JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate \[=A\])                                 [MP PET 1996]

    A)              \[\frac{{{\varepsilon }_{0}}A}{\left( \frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}}+\frac{{{d}_{3}}}{{{K}_{3}}} \right)}\]

    B)               \[\frac{{{\varepsilon }_{0}}A}{\left( \frac{{{d}_{1}}+{{d}_{2}}+{{d}_{3}}}{{{K}_{1}}+{{K}_{2}}+{{K}_{3}}} \right)}\]

    C)              \[\frac{{{\varepsilon }_{0}}A({{K}_{1}}{{K}_{2}}{{K}_{3}})}{{{d}_{1}}{{d}_{2}}{{d}_{3}}}\]

    D)                    \[{{\varepsilon }_{0}}\left( \frac{A{{K}_{1}}}{{{d}_{1}}}+\frac{A{{K}_{2}}}{{{d}_{2}}}+\frac{A{{K}_{3}}}{{{d}_{3}}} \right)\]

    Correct Answer: A

    Solution :


You need to login to perform this action.
You will be redirected in 3 sec spinner