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JIPMER Jipmer Medical Solved Paper-2013

  • question_answer
    A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each pate is \[A\,{{m}^{2}}\] and the separation is d metre. The dielectric constants are\[{{K}_{1}}\]and\[{{K}_{2}}\]respectively. Its capacitance in farad will be

    A)  \[\frac{2{{\varepsilon }_{0}}A}{d}\left( \frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}{{K}_{2}}} \right)\]    

    B)  \[\frac{2{{\varepsilon }_{0}}A}{d}\left( \frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}} \right)\]

    C)  \[\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{{{K}_{1}}+{{K}_{2}}}{2{{K}_{1}}{{K}_{2}}} \right)\]    

    D)  \[\frac{{{\varepsilon }_{0}}A{{K}_{1}}{{K}_{2}}}{2\,({{d}_{2}}{{K}_{1}}+{{d}_{1}}{{K}_{2}})}\]

    Correct Answer: B

    Solution :

    This is equivalent to the series combination of two capacitor \[{{C}_{equivalent}}=\frac{{{\varepsilon }_{0}}A}{\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}}=\frac{{{\varepsilon }_{0}}A}{t}\left( \frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}} \right)\] \[=\frac{2{{\varepsilon }_{0}}A}{d}\left( \frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}} \right)\]


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