91Ó°ÊÓ

Consider the formula for the gauss law for the electric displacement as follows:

$\mathbf{∫}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{D}}\mathbf{.}\mathit{d}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{a}}\mathbf{=}\mathit{Q}$

Here, D is the electric displacement, $d\stackrel{â‡\P Ä}{a}$ is the area of element and Q is the charge that is enclosed.

Consider the expression for charge displacement as follows:

$\begin{array}{rcl}∫\mathit{D}\mathbf{.}\mathit{d}\mathit{a}& =& D\left(2\mathrm{π}sl\right)\\ D\left(2\mathrm{π}sl\right)& =& Q\\ D& =& \frac{Q}{2\mathrm{π}sl}\end{array}$

Consider the electric field for the range a < s < b is as follows:

role="math" localid="1658728408200" $\begin{array}{rcl}E& =& \frac{D}{{\mathrm{Î\mu }}_0}\\ & =& \frac{Q}{2_{{\mathrm{Ï€Î\mu }}_{0}sl}}\end{array}$

Consider the electric field for the range b < r < c is as follows:

$\begin{array}{rcl}E& =& \frac{D}{{\mathrm{Î\mu }}_0{\mathrm{Î\mu }}_r}\\ & =& \frac{Q}{2\mathrm{Ï€Î\mu }sl}\end{array}$

Solve for the potential difference as:

role="math" localid="1658728840392" $\begin{array}{rcl}V& =& -{∫}_c^{a}Edl\\ & =& {∫}_c^a\left(\frac{Q}{2{\mathrm{Ï€Î\mu }}_0\mathrm{l}}\right)\frac{ds}{s}+{∫}_c^a\left(\frac{Q}{2\mathrm{Ï€Î\mu \pm ô}}\right)\frac{ds}{s}\\ & =& \frac{Q}{2{\mathrm{Ï€Î\mu }}_0\mathrm{l}}\left[{\left[Ins\right]}_a^b+\frac{{\mathrm{Î\mu }}_0}{\mathrm{Î\mu }}{\left[Ins\right]}_b^c\right]\\ & =& \frac{Q}{2{\mathrm{Ï€Î\mu }}_0\mathrm{l}}\left[In\frac{b}{a}+\frac{{\mathrm{Î\mu }}_0}{\mathrm{Î\mu }}In\frac{c}{b}\right]\end{array}$

Solve further as:

$V=\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{Q}{2{\mathrm{Ï€Î\mu }}_0\mathrm{l}}\end{array}\begin{array}{rcl}& & \left[In\frac{b}{a}+\frac{1}{{\mathrm{Î\mu }}_r}In\frac{c}{b}\right]\end{array}$

Consider the formula for the capacitance per unit length:

$\frac{c}{l}=\frac{Q}{VI}$

Substitute the values and solve as:

role="math" localid="1658729102645" $\begin{array}{rcl}\frac{C}{I}& =& \frac{Q}{I\left[{\displaystyle \frac{Q}{2{\mathrm{Ï€Î\mu }}_0\mathrm{l}}}\left[In{\displaystyle \frac{b}{a}}+{\displaystyle \frac{1}{{\mathrm{Î\mu }}_r}}In{\displaystyle \frac{c}{b}}\right]\right]}\\ & =& \frac{2{\mathrm{Ï€Î\mu }}_0\mathrm{I}}{In{\displaystyle \frac{b}{a}}+{\displaystyle \frac{1}{{\mathrm{Î\mu }}_r}}In{\displaystyle \frac{c}{b}}}\end{array}$

Therefore, the capacitance per unit length is $\begin{array}{rcl}& & \frac{2{\mathrm{Ï€Î\mu }}_0\mathrm{I}}{In{\displaystyle \frac{b}{a}}+{\displaystyle \frac{1}{{\mathrm{Î\mu }}_r}}In{\displaystyle \frac{c}{b}}}\end{array}$.