A coaxial cable used in a transmission line has an inner radius of 0.10 mmand an outer radius of 0.60 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with polystyrene.

a) Length of the transmission line, L = 1m

b) Inner radius of the transmission line, a = 0.10 mm

c) Outer radius of the transmission line, b = 0.60 mm

d) Dielectric constant of polystyrene, k = 2.6

Capacitance per unit length of (cylindrical capacitor) coaxial cable can be calculated using the formula of capacitance for a cylindrical capacitor with dielectric.

Formulae:

The electric flux passing through a surface,$\varphi \overrightarrow{E}.d\overrightarrow{s}=\frac{q_{enclosed}}{{{\displaystyle \stackrel{,}{o}}}_0}$ …(i)

The potential difference due to the electric field,$V=-{\int }_b^a\overrightarrow{E}.d\overrightarrow{r}$ …(ii)

The capacitance due to stored charge,$C=\frac{q}{V}$ …(iii)

Where

${\epsilon }_0=8.85\times {10}^{-12}\frac{C^2}{N.m^2}$is permittivity of free space

Let us consider a coaxial cylinder of inner radius and outer radius b . Let q and -q be the total charges on inner and outer cylinder respectively. Now using Gauss’s law, we can find out the electric field using equation (i) and the surface area of the cylinder at the point r from the axis of the inner cylinder is given as:

$$\begin{gathered} E\left(2\pi rL\right)=\frac{q}{{{\displaystyle \stackrel{,}{o}}}_0} \\ E=\frac{q}{\left(2\pi rL\right){\stackrel{,}{o}}_0} \end{gathered}$$

Now, the potential difference at distance r using the above expression in equation (ii) is given as:

$$\begin{gathered} V=\frac{-q}{\left(2\pi rL\right){\stackrel{,}{o}}_0}{\int }_b^a\frac{dr}{r} \\ =\frac{-q}{\left(2\pi rL\right){\stackrel{,}{o}}_0}\ln \left(\frac{b}{a}\right) \end{gathered}$$

Thus, the value of the capacitance using the above value of potential difference is given using equation (iii) as follows:

$C=\frac{2\pi r{\stackrel{,}{o}}_{0}L}{in\left(\frac{b}{a}\right)}$

For the dielectric medium, we need to only multiply this by, so we have the required expression of the new capacitance per meter as follows:

$$\begin{gathered} \mathrm{C}=\mathrm{K}\frac{2\mathrm{\pi r}{\stackrel{,}{\mathrm{o}}}_0\mathrm{L}}{\mathrm{in}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)} \\ =\mathrm{K}.\frac{2\mathrm{\pi r}{\stackrel{,}{\mathrm{o}}}_0}{\mathrm{in}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)} \\ =\frac{2.6\times 2\mathrm{\pi }\left(8.85\times {10}^{-12}{\displaystyle \frac{{\mathrm{C}}^2}{\mathrm{N}.{\mathrm{m}}^2}}\right)}{\ln \left({\displaystyle \frac{0.60\mathrm{mm}}{0.10\mathrm{mm}}}\right)} \\ =81\mathrm{pF}.{\mathrm{m}}^{-1} \end{gathered}$$

Hence, the required value of capacitance per meter is$81\mathrm{pF}.{\mathrm{m}}^{-1}$ .