Question

An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume. Calculate the charge enclosed by a concentric spherical surface with the following radius.(a) r = 1.00 cm(b) r = 6.50 cm

Answer 1

Explanation:

(a)   Formula to calculate the density is as follows.

            $\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

                       = $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

                     = $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

            $q_{in} = \rho(\frac{4}{3} \pi r^{3})$

                      = $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

                      = $10.106 \times 10^{-8}$ C

or,                   = 101.06 nC

(b)  For r = 6.50 cm, the value of charge will be calculated as follows.

                $q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

                          = $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

                          = 7.454 $\mu C$