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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Mock Test - 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

    A) \[11.2\mu F\]     

    B) \[15.6\mu F\]

    C) \[19.2\mu F\]     

    D) \[22.4\mu F\]

    Correct Answer: A

    Solution :

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


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