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J & K CET Medical J & K - CET Medical Solved Paper-2010

  • question_answer
    A capacitor is charged by a battery and the energy stored is U. The battery is now removed and the separation distance between the plates is doubled. The energy stored now is

    A) \[\frac{u}{2}\]                                  

    B)  U

    C)  2 U                                       

    D)  4U

    Correct Answer: C

    Solution :

                     Energy stored \[U=\frac{1}{2}qV\] As the distance d is increased between the two plates. Now, stored energy, \[U=\frac{1}{2}qV\] \[=\frac{1}{2}q\left[ \frac{q}{C} \right]\] \[=\frac{1}{2}\frac{{{q}^{2}}d}{{{\varepsilon }_{0}}A}\] \[U\propto d\] Hence, \[U=2U\]


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