A 60; pF capacitor is fully charged by a 20; | vedclass

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Question

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

Vedclass · Accepted Answer

$6$

Vedclass · Answer

$\Delta \mathrm{Q}_{\mathrm{L}}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}-\left[\frac{(\mathrm{Q} / 2)^{2}}{2 \mathrm{C}} 	imes 2ight]=\frac{\mathrm{Q}^{2}}{4 \mathrm{C}}$

$=\frac{1}{4} \mathrm{CV}^{2}$

$=\frac{1}{4} 	imes 60 	imes 10^{-12} 	imes 4 	imes 10^{2}$

$=6 \mathrm{nJ}$

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 $nJ$)

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