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

The conducting sphere A is connected to the spherical conducting shell B by a thin wire. Since the shells are conductors, the potential on sphere A and shell B will be the same due to the flow of charge between them to maintain the same potential. When the two spheres are connected by a wire, charge will flow from sphere A to shell B until both spheres reach the same potential. The total charge is redistributed between the two spheres. The potential on sphere A and shell B is calculated using the formula for the potential of a spherical conductor: \[ V = \frac{Q}{4 \pi \varepsilon_0 r} \] Where: - \( Q \) is the charge on the conductor, - \( r \) is the radius of the conductor. Since both spheres reach the same potential, the charge is distributed in such a way that: \[ V_A = V_B = \frac{Q_{\text{total}}}{4 \pi \varepsilon_0 (r + R)} \] Thus, the final potential on both sphere A and shell B is the same, and it depends on the total charge and the sum of the radii of the two conductors.