JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • 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)  zero

    B)  \[q({{Q}_{1}}-{{Q}_{2}})(\sqrt{2}-1)/\sqrt{2}4\pi {{\varepsilon }_{0}}R\]

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

    D)  \[q({{Q}_{1}}+{{Q}_{2}})(\sqrt{2}+1)/\sqrt{2}\,4\pi {{\varepsilon }_{0}}R\]

    Correct Answer: B

    Solution :

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

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