Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2007

  • question_answer
    Capacitor of 12 pF capacitance is connected to 50V battery then electrostatic potential energy will be

    A) \[1.5\times {{10}^{-8}}\text{J}\]                               

    B) \[2.5\times {{10}^{-7}}\text{J}\]

    C) \[3.5\times {{10}^{-5}}\text{J}\]                               

    D) \[4.5\times {{10}^{-2}}\text{J}\]

    Correct Answer: A

    Solution :

    Given, \[C=12pF=12\times {{10}^{-12}}F,\] \[V=50volt\] \[\therefore \]Potential energy \[(U)=\frac{1}{2}C{{V}^{2}}\]                 \[=\frac{1}{2}\times 12\times {{10}^{-12}}\times {{(50)}^{2}}\]                 \[=6\times {{10}^{-12}}\times 2500\]                 \[=1.5\times {{10}^{-8}}J\]


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