A capacitor of capacitance $50 pF$ is charged by 100 $V$ source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is $nJ$ .

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answer is 125.

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Detailed Solution

$Energy loss = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2} = \frac{1}{2} \frac{50 \times 50 \times 10^{- 12} \times 10^{- 12}}{(50 + 50) 10^{- 12}} (100 - 0)^{2} = 125 n J$

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