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Manipal Medical Manipal Medical Solved Paper-2013

  • question_answer
    Two plates of a parallel plate capacitor of capacity\[50\mu F\]are charged by a battery to a potential of 100 V. The battery remains connected and the plates are separated from each other so that the distance between them is doubled. The energy spent by the battery in doing so, will be

    A)  \[12.5\times {{10}^{-2}}J\]

    B)  \[-25\times {{10}^{-2}}J\]

    C)  \[25\times {{10}^{-2}}J\]

    D)  \[-12.5\times {{10}^{-1}}J\]

    Correct Answer: A

    Solution :

     We know that when separation between the plates is doubled the capacitance becomes one half i.e., \[C=25\,\mu F\] The energy spend by the battery is given by \[=qV=(CV)V=C{{V}^{2}}\] \[=25\times 10-6\times {{(100)}^{2}}\] \[=25\times {{10}^{-2}}\]


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