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  4. A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)

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Q. A $60\, pF$ capacitor is fully charged by a $20\, V$ supply. It is then disconnected from the supply and is connected to another uncharged $60\, pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)

JEE MainJEE Main 2020Electrostatic Potential and Capacitance

Solution:

$\Delta Q_{L} = \frac{Q^{2}}{2C}-\left[\frac{\left(Q/ 2\right)^{2}}{2C}\times2\right] = \frac{Q^{2}}{4C}$
$= \frac{1}{4}CV^{2}$
$= \frac{1}{4}× 60 × 10^{-12} × 4 × 10^{2}$
$= 6nJ$

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