Assuming the potential at point B is zero, determine the potentials...

Given the potential at point B is $V_B = 0\text{ V}$.

The battery is connected across points A and B with its positive terminal (longer line) towards A and negative terminal (shorter line) towards B. Thus, the potential at point A is $V_A = 50\text{ V}$.

In the upper branch, the $4\ \mu\text{F}$ and $8\ \mu\text{F}$ capacitors are connected in series across A and B. The potential difference across the $8\ \mu\text{F}$ capacitor is $V_C - V_B$.

Using the voltage divider rule for capacitors in series ($V_1 = V \times \dfrac{C_2}{C_1 + C_2}$), the voltage across the $8\ \mu\text{F}$ capacitor is:
$V_C - V_B = (V_A - V_B) \times \dfrac{4}{4 + 8}$

$V_C - 0 = 50 \times \dfrac{4}{12} = \dfrac{50}{3}\text{ V}$

In the lower branch, the $3\ \mu\text{F}$ and $6\ \mu\text{F}$ capacitors are connected in series across A and B. The potential difference across the $6\ \mu\text{F}$ capacitor is $V_D - V_B$.

Similarly, the voltage across the $6\ \mu\text{F}$ capacitor is:
$V_D - V_B = (V_A - V_B) \times \dfrac{3}{3 + 6}$

$V_D - 0 = 50 \times \dfrac{3}{9} = \dfrac{50}{3}\text{ V}$

Therefore, the potentials at points C and D are both $\dfrac{50}{3}\text{ V}$.