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

  • question_answer
    The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively \[{{C}_{0}}\] and \[{{W}_{0}}\]. If the air is replaced by glass (dielectric constant = 5) between the plates, the capacity of the plates and the energy stored in it will respectively be

    A) \[5{{C}_{0,}}5{{W}_{0}}\]     

    B) \[5{{C}_{0,}}\frac{{{W}_{0}}}{5}\]

    C) \[\frac{{{C}_{0}}}{5},5{{W}_{0}}\]    

    D) \[\frac{{{C}_{0}}}{5},\frac{{{W}_{0}}}{5}\]

    Correct Answer: B

    Solution :

    [b] When a dielectric K is introduced in a parallel plate condenser its capacity becomes K times. Hence \[C'=5{{C}_{0}}\] Energy stored \[{{W}_{0}}=\frac{{{q}^{2}}}{2{{C}_{0}}}\] \[\therefore W'=\frac{{{q}^{2}}}{2C'}=\frac{{{q}^{2}}}{2\times 5{{C}_{0}}}\Rightarrow W'=\frac{{{W}_{0}}}{5}\]


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