Hence, the capacitance per unit length is given by,