A 2 µC electric charge experiences an electric potential energy of 15.43 J when placed 3 mm from another charge. Determine the value of this second charge.

To determine the value of the second charge, we can use the equation for electric potential energy:

$$PE = \frac{k \cdot |q_1 \cdot q_2|}{r}$$

Where:
- $PE$ is the electric potential energy
- $k$ is the Coulomb's constant ($9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$)
- $q_1$ and $q_2$ are the charges in coulombs
- $r$ is the distance between the charges in meters

Plugging in the given values, we have:

$$15.43 \, \text{J} = \frac{(9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot (2 \times 10^{-6} \, \text{C}) \cdot q_2}{0.003 \, \text{m}}$$

Simplifying, we find:

$$q_2 = \frac{15.43 \, \text{J} \cdot 0.003 \, \text{m}}{(9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot (2 \times 10^{-6} \, \text{C})}$$

Calculating the value, we get:

$$q_2 \approx 25.7167 \, \text{C}$$

Answer: The value of the second charge is approximately 25.7167 C.