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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2003

  • question_answer
    Capacitance of a parallel plate capacitor becomes \[\frac{4}{3}\]times its original value, if a dielectric slab of thickness t = d/2 is inserted between the plates [d is the separation between the plates]. The dielectric constant of the slab is:

    A)  4

    B)  8

    C)  2

    D)  6

    Correct Answer: C

    Solution :

    The capacitance of air capacitor \[C=\frac{{{\varepsilon }_{0}}A}{d}\]                When a dielectric slab of thickness \[t=d/2\] is inserted between plates, the capacity becomes                 \[C=\frac{A{{\varepsilon }_{0}}}{d-d/2\left( 1-\frac{1}{K} \right)}\]                 \[\frac{4}{3}\frac{A{{\varepsilon }_{0}}}{d}=\frac{{{\varepsilon }_{0}}A}{d-d/2\left( 1-\frac{1}{K} \right)}\]                \[3d-4d\left( 1-\frac{1}{2}+\frac{1}{2K} \right)\]                \[3=4\left( \frac{1}{2}+\frac{1}{2K} \right)\]  or   \[\frac{4}{2K}=3-2\]                 or       \[K=2\]


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