A capacitor C is fully charged with voltage V | vedclass

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Question

$\frac{ CV _{0}- q }{ C }=\frac{ q }{ C / 2}=\frac{2 q }{ C }$

$V _{0}=\frac{3 q }{ C } \Rightarrow q =\frac{ CV _{0}}{3}$

$U _{ i }=\frac{1}{2}

Vedclass · Accepted Answer

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Vedclass · Answer

$\frac{ CV _{0}- q }{ C }=\frac{ q }{ C / 2}=\frac{2 q }{ C }$

$V _{0}=\frac{3 q }{ C } \Rightarrow q =\frac{ CV _{0}}{3}$

$U _{ i }=\frac{1}{2} CV _{0}^{2}$

$U _{ f }=\frac{\left(\frac{2 CV _{0}}{3}\right)^{2}}{2 C }+\frac{\left(\frac{ CV _{0}}{3}\right)^{2}}{2\left(\frac{ C }{2}\right)}$

$=\frac{1}{2} CV _{0}^{2}\left[\frac{4}{9}+\frac{2}{9}\right]=\frac{1}{2} CV _{0}^{2}\left(\frac{2}{3}\right)$

Heat loss $=\frac{1}{2} CV _{0}^{2}-\left(\frac{2}{3}\right)\left(\frac{1}{2} CV _{0}^{2}\right)$

$=\frac{1}{6} CV _{0}^{2}$

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