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.

Answer 1

Since sphere A is initially at potential V, its charge is:

Q_A = `V_r/k`

Where, k = 9 × 10⁹ Nm² C⁻²

The outer shell B is initially uncharged, so its charge:

Q_B = 0

When A and B are connected by a thin wire, the charge redistributes such that both reach the same final potential V_f.

The potential of a conducting sphere of radius r with charge `Q_A^'` is:

V_A = `(kQ_A^')/r`

The potential of the outer shell B with charge `Q_B^'` is:

V_B = `(kQ_B^')/R`

Since, they are connected, V_A = V_B = V_f and charge is conserved:

`Q_A^' = Q_B^' = Q_A`

Substituting for potentials:

`(kQ_A^')/r = (kQ_B^')/R = V_f`

Solving for `Q_A^'` and `Q_B^'`: 

`Q_A^' = (V_f r)/k` and

`Q_B^' = (V_fR)/k`

Using charge conservation:

`(V_f r)/k + (V_f R)/k = (Vr)/k`

V_f(r + R) = Vr

V_f = `(Vr)/((r + R))`