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

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

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

    B) \[\frac{{{\varepsilon }_{0}}A}{t}.\frac{{{k}_{1}}+{{k}_{2}}}{2}\]

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

    D) \[\frac{{{\varepsilon }_{0}}A}{t}.\frac{{{k}_{1}}-{{k}_{2}}}{2}\]

    Correct Answer: B

    Solution :

    [b] The two capacitors are in parallel so \[C=\frac{{{\varepsilon }_{0}}A}{t\times 2}\left( {{k}_{1}}+{{k}_{2}} \right)\]


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