An insulated conductor initiallly free from charge is charged by repeated contacts with a plate which after each contact is replenished to a charge `Q`. If `q` is the charge on the conductor after first operation prove that the maximum charge which can be given to the conductor in this way is `(Qq)/(Q-q)`

Explanation

Let `C_1` be the capacity of plate and `C_2` that of the conductor. After first contact charge on conductor is `q`. Therefore, charge on plate will remain `Q-q`. As the charge redistributes in the ratio of capacities.
`(Q-q)/q=C_1/C_2`………i
Let `q_n` be the maximum charge which can be given to the conductor. The flow of charge from the plate to the conductor will stop when
`V_("conductor")=V_(plate)`
`q_m/C_2=Q/C_1implies q_m=(C_2/C_1)Q`
Substituting `C_2/C_1` from eqn i we get
`q_m=(Qq)/(Q-q)`