• # question_answer Two identical thin rings, each of radius R metres, are coaxially placed at a distance R metres 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 the other is        A)  zeroB)  $q({{Q}_{1}}-{{Q}_{2}})(\sqrt{2}-1)/\sqrt{2}4\pi {{\varepsilon }_{0}}R$C)  $q\sqrt{2}({{Q}_{1}}+{{Q}_{2}})/4\pi {{\varepsilon }_{0}}R$D)  $q({{Q}_{1}}+{{Q}_{2}})(\sqrt{2}+1)/\sqrt{2}\,4\pi {{\varepsilon }_{0}}R$

${{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{Q}_{1}}}{R}+\frac{{{Q}_{2}}}{\sqrt{2}R} \right];$ ${{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{Q}_{2}}}{R}+\frac{{{Q}_{1}}}{\sqrt{2}R} \right]$ ${{V}_{A}}-{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}R}\left[ {{Q}_{1}}+\frac{{{Q}_{2}}}{\sqrt{2}}-{{Q}_{2}}-\frac{{{Q}_{1}}}{\sqrt{2}} \right]$ Work done $=Q\times V=q\times ({{V}_{A}}-{{V}_{B}})$ $=\frac{q}{4\pi {{\varepsilon }_{0}}R}\left[ {{Q}_{1}}+\frac{{{Q}_{2}}}{\sqrt{2}}-{{Q}_{2}}-\frac{{{Q}_{1}}}{\sqrt{2}} \right]$ $=\frac{q}{4\pi {{\varepsilon }_{0}}R}\times \frac{1}{\sqrt{2}}\left[ \sqrt{2}{{Q}_{1}}+{{Q}_{2}}-\sqrt{2}{{Q}_{2}}-{{Q}_{1}} \right]$ $=\frac{q({{Q}_{1}}-{{Q}_{2}})\,(\sqrt{2}-1)}{(\sqrt{2}\,4\pi {{\varepsilon }_{0}}\,R)}$