Question

Two identical thin rings, each of radius $10cm$ carrying charges $10C$ and $5C$ are coaxially placed at a distance $10cm$ apart. The work done in moving a charge $q$ from the centre of the first ring to that of the second is

• A. $\frac{q}{8\pi {\epsilon }_0}\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)$
• B. $\frac{q}{8\pi {\epsilon }_0}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
• C. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)$
• D. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
• E. $\frac{q}{4\pi {\epsilon }_0}\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)$

Answer 1

The correct option is B $\frac{q}{8\pi {\epsilon }_0}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$

Work done $W=q(V_2-V_1)$
Now, $V_1=\frac{Q_1}{4\pi {\epsilon }_0{R}_1}+\frac{Q_2}{4\pi {\epsilon }_{0}R\sqrt{2}}$
$=\frac{10}{4\pi {\epsilon }_0\times 10}+\frac{5}{4\pi {\epsilon }_{0}10\sqrt{2}}$
$V_2=\frac{Q_2}{4\pi {\epsilon }_{0}R}+\frac{Q_1}{4\pi {\epsilon }_{0}R\sqrt{2}}$
$=\frac{5}{4\pi {\epsilon }_0\times 10}+\frac{10}{4\pi \epsilon 10\sqrt{2}}$
$V_2-V_1=\frac{5}{4\pi {\epsilon }_{0}10\sqrt{2}}-\frac{5}{4\pi {\epsilon }_0\times 10}$
$=\frac{5}{4\pi {\epsilon }_{0}10}[\frac{1}{\sqrt{2}}-1]=\frac{1}{8\pi {\epsilon }_0}\left[\frac{\sqrt{2}-1}{\sqrt{2}}\right]$
$\therefore W=\frac{q}{8\pi {\epsilon }_0}\left[\frac{\sqrt{2}-1}{\sqrt{2}}\right]$