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

  • question_answer
    What is the magnitude of a point charge due to which the electric field \[30\,cm\] away has the magnitude \[2\,newton/coulomb\] \[[1/4\pi {{\varepsilon }_{0}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}]\]                                                                                                       [MP PMT 1996]

    A)                    \[2\times {{10}^{-11}}coulomb\]

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

    C)                    \[5\times {{10}^{-11}}coulomb\]                             

    D)            \[9\times {{10}^{-11}}coulomb\]

    Correct Answer: A

    Solution :

                         Electric field due to a point charge \[E=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \ \[q=E\times 4\pi {{\varepsilon }_{0}}{{r}^{2}}=2\times \frac{1}{9\times {{10}^{9}}}\times {{\left( \frac{30}{100} \right)}^{2}}\]= 2 ´ 10?11 C


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