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

  • question_answer
    What is the magnitude of a point charge which produces an electric field of 2 N/coulomb at a distance of 60 cm (\[1/4\pi {{\varepsilon }_{0}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}\]) [MP PET 2000; RPET 2001]

    A)            \[8\times {{10}^{-11}}\]C

    B)                                      \[2\times {{10}^{-12}}\]C

    C)            \[3\times {{10}^{-11}}\]C

    D)                                      \[6\times {{10}^{-10}}\]C

    Correct Answer: A

    Solution :

                 \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{{{r}^{2}}}\]Þ \[2=9\times {{10}^{9}}\times \frac{Q}{{{(0.6)}^{2}}}\]Þ Q = 8 ´ \[{{10}^{-11}}C\]


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