Answer
(a) $C = 11.0~\mu F$
(b) $Q = 2.64\times 10^{-4}~C$
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:
$C = 2.0~\mu F+6.0~\mu F+3.0~\mu F$
$C = 11.0~\mu F$
(b) We can find the charge on the $6.0~\mu F$ capacitor:
$Q = C~V$
$Q = (6.0 \times 10^{-6} F)(44.0~V)$
$Q = 2.64\times 10^{-4}~C$
