Question

A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 5 mm , and the outer one a radius 11 mm . The common length of the cylinders is 160 m . What is the potential energy stored in this capacitor when a potential difference 6 V is

Answer 1

Answer:

The  potential energy is $PE = 2.031 *10^{-7} \ J$

Explanation:

From the question we are told that

    The inner radius is  $r_i = 5 \ mm = 0.005 \ m$

      The outer radius is  $r_o = 11 \ mm = 0.011 \ m$

     The  common length is  $l = 160 \ m$

      The  potential  difference is   $V = 6 \ V$

Generally the capacitance of the cylindrical capacitor is mathematically represented as

       $C = \frac{2 \pi * k * \epsilon_o }{ ln \frac{ r_o }{r_i} } * l$

Where  $\epsilon _o$ is the permitivity of free space with the values $\epsilon _o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

and  k  is the dielectric constant  of the dielectric material here the  dielectric material is free space so  k  =   1

     Substituting values

             $C = \frac{2* 3.142 * 1 * 8.85*10^{-12} }{ ln \frac{ 0.011}{0.005} } * 160$

             $C = 1.129 *10^{-8} \ F$

The potential energy stored is mathematically represented as

       $PE = \frac{1}{2} * C * V ^2$

substituting values

      $PE = 0.5 * 1.129 *10^{-8} * (6)^2$

      $PE = 2.031 *10^{-7} \ J$