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

  • question_answer
    Two positive charges of 20 \[coulomb\] and \[Q\ coulomb\]are situated at a distance of \[60\,cm\]. The neutral point between them is at a distance of \[20\,cm\] from the \[20\,coulomb\] charge. Charge \[Q\] is

    A)                    \[30\,C\]

    B)                                      \[40\,C\]

    C)                    \[60\,C\]

    D)                                      \[80\,C\]

    Correct Answer: D

    Solution :

                         At neutral point \[k\times \frac{20}{{{(20\times {{10}^{-2}})}^{2}}}=k\times \frac{Q}{{{(40\times {{10}^{-2}})}^{2}}}\] Þ Q = 80 C


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