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

  • question_answer
    Two metal pieces having a potential difference of \[800\,V\] are  \[0.02\,m\] apart horizontally. A particle of mass \[1.96\times {{10}^{-15}}kg\]is suspended in equilibrium between the plates. If \[e\] is the elementary charge, then charge on the particle is                                           [MP PET 1999]

    A)                    \[e\]

    B)                                      \[3e\]

    C)                    \[6e\]

    D)                                      \[8e\]

    Correct Answer: B

    Solution :

                         For equilibrium mg = qE \ \[1.96\times {{10}^{-15}}\times 9.8=q\times \,\left( \frac{800}{0.02} \right)\] Þ \[q=\frac{1.96\times {{10}^{-15}}\times 9.8\times 0.02}{800}\] Þ \[n\times 1.6\times {{10}^{-19}}=\frac{1.96\times {{10}^{-15}}\times 9.8\times 0.02}{800}\]Þ n = 3.


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