Q. Two identical thin rings each of radius $R$ meters are coaxially placed at a diatance $R$ meters apart. If $Q_{1}$ coulomb and $Q_{2}$ coulomb 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 other is
A) Zero
B) $\frac{q(Q_{1} - Q_{2})(\sqrt{2} - 1)}{\sqrt{2}.4\pi\varepsilon_{0}R}$
C) $\frac{q\sqrt{2}(Q_{1} + Q_{2})}{4\pi\varepsilon_{0}R}$
D) $\frac{q(Q_{1} + Q_{2})(\sqrt{2} + 1)}{\sqrt{2}.4\pi\varepsilon_{0}R}$
Explanation
(b) $W = q(V_{O_{2}} - V_{O_{1}})$
[image:1_image271.png]
where $V_{O_{1}} = \frac{Q_{1}}{4\pi\varepsilon_{0}R} + \frac{Q_{2}}{4\pi\varepsilon_{0}R\sqrt{2}}$
$and\ V_{O_{2}} = \frac{Q_{2}}{4\pi\varepsilon_{0}R} + \frac{Q_{1}}{4\pi\varepsilon_{0}R\sqrt{2}}$
$\Rightarrow V_{O_{2}} - V_{O_{1}} = \frac{\left( Q_{2} - Q_{1} \right)}{4\pi\varepsilon_{0}R}\left\lbrack 1 - \frac{1}{\sqrt{2}} \right\rbrack$
So, $W = \frac{q.\left( Q_{2} - Q_{1} \right)}{4\pi\varepsilon_{0}R}\frac{(\sqrt{2} - 1)}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.