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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 is charged to a potential difference of 500 V and then insulated. A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes 75V. The dielectric constant of plastic is

    A) 10/3                 

    B)        5

    C) 20/3                 

    D)        10

    Correct Answer: C

    Solution :

    [c] \[{{V}_{0}}=\frac{q}{{{C}_{0}}}\] \[V=\frac{q}{C}\Rightarrow \frac{V}{{{V}_{0}}}=\frac{{{C}_{0}}}{C}\Rightarrow \frac{{{C}_{0}}}{C}=\frac{500}{75}=\frac{20}{3}\]


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