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

  • question_answer
    The distance between a proton and electron both having a charge \[1.6\times {{10}^{-19}}coulomb\], of a hydrogen atom is \[{{10}^{-10}}metre\]. The value of intensity of electric field produced on electron due to proton will be                   [MP PET 1996]

    A)                    \[2.304\times {{10}^{-10}}N/C\]

    B)                                      \[14.4\,V/m\]

    C)                    \[16\,V/m\]

    D)                                      \[1.44\times {{10}^{11}}N/C\]

    Correct Answer: D

    Solution :

                 \[E=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]\[=9\times {{10}^{9}}\times \frac{1.6\times {{10}^{-19}}}{{{({{10}^{-10}})}^{2}}}=1.44\,\times {{10}^{11}}N/C\]


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