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 $nJ$)
A$5$
B$6$
C$7$
D$8$
JEE MAIN 2020, Medium
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B$6$
b $\Delta \mathrm{Q}_{\mathrm{L}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}-\left[\frac{(\mathrm{Q} / 2)^{2}}{2 \mathrm{C}} \times 2\right]=\frac{\mathrm{Q}^{2}}{4 \mathrm{C}}$
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