To solve the problem of finding the charge stored on capacitor $C_3$, we can break it down step by step based on the circuit provided.

Given:

• $C_1 = 5.15 \, \mu\text{F}$
• $C_2 = 11.1 \, \mu\text{F}$
• $C_3 = 2.35 \, \mu\text{F}$
• Potential difference between points $a$ and $b$, $V_{ab} = 63.0 \, \text{V}$

From the circuit diagram, $C_1$, $C_2$, and $C_3$ are arranged in a specific configuration:

• $C_1$ and $C_3$ are in series.
• This combination is in parallel with $C_2$.

Step 1: Find the equivalent capacitance of $C_1$ and $C_3$ (series combination).

For capacitors in series, the equivalent capacitance $C_{13}$ is given by: $\frac{1}{C_{13}} = \frac{1}{C_1} + \frac{1}{C_3}$ Substituting the values: $\frac{1}{C_{13}} = \frac{1}{5.15} + \frac{1}{2.35}$ $C_{13} = \frac{1}{\left(\frac{1}{5.15} + \frac{1}{2.35}\right)} \approx 1.634 \, \mu\text{F}$

Step 2: Find the total equivalent capacitance $C_{eq}$ of the circuit.

Now, the equivalent capacitance $C_{eq}$ of the whole system is the parallel combination of $C_{13}$ and $C_2$: $C_{eq} = C_{13} + C_2 = 1.634 \, \mu\text{F} + 11.1 \, \mu\text{F} = 12.734 \, \mu\text{F}$

Step 3: Calculate the total charge stored in the system.

The total charge $Q_{total}$ is given by: $Q_{total} = C_{eq} \cdot V_{ab}$ Substituting the values: $Q_{total} = 12.734 \times 10^{-6} \, \text{F} \times 63.0 \, \text{V} \approx 0.8022 \, \text{mC} = 802.2 \, \mu\text{C}$

Step 4: Find the voltage across $C_1$ and $C_3$.

Since $C_1$ and $C_3$ are in series, they have the same charge. To find the charge on $C_3$, we first need to find the voltage across the series combination $C_{13}$.

The voltage across $C_2$ is given by: $V_{C2} = \frac{Q_{total}}{C_2} = \frac{802.2 \, \mu\text{C}}{11.1 \, \mu\text{F}} \approx 72.28 \, \text{V}$

Now, the voltage across the series combination ( C_3