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Two identical thin rings each of radius $R$ meters are coaxially placed at a distance $R$ 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
Zero
$\frac{{q({Q_1} - {Q_2})(\sqrt 2 - 1)}}{{\sqrt 2 .4\pi {\varepsilon _0}R}}$
$\frac{{q\sqrt 2 ({Q_1} + {Q_2})}}{{4\pi {\varepsilon _0}R}}$
$\frac{{q({Q_1} + {Q_2})(\sqrt 2 + 1)}}{{\sqrt 2 .4\pi {\varepsilon _0}R}}$
Solution
(b)$W = q\,({V_{{O_2}}} – {V_{{O_1}}})$
where ${V_{{O_1}}} = \frac{{{Q_1}}}{{4\pi {\varepsilon _0}R}} + \frac{{{Q_2}}}{{4\pi {\varepsilon _0}R\sqrt 2 }}$
and ${V_{{O_2}}} = \frac{{{Q_2}}}{{4\pi {\varepsilon _0}R}} + \frac{{{Q_1}}}{{4\pi {\varepsilon _0}R\sqrt 2 }}$
$==>$ ${V_{{O_2}}} – {V_{{O_1}}} = \frac{{({Q_2} – {Q_1})}}{{4\pi {\varepsilon _0}R}}\left[ {1 – \frac{1}{{\sqrt 2 }}} \right]$
So, $W = \frac{{q.({Q_2} – {Q_1})}}{{4\pi {\varepsilon _0}R}}\frac{{(\sqrt 2 – 1)}}{{\sqrt 2 }}$
