Consider the circuit shown in the figure. The switch S is open for...

When the switch $S$ is open, the two capacitors are connected in series across the battery.

The equivalent capacitance of the circuit is $C_{eq} = \dfrac{C \times C}{C + C} = \dfrac{C}{2}$.

The initial energy stored in the capacitors is:

$U_i = \dfrac{1}{2} C_{eq} \mathcal{E}^2 = \dfrac{1}{2} \left(\dfrac{C}{2}\right) \mathcal{E}^2 = \dfrac{1}{4} C \mathcal{E}^2$

When the switch $S$ is closed, the capacitor on the right is short-circuited because both of its plates are connected to the same potential (the negative terminal of the battery). Therefore, it stores no energy.

The capacitor on the left is now connected directly across the battery, so the potential difference across it is $\mathcal{E}$.

The final energy stored in the circuit is:

$U_f = \dfrac{1}{2} C \mathcal{E}^2$

The change in the energy stored in the capacitors is:

$\Delta U = U_f - U_i = \dfrac{1}{2} C \mathcal{E}^2 - \dfrac{1}{4} C \mathcal{E}^2 = \dfrac{1}{4} C \mathcal{E}^2$

Answer: $\dfrac{C\mathcal{E}^2}{4}$