$\because 1 \mu \mathrm{F} \& 2 \mu \mathrm{F}$ are in parallel.



$\therefore$ The equivalent capacitance of the series combination is, $C_{\mathrm{eq}}$ is$=\frac{3C}{C+3}$. So, the total charge supplied by the battery is, 
$\text{⇒}\mathit{\text{Q}}=C_{eq}E=\frac{3CE}{C+3}$.$\therefore$ The potential difference across the parallel combination of $1 \mu \mathrm{F}$ and $2\mu \mathrm{F}$ is, $∆V=\frac{Q}{3}=\frac{CE}{C+3}$. So charge on $2 \mu \mathrm{F}$ capacitor is, $Q_2=C_2∆V=\frac{2CE}{C+3}$$\Rightarrow \frac{Q_2}{2E}=\frac{C}{C+3}\Rightarrow \frac{Q_2}{2E}=\frac{C+3-3}{C+3}$

$\Rightarrow \frac{Q_2}{2E}=1-\frac{3}{C+3}$⇒ So the graph is a hyperbola. With downward curve line. i.e.