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

  • question_answer
    Two parallel plates separated by a distance of \[5mm\] are kept at a potential difference of \[50\,V.\] A particle  of mass \[{{10}^{-15}}kg\] and charge \[{{10}^{-11}}C\] enters in it with a velocity \[{{10}^{7}}m/s.\] The acceleration of the particle will be  [MP PMT 1997]

    A)            \[{{10}^{8}}m/{{s}^{2}}\]

    B)                                      \[5\times {{10}^{5}}m/{{s}^{2}}\]

    C)            \[{{10}^{5}}m/{{s}^{2}}\]

    D)                                      \[2\times {{10}^{3}}m/{{s}^{2}}\]

    Correct Answer: A

    Solution :

                 \[a=\frac{qE}{m}=\frac{q}{m}\left( \frac{V}{d} \right)\]\[=\frac{{{10}^{-11}}}{{{10}^{-15}}}\times \frac{50}{5\times {{10}^{-3}}}={{10}^{8}}m/se{{c}^{2}}\]


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