The circuit effectively consists of 5 identical capacitors (The right middle capacitor is short circuited), each with capacitance $C$. Initially, the switch is open and only capacitor $A$ (Top-Left) is charged with an initial charge $q_0$. All other capacitors are uncharged.

When the switch is closed, it connects the left terminal of capacitor $A$ to the rest of the network. We can model this as a redistribution of charge from capacitor $A$ to an equivalent capacitor representing the rest of the circuit.

We view the circuit from the terminals of capacitor $A$ (Left Node and Top-Middle Node).

• Right Branch: The Top-Right capacitor and the Right-Vertical capacitor are in series. $$ C_{right} = \frac{C \times C}{C + C} = \frac{C}{2} $$
• Middle Section: This series combination is in parallel with the Middle-Vertical capacitor. $$ C_{mid\_eff} = C + \frac{C}{2} = \frac{3C}{2} $$ This equivalent capacitance ($3C/2$) is connected between the Top-Middle Node and the Bottom Node.
• Left Bottom Branch: This entire $3C/2$ equivalent is in series with the Bottom-Left capacitor (since the path goes from the switch, through Bottom-Left, through the network, to the right plate of A). $$ C_{ext} = \frac{C \times \frac{3C}{2}}{C + \frac{3C}{2}} = \frac{1.5 C^2}{2.5 C} = 0.6 C $$

The initial charge $q_0$ on capacitor $A$ redistributes between capacitance $A$ ($C$) and the equivalent capacitance of the rest of the circuit ($C_{ext} = 0.6C$). Since they are connected in parallel (forming a closed loop), they share the same potential $V$.

The final charge $q_A$ on capacitor $A$ is:

Given that $q_A = 5.0 \, \mu\text{C}$: