JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Grouping of Capacitors

  • question_answer
    A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is \[A\ metr{{e}^{2}}\]and the separation is\[t\ metre\]. The dielectric constants are \[{{k}_{1}}\] and \[{{k}_{2}}\] respectively. Its capacitance in farad will be                                                          [MNR 1985; DCE 1999; AIIMS 2001]

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

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

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

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

    Correct Answer: B

    Solution :

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


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