In the following circuit diagram, potential difference across $4 \mu \mathrm{~F}$ capacitor is

1. Parallel Combination: The \(2 \mu F\) and \(6 \mu F\) capacitors are in parallel, so their equivalent capacitance: \(C_{e q}=2+6=8 \mu F\) 2. Series Combination: The \(4 \mu F^{\prime}\) capacitor is in series with the \(8 \mu F\) equivalent capacitor. The total capacitance is: \(\frac{1}{C_{t o t a l}}=\frac{1}{4}+\frac{1}{8}=\frac{3}{8} \Longrightarrow C_{t o t a l}=\frac{8}{3} \mu F\) 3. Charge Calculation: The total charge in the circuit is: \(Q=C_{t o t a l} \cdot V=\frac{8}{3} \cdot 12=32 \mu C\) 4. Voltage Across \(4 \mu F\) Capacitor: Since the charge is the same in series, the voltage across the \(4 \mu F\) capacitor is: \(V=\frac{Q}{C}=\frac{32}{4}=8 \mathrm{~V}\) Final answer: \(\mathrm{8} \mathrm{V}\).