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JEE Main & Advanced Sample Paper JEE Main Sample Paper-44

  • question_answer
    A parallel plate capacitor of capacity \[100\mu \]F is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated, so that the distance between them becomes double the original distance, the additional energy given to the battery of the capacitor in joules is

    A)  \[\frac{125\times {{10}^{-3}}}{2}\]                     

    B)  \[\frac{12.5\times {{10}^{-3}}}{2}\]

    C)  \[\frac{1.25\times {{10}^{-3}}}{2}\]                    

    D)  \[\frac{0.125\times {{10}^{-3}}}{2}\]

    Correct Answer: A

    Solution :

     Initially energy stored in capacitor \[{{E}_{1}}=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}\,\times 100\times {{10}^{-6}}\times 50\times 50\] \[\Rightarrow \]            \[{{E}_{1}}=125\times {{10}^{-3}}J\] When the distance is doubled, the capacitance decreases to 1/2, i.e.,       \[C'=50\,\mu F\] \[\therefore \]    Final energy,  \[{{E}_{2}}=\frac{1}{2}C'{{V}^{2}}=\frac{1}{2}\times 50\times {{10}^{-6}}\times {{(50)}^{2}}\] \[=\frac{125\times {{10}^{-3}}}{2}\] Additional energy, \[E={{E}_{1}}-{{E}_{2}}=\frac{125\times {{10}^{-3}}}{2}J\]


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