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CMC Medical CMC-Medical Ludhiana Solved Paper-2014

  • question_answer
    The potential energy of a charged parallel plate capacitor is\[{{U}_{0}}.\] If a slab of dielectric constant \[K\] is increased 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 \[V=\frac{V}{K}\]      \[U=\frac{{{U}_{0}}}{K}\]


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