Question

A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm) which has a net charge of –4.0 nC. Determine the resulting charge density on the inner surface of the conducting sphere.

Answer 1

Explanation:

The given data is as follows.

             q = 6.0 nC = $6 \times 10^{-9} C$

         inner radius (r) = 1.0 cm = 0.01 m   (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

            $\sigma = \frac{q_{in}}{A}$ .......... (1)

Since, area of the sphere is as follows.

               A = $4 \pi r^{2}$ ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

      $\sigma = \frac{q_{in}}{4 \pi r^{2}}$

                   = $\frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}$            

                   = $0.477 \times 10^{-5}$

or,               = 4.77 $\mu C/m^{2}$

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 $\mu C/m^{2}$.