Question

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)

Answer 1

Given:       $q_(initial)$  = $0$

To Find: Prove that the maximum charge is  $\frac{Qq}{Q-q}$

Solution:

At each contact, the potential both becomes same. At the first contact, charges on the conductor and the plate are $Q-q$ and $q$ respectively.

So

                $V_1 = \frac{Q-q}{C_1} = \frac{q}{C_2} \\\frac{C_1}{C_2} = \frac{Q-q}{q} ................... (i)$

In the second contact, the charge transferred to the plate is $q_2$.

Then

              ⇒   $V_2 =\frac{Q - q_2}{C_1} =\frac{ q+q_2}{C_2} \\\\ \frac{Q-q}{q} = \frac{Q-q_2}{q+q_2}$

             ⇒  $(q+q_2)(Q-q)=q(Q-q_2)$

             ⇒  $qQ-q^2+q_2Q-q_2q = qQ-qq_2$

             ⇒  $q_2Q=q^2$

              ⇒  $q^2= \frac{q^2}{Q}$

The total charge transferred in a large number of contacts is

            $q_(max) = q+ \frac{q^2}{Q} + \frac{q^3}{Q_2} + \frac{q^4}{Q_3}+ ... + \infty$

                     ⇒   $\frac{q}{1-\frac{q}{Q} } = \frac{q Q}{Q-q}$

Thus,   The maximum charge is  $\frac{Qq}{Q-q}$

Hence proved.