Question bank/JEE Main/Physics

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

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[IMAGE] V_0 = 20 V
Heat loss = Ui - Uf
$= \frac{1}{2}CV_{0}^{2} - 2\left\lbrack \frac{1}{2}C\left( \frac{V_{0}}{2} \right)^{2} \right\rbrack$
$= \frac{CV_{0}^{2}}{4}$
$= \frac{\left( 60 \times 10^{- 12} \right)(20)^{2}}{4}J$
= 6 × 10^-9 J = 6 nJ

Source: Grass-G / P-12 Physics paper 2 Nov.docx