Question

A thin spherical shell of radius 15 cm carries 4.8 μC, distributed uniformly over its surface. At the center of the shell is a point charge. Part A If the electric field at (just barely outside) the surface of the sphere is 750 kN/C and points outward, what is the charge of the point charge?

Answer 1

Explanation:

It is known that on the surface of the shell the electric field will be the sum of electric field due to the shell and due to the point charge.

          $E_{shell} + E_{p.cha} = 750 \times 10^{3} N/C$

    $\frac{kq_{shell}}{r^{2}} + \frac{kq_{p.ch}}{r^{2}} = 750 \times 10^{3} N/C$

    $(q_{shell} + q_{p.cha}) = \frac{750 \times 10^{3} N/C}r^{2}{k}$

    $4.8 \times 10^{-6} C + q_{p.cha} = \frac{(750 \times 10^{3}N/C)(15 \times 10^{-2}m)^{2}}{9 \times 10^{9} Nm^{2}/C^{2}}$

     $q_{p.cha} = -2.93 \times 10^{-6} C (\frac{1.0 \muC}{10^{-6}C})$

                = $-3.0 \muC$

Thus, we can conclude that charge of the point charge is $-3.0 \muC$.