## University Physics with Modern Physics (14th Edition)

(a) $Q_1 =156 \mathrm{~\mu C}$ and $Q_2 = 260 \mathrm{~\mu C}$ (b) $V_1 = V_2 = 52 \,\text{V}$
(a) As both capacitors are in parallel, therefore, they have the same potential even they have different capacitances. So the charge for each capacitor is calculated as next: For $C_1$ $$Q_1 = C_1 V = (3\mathrm{~\mu F} ) (52 \,\text{V}) = \boxed{156 \mathrm{~\mu C}}$$ For $C_2$ $$Q_2 = C_2 V = (5\mathrm{~\mu F} ) (52 \,\text{V}) = \boxed{260 \mathrm{~\mu C}}$$ (b) In parallel, the combination of the two capacitors has the same potential for each capacitor $$\boxed{V_1 = V_2 = 52 \,\text{V}}$$