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It seems you provided an image containing a circuit problem with multiple parts related to the calculation of equivalent capacitance, charge, and potential difference across capacitors. I will now work through each part of the question.

### (a) Equivalent Capacitance (in μF)

The first step is to find the equivalent capacitance of all the capacitors in the circuit:

- The two capacitors on the left and right, each with \( C_1 = 49.60 \, \mu F \), are in series with the parallel combination of the capacitors \( C_2 = 43.60 \, \mu F \) and \( C_3 = 6.00 \, \mu F \).

1. **Parallel Combination**: 
   The equivalent capacitance of the two capacitors in parallel is:
   \[
   C_{parallel} = C_2 + C_3 = 43.60 \, \mu F + 6.00 \, \mu F = 49.60 \, \mu F
   \]

2. **Series Combination**:
   Now, the total equivalent capacitance is the series combination of three capacitors, two \( C_1 = 49.60 \, \mu F \) on the sides, and the equivalent \( C_{parallel} = 49.60 \, \mu F \):
   \[
   \frac{1}{C_{total}} = \frac{1}{49.60} + \frac{1}{49.60} + \frac{1}{49.60}
   \]
   Solving for \( C_{total} \):
   \[
   C_{total} = \frac{49.60}{3} = 16.53 \, \mu F
   \]

Thus, the equivalent capacitance is \( C_{total} = 16.53 \, \mu F \).

---

### (b) Charge Stored on Each Capacitor (in μC)

To find the charge on each capacitor, we use the relationship:
\[
Q = C \times V
\]
where \( V = 9.00 \, V \) is the total potential difference provided by the battery.

1. **Charge on Equivalent Capacitance**:
   The total charge stored by the equivalent capacitance is:
   \[
   Q_{total} = C_{total} \times V = 16.53 \, \mu F \times 9.00 \, V = 148.77 \, \mu C
   \]
   Since the capacitors are in series, the charge on each capacitor is the same.

2. **Charge on Each Capacitor**:
   Each capacitor in series stores the same charge, so:
   \[
   Q_{C1} = Q_{C2} = Q_{C3} = Q_{total} = 148.77 \, \mu C
   \]

---

### (c) Potential Difference Across Each Capacitor (in V)

We now calculate the potential difference across each capacitor.

1. **For the 49.60 μF Capacitors (C1)**:
   The potential difference across each of the 49.60 μF capacitors is:
   \[
   V = \frac{Q}{C} = \frac{148.77 \, \mu C}{49.60 \, \mu F} = 3.00 \, V
   \]

2. **For the Parallel Capacitors (C2 = 43.60 μF and C3 = 6.00 μF)**:
   The total potential difference across these capacitors is the remaining:
   \[
   V_{C2 \, and \, C3} = 9.00 \, V - 3.00 \, V = 6.00 \, V
   \]
   Since they are in parallel, both capacitors will have this potential difference of 6.00 V.

Thus, the potential differences are:
- \( V_{C1} = 3.00 \, V \)
- \( V_{C2} = 6.00 \, V \)
- \( V_{C3} = 6.00 \, V \)

---

Would you like more details on any of these calculations? Here are five related questions to extend this understanding:

1. How does the configuration of series and parallel capacitors affect the total capacitance?
2. Why is the charge the same on all capacitors in series?
3. How does the potential difference split between capacitors in series and parallel?
4. What happens to the total capacitance if one of the parallel capacitors' value changes?
5. How would the total charge change if the battery voltage increased?

**Tip**: For capacitors in series, the total capacitance is always smaller than the smallest capacitor in the series.

The circuit in the figure contains a 9.00 V battery and four capacitors. Find the equivalent capacitance, the charge stored by each capacitor, and the potential difference across each capacitor.

This problem demonstrates how to calculate the equivalent capacitance, the charge stored by each capacitor, and the potential difference across each capacitor in a circuit containing four capacitors and a 9.00 V battery. Key concepts include series and parallel capacitors, capacitance formulas, and electric charge distribution.

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