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

  • question_answer
    Three plates of common surface area A are connected as shown. The effective capacitance will be          [Orissa PMT 2004]

    A)            \[\frac{{{\varepsilon }_{0}}A}{d}\]

    B)            \[\frac{3{{\varepsilon }_{0}}A}{d}\]

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

    D)                    \[\frac{2{{\varepsilon }_{0}}A}{d}\]

    Correct Answer: D

    Solution :

               The given circuit is equivalent to parallel combination of two identical capacitors, each having capacitance \[C=\frac{{{\varepsilon }_{0}}A}{d}\]. Hence \[{{C}_{eq}}=2C=\frac{2{{\varepsilon }_{0}}A}{d}\]


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