Question

A thin spherical conducting shell of radius \( R \) has a charge \( q \). A point charge \( Q \) is placed at the center of the shell. Find:
(i) The charge density on the outer surface of the shell.
(ii) The potential at a distance \( R/2 \) from the center of the shell.

Hint:

For a conducting shell, the electric field inside is zero, and the potential remains constant throughout the interior.

Answer 1

Charge Density and Potential in a Conducting Shell 
 

1: Understanding the Charge Distribution
- The conducting shell redistributes its charge to maintain electrostatic equilibrium.
- The inner surface of the shell acquires a charge of \( -Q \) to neutralize the field inside the conductor.
- The outer surface must therefore carry a charge: \[ q_{\text{outer}} = q + Q \] 
2: Surface Charge Density on Outer Surface The surface charge density \( \sigma \) is given by: \[ \sigma = \frac{\text{Charge on outer surface}}{\text{Surface area of the sphere}} \] \[ \sigma = \frac{q + Q}{4\pi R^2} \] Thus, the charge density on the outer surface is: \[ \boxed{\sigma = \frac{q + Q}{4\pi R^2}} \] 
3: Potential at \( R/2 \) from the Center
- Inside a conducting shell, the potential is uniform and equal to the potential at the surface.
- The potential at the surface is given by: \[ V = \frac{1}{4\pi \epsilon_0} \left( \frac{Q}{R} + \frac{q}{R} \right) \] \[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q + q}{R} \] Since the entire region inside the shell has the same potential, the potential at \( R/2 \) is the same as at the surface: \[ \boxed{V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q + q}{R}} \] 
4: Conclusion
- Charge density on the outer surface: \( \frac{q + Q}{4\pi R^2} \)
- Potential at \( R/2 \): \( \frac{1}{4\pi \epsilon_0} \cdot \frac{Q + q}{R} \)