Question

problem 1
two concentric spheres are shown in the figure. the inner sphere is a solid non - conductor and carries a charge of 45.0 nc uniformly distributed over its outer surface. the outer sphere is a conducting shell that carries a net charge of - 8.00 nc. no other charges are present. the radii shown in the figure have the values r1 = 10.0 cm, r2 = 20.0 cm, and r3 = 30.0 cm. (k = 8.99×10^9 n·m²/c²)
(a) find the total excess charge on the inner and outer surfaces of the conducting sphere. explain your reason.
(b) find the magnitude and direction of the electric field at the following distances r from the center of the inner sphere: (i) r = 9.0 cm, (ii) r = 15.0 cm. (give information, write equations, and answer)

Explanation:

Step1: Use Gauss's law for conductors in electrostatic equilibrium.

In electrostatic equilibrium, the electric - field inside a conductor is zero. Consider a Gaussian surface inside the conducting shell. The net flux through this Gaussian surface is zero. By Gauss's law $\varPhi_E=\frac{Q_{enclosed}}{\epsilon_0}=0$. Let the charge on the inner sphere be $Q_1 = 45.0\ \mu C$ and the charge on the conducting shell be $Q_2=- 6.00\ \mu C$.

Step2: Determine the charge on the inner surface of the shell.

To make the electric - field inside the conducting shell zero, the charge induced on the inner surface of the shell $q_{inner}$ must be such that the net charge enclosed by a Gaussian surface just inside the shell is zero. So, $q_{inner}=-Q_1=- 45.0\ \mu C$.

Step3: Determine the charge on the outer surface of the shell.

The total charge on the shell is $Q_2$. Let the charge on the outer surface be $q_{outer}$. We know that $Q_2 = q_{inner}+q_{outer}$. So, $q_{outer}=Q_2 - q_{inner}$. Substituting $Q_2=-6.00\ \mu C$ and $q_{inner}=-45.0\ \mu C$, we get $q_{outer}=-6.00\ \mu C-(-45.0\ \mu C)=39.0\ \mu C$.

Answer:

a) Charge on the inner surface: $-45.0\ \mu C$
b) Charge on the outer surface: $39.0\ \mu C$