Question

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air. The potential at the outer conductor is 1.5 kV relative to the inner conductor. What is line charge density on the positively charged conductor

Answer 1

Answer:

The value is  $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

   The radius of the inner conductor  is  $r_1 = 1 \ mm = 0.001 \ m$

    The radius of the outer conductor is  $r_2 = 3 \ mm = 0.003 \ m$

    The potential at the outer conductor is  $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

      $C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value  $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=>   $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=>   $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential  of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line  charge density of the outer  conductor is mathematically represented as

      $\rho_l = C * V$

=>   $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=>   $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

        $\rho_s = \frac{\rho_l }{2 \pi * r_2 }$    here $2 \pi r = (circumference \ of \ outer \ conductor )$

=>    $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=>    $\rho_s = 4.026 *10^{-6} \ C/m^2$