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Haryana PMT Haryana PMT Solved Paper-2000

  • question_answer
    The capacitance of a parallel plate air capacitor is 10 \[\mu \]F. If now the overlapping area of plates is doubled and separation between the plates is halved, the new capacitance will be :

    A)  60\[\text{ }\!\!\mu\!\!\text{ F}\]                           

    B)  10\[\text{ }\!\!\mu\!\!\text{ F}\]

    C)  20 \[\text{ }\!\!\mu\!\!\text{ F}\]                                          

    D)  40 \[\text{ }\!\!\mu\!\!\text{ F}\]

    Correct Answer: D

    Solution :

                    Using the relation \[C=\frac{{{\varepsilon }_{0}}A}{d}\] (Initially) \[C=10\mu F\] Now,    \[C=\frac{{{\varepsilon }_{0}}A}{d}\]                 \[=\frac{{{\varepsilon }_{0}}2A}{d/2}=\frac{4{{\varepsilon }_{0}}A}{d}\] \[\left[ Here,\,\,\,\,\,A=2A,\,d=\frac{d}{2} \right]\] So,          \[\frac{C}{C}=4\] or        \[C=4\,\,C=4\times 10=40\mu F\]


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