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JIPMER Jipmer Medical Solved Paper-2003

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

    A)  \[12.5\times {{10}^{-2}}\text{J}\]    

    B)         \[35\times {{10}^{-2}}\text{J}\]               

    C)  \[42.5\times {{10}^{-2}}\text{J}\]    

    D)  \[25\times {{10}^{-2}}\text{J}\]

    Correct Answer: D

    Solution :

    When separation between the plates is doubled the capacitance becomes one half. It means \[C=25\mu F\] Now energy spent by the battery \[=qV=(CV)\,V\] \[=C{{V}^{2}}=25\times {{10}^{-6}}\times {{(100)}^{2}}\] \[=25\times {{10}^{-2}}J\]


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