JIPMER Jipmer Medical Solved Paper-2006

  • question_answer
    Two equal -ve charges -q are fixed at the point \[(0,a)\] and \[(0,a)\] on the y-axis. A positive charge Q is released from rest at the point \[(2a,0)\]on the x-axis. The charge will;

    A) execute SHM about the origin

    B)        move to the origin and remain at rest

    C)        move to infinity

    D)        execute oscillatory but not SHM

    Correct Answer: D

    Solution :

    Component of force on charge + Q at P, along x-axis, \[{{F}_{x}}=\frac{2Qq}{4\pi {{\varepsilon }_{0}}({{a}^{2}}+{{x}^{2}})}\cos \theta \]      \[=\frac{2Qq}{4\pi {{\varepsilon }_{0}}({{a}^{2}}+{{x}^{2}})}\times \frac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}\]      \[=\frac{2Qqx}{4\pi {{\varepsilon }_{0}}{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}\] Which is not directly proportional to x. So, motion is oscillatory but not SHM.


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