Question

A coaxial cable used in a transmission line has an inner radius of 0.12 mm and an outer radius of 0.76 mm. Calculate the capacitance per meter for the cable. Assume that the space between the conductors is filled with a material with a dielectric constant of 2.9.

Answer 1

The capacitance of the cylindrical capacitor is given by \(\mathrm{C}=\mathrm{K} \mathrm{C}_{0}=\frac{2 \pi \mathrm{K} \varepsilon_{0} \mathrm{~L}}{\ln (b / a)}\)

Where \(\mathrm{C}_{0}\) is the capacitance with out the dielectic, \(\mathrm{K}\) is the di electric constant, \(L\) is the length, a is the inner radius, and \(b\) is the outer radius.

The capacitance per unit length of the cable is

$$ \frac{\mathrm{C}}{\mathrm{L}}=\frac{2 \pi \mathrm{K} \varepsilon_{0}}{\ln (\mathrm{b} / \mathrm{a})} $$

inner radius \(a=0.12 \mathrm{~mm}\)

and the outer radius \(b=0.76 \mathrm{~mm}\)

di-electric constant \(\mathrm{K}=2.9\)

$$ \begin{aligned} \text { Hence } & \frac{\mathrm{C}}{\mathrm{L}}=\frac{2 \pi(2.9)\left(8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)}{\ln (0.76 \mathrm{~mm} / 0.12 \mathrm{~mm})} \\ \Rightarrow \frac{\mathrm{C}}{\mathrm{L}} &=\frac{161.1762 \times 10^{-12}}{1.8458} \\ \Rightarrow \frac{\mathrm{C}}{\mathrm{L}} &=87.32 \times 10^{-12} \mathrm{~F} / \mathrm{m} \\ &=87.32 \mathrm{pF} / \mathrm{m} \approx 87 \mathrm{pF} / \mathrm{m} \end{aligned} $$