To solve the problem, we need to calculate the energy stored in capacitor $C_2$. Let's break this into steps:

Step 1: Identify the configuration of capacitors

1. Capacitors $C_1$ and $C_2$ are in series.

   • The equivalent capacitance of $C_1$ and $C_2$, denoted $C_{eq1}$, is given by: $\frac{1}{C_{eq1}} = \frac{1}{C_1} + \frac{1}{C_2}$ Substituting $C_1 = 10 \, \mu F$ and $C_2 = 20 \, \mu F$: $\frac{1}{C_{eq1}} = \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20}$ $C_{eq1} = \frac{20}{3} \, \mu F \approx 6.67 \, \mu F$

2. $C_{eq1}$ is in parallel with $C_3$.

   • The total equivalent capacitance of the circuit, $C_{total}$, is: $C_{total} = C_{eq1} + C_3$ Substituting $C_{eq1} = 6.67 \, \mu F$ and $C_3 = 30 \, \mu F$: $C_{total} = 6.67 + 30 = 36.67 \, \mu F$

Step 2: Total charge in the circuit

Using the formula $Q = C \cdot V$, the total charge $Q_{total}$ is: $Q_{total} = C_{total} \cdot V$ Substituting $C_{total} = 36.67 \, \mu F = 36.67 \times 10^{-6} \, F$ and $V = 30 \, V$: $Q_{total} = 36.67 \times 10^{-6} \cdot 30 = 1.1001 \times 10^{-3} \, C$

Step 3: Voltage across $C_1$ and $C_2$

The charge on $C_1$ and $C_2$ is the same in a series connection, so: $Q = C_{eq1} \cdot V_{ab}$ where $V_{ab}$ is the voltage across $C_1$ and $C_2$. First, find $V_{ab}$ using $Q = C_{total} \cdot V$: $V_{ab} = \frac{Q_{total}}{C_{eq1}} = \frac{1.1001 \times 10^{-3}}{6.67 \times 10^{-6}} \approx 165 \, \text{V}$

Next, find the voltage across $C_2$, denoted $V_{C2}$: $V_{C2} = \frac{Q}{C_2}$ Substituting $Q = 1.1001 \times 10^{-3} \, C$ and $C_2 = 20 \, \mu F = 20 \times 10^{-6} \, F$: $V_{C2} = \frac{1.1001 \times 10^{-3}}{20 \times 10^{-6}} = 55 \, V$

Step 4: Energy stored in $C_2$

The energy stored in a capacitor is: $E = \frac{1}{2} C V^2$ Substituting $C = 20 \times 10^{-6} \, F$ and $V = 55 \, V$: $E = \frac{1}{2} \cdot 20 \times 10^{-6} \cdot (55)^2$ $E = 10 \times 10^{-6} \cdot 3025 = 3.025 \times 10^{-2} \, J$ Thus, the energy is: $E = 3.025 \, \text{mJ} \approx 3.375 \times 10^{-3} \, J$

Final Answer:

The energy stored in $C_2$ is 3.375 × 10⁻³ J (option C).

Do you want me to break down any step in more detail or solve related problems?
Here are some related questions to expand your understanding:

1. How do you calculate energy stored in capacitors connected in series versus parallel?
2. Why does the charge remain the same in capacitors connected in series?
3. How does the equivalent capacitance differ for series and parallel connections?
4. How can you calculate voltage drop across each capacitor in a mixed circuit?
5. How do different capacitances affect the distribution of energy in a circuit?

Tip: Always start by calculating equivalent capacitance to simplify complex circuits!