Answer
a) $8\;\rm V$
b) $3.7\;\rm V$
c) $8\;\rm V$
d) $4.8\times10^{-6}\;\rm C$
Work Step by Step
a)
The potential at point a when the switch is open is the same potential difference of 4.4 $\Omega$-resistor.
$$V_a=4.4I\tag 1 $$
Noting that the two resistors are in series as long as the switch is open, so they got the same current. Also, After a long time when the two capacitors are fully charged, there is no current passes through the right loop.
Applying Kirchhoff's second law on the left loop clockwise starting from the zero voltage end of the battery.
$$24- 8.8I-4.4I=0$$
Solving for $I$;
$$I=1.82\;\rm A$$
Plugging into (1);
$$V_a=4.4\cdot 1.82=\color{red}{\bf 8}\;\rm V$$
b)
Since the two capacitors are fully charged and no current passes through the right loop, and since the voltage of this branch that contains both of them is equal to that of the battery because their wire is in parallel to the battery, the voltage at point b is the voltage across the second capacitor.
$$V_b=\dfrac{Q_2}{C_2}=\dfrac{Q_2}{0.36\times10^{-6}}\tag 2$$
Now we need to find the charge stored in the second capacitor. Noting that the two capacitors are in series which means that they have the same amount of charge stored. And also, the equivalent capacitor must have the same amount of charge.
$$C_{eq}=\left(\dfrac{1}{C_1}+\dfrac{1}{C_2}\right)^{-1}=\left(\dfrac{1}{0.48\times10^{-6}}+\dfrac{1}{0.36\times10^{-6}}\right)^{-1}$$
$$C_{eq}=\bf2.06\times10^{-7}\;\rm F$$
We also know that $$Q=CV$$
Hence,
$$Q_{eq}=Q_1=Q_2=C_{eq}V=2.06\times10^{-7}\cdot 24=\bf 4.94\times10^{-6}\;\rm C$$
Plugging into (2);
$$V_b =\dfrac{ 4.94\times10^{-6}}{0.36\times10^{-6}}=\color{red}{\bf13.7}\;\rm V$$
c)
After a long time when the two capacitors are fully charged after closing the switch, there will be no current passes through the left loop again.
And hence, the voltage at point a will be the same again as when the switch was open (8 V).
And since point a is connected to point b by a wire with no resistance, so the two points must have the same voltage.
Thus,
$$V_b =\color{red}{\bf 8}\;\rm V$$
d)
First of all, we need to find the charge in each capacitor.
$$Q_1=C_1V_1=0.48\times10^{-6}\cdot (24-8)=\bf 7.68\times 10^{-6}\;\rm C$$
$$Q_2=C_2V_2=0.36\times10^{-6}\cdot 8=\bf 2.88\times 10^{-6}\;\rm C$$
We know that the negative plate of the second capacitor is facing the positive plate of the first one.
Thus, the total amount of charge at point b is given by
$$Q_{b}=-Q_1+Q_2=-7.68\times10^{-6}+2.88\times 10^{-6}=\bf \color{red}{\bf-4.8\times10^{-6}}\;\rm C$$
Thus the amount of charge passed through the switch is
$$Q= \color{red}{\bf 4.8\times10^{-6}}\;\rm C$$
