Two identical spherical conductors B and C carry equal charges Q each. A third identical conductor A, initially uncharged, is touched first to B and then to C, and is then removed. What is the charge on A?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3Q/4 \)
When two identical conductors touch, their total charge is shared equally between them, and by conservation of charge the total stays the same.
When A (charge 0) touches B (charge \( Q \)), each gets:
\( \displaystyle \frac{0+Q}{2}=\frac{Q}{2} \)
A now carries \( Q/2 \) and touches C (charge \( Q \)). Each then gets:
\( \begin{aligned} q_A &= \frac{Q/2+Q}{2} \\ &= \frac{3Q/2}{2} \\ &= \frac{3Q}{4} \end{aligned} \)
Therefore, A is left with a charge of \( 3Q/4 \) when it is removed.