Two identical thin rings, each of radius R, are coaxially placed a distance R apart. If Q1 and Q2 are respectively the charges uniformly spread on the two ring to that of the other is :
Answer & explanation
Correct answer: option 2
$V_{C_1} =V_{Q_1}+V_{Q_2}$
$=\frac{1}{4 \pi \varepsilon_0} \frac{Q_1}{R}+\frac{1}{4 \pi \varepsilon_0} \frac{Q_2}{R \sqrt{2}}$
$=\frac{1}{4 \pi \varepsilon_0 R}\left(Q_1+\frac{Q_2}{\sqrt{2}}\right)$
Similarly $\quad V_{C_2}=\frac{1}{4 \pi \varepsilon_0 R}\left(Q_2+\frac{Q_1}{\sqrt{2}}\right)$
∴ $\Delta V=V_{C_1}-V_{C_2}$
$=\frac{1}{4 \pi \varepsilon_0 R}\left[\left(Q_1-Q_2\right)-\frac{1}{\sqrt{2}}\left(Q_1-Q_2\right)\right]$
$=\frac{Q_1-Q_2}{\sqrt{2}\left(4 \pi \varepsilon_0 R\right)}(\sqrt{2}-1)$
$W=q \Delta V=q\left(Q_1-Q_2\right)(\sqrt{2}-1) / \sqrt{2}\left(4 \pi \varepsilon_0 R\right)$