JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V0. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is                                           [Kerala PET 2002]

    A)            \[\frac{3{{\varepsilon }_{0}}AV_{0}^{2}}{d}\]

    B)                                      \[\frac{{{\varepsilon }_{0}}AV_{0}^{2}}{2d}\]

    C)            \[\frac{{{\varepsilon }_{0}}AV_{0}^{2}}{3d}\]           

    D)            \[\frac{{{\varepsilon }_{0}}AV_{0}^{2}}{d}\]

    Correct Answer: D

    Solution :

                       Work done \[W={{U}_{f}}-{{U}_{i}}\]                    \[{{U}_{i}}=\frac{1}{2}C{{V}_{0}}^{2}\text{ and }{{\text{U}}_{\text{f}}}=\frac{1}{2}\frac{(C)}{3}.{{(3{{V}_{0}})}^{2}}\]\[=3\times \frac{1}{2}C{{V}_{0}}^{2}\]                    So \[W=\frac{{{\varepsilon }_{0}}A{{V}_{0}}^{2}}{d}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner