Let's break this problem down step by step.

Given:

• $C_1 = 1000$ pF
• $C_2 = 100$ pF
• $C_3 = 500$ pF
• Voltage supply: $4.5V$

Step 1: Identify Series and Parallel Combinations

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

Step 2: Calculate Equivalent Capacitance of $C_2$ and $C_3$ (Parallel)

For capacitors in parallel, the total capacitance is: $C_{\text{parallel}} = C_2 + C_3$

$C_{\text{parallel}} = 100 + 500 = 600 \text{ pF}$

Step 3: Calculate Total Capacitance $C_{\text{total}}$ (Series Combination with $C_1$)

For capacitors in series, the total capacitance is: $\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_{\text{parallel}}}$

$\frac{1}{C_{\text{total}}} = \frac{1}{1000} + \frac{1}{600}$

Finding a common denominator:

$\frac{1}{1000} = 0.001, \quad \frac{1}{600} = 0.001667$

$\frac{1}{C_{\text{total}}} = 0.001 + 0.001667 = 0.002667$

$C_{\text{total}} = \frac{1}{0.002667} = 375 \text{ pF}$

Step 4: Calculate Total Voltage Across Each Capacitor

Since $C_1$ is in series with the $C_2 - C_3$ combination, they share the same charge.

Using the formula for voltage in series:

$V_1 = Q / C_1, \quad V_{\text{parallel}} = Q / C_{\text{parallel}}$

First, find total charge:

$Q_{\text{total}} = C_{\text{total}} \times V_{\text{total}}$

$Q_{\text{total}} = (375 \times 10^{-12}) \times 4.5$

$Q_{\text{total}} = 1.6875 \times 10^{-9} \text{ C}$

Now, voltage across $C_1$:

$V_1 = \frac{Q_{\text{total}}}{C_1} = \frac{1.6875 \times 10^{-9}}{1000 \times 10^{-12}}$

$V_1 = 1.6875V$

Voltage across the parallel combination:

$V_{\text{parallel}} = \frac{Q_{\text{total}}}{C_{\text{parallel}}} = \frac{1.6875 \times 10^{-9}}{600 \times 10^{-12}}$

$V_{\text{parallel}} = 2.8125V$

Since $C_2$ and $C_3$ are in parallel, they have the same voltage:

$V_2 = V_3 = 2.8125V$

Step 5: Calculate Charge on $C_2$ and $C_3$

Since $C_2$ and $C_3$ are in parallel, their charge is:

$Q_2 = C_2 \times V_2$

$Q_2 = (100 \times 10^{-12}) \times 2.8125$

$Q_2 = 2.8125 \times 10^{-10} \text{ C}$

Similarly, for $C_3$:

$Q_3 = C_3 \times V_3$

$Q_3 = (500 \times 10^{-12}) \times 2.8125$

$Q_3 = 1.40625 \times 10^{-9} \text{ C}$

Final Answers:

1. Total Capacitance: $C_{\text{total}} = 375$ pF
2. Total Charge: $Q_{\text{total}} = 1.6875 \times 10^{-9}$ C
3. Voltage Across $C_1$: $V_1 = 1.6875V$
4. Voltage Across $C_2$ and $C_3$: $V_2 = V_3 = 2.8125V$
5. Charge on $C_2$: $Q_2 = 2.8125 \times 10^{-10}$ C
6. Charge on $C_3$: $Q_3 = 1.40625 \times 10^{-9}$ C

Would you like me to clarify any steps? Let me know if you need further details!

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Follow-up Questions:

1. How do you determine whether capacitors are in series or parallel?
2. What happens if an additional capacitor is added in series?
3. How does increasing the supply voltage affect the charge and voltage distribution?
4. Why do capacitors in parallel have the same voltage?
5. What real-life applications use capacitor networks like this?

Tip: Always check the units carefully when working with capacitance, voltage, and charge to avoid calculation errors!