Question

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/eo, with Qin/e, where ε is the permittivity of the material. (Technically, Eo is called the vacuum permittivity.) Suppose that a 70 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C.
What is the permittivity of rubber?

Answer 1

Answer:

The permittivity of rubber is  $\epsilon = 8.703 *10^{-11}$

Explanation:

From the question we are told that

     The  magnitude of the point charge is  $q_1 = 70 \ nC = 70 *10^{-9} \ C$

      The diameter of the rubber shell is  $d = 32 \ cm = 0.32 \ m$

       The Electric field inside the rubber shell is  $E = 2500 \ N/ C$

The radius of the rubber is  mathematically evaluated as

              $r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m$

Generally the electric field for a point  is in an insulator(rubber) is mathematically represented as

         $E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}$

Where $\epsilon$ is the permittivity of rubber

    =>     $E * \epsilon * 4 * \pi * r^2 = Q$

   =>      $\epsilon = \frac{Q}{E * 4 * \pi * r^2}$

substituting values

            $\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}$

            $\epsilon = 8.703 *10^{-11}$