This problem involves electric potential energy and conservation of energy principles in an electrostatic context. Let's break it down and solve:

Problem:

• $q = -2 \, \mu C = -2 \times 10^{-6} \, C$
• $Q = 6 \, \mu C = 6 \times 10^{-6} \, C$
• Initial distance from $Q$: $r_1 = 2 \, m$
• Final distance from $Q$: $r_2 = 1 \, m$

Goal: Find the kinetic energy of charge $q$ at $r_2 = 1 \, m$.

Method:

1. Use Conservation of Energy: $\text{Initial Potential Energy} + \text{Initial Kinetic Energy} = \text{Final Potential Energy} + \text{Final Kinetic Energy}.$ Since $q$ starts from rest, its initial kinetic energy is zero: $U_1 = U_2 + K.$ The kinetic energy at $r_2$ is then: $K = U_1 - U_2.$

2. Calculate the Potential Energy: The potential energy between two charges is: $U = k_e \frac{qQ}{r},$ where $k_e = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2$ (Coulomb constant).

   • Initial Potential Energy: $U_1 = k_e \frac{qQ}{r_1}.$

   • Final Potential Energy: $U_2 = k_e \frac{qQ}{r_2}.$

3. Find the Kinetic Energy: Subtract $U_2$ from $U_1$: $K = k_e qQ \left(\frac{1}{r_1} - \frac{1}{r_2}\right).$

Solution:

1. Plug in the values: $q = -2 \times 10^{-6} \, C, \quad Q = 6 \times 10^{-6} \, C, \quad r_1 = 2 \, m, \quad r_2 = 1 \, m.$

2. Compute $K$: $K = (9 \times 10^9) \cdot (-2 \times 10^{-6}) \cdot (6 \times 10^{-6}) \cdot \left(\frac{1}{2} - \frac{1}{1}\right).$

3. Simplify: $K = (9 \times 10^9) \cdot (-12 \times 10^{-12}) \cdot \left(\frac{1}{2} - 1\right).$ $K = (9 \times 10^9) \cdot (-12 \times 10^{-12}) \cdot \left(-\frac{1}{2}\right).$ $K = (9 \times 10^9) \cdot 6 \times 10^{-12}.$ $K = 54 \times 10^{-3} = 0.054 \, \text{J}.$

Final Answer:

The kinetic energy of $q$ at $r = 1 \, m$ is: $K = 0.054 \, \text{J}.$

Would you like me to provide more details or analyze further? Here are 5 related questions:

1. How is the direction of motion determined in problems with positive and negative charges?
2. What would happen to the result if $q$ were positive?
3. How do you derive the formula for potential energy between charges?
4. What is the physical significance of the change in potential energy?
5. How would the result differ if $Q$ were a different charge distribution (e.g., a ring)?

Tip: Always double-check the sign of charges when dealing with potential energy, as it determines whether the force is attractive or repulsive!