A conducting sphere of radius R carrying charge Q is joined to an uncharged conducting sphere of radius 2R. The charge flowing between them is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{2Q}{3} \)
When a wire joins two conductors, charge flows until both are at the same potential. The total charge is conserved. A sphere of radius \( a \) carrying charge \( q \) has potential \( kq/a \).
Let the small sphere keep \( Q_1 \) and the large sphere receive \( Q_2 \). Equal potentials give:
\( \displaystyle \frac{kQ_1}{R}=\frac{kQ_2}{2R} \)
\( \Rightarrow\; Q_2=2Q_1 \)
Conservation of charge gives:
\( Q_1+Q_2=Q \)
\( \begin{aligned} Q_1+2Q_1 &= Q \\ Q_1 &= \frac{Q}{3} \end{aligned} \)
\( \begin{aligned} Q_2 &= Q-Q_1 \\ &= Q-\frac{Q}{3} \\ &= \frac{2Q}{3} \end{aligned} \)
The large sphere was uncharged at first, so the charge that flows through the wire is \( 2Q/3 \).