Answer:
sigma = 3.4*10^-4 C / m^2
Explanation:
Given:
- The net charge of the conducting sphere Q_s = 1.7 uC
- The net charge in the shell Q_shell = 9.6 uC
- Outer radius of the shell r = 4.3 cm
Find:
Determine the surface charge density on the outer surface of the shell.
Solution:
- The concept here is that the net charge on the conducting sphere must be equal to charge on inner shell. Hence, Q_s = Q_in
Q_shell = Q_in + Q_out
Q_out = Q_shell - Q_in
Q_out = 9.6 - 1.7 = 7.9 uC
- Now we can calculate the charge density of the outer shell by:
sigma = Q_out / A_out
sigma = (7.9*10^-6) / 4*pi*(0.043)^2
sigma = 3.4*10^-4 C / m^2