question_answer

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 qfrom the centre of one ring to that of other is                                                            [MP PMT 1999; AMU (Engg.) 1999]

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)            \[\frac{q({{Q}_{1}}+{{Q}_{2}})(\sqrt{2}+1)}{\sqrt{2}.4\pi {{\varepsilon }_{0}}R}\]

Correct Answer: B

Solution :

           \[W=q\,({{V}_{{{O}_{2}}}}-{{V}_{{{O}_{1}}}})\] where \[{{V}_{{{O}_{1}}}}=\frac{{{Q}_{1}}}{4\pi {{\varepsilon }_{0}}R}+\frac{{{Q}_{2}}}{4\pi {{\varepsilon }_{0}}R\sqrt{2}}\] and \[{{V}_{{{O}_{2}}}}=\frac{{{Q}_{2}}}{4\pi {{\varepsilon }_{0}}R}+\frac{{{Q}_{1}}}{4\pi {{\varepsilon }_{0}}R\sqrt{2}}\] Þ \[{{V}_{{{O}_{2}}}}-{{V}_{{{O}_{1}}}}=\frac{({{Q}_{2}}-{{Q}_{1}})}{4\pi {{\varepsilon }_{0}}R}\left[ 1-\frac{1}{\sqrt{2}} \right]\] So, \[W=\frac{q.({{Q}_{2}}-{{Q}_{1}})}{4\pi {{\varepsilon }_{0}}R}\frac{(\sqrt{2}-1)}{\sqrt{2}}\]