ln the circuit shown in figure the potential difference across the 4.5 $\mathrm{\mu }$F capacitor is

Let us calculate the total capacitance of the circuit. The capacitors of capacitances 3 $\mathrm{\mu }$F and 6 $\mathrm{\mu }$F arc in parallel, Their capacitance C' is given by ${\mathrm{C}}^{\mathrm{\prime }}=3\mathrm{\mu F}+6\mathrm{\mu F}=9\mathrm{\mu F}$
The capacitor of capacitance C' is in series with the capacitor of capacitance 4.5 $\mathrm{\mu }$F. Therefore, the capacitance of the system is given by $\frac{1}{\mathrm{C}}=\frac{1}{9}+\frac{1}{4\cdot 5}=\frac{1}{3}$
or  $\mathrm{C}=3\mathrm{\mu F}$
The charge flowing through the circuit
$=3\times 12=36\mathrm{\mu C}$
$\therefore$Potential difference a cross 4.5 $\mathrm{\mu }$F capacitor
$=\frac{\mathrm{q}}{\mathrm{C}}=\frac{36}{4\cdot 5}=8 \mathrm{volt}$