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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    A parallel plate capacitor of capacity \[{{C}_{0}}\] is charged to a potential \[{{V}_{0}}\] (i)       The energy stored in the capacitor when the battery is disconnected and the separation is doubled \[{{E}_{1}}\] (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is \[{{E}_{2}}.\] Then \[{{E}_{1}}/{{E}_{2}}\] value is                                    [EAMCET 2003]

    A)            4

    B)            3/2

    C)            2    

    D)            1/2

    Correct Answer: A

    Solution :

                          Let \[E=\frac{1}{2}{{C}_{0}}{{V}_{0}}^{2}\,\text{then}\,\text{ }{{E}_{\text{1}}}=2E\] and \[{{E}_{2}}=\frac{E}{2}\] So \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{4}{1}\]


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