
\because 1 \mu F & 2 \mu F are in parallel.

∴ The equivalent capacitance of the series combination is, Ceq is=C+33C. So, the total charge supplied by the battery is,
⇒Q=CeqE=C+33CE.∴ The potential difference across the parallel combination of 1μF and 2μF is, ΔV=3Q=C+3CE. So charge on 2μF capacitor is, Q2=C2ΔV=C+32CE⇒2EQ2=C+3C⇒2EQ2=C+3C+3−3
⇒2EQ2=1−C+33⇒ So the graph is a hyperbola. With downward curve line. i.e.





