A battery of emf 18 V and negligible internal resistance is connected across two parallel branches. Branch 1 consists of 6 Ω followed by 3 Ω in…

In the steady state no current flows through the capacitor, so the two branches are independent. Branch 1 carries $I_1=\dfrac{18}{6+3}=2\ \mathrm{A}$. Taking the negative terminal as $0\ \mathrm{V}$, the junction between the $6\ \Omega$ and $3\ \Omega$ resistors is at $18-6\times2=6\ \mathrm{V}$. Branch 2 carries $I_2=\dfrac{18}{2+4}=3\ \mathrm{A}$, so its mid-junction is at $18-2\times3=12\ \mathrm{V}$. The capacitor therefore has a potential difference $|12-6|=6\ \mathrm{V}$ across it, and its charge is $Q=CV=1.5\ \mu\mathrm{F}\times6\ \mathrm{V}=9\ \mu\mathrm{C}$. Remember that the capacitor carries no steady current, so it does not disturb the potentials of the two junctions; the sign of the charge is irrelevant to the magnitude asked, and the plate connected to the higher-potential junction is positive.