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AMU Medical AMU Solved Paper-2010

  • question_answer
    A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness \[\left( \frac{3}{4} \right)\] d, where d is the separation of the plates. The ratio of the capacitance C (in the presence of the dielectric) to the capacitance \[{{C}_{o}}\](in the absence of the dielectric) is

    A)  \[\frac{3K}{K+3}\]                          

    B)  \[\frac{3}{4}K\]

    C)  \[\frac{4K}{K+3}K\]                       

    D)  \[\frac{4}{3}K\]

    Correct Answer: C

    Solution :

                     The capacitance of a parallel plate capacitor in the absence of the dielectric is                 \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}\] If a dielectric slab is partially filled between the plates of capacitor                 \[C=\frac{{{\varepsilon }_{0}}A}{d-t+\frac{t}{K}}\]                 \[C=\frac{{{\varepsilon }_{0}}A}{\left( d-\frac{3}{4}d \right)+\left( \frac{3d}{4K} \right)}\]                 \[C=\frac{{{\varepsilon }_{0}}A}{\frac{d}{4}+\frac{3d}{4K}}\]                 \[=\frac{4K{{\varepsilon }_{0}}A}{d(K+3)}\] \[\therefore \]  \[\frac{C}{{{C}_{0}}}=\frac{4K{{\varepsilon }_{0}}A}{d(K+3)}\times \frac{d}{{{\varepsilon }_{0}}A}\]                 \[=\frac{4K}{(K+3)}\]


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