Solution:

To solve this problem, let's analyze the situation step by step:

1. Initial Charges on Capacitors

The charge on each capacitor is given by $Q = CV$:

• For the first capacitor ($C_1 = 2 \, \mu\text{F}, V_1 = 10 \, \text{V}$):

  $$Q_1 = C_1 V_1 = 2 \times 10^{-6} \cdot 10 = 20 \, \mu\text{C}$$

• For the second capacitor ($C_2 = 3 \, \mu\text{F}, V_2 = 20 \, \text{V}$):

  $$Q_2 = C_2 V_2 = 3 \times 10^{-6} \cdot 20 = 60 \, \mu\text{C}$$

2. Connecting Capacitors with Opposite Polarity

When connected with opposite polarity, the charges on the two capacitors partially cancel each other. The net charge is:

$$Q_{\text{net}} = Q_2 - Q_1 = 60 \, \mu\text{C} - 20 \, \mu\text{C} = 40 \, \mu\text{C}$$

The equivalent capacitance of the two capacitors in parallel is:

$$C_{\text{eq}} = C_1 + C_2 = 2 \, \mu\text{F} + 3 \, \mu\text{F} = 5 \, \mu\text{F}$$

3. Final Energy Stored in the System

The energy stored in a capacitor is given by:

$$U = \frac{1}{2} \frac{Q_{\text{net}}^2}{C_{\text{eq}}}$$

Substitute the values:

$$U = \frac{1}{2} \cdot \frac{(40 \times 10^{-6})^2}{5 \times 10^{-6}}$$ $$U = \frac{1}{2} \cdot \frac{1600 \times 10^{-12}}{5 \times 10^{-6}}$$ $$U = \frac{1}{2} \cdot 320 \times 10^{-6} = 160 \times 10^{-6} \, \text{J}$$ $$U = 540 \, \mu\text{J}$$

Final Answer:

The final energy stored in the system is 540 μJ.