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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electrostatic Potential and Capacitance

  • question_answer
    For the configuration of media permittivities \[{{\varepsilon }_{0}},\varepsilon \] and \[{{\varepsilon }_{0}}\]between parallel plated each of area A, as shown in Fig. the equivalent capacitance

    A) \[{{\varepsilon }_{0}}A/d\]

    B) \[\varepsilon {{\varepsilon }_{0}}A/d\]

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

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

    Correct Answer: D

    Solution :

    [d] \[{{C}_{eq}}=\frac{{{\varepsilon }_{0}}}{\frac{d}{{{K}_{1}}}+\frac{d}{{{K}_{2}}}+\frac{d}{{{K}_{3}}}}\] Here, \[{{K}_{1}}={{K}_{3}}=1,{{K}_{2}}=\varepsilon /{{\varepsilon }_{0}}\]


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