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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Self Evaluation Test - Electrostatic Potential and Capacitance

  • question_answer
    A parallel plate capacitor with air between the plates has a capacitance of 8 pF. Calculate the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant. \[\left( {{\varepsilon }_{r}}=6 \right)\]

    A) 72 pF

    B) 81 pF   

    C) 84 pF

    D) 96 PF

    Correct Answer: D

    Solution :

    [d] Capacity of parallel plate capacitor \[C=\frac{{{\varepsilon }_{r}}{{\varepsilon }_{0}}A}{d}\text{         }\left( \text{For air }{{\varepsilon }_{r}}=i \right)\] So, \[\frac{{{\varepsilon }_{0}}A}{d}=8\times {{10}^{-12}}\] If \[d\to \frac{d}{2}\]and \[{{\varepsilon }_{r}}\to 6\]then new capacitance \[C'=6\times \frac{{{\varepsilon }_{0}}A}{d/2}=12\frac{{{\varepsilon }_{0}}A}{d}=12\times 8pF=96pF\]


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