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

  • question_answer
    An electron enters between two horizontal plates separated by 2mm and having a potential difference of 1000V. The force on electron is                                                    [JIPMER 1999]

    A)            \[8\times {{10}^{-12}}\]N

    B)                                      \[8\times {{10}^{-14}}\]N

    C)            \[8\times {{10}^{9}}\] N

    D)                                      \[8\times {{10}^{14}}\] N

    Correct Answer: B

    Solution :

                         Force on electron  F = QE \[=Q\,\left( \frac{V}{d} \right)\] Þ \[F=(1.6\times {{10}^{-19}})\,\left( \frac{1000}{2\times {{10}^{-3}}} \right)=8\times {{10}^{-14}}N\]


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