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12th Class Physics Electrostatics & Capacitance Question Bank MCQ - Electrostatic Potential and Capacitance

  • question_answer
    A slab of material dielectric constant K has the same area A as the plates of a parallel plate capacitor, and has thickness \[\left( \frac{3}{4}d \right)\], where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is:

    A) \[C=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{K+3}{4K} \right)\]

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

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

    D) \[C=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{4K}{K+3} \right)\]

    Correct Answer: D

    Solution :

    (d) \[C=\frac{{{\varepsilon }_{0}}A}{d}\left( \frac{4K}{K+3} \right)\] Here, thickness of the slab, \[t=\frac{3}{4}\,d\] Capacitance, \[C=\frac{{{\varepsilon }_{0}}A}{d-t\left( 1-\frac{1}{K} \right)}\] \[=\frac{{{\varepsilon }_{0}}A}{d-\frac{3}{4}d\left( 1-\frac{1}{K} \right)}=\frac{{{\varepsilon }_{0}}A}{\frac{d}{4}+\frac{3}{4}\frac{d}{K}}=\frac{{{\varepsilon }_{0}}A}{\frac{d}{4}\left( 1+\frac{3}{K} \right)}\] \[=\frac{{{\varepsilon }_{0}}A}{d}\frac{4K}{\left( K+3 \right)}\]


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