Electrostatics Ques 165
- Two identical thin rings, each of radius $R$, are coaxially placed 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 that of the other is
(1992, 2M)
(a) zero
(b) $\frac{q\left(Q_{1}-Q_{2}\right)(\sqrt{2}-1)}{\sqrt{2}\left(4 \pi \varepsilon_{0} R\right)}$
(c) $\frac{q \sqrt{2}\left(Q_{1}+Q_{2}\right)}{\left(4 \pi \varepsilon_{0} R\right)}$
(d) $q\left(Q_{1} / Q_{2}\right)(\sqrt{2}+1) \sqrt{2}\left(4 \pi \varepsilon_{0} R\right)$
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Answer:
Correct Answer: 165.(b)
Solution:
Formula:
Work Done By An Electric Force:
- $V_{C_{1}}=V_{Q_{1}}+V_{Q_{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1}}{R}+\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{2}}{R \sqrt{2}}$
$$ =\frac{1}{4 \pi \varepsilon_{0} R} \quad Q_{1}+\frac{Q_{2}}{\sqrt{2}} $$
Similarly $V_{C_{2}}=\frac{1}{4 \pi \varepsilon_{0} R} \quad Q_{2}+\frac{Q_{1}}{\sqrt{2}}$
$\therefore \Delta V=V_{C_{1}}-V_{C_{2}}=\frac{1}{4 \pi \varepsilon_{0} R}\left(Q_{1}-Q_{2}\right)-\frac{1}{\sqrt{2}}\left(Q_{1}-Q_{2}\right)$
$$ \begin{aligned} & =\frac{Q_{1}-Q_{2}}{\sqrt{2}\left(4 \pi \varepsilon_{0} R\right)}(\sqrt{2}-1) \\ W & =q \Delta V=q\left(Q_{1}-Q_{2}\right)(\sqrt{2}-1) / \sqrt{2}\left(4 \pi \varepsilon_{0} R\right) \end{aligned} $$
BETA
