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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    Capacitance of a parallel plate capacitor becomes 4/3 times its original value if a dielectric slab of thickness t = d/2 is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is         [KCET 2003]

    A)            8    

    B)            4

    C)            6    

    D)            2

    Correct Answer: D

    Solution :

     \[{{C}_{air}}=\frac{{{\varepsilon }_{0}}A}{d}\], with dielectric slab C¢= \[\frac{{{\varepsilon }_{0}}A}{\left( d-t+\frac{t}{K} \right)}\]                    Given \[{C}'=\frac{4}{3}C\Rightarrow \]\[\frac{{{\varepsilon }_{0}}A}{\left( d-t+\frac{t}{K} \right)}=\frac{4}{3}\times \frac{{{\varepsilon }_{0}}A}{d}\]                    \[\Rightarrow K=\frac{4t}{4t-d}=\frac{4(d/2)}{4[(d/2)-d]}=2\]


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