Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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14CV12−V22

14CV12+V22

14CV1−V22

14CV1+V22

answer is C.

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Detailed Solution

Common potential V=CV1+CV2C+C=V1+V22 Loss =12CV12+12CV22−12(C+C)V2 =14CV1−V22

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