Combination of Capacitors question from JEE Main 2015 (11 Apr, Shift 1)

Answer: (a)

1. Switch open: each branch is $2$ and $3$ in series, $Q = \frac{6}{5}\times 10 = 12\ \mu\text{C}$. Nodes a and b each hold net charge $-12 + 12 = 0$.
2. Switch closed: left pair $2 + 3 = 5$, right pair $3 + 2 = 5$, so the junction sits at 5 V and each capacitor has 5 V.
3. Node a now: $-2\times5 + 3\times5 = +5\ \mu\text{C}$; node b: $-3\times5 + 2\times5 = -5\ \mu\text{C}$.
4. So $5\ \mu\text{C}$ moves through S from b to a.
5. So the answer is (a) 5 $\mu$C from b to a.