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 $\dots \; nJ$.
JEE MAIN 2022, Diffcult
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Energy loss $=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}$
$=\frac{1}{2} \frac{50 \times 50 \times 10^{-12} \times 10^{-12}}{(50+50) 10^{-12}}(100-0)^{2}=125 \; n J$
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