Answer
(a) $23\mu F$
(b) the $15\mu F$ capacitor
(c) $Q_{7.5\mu F}=110\mu C, Q_{15\mu F}=230\mu C$
Work Step by Step
(a) We know that the equivalent capacitance is given as
$C=C_1+C_2$
$C=7.5\times 10^{-6}F+15\times 10^{-6}F=23\times 10^{-6}F=23\mu F$
(b) We know that $Q=CV$. This equation shows that the charge stored on the capacitor depends on the capacitance of the capacitor. Therefore, the larger capacitance $C_2=15\mu F$ will store more charge.
(c) We know that $Q=C_1V$
$\implies Q=(7.5\times 10^{-6}F)(15.0V)$
$Q=110\mu F$
Similarly $Q=C_2V$
$\implies Q=(15\times 10^{-6}F)(15.0V)$
$Q=230\mu F$
