Answer
$V_1=11.48V$
$V_2=6.26V$
$V_3=5.22V$
$Q_1=11.5\mu C$
$Q_2=12.5\mu C$
$Q_3=12.5\mu C$
Work Step by Step
$Q=CV$
$Q_{total}=(1.0\mu F)(24V)=24\mu C$
$Q_2=Q_3$
$C_2V_2=C_3V_3$
$V_2=\frac{C_3V_3}{C_2}=1.2V_3$
$V_1=V_2+V_3=2.2V_3$
$Q_{total}=Q_1+Q_2$
$24\mu C=(1\mu F)V_1+(2\mu F)(0.55V_1)$
$V_1=11.48V$
$V_3=5.22V$
$V_2=6.26V$
$Q_1=C_1V_1=(1.0\mu F)(11.48V)=11.5\mu C$
$Q_2=C_2V_2=(2.0\mu F)(6.26V)=12.5\mu C$
$Q_3=C_3V_3=(2.4\mu F)(5.22V)=12.5\mu C$
