The potential difference across the $4 \mu \mathrm{~F}$ capacitor in the following circuit is

The potential difference across the $4 \mu \mathrm{F}$ capacitor is determined by simplifying the circuit and analyzing charge distribution. The $2 \mu \mathrm{F}$ and $4 \mu \mathrm{F}$ capacitors are in parallel, with equivalent capacitance $C_{\text{parallel}} = 2 \mu \mathrm{F} + 4 \mu \mathrm{F} = 6 \mu \mathrm{F}$. This combination is in series with the other $4 \mu \mathrm{F}$ capacitor. The equivalent capacitance is found from $\frac{1}{C_{\text{eq}}} = \frac{1}{4 \mu \mathrm{F}} + \frac{1}{6 \mu \mathrm{F}} = \frac{5}{12 \mu \mathrm{F}}$, giving $C_{\text{eq}} = \frac{12}{5} \mu \mathrm{F} = 2.4 \mu \mathrm{F}$. With a $9 \mathrm{V}$ battery, the total charge is $Q = (2.4 \mu \mathrm{F}) \times 9 \mathrm{V} = 21.6 \mu \mathrm{C}$. In series, this charge is the same on the $4 \mu \mathrm{F}$ capacitor. The potential difference across it is $V_1 = \frac{21.6 \mu \mathrm{C}}{4 \mu \mathrm{F}} = 5.4 \mathrm{V}$. Final answer: $\boxed{5.4 \mathrm{V}}$