**Question:**Four capacitors having capacitances of 2 µF, 6 µF, 14 µF, and 5 µF are connected in parallel and this combination is across an ideal voltage source. If the charge on the 6 µF capacitor is 180 µC, what is the total charge on this circuit?- ○ 720 µC- ○ 120 µC- ○ 810 µC- ○ 4860 µC- ○ 1080 µC- ○ 180 µC**Explanation:**Capacitors in parallel share the same voltage, \( V \). The charge on any capacitor in parallel is given by \( Q = C \times V \). Here, the charge \( Q \) on the 6 µF capacitor is 180 µC, so:\[ V = \frac{Q}{C} = \frac{180 \, \mu C}{6 \, \mu F} = 30\, \text{V} \]The total charge for capacitors in parallel is the sum of individual charges:\[ Q_{\text{total}} = (2 + 6 + 14 + 5) \, \mu F \times 30 \, \text{V} \]\[ = 27 \, \mu F \times 30 \, \text{V} \]\[ = 810 \, \mu C \]Therefore, the total charge on the circuit is 810 µC.

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