Question

A small conducting sphere A of radius \( r \) charged to a potential \( V \), is enclosed by a spherical conducting shell B of radius \( R \). If A and B are connected by a thin wire, calculate the final potential on sphere A and shell B.

Hint:

When two conductors are connected by a wire, they must be at the same potential, which is achieved by redistributing the charge between them.

Answer 1

Final Potential on Sphere A and Shell B

Given:

• Radius of sphere A: \( r \)
• Radius of shell B: \( R \)
• Initial potential on sphere A: \( V \)
• Spheres A and B are connected by a thin wire.

Solution:

1. Initial Setup:

Sphere A initially possesses a charge \( Q_A \) corresponding to potential \( V \). Shell B is initially uncharged. Upon connection via a wire, charge redistribution occurs until both conductors reach an equipotential state, as the wire enforces zero potential difference.

2. Charges and Potentials:

The potential of a charged sphere is given by: \[ V = \frac{kQ}{r} \] where \( k \) is Coulomb's constant \( \left( k = \frac{1}{4 \pi \epsilon_0} \right) \), \( Q \) is the charge, and \( r \) is the radius.

3. Charge on Sphere A:

The initial charge on sphere A is: \[ Q_A = \frac{V r}{k} \]

4. Final Situation:

After connection, charge flows until both sphere A and shell B attain a common final potential, denoted as \( V_f \).

5. Total Charge:

Charge conservation dictates that the total charge remains constant. Initially, this total charge is solely on sphere A: \[ Q_{\text{total}} = Q_A = \frac{V r}{k} \]

6. Potential on Sphere A and Shell B:

The final potentials are: \[ V_A = \frac{k Q_A'}{r} \] and \[ V_B = \frac{k Q_B'}{R} \], where \( Q_A' \) and \( Q_B' \) are the final charges. Due to equipotential, \( V_A = V_B = V_f \). A common misconception is that \( V_f = \frac{k Q_A}{R} \) or \( V_f = \frac{k Q_A}{r} \).

7. Final Potential Calculation:

Utilizing charge conservation \( Q_A' + Q_B' = Q_A \) and equipotential \( \frac{Q_A'}{r} = \frac{Q_B'}{R} \), and substituting into the potential equation leads to the correct final potential. The formula \( V_f = \frac{k Q_A}{r + R} \) is incorrect for this configuration. The correct derivation leads to: \[ V_f = \frac{V r}{r + R} \]

Conclusion:

The final common potential on sphere A and shell B is \( V_f \): \[ V_f = \boxed{\frac{V r}{r + R}} \]