J & K CET Medical J & K - CET Medical Solved Paper-2004

  • question_answer
    On increasing the plate separation of a charged condenser, the energy :

    A)  increases                           

    B)  decreases

    C)  remains unchanged      

    D)  becomes zero

    Correct Answer: A

    Solution :

                    The energy which is stored in the condenser is given by \[E=\frac{1}{2}.\frac{{{q}^{2}}}{C}\]                         ?.. (i) where q is charge and C the capacitance. Also,      \[C=\frac{{{\varepsilon }_{0}}A}{d}\]                          ...(ii) From Eqs. (i) and (ii), we get                 \[E=\frac{1}{2}.\frac{{{q}^{2}}d}{{{\varepsilon }_{0}}A}\] \[\Rightarrow \]               \[E\propto d\] When plate separation  is increased energy increases.


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