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 centre of the other is

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

Answer 1

The correct option is B $\frac{q}{4\pi {\epsilon }_0\sqrt{2}R}(Q_1-Q_2)(\sqrt{2}-1)$

From the figure we can conclude that, potential at $C_1$ is due to charge $Q_1$ and $Q_2$,

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

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

Similarly , potential at $C_2$ due to $Q_1$ and $Q_2$ can be written as,

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


Therefore the work done, $W$ in moving charge $q$ from ${\text{C}}_1$ to ${\text{C}}_2$ is,

$W=q(V_1-V_2)$

Substituting the values in the above equation,

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

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

Hence, option (b) is the correct option.