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

  • question_answer
    The electric field due to a charge at a distance of 3 m from it is 500 N/coulomb. The magnitude of the charge is \[\left[ \frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}\frac{N-{{m}^{2}}}{coulom{{b}^{2}}} \right]\]                                               [MP PMT 2000]

    A)            2.5 micro-coulomb            

    B)            2.0 micro-coulomb

    C)            1.0 micro-coulomb            

    D)            0.5 micro-coulomb

    Correct Answer: D

    Solution :

                         \[E=9\times {{10}^{9}}\times \frac{Q}{{{r}^{2}}}\,\Rightarrow 500=9\times {{10}^{9}}\times \frac{Q}{{{(3)}^{2}}}\]Þ Q = 0.5 mC


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