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Haryana PMT Haryana PMT Solved Paper-2003

  • question_answer
    What is the magnitude of the point charge due to which the electric field 30 cm away has    the    magnitude    of   2N/C\[\left( \frac{1}{4{{\pi }_{\varepsilon 0}}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}} \right):\]

    A)  \[4\times {{10}^{-11}}C\]                            

    B)  \[2\times {{10}^{-11}}C\]

    C)  \[5.4\times {{10}^{-11}}C\]                        

    D)  \[7.5\times {{10}^{-11}}C\]

    Correct Answer: B

    Solution :

                    Electric field due to point charge \[E=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] So,         \[q=E\times 4\pi \omega {{r}^{2}}\]                 \[=2\times \frac{1}{9\times {{10}^{9}}}\times {{\left( \frac{30}{100} \right)}^{2}}\] \[=2\times {{10}^{-11}}\] coulomb            


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