2. Electric Potential and Capacitance
medium

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)

Solution

$\Delta \mathrm{Q}_{\mathrm{L}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}-\left[\frac{(\mathrm{Q} / 2)^{2}}{2 \mathrm{C}} \times 2\right]=\frac{\mathrm{Q}^{2}}{4 \mathrm{C}}$

$=\frac{1}{4} \mathrm{CV}^{2}$

$=\frac{1}{4} \times 60 \times 10^{-12} \times 4 \times 10^{2}$

$=6 \mathrm{nJ}$

Standard 12
Physics

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