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 a 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
Answer: (C) $\dfrac5{13}$
With $K$ closed: $E_1 = \frac12(2C)V^2 = CV^2$.
After opening $K$, the capacitor still joined to the source stays at $V$; with the dielectric its energy is $\frac12(5C)V^2 = 2.5CV^2$. The other capacitor is isolated with charge $CV$; with the dielectric its energy is $\dfrac{(CV)^2}{2(5C)} = 0.1CV^2$.
$$E_2 = 2.6CV^2,\qquad\frac{E_1}{E_2} = \frac{1}{2.6} = \frac{5}{13}$$
Solution by Sreeraj P, M.Sc Physics