JEE Main & Advanced Physics Electrostatics & Capacitance NEET PYQ - Electrostatics and Capacitance

  • question_answer
    A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference V is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is    [NEET (Re) 2015]

    A) \[\frac{{{C}^{2}}{{V}^{2}}}{2d}\]

    B) \[\frac{C{{V}^{2}}}{2d}\]

    C) \[\frac{C{{V}^{2}}}{d}\]

    D) \[\frac{{{C}^{2}}{{V}^{2}}}{2{{d}^{2}}}\]

    Correct Answer: D

    Solution :

    [d] Force between plates of parallel capacitor,
                \[F=qE=q\left[ \frac{\sigma }{2{{\varepsilon }_{0}}} \right]\]
    \[\because \] Surface charge density \[\sigma =\frac{\sigma }{A}\]
    \[\therefore \]      \[F=q\left[ \frac{q}{2A{{\varepsilon }_{0}}} \right]\Rightarrow F=\frac{{{q}^{2}}}{2A{{\varepsilon }_{0}}}\]
    So, net charge across a capacitor, \[q=CV\]
    \[F=\frac{{{C}^{2}}{{V}^{2}}}{2A{{\varepsilon }_{0}}}\]          \[\left[ C=\frac{A{{\varepsilon }_{0}}}{d} \right]\]
    \[F=\frac{\left( \frac{A{{\varepsilon }_{0}}}{d} \right)\times C{{V}^{2}}}{2A{{\varepsilon }_{0}}}=\frac{C{{V}^{2}}}{2d}\]


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