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 $$\dfrac{E_1}{E_2}$$ will be

Initially switch KKK is closed, so both identical capacitors C are in parallel across voltage V.

Equivalent capacitance:

$$C_{eq}=2C$$

Initial energy:

$$E_1=\frac{1}{2}(2C)V^2$$

$$E_1=CV^2$$

Now switch is opened and dielectric constant

k=5

is inserted.

Left capacitor remains connected to battery, so voltage remains constant.

New capacitance:

$$C_1'=5C$$

Energy in left capacitor:

$$E_L=\frac{1}{2}(5C)V^2$$

$$=\frac{5}{2}CV^2$$

Right capacitor becomes isolated, so charge remains constant.

Initial charge on it was

Q=CV

After dielectric insertion,

$$C_2'=5C$$

Energy:

$$E_R=\frac{Q^2}{2C_2'}=\frac{(CV)^2}{2(5C)}=2(5C)(CV)2​$$

$$=\frac{1}{10}CV^2$$

Total final energy:

$$E_2=\frac{5}{2}CV^2+\frac{1}{10}CV^2$$

$$=\frac{26}{10}CV^2$$

$$=\frac{13}{5}CV^2$$

Therefore

$$\frac{E_1}{E_2}=\frac{CV^2}{(13/5)CV^2}$$

$$=\frac{5}{13}$$