BHU PMT BHU PMT Solved Paper-2001

  • question_answer
    If the distance between parallel plates of a Capacitor is halved and dielectric constant is Doubled then the capacitance will:

    A)  remain the same            

    B)  increase 4 times

    C)  increase 2 times             

    D)  decrease 2 times

    Correct Answer: B

    Solution :

    Key-Idea: In order to obtain high capacitance, plats of large area should be taken and kept close to each other. The capacitance of a parallel plate capacitor of plate area A and the distance \[d\], between the plated is given by \[C=\frac{K{{\varepsilon }_{o}}A}{d}\] Given, \[{{d}_{1}}=d,\,\,{{d}_{2}}=\frac{d}{2},\,{{K}_{1}}=K,\,\,{{K}_{2}}=2K\] \[\therefore \]                  \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{{{K}_{1}}}{{{d}_{1}}}\times \frac{{{d}_{2}}}{{{K}_{2}}}\]                                                        \[=\frac{K}{d}\times \frac{d}{2\times 2\,K}\]                                                 \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{1}{4}\] \[\Rightarrow \]                               \[{{C}_{2}}=4\,{{C}_{1}}\] Hence, Capacitance increases four times.


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