RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
                      The uniform electric field in the space between the plates of a parallel plate capacitor of plate separation d and plate areas A is E. The energy of this charged capacitor is

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

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

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

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

    Correct Answer: C

    Solution :

    Energy stored\[=\frac{1}{2}C{{V}^{2}}\]                 \[=\frac{1}{2}\left( \frac{{{\varepsilon }_{0}}A}{d} \right)({{V}^{2}})\]                 \[=\frac{1}{2}{{\left( \frac{V}{d} \right)}^{2}}Ad\]                 \[=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}Ad\]


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