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AFMC AFMC Solved Paper-2009

  • question_answer
    The potential energy of a charged parallel plate capacitor is U0. If a slab of dielectric constant K is inserted between the plates, then the new potential energy will be

    A) \[\frac{{{U}_{0}}}{K}\]                                  

    B) \[{{U}_{0}}{{K}^{2}}\]

    C) \[\frac{{{U}_{0}}}{{{K}^{2}}}\]                                   

    D) \[{{U}_{0}}^{2}\]

    Correct Answer: A

    Solution :

    When a dielectric slab of dielectric constant K is inserted between the plates of charged parallel plate capacitor, then capacity becomes C' = KC potential becomes \[V'=\frac{V}{K}\] and energy \[U'=\frac{U}{K}\]


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