Answer
The charge that passes through point P from the battery is $~~7.20~\mu C$
Work Step by Step
We can find the equivalent capacitance of $C_1, C_2,$ and $C_3$:
$\frac{1}{C_{eq}} = \frac{1}{8.00~\mu F}+\frac{1}{6.00~\mu F}+\frac{1}{8.00~\mu F}$
$C_{eq} = 2.40~\mu F$
We can find the charge stored on $C_{eq}$:
$q = C_{eq}~V$
$q = (2.40~\mu F)(12.0~V)$
$q = 28.8~\mu C$
When the switch is closed, $C_2$ and $C_4$ are in parallel.
The equivalent capacitance of these two capacitors is $6.00~\mu F+6.00~\mu F = 12.0~\mu F$
We can find the equivalent capacitance of $C_1, C_2, C_3,$ and $C_4$:
$\frac{1}{C_{eq}} = \frac{1}{8.00~\mu F}+\frac{1}{12.0~\mu F}+\frac{1}{8.00~\mu F}$
$C_{eq} = 3.00~\mu F$
We can find the charge stored on $C_{eq}$:
$q = C_{eq}~V$
$q = (3.00~\mu F)(12.0~V)$
$q = 36.0~\mu C$
We can find the charge that passes through point P from the battery:
$q = 36.0~\mu C-28.8~\mu C = 7.20~\mu C$
The charge that passes through point P from the battery is $~~7.20~\mu C$
