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

  • question_answer
    A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants \[{{k}_{1}},{{k}_{2}}\] and \[{{k}_{3}}\] as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by                                               [IIT-JEE Screening 2000]

    A)            \[\frac{1}{k}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}}+\frac{1}{2{{k}_{3}}}\]

    B)                                      \[\frac{1}{k}=\frac{1}{{{k}_{1}}+{{k}_{2}}}+\frac{1}{2{{k}_{3}}}\]

    C)            \[k=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}+2{{k}_{3}}\]         

    D)            \[k={{k}_{1}}+{{k}_{2}}+2{{k}_{3}}\]

    Correct Answer: B

    Solution :

                       \[{{C}_{1}}=\frac{{{K}_{1}}{{\varepsilon }_{0}}\frac{A}{2}}{\left( \frac{d}{2} \right)}=\frac{{{K}_{1}}{{\varepsilon }_{0}}A}{d}\]                               \[{{C}_{2}}=\frac{{{K}_{2}}{{\varepsilon }_{0}}\frac{A}{2}}{\left( \frac{d}{2} \right)}=\frac{{{K}_{2}}{{\varepsilon }_{0}}A}{d}\] and \[{{C}_{3}}=\frac{{{K}_{3}}{{\varepsilon }_{0}}A}{\left( \frac{d}{2} \right)}=\frac{2{{K}_{3}}{{\varepsilon }_{0}}A}{d}\]                    \[\frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{1}}+{{C}_{2}}}+\frac{1}{{{C}_{3}}}\]\[=\frac{1}{\frac{{{\varepsilon }_{0}}A}{d}({{K}_{1}}+{{K}_{2}})}+\frac{1}{\frac{{{\varepsilon }_{0}}}{d}\times 2{{K}_{3}}}\]                    \[\frac{1}{{{C}_{eq}}}=\frac{d}{{{\varepsilon }_{0}}A}\left[ \frac{1}{{{K}_{1}}+{{K}_{2}}}+\frac{1}{2{{K}_{3}}} \right]\]                    \[{{C}_{eq}}={{\left[ \frac{1}{{{K}_{1}}+{{K}_{2}}}+\frac{1}{2{{K}_{3}}} \right]}^{-1}}.\frac{{{\varepsilon }_{0}}A}{d}\]                    So \[{{K}_{eq}}={{\left[ \frac{1}{{{K}_{1}}+{{K}_{2}}}+\frac{1}{2{{K}_{3}}} \right]}^{-1}}\]


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