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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is \[2\mu \,F\]. The separation is reduced to half and it is filled with a dielectric substance of value 2.8. The final capacity of the capacitor is [CBSE PMT 2000]

    A)            \[11.2\mu F\]                      

    B)            \[15.6\mu F\]

    C)            \[19.2\mu F\]                      

    D)            \[22.4\mu F\]

    Correct Answer: A

    Solution :

               \[C=\frac{{{\varepsilon }_{0}}KA}{d}\] Þ \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{{{K}_{1}}}{{{K}_{2}}}\times \frac{{{d}_{2}}}{{{d}_{1}}}\]                    \[\frac{2}{{{C}_{2}}}=\frac{1}{2.8}\times \frac{(0.4/2)}{(0.4)}\] Þ \[{{C}_{2}}\,=\,11.2\,\mu \,F\]


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