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Question

Energy in each capacitor is $\frac{1}{2} cV ^2$

when switch is opened

$Q$ on $2^{nd}$ capacitor is $cv$ even after introduction of dielectric $Q

Vedclass · Accepted Answer

$3 : 5$

Vedclass · Answer

Energy in each capacitor is $\frac{1}{2} cV ^2$

when switch is opened

$Q$ on $2^{nd}$ capacitor is $cv$ even after introduction of dielectric $Q$ remain same but

$C$ will become $KC$

Energy $=\frac{ Q ^2}{2 C ^2}=\frac{( cv )^2}{2\left( K _{ l }ight)}=\frac{ cv ^2}{2 k }$

Energy is set is $\frac{1}{2} cv ^2$

same as before

after dielectric insertion

$c$ become $kc$

Energy $=\frac{1}{2}( kc ) v ^2$

$=\frac{ kcv ^2}{2}$

Total energy initially $=\frac{1}{2} c v^2+\frac{1}{2} c v^2=c^2 \cdots(1)$

Total energy after opening switch $= k \frac{ cv ^2}{2}+\frac{ cv ^2}{2 k }$

$=\frac{c v^2}{2}\left( k +\frac{1}{ k }ight)$

put $k=3$

$=\frac{ cV ^2(10)}{2 	imes 3}=\frac{5 CV ^2}{3} \ldots(2)$

$(1)$ $\div(2)$

$=3 / 5$

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