Question

A conductor initially 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, then the maximum charge which can be given to the conductor in this way will be:
(When 2 conductors are in contact, potential on them becomes equal)

• A. $\frac{3Qq}{2Q+q}$
• B. $\frac{2Qq}{Q+q}$
• C. $\frac{Qq}{Q+q}$
• D. $\frac{Qq}{Q-q}$

Answer 1

The correct option is D $\frac{Qq}{Q-q}$

Let $C_1$ be the capacitance of the plate and $C_2$ be the capacitance of conductor.

After first contact,


Here charge $q$ has been transferred to the neutral conductor and plate is replenished back to charge $Q$ again.

When conductors are brought in contact,

$V_1=V_2$

By using definition of capacitance

$\Rightarrow \frac{Q-q}{C_1}=\frac{q}{C_2}$

$\Rightarrow \frac{Q-q}{q}=\frac{C_1}{C_2}........\left(1\right)$

Let $q_{max}$ be the maximum charge given to the conductor. Hence, flow of charge from plate to conductor will stop at this point and plate and conductor will be at the same potential.

$\therefore V_{conductor}=V_{plate}$

$\Rightarrow \frac{q_{max}}{C_2}=\frac{Q}{C_1}$

[$\because$ plate will be having charge $Q$ each time]

$\Rightarrow q_{max}=\left(\frac{C_2}{C_1}\right)Q....\left(2\right)$

From Eqs. $\left(1\right)\&\left(2\right)$

$\Rightarrow q_{max}=\frac{1}{\left(\frac{Q-q}{q}\right)}.Q$

$\therefore q_{max}=\frac{Qq}{Q-q}$

Hence, option $\left(d\right)$ is correct.

$\begin{array}{c}\text{Why this question?}\\ \text{Tip: Whenever you see a problem}\\ \text{involving charged conductors brought}\\ \text{in contact, just remember to apply}\\ \text{"Concept of equating the potential of}\\ \text{both conductors". Also, the charges will}\\ \text{distribute in ratio of capacitance!!!}\end{array}$