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CET Karnataka Engineering CET - Karnataka Engineering 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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