Question

A coaxial cable contains an insulating material of conductivity $\sigma$ .If the radius of the central wire is a and that of the sheet is b, then the conductance per unit length is

• A. $\frac{\ln \left(\frac{b}{a}\right)}{2\pi \sigma V_0}$
• B. $\frac{4\pi \sigma }{\ln \left(\frac{b}{a}\right)}$
• C. $\frac{2\pi \sigma }{\ln \left(\frac{a}{b}\right)}$
• D. $\frac{2\pi \sigma }{\ln \left(\frac{b}{a}\right)}$

Answer 1

The correct option is D $\frac{2\pi \sigma }{\ln \left(\frac{b}{a}\right)}$

Consider the length of the coaxial cable as L

Let $V_0$ be the potential difference between the inner and outer conductors so that,

$V_{(\rho =a)}=0$

and $V_{(\rho =b)}=V_0$

$\overrightarrow{J}=\sigma \overrightarrow{E}=\frac{-\sigma V_0}{\rho \ln \frac{b}{a}}{\hat{a}}_p$

$\overrightarrow{dS}=-\rho d\varphi dz{\hat{a}}_p$

$I=\int \overrightarrow{J}.\overrightarrow{dS}={\int }_{\Phi =0}^{2\pi }{\int }_{z=0}^L\frac{V_0\sigma }{\rho \ln \frac{b}{a}}\rho d\varphi dZ=\frac{2\pi L\sigma V_0}{\ln \frac{b}{a}}$

Resistance per unit lengh , $\frac{R}{L}=\frac{V_0}{I}=\frac{\ln \left(\frac{b}{a}\right)}{2\pi \sigma }$

The conductance per unit length is,

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