A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E _{1}$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _{2}$. The ratio $E _{1} / E _{2}$ will be :
JEE MAIN 2022, Diffcult
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$(1)$ Switch is closed
$C _{ eq }=2 C$
Energy $E_{1}=\frac{1}{2} C_{\text {oq }} V^{2}$
$=\frac{1}{2} 2 C ^{2} \times V ^{2}$
$E _{1}= CV ^{2}$
$(ii)$ When switch is opened charge on right capacitor remain CV while potential on left capacitor remain same
Dielectric $K =5$
$C ^{\prime}= KC$
$C ^{\prime}=5\,C$
$E _{2}=\frac{1}{2}(5 C ) V ^{2}+\frac{( CV )^{2}}{2(5 C )}$
$E _{2}=\frac{5 CV ^{2}}{2}+\frac{ CV ^{2}}{10}$
$E _{2}=\frac{13 CV ^{2}}{5}$
$\frac{ E _{1}}{ E _{2}}=\frac{ CV ^{2}}{\frac{13 CV ^{2}}{5}}=\frac{5}{13}$
$\frac{ E _{1}}{ E _{2}}=\frac{5}{13}$
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