Question

Two identical thin rings each of radius $R$ are coaxially placed at a distance $R$ apart. If $Q_1$ and $Q_2$ are respectively the charges uniformly spread on the two rings, the work done in moving a charge $q$ from the centre of one ring to the other is :

• A. $\frac{q\left(Q_1/Q_2\right)(\sqrt{2}+1)}{4\pi {\epsilon }_0\sqrt{2}R}$
• B. $\frac{q(Q_1-Q_2)(\sqrt{2}-1)}{4\pi {\epsilon }_0\sqrt{2}R}$
• C. $\frac{q\sqrt{2}(Q_1+Q_2)}{4\pi {\epsilon }_{0}R}$
• D. zero

Answer 1

Answer: (B)

$\frac{q(Q_1-Q_2)(\sqrt{2}-1)}{4\pi {\epsilon }_0\sqrt{2}R}$

$\because$ work done $W=q(V_1-V_2)$

potential at A is

$V_1=\frac{1}{4\pi {\epsilon }_0}(\frac{Q_1}{R}+\frac{Q_2}{\sqrt{2}R})$

potential at B is

$V_2=\frac{1}{4\pi {\epsilon }_0}(\frac{Q_2}{R}+\frac{Q_1}{\sqrt{2}R})$

$W=q(V_1-V_2)W=\frac{q}{4\pi {\epsilon }_0}[(\frac{Q_1}{R}+\frac{Q_2}{\sqrt{2}R})-(\frac{Q_2}{R}+\frac{Q_1}{\sqrt{2}R})]W=\frac{q(Q_1-Q_2)(\sqrt{2}-1)}{4\pi {\epsilon }_0\sqrt{2}R}$