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AMU Medical AMU Solved Paper-2012

  • question_answer
    In a parallel-plate capacitor with plate area A and charge Q, the force on one plate because of the charge on the other is equal to

    A)  \[\frac{{{Q}^{2}}}{{{\varepsilon }_{0}}{{A}^{2}}}\]                           

    B)  \[\frac{{{Q}^{2}}}{2{{\varepsilon }_{0}}{{A}^{2}}}\]

    C)  \[\frac{{{Q}^{2}}}{{{\varepsilon }_{0}}A}\]                                          

    D)  \[\frac{{{Q}^{2}}}{2{{\varepsilon }_{0}}A}\]

    Correct Answer: D

    Solution :

                     Force between the plates of a parallel plate capacitor\[\left| F \right|=\frac{{{V}^{2}}A}{2\,{{\varepsilon }_{0}}}=\frac{{{Q}^{2}}}{2\,{{\varepsilon }_{0}}A}=\frac{C{{V}^{2}}}{2\,d}\]


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