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A 400 pF capacitor is charged with a 100 V battery. After disconnecting battery this capacitor is connected with another 400 pF capacitor. Then find out energy loss.

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$1 x 10^{- 6} J$

$2 x 10^{- 6} J$

$3 x 10^{- 6} J$

$4 x 10^{- 6} J$

answer is A.

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Detailed Solution

Energy loss = $\frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} - V_{2})}^{2}$

$= \frac{400 x 10^{- 12}}{4} {(100 - 0)}^{2} J = 1 x 10^{- 6} J$

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