Answer
(a) $C = 23.0~\mu F$
(b) $Q = 3.68\times 10^{-4}~C$
(c) $Q = 4.80\times 10^{-5}~C$
Work Step by Step
(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:
$Q = C~V$
$Q = (3.0 \times 10^{-6} F)(16.0~V)$
$Q = 4.80\times 10^{-5}~C$
