Question

g What are the magnitude and direction of the electric field at a distance 43.5 mmmm from the center of the shells

Answer 1

Complete Question

Two spherical shells have a common center. A -1.6 x 10-6 C charge is spread uniformly over the inner shell, which has a radius of 0.030 m. A +5.1 x 10-6 C charge is spread uniformly over the outer shell, which has a radius of 0.15 m.What are the magnitude and direction of the electric field at a distance 43.5 mm from the center of the shells.

Answer:

The magnitude is  $E = 7.6021*10^{6} \N/C$

The direction is radially inward toward the center

Explanation:

From the question we are told that

   The charge on the inner shell is  $q_i = -1.6*10^{-6} \ C$

    The  radius of the inner shell is  $c_1 = 0.030 \ m$

   The  charge on the outer shell is  $q_o = 5.1*10^{-6}\ C$

     The  radius of the outer shell is  $c_2 = 0.15\ m$

   The distance considered is  $r = 43.5 \ mm = 0.0435 \ m$

Generally the electric field at the position considered is mathematically represented as

          $E = \frac{Q_c }{4\pi r^2 \epsilon_o }$

Here  $Q_c$ is the charge  which is enclosed by the distance considered which in this case is the charge on the inner shell

  So  $Q_c =q_i = -1.6*10^{-6} \ C$

Hence

          $E = \frac{-1.6 *10^{-6} }{4* 3.142 *0.0435^2* 8.85*10^{-12} }$

 =>      $E = -7.6021*10^{6} \N/C$

The negative sign is show that the direction of the field is radially inward toward the center