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

  • question_answer
    A parallel plate capacitor has plates of area \[A\] and separation \[d\] and is charged to a potential difference\[V\]. The charging battery is then disconnected and the plates are pulled apart until their separation is 2\[d\]. What is the work required to separate the plates?

    A) \[2{{\varepsilon }_{0}}A{{V}^{2}}/d\]             

    B) \[{{\varepsilon }_{0}}A{{V}^{2}}/d\]

    C) \[3{{\varepsilon }_{0}}A{{V}^{2}}/d\] 

    D) \[{{\varepsilon }_{0}}A{{V}^{2}}/2d\]

    Correct Answer: D

    Solution :

    [d] \[W={{U}_{2}}-{{U}_{1}}=\frac{{{q}^{2}}}{2}\left[ \frac{1}{{{C}_{2}}}-\frac{1}{{{C}_{1}}} \right]\] \[{{C}_{1}}=\frac{{{\varepsilon }_{0}}A}{d},{{C}_{2}}=\frac{{{C}_{1}}}{2}=\frac{{{\varepsilon }_{0}}A}{2d}\] \[q={{C}_{1}}V=\frac{{{\varepsilon }_{0}}AV}{d}\] Solve to get \[W=\frac{1}{2}\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{d}\] Alternatively: \[W=Fd=\frac{{{Q}^{2}}}{2A{{\varepsilon }_{0}}}d=\frac{{{C}^{2}}_{1}{{V}^{2}}}{2{{\varepsilon }_{0}}A}d=\frac{1}{2}\frac{{{\varepsilon }_{0}}A{{V}^{2}}}{d}\]


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