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

  • question_answer
    Two dielectric slabs of constant \[{{K}_{1}}\] and \[{{K}_{2}}\] have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor                                                        [MNR 1985; MP PET 1999; DCE 2002]         

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

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

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

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

    Correct Answer: D

    Solution :

               The two capacitors formed by the slabs may assumed to be in series combination.


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