Three capacitors of capacitances \( 2\, \mu\text{F} \), \( 3\, \mu\text{F} \), and \( 6\, \mu\text{F} \) are connected in series across a \( 120\text{ V} \) DC source. The potential difference across the \( 2\, \mu\text{F} \) capacitor is:

Equivalent capacitance \( \frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1\, \mu\text{F}^{-1} \implies C_{\text{eq}} = 1\, \mu\text{F} \). Total charge \( Q = C_{\text{eq}} V = 1\, \mu\text{F} \times 120\text{ V} = 120\, \mu\text{C} \). Voltage across \( 2\, \mu\text{F} \) is \( V_1 = \frac{Q}{C_1} = \frac{120\, \mu\text{C}}{2\, \mu\text{F}} = 60\text{ V} \).