Determine the potential difference between points A and B , and...

In steady state, the capacitors are fully charged and act as open circuits. Therefore, no direct current flows through the circuit ($I = 0$).

Since the current is zero, there is no voltage drop across the $20\ \Omega$ and $10\ \Omega$ resistors. The potential difference across the capacitor network, between points A and C, is equal to the electromotive force of the battery:

$V_A - V_C = 100\text{ V}$

Let us analyze the connections of the capacitors in the network. The left vertical wire is entirely at potential $V_A$, and the right vertical wire is entirely at potential $V_C$.

The top $3\ \mu\text{F}$ capacitor is connected between node A and node B. The middle horizontal wire is connected to the left vertical wire (node A). The vertical $3\ \mu\text{F}$ capacitor connects this middle wire to node B. Thus, both $3\ \mu\text{F}$ capacitors are connected in parallel between nodes A and B. Their equivalent capacitance is:

$C_{AB} = 3\ \mu\text{F} + 3\ \mu\text{F} = 6\ \mu\text{F}$

Similarly, the top $1\ \mu\text{F}$ capacitor is connected between node B and node C. The middle horizontal wire is connected to the right vertical wire (node C). The vertical $1\ \mu\text{F}$ capacitor connects this middle wire to node B. Thus, both $1\ \mu\text{F}$ capacitors are connected in parallel between nodes B and C. Their equivalent capacitance is:

$C_{BC} = 1\ \mu\text{F} + 1\ \mu\text{F} = 2\ \mu\text{F}$

The middle $1\ \mu\text{F}$ capacitor is connected directly between node A and node C. It is in parallel with the series combination of $C_{AB}$ and $C_{BC}$, and does not affect the potential at node B.

The equivalent capacitors $C_{AB}$ and $C_{BC}$ are in series across the potential difference of $100\text{ V}$. In a series combination, the potential difference across a capacitor is inversely proportional to its capacitance.

The potential difference between A and B is:

$V_{AB} = V_{AC} \times \dfrac{C_{BC}}{C_{AB} + C_{BC}}$

$V_{AB} = 100 \times \dfrac{2}{6 + 2} = 100 \times \dfrac{2}{8} = 25\text{ V}$

The potential difference between B and C is:

$V_{BC} = V_{AC} \times \dfrac{C_{AB}}{C_{AB} + C_{BC}}$

$V_{BC} = 100 \times \dfrac{6}{6 + 2} = 100 \times \dfrac{6}{8} = 75\text{ V}$

Answer: $25 \text{ V}$ and $75 \text{ V}$