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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Electric Field and Potential

  • question_answer
    A particle \[A\] has charge \[+q\] and a particle \[B\] has charge \[+\,4q\] with each of them having the same mass \[m\]. When allowed to fall from rest through the same electric potential difference, the ratio of their speed \[\frac{{{v}_{A}}}{{{v}_{B}}}\] will become                                                                                                                                 [BHU 1995; MNR 1991; UPSEAT 2000; Pb PET 2004]                   

    A)                    \[2:1\]

    B)                                                 \[1:2\]

    C)                                                               \[1:4\]

    D)                                                 \[4:1\]

    Correct Answer: B

    Solution :

                         Using \[v=\sqrt{\frac{2QV}{m}}\] Þ \[v\propto \sqrt{Q}\]Þ \[\frac{{{v}_{A}}}{{{v}_{B}}}=\sqrt{\frac{{{Q}_{A}}}{{{Q}_{B}}}}=\sqrt{\frac{q}{4q}}=\frac{1}{2}\]


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