Question

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

• A. $\frac{q}{8\pi\varepsilon_{0}} \left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)$
• B. $\frac{q}{8\pi\varepsilon_{0}} \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$
• C. $\frac{q}{4\pi\varepsilon_{0}} \left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)$
• D. $\frac{q}{4\pi\varepsilon_{0}} \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$

Answer 1

The Correct Option is B

Work done $W=q\left(V_{2}-V_{1}\right)$

$V_{1} =\frac{Q_{1}}{4 \pi \varepsilon_{0} R_{1}}+\frac{Q_{2}}{4 \pi \varepsilon_{0} R \sqrt{2}} $
$=\frac{10}{4 \pi \varepsilon_{0} \times 10}+\frac{5}{4 \pi \varepsilon_{0} 10 \sqrt{2}} $
$V_{2} =\frac{Q_{2}}{4 \pi \varepsilon_{0} R}+\frac{Q_{1}}{4 \pi \varepsilon_{0} R \sqrt{2}} $
$=\frac{5}{4 \pi \varepsilon_{0} \times 10}+\frac{10}{4 \pi \varepsilon 10 \sqrt{2}} $
$V_{2}-V_{1} =\frac{5}{4 \pi \varepsilon_{0} 10 \sqrt{2}}-\frac{5}{4 \pi \varepsilon_{0} \times 10} $
$=\frac{5}{4 \pi \varepsilon_{0} 10}\left[\frac{1}{\sqrt{2}}-1\right]=\frac{1}{8 \pi \varepsilon_{0}}\left[\frac{\sqrt{2}-1}{\sqrt{2}}\right] $
$\therefore W =\frac{q}{8 \pi \varepsilon_{0}}\left[\frac{\sqrt{2}-1}{\sqrt{2}}\right]$