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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    A capacitor of capacity \[C\] is connected with a battery of potential \[V\] in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential \[V\] again, the energy given by the battery will be                                                                                            [MP PET 1989]

    A)                    \[C{{V}^{2}}/4\]

    B)                                      \[C{{V}^{2}}/2\]

    C)                    \[3C{{V}^{2}}/4\]        

    D)            \[C{{V}^{2}}\]

    Correct Answer: D

    Solution :

     Extra charge Q = (2CV ? CV) = CV flows through potential V of the battery. Thus W = QV = \[C{{V}^{2}}\]


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