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

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

The equivalent capacitance is $C_{eq1} = \dfrac{C \times C}{C + C} = \dfrac{C}{2}$.

The initial charge supplied by the battery is $Q_i = C_{eq1}\mathcal{E} = \dfrac{C\mathcal{E}}{2}$.

When the switch $S$ is closed, the right capacitor is short-circuited and the left capacitor is directly connected across the battery.

The new equivalent capacitance is $C_{eq2} = C$.

The final charge supplied by the battery is $Q_f = C_{eq2}\mathcal{E} = C\mathcal{E}$.

The extra charge that flows through the battery is $\Delta Q = Q_f - Q_i = C\mathcal{E} - \dfrac{C\mathcal{E}}{2} = \dfrac{C\mathcal{E}}{2}$.

The work done by the battery is $W = \Delta Q \times \mathcal{E} = \left(\dfrac{C\mathcal{E}}{2}\right)\mathcal{E} = \dfrac{C\mathcal{E}^2}{2}$.