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

  • question_answer
    The potential at a point, due to a positive charge of \[100\mu C\] at a distance of 9m, is                                                 [KCET (Med.) 2000]

    A)            \[{{10}^{4}}\]V                    

    B)            \[{{10}^{5}}\]V

    C)            \[{{10}^{6}}\]V                    

    D)            \[{{10}^{7}}\]V

    Correct Answer: B

    Solution :

                 By using \[V=9\times {{10}^{9}}\times \frac{Q}{r}\]\[=9\times {{10}^{9}}\times \frac{100\times {{10}^{-6}}}{9}={{10}^{5}}V\]


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