Question 6.9: A coaxial cable contains an insulating material of conductiv...

Consider length L of the coaxial cable as shown in Figure 6.14. Let $V_{o}$ be the potential difference between the inner and outer conductors so that $V(\rho=a)=0$ and $V(\rho=b)=V_{o}$, which allows V and E to be found just as in part (a) of Example 6.8. Hence

$J=\sigma E=\frac{-\sigma V_{o}}{\rho \ln\frac{b}{a}}a_{\rho},       dS=-\rho d\phi dza_{\rho}$

$I=\int_{S}J\cdot dS=\int_{\phi=0}^{2\pi}\int_{z=0}^{L}\frac{V_{o}\sigma}{\rho\ln\frac{b}{a}}\rho dzd\phi=\frac{2\pi L\sigma V_{o}}{\ln\frac{b}{a}}$

The resistance of the cable of length is given by

$R=\frac{V_{o}}{I}\cdot\frac{+}{L}=\frac{\ln^{\frac{b}{a}}}{-\pi \sigma}$

and the conductance per unit length is

$G=\frac{1}{R}=\frac{2\pi\sigma}{\ln(\frac{b}{a})}$

as required.