In a circuit, a 50 V battery is connected across points A and B,...

The circuit can be analyzed by finding the potentials at points C and D. Let the potential at B be $0\text{ V}$, then the potential at A is $50\text{ V}$.

In the upper branch, the $4\mu\text{F}$ and $8\mu\text{F}$ capacitors are in series. The equivalent capacitance is:
$C_{upper} = \dfrac{4 \times 8}{4 + 8} = \dfrac{32}{12} = \dfrac{8}{3}\mu\text{F}$

The charge on the upper branch is:
$Q_{upper} = C_{upper} \times V = \dfrac{8}{3} \times 50 = \dfrac{400}{3}\mu\text{C}$

The potential difference across the $4\mu\text{F}$ capacitor is:
$V_A - V_C = \dfrac{Q_{upper}}{4\mu\text{F}} = \dfrac{400/3}{4} = \dfrac{100}{3}\text{ V}$

In the lower branch, the $3\mu\text{F}$ and $6\mu\text{F}$ capacitors are in series. The equivalent capacitance is:
$C_{lower} = \dfrac{3 \times 6}{3 + 6} = \dfrac{18}{9} = 2\mu\text{F}$

The charge on the lower branch is:
$Q_{lower} = C_{lower} \times V = 2 \times 50 = 100\mu\text{C}$

The potential difference across the $3\mu\text{F}$ capacitor is:
$V_A - V_D = \dfrac{Q_{lower}}{3\mu\text{F}} = \dfrac{100}{3}\text{ V}$

Since $V_A - V_C = V_A - V_D = \dfrac{100}{3}\text{ V}$, we have $V_C = V_D$.

Alternatively, the circuit forms a Wheatstone bridge. The ratio of capacitances in the upper branch is $\dfrac{4}{8} = \dfrac{1}{2}$, and in the lower branch is $\dfrac{3}{6} = \dfrac{1}{2}$. Since the ratios are equal, the bridge is balanced, meaning points C and D are at the same potential.

Because the potential difference between C and D is zero ($V_C - V_D = 0$), no charge will appear on any additional capacitor connected between these two points. Thus, the charge is $0$.