A $60\; pF$ capacitor is fully charged by a $20\; \mathrm{V}$ supply. It is then disconnected from the supply and is connected to another uncharged $60 \;pF$ capactior is parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$)
JEE MAIN 2020, Medium
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$\Delta \mathrm{Q}_{\mathrm{L}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}-\left[\frac{(\mathrm{Q} / 2)^{2}}{2 \mathrm{C}} \times 2\right]=\frac{\mathrm{Q}^{2}}{4 \mathrm{C}}$
$=\frac{1}{4} \mathrm{CV}^{2}$
$=\frac{1}{4} \times 60 \times 10^{-12} \times 4 \times 10^{2}$
$=6 \mathrm{nJ}$
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