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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2013

  • question_answer
    The potential energy of a charged parallel plate capacitor is\[{{U}_{0}}\]. If a slab of dielectric constant k is inserted between the plates, then the new potential energy will be

    A)  \[\frac{{{v}_{0}}}{k}\]                                   

    B)  \[{{v}_{0}}{{k}^{2}}\]

    C)  \[\frac{{{v}_{0}}}{{{k}^{2}}}\]                                    

    D)  \[v_{0}^{2}\]

    Correct Answer: A

    Solution :

                    We know that, the potential energy of a charged parallel plate capacitor \[{{U}_{0}}=\frac{1}{2}CV_{0}^{2}\]                                ...(i) Now,      \[U=\frac{1}{2}C{{V}^{2}}\]                             ...(ii) We know,           \[V=\left( \frac{{{V}_{0}}}{K} \right)\] From Eq. (ii) \[U=\frac{1}{2}\frac{CV_{0}^{2}}{{{K}^{2}}}=\frac{{{V}_{0}}}{{{K}^{2}}}\]


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