Let's solve the given problem step by step.

Step 1: Given Information

• The capacitance of each capacitor $C = 0.045 \, \text{F}$.
• The total equivalent capacitance $C_{\text{eq}}$ is given by the formula:

$C_{\text{eq}} = \frac{1}{\left(\frac{1}{2C + 3C}\right) + \left(\frac{1}{5C}\right)}$

Step 2: Simplify the Expressions Inside the Formula

1. First Term: $2C + 3C = 5C$, so the first term becomes $\frac{1}{5C}$.
2. Second Term: The second term is already in the form $\frac{1}{5C}$.

Thus, the equation becomes:

$C_{\text{eq}} = \frac{1}{\left( \frac{1}{5C} \right) + \left( \frac{1}{5C} \right)}$

Step 3: Combine the Terms

Since both terms are $\frac{1}{5C}$, we can combine them:

$C_{\text{eq}} = \frac{1}{\frac{2}{5C}}$

Step 4: Simplify Further

Now simplify the right-hand side:

$C_{\text{eq}} = \frac{5C}{2}$

Step 5: Substitute the Value of $C$

Now substitute $C = 0.045 \, \text{F}$ into the equation:

$C_{\text{eq}} = \frac{5 \times 0.045}{2} = \frac{0.225}{2} = 0.1125 \, \text{F}$

Final Answer:

The equivalent capacitance $C_{\text{eq}}$ is $0.1125 \, \text{F}$.

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Additional Related Questions:

1. How would the equivalent capacitance change if the capacitors were arranged in a different configuration (e.g., series instead of parallel)?
2. What is the effect of increasing the capacitance of each individual capacitor on the overall equivalent capacitance?
3. How can you calculate the total energy stored in this capacitor network if it is charged to a potential difference of 10 V?
4. What happens to the equivalent capacitance if the number of capacitors increases in a similar configuration?
5. How would the result change if the capacitors had different capacitances?

Tip: When solving problems involving capacitors in parallel and series, always simplify step-by-step and carefully track the relationships between the individual and equivalent capacitances!