Question

Two charged concentric spheres have radii of 0.008 m and 0.018 m. The charge on the inner sphere is 3.62 10-8 C and that on the outer sphere is 1.62 10-8 C. Find the magnitude of the electric field (in N/C) at 0.012 m.

Answer 1

Answer:

The electric field is  $E = 2.2625 *10^{6} \ N/C$

Explanation:

From the question we are told that

       The radius of the inner sphere  is  $r_1 = 0.008\ m$

        The radius of the outer sphere is $r _2 = 0.018 \ m$

       The charge on the inner sphere is  $q_1 = 3.62 *10^{-8} \ C$

        The charge on the outer sphere is  $q_2 = 1.62 *10^{-8} \ C$

        The position from the  origin is $d = 0.012 \ m$

Generally the electric field is mathematically represented as

        $E = \frac{k (q_1 )}{ r^2}$

The reason for using  $q_1$ for the calculation is due to the fact that the position considered is greater than the $r_1$ but less than $r_2$

 Here k is the Coulomb constant with value   $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A{-2}$

    So  

         $E = \frac{9*10^9 (3.62 *10^{-8}}{0.012^2}$

         $E = 2.2625 *10^{6} \ N/C$