Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If Q₁ coulomb and Q₂ 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

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$\large \frac{{q({Q_1} + {Q_2})(\sqrt 2 + 1)}}{{\sqrt 2 .4\pi {\varepsilon _0}R}}$

zero

$\large \frac{{q({Q_1} - {Q_2})(\sqrt 2 - 1)}}{{\sqrt 2 .4\pi {\varepsilon _0}R}}$

$\large \frac{{q\sqrt 2 ({Q_1} + {Q_2})}}{{4\pi {\varepsilon _0}R}}$

answer is B.

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Detailed Solution

$\large W = q\,({V_{{O_2}}} - {V_{{O_1}}})$
Where $\large {V_{{O_1}}} = \frac{{{Q_1}}}{{4\pi {\varepsilon _0}R}} + \frac{{{Q_2}}}{{4\pi {\varepsilon _0}R\sqrt 2 }}$
and $\large {V_{{O_2}}} = \frac{{{Q_2}}}{{4\pi {\varepsilon _0}R}} + \frac{{{Q_1}}}{{4\pi {\varepsilon _0}R\sqrt 2 }}$
⇒ $\large {V_{{O_2}}} - {V_{{O_1}}} = \frac{{({Q_2} - {Q_1})}}{{4\pi {\varepsilon _0}R}}\left[ {1 - \frac{1}{{\sqrt 2 }}} \right]$
So, $\large W = \frac{{q.({Q_2} - {Q_1})}}{{4\pi {\varepsilon _0}R}}\frac{{(\sqrt 2 - 1)}}{{\sqrt 2 }}$

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