Question

A certain coaxial cable consists of a copper wire, radius a, surrounded by a concentric copper tube of inner radius c (Fig. 4.26). The space between is partially filled (from b out to c) with material of dielectric constant ${\mathbf{\in }}_{\mathbf{r}}$, as shown. Find the capacitance per unit length of this cable.

Short answer

The capacitance per unit length is $\frac{2\mathrm{\pi }{\in }_0\mathrm{I}}{In{\displaystyle \frac{b}{a}}+{\displaystyle \frac{1}{{\in }_r}}In{\displaystyle \frac{c}{b}}}$.

Step-by-step solution

Define the formulas

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

$\mathbf{\int }\stackrel{\mathbf{\rightharpoonup }}{\mathbf{D}}\mathbf{.}\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}\mathbf{=}\mathit{Q}$

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

Solve for the capacitance per unit length as:

Consider the expression for charge displacement as follows:

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

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

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

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

$\begin{array}{rcl}E& =& \frac{D}{{\epsilon }_0{\epsilon }_r}\\ & =& \frac{Q}{2\mathrm{\pi \epsilon }sl}\end{array}$

Solve for the potential difference as:

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

Solve further as:

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

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" $\begin{array}{rcl}\frac{C}{I}& =& \frac{Q}{I\left[{\displaystyle \frac{Q}{2{\mathrm{\pi \epsilon }}_0\mathrm{l}}}\left[In{\displaystyle \frac{b}{a}}+{\displaystyle \frac{1}{{\epsilon }_r}}In{\displaystyle \frac{c}{b}}\right]\right]}\\ & =& \frac{2{\mathrm{\pi \epsilon }}_0\mathrm{I}}{In{\displaystyle \frac{b}{a}}+{\displaystyle \frac{1}{{\epsilon }_r}}In{\displaystyle \frac{c}{b}}}\end{array}$

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