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

  • question_answer
    The energy required to charge a parallel plate condenser of plate separation a and plate area of cross-section A such that the uniform electric field between the plates is E, is   [AIPMPT (S) 2008]

    A) \[\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}/Ad\]

    B) \[{{\varepsilon }_{0}}{{E}^{2}}/Ad\]

    C) \[{{\varepsilon }_{0}}{{E}^{2}}Ad\]

    D) \[\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}Ad\]

    Correct Answer: C

    Solution :

    [c] Energy given by the cell \[E=C{{V}^{2}}\]         
    Here, \[C=\] capacitance of condenser \[=\frac{A{{\varepsilon }_{0}}}{d}\]
    \[V=\] potential difference across the plates \[=Ed\]
    Therefore, \[E=\left( \frac{A{{\varepsilon }_{0}}}{d} \right){{(Ed)}^{2}}\]
    \[=A\,{{\varepsilon }_{0}}{{E}^{2}}d\]


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