JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    In a capacitor of capacitance \[20\mu \,F\], the distance between the plates is 2mm. If a dielectric slab of width 1mm and dielectric constant 2 is inserted between the plates, then the new capacitance is                                     [BHU 2000]

    A)            \[2\mu \,F\]                         

    B)            \[15.5\mu \,F\]

    C)            \[26.6\mu \,F\]

    D)                                      \[32\mu \,F\]

    Correct Answer: C

    Solution :

               \[C=\frac{{{\varepsilon }_{0}}A}{d}\] and \[C'=\frac{{{\varepsilon }_{0}}A}{\left( d-t+\frac{t}{K} \right)}\]Þ \[\frac{C}{C'}=\frac{\left( d-t+\frac{t}{K} \right)}{d}\] \[\Rightarrow \frac{20}{C'}=\frac{\left( 2\times {{10}^{-3}}-1\times {{10}^{-3}}+\frac{1\times {{10}^{-3}}}{2} \right)}{2\times {{10}^{-3}}}\]Þ\[C'=26.6\mu F\]


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