## College Physics (4th Edition)

(a) $C = 23.0~\mu F$ (b) $Q = 3.68\times 10^{-4}~C$ (c) $Q = 4.80\times 10^{-5}~C$
(a) Since the capacitors are connected in parallel, we can add the capacitance of each capacitor together to find the equivalent capacitance: $C = 4.0~\mu F+2.0~\mu F+3.0~\mu F+9.0~\mu F+5.0~\mu F$ $C = 23.0~\mu F$ (b) We can find the charge on a $23.0~\mu F$ capacitor: $Q = C~V$ $Q = (23.0 \times 10^{-6} F)(16.0~V)$ $Q = 3.68\times 10^{-4}~C$ (c) We can find the charge on the $3.0~\mu F$ capacitor: $Q = C~V$ $Q = (3.0 \times 10^{-6} F)(16.0~V)$ $Q = 4.80\times 10^{-5}~C$