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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively \[{{C}_{o}}\] and \[{{W}_{o}}\]. 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)                    \[5{{C}_{o}},\ 5{{W}_{o}}\]

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

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

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

    Correct Answer: B

    Solution :

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


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