In the given circuit, charge Q _ 2 on the 2 capacitor changes as C...

$∵ 1\muF \text{ and } 2\muF$ are in parallel.



$∴$ The equivalent capacitance of the series combination is, ${C}_{eq}$ is$= \dfrac{3C}{C + 3}$. So, the total charge supplied by the battery is,
$\text{⇒}\text{Q} = {C}_{eq}E = \dfrac{3CE}{C + 3}$.$∴$ The potential difference across the parallel combination of $1\muF$ and $2\muF$ is, $∆ V = \dfrac{Q}{3} = \dfrac{CE}{C + 3}$. So charge on $2\muF$ capacitor is, ${Q}_{2} = {C}_{2} ∆ V = \dfrac{2CE}{C + 3}$$⇒ \dfrac{{Q}_{2}}{2E} = \dfrac{C}{C + 3} ⇒ \dfrac{{Q}_{2}}{2E} = \dfrac{C + 3 - 3}{C + 3}$

$⇒ \dfrac{{Q}_{2}}{2E} = 1 - \dfrac{3}{C + 3}$→ So the graph is a hyperbola. With downward curve line. i.e.