\(C _{ eq }=2 C\)

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 \(K =5\)

\(C ^{\prime}= KC\)

\(C ^{\prime}=5\,C\)

\(E _{2}=\frac{1}{2}(5 C ) V ^{2}+\frac{( CV )^{2}}{2(5 C )}\)

\(E _{2}=\frac{5 CV ^{2}}{2}+\frac{ CV ^{2}}{10}\)

\(E _{2}=\frac{13 CV ^{2}}{5}\)

\(\frac{ E _{1}}{ E _{2}}=\frac{ CV ^{2}}{\frac{13 CV ^{2}}{5}}=\frac{5}{13}\)

\(\frac{ E _{1}}{ E _{2}}=\frac{5}{13}\)