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12th Class Physics Electrostatics & Capacitance Question Bank Case Based (MCQs) - Electrostatic Potential and Capacitance

  • question_answer
    A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacity is:                                            

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

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

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

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

    Correct Answer: C

    Solution :

    (c) \[\frac{{{\varepsilon }_{0}}A}{d-b}\] As capacitance, \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}\] \[\therefore \]After inserting copper plate, \[C=\frac{{{\varepsilon }_{0}}A}{d-b}\]


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