Given:
- Charge \(q_1 = +2 \times 10^{-6} \, \text{C}\)
- Charge \(q_2 = -2 \times 10^{-6} \, \text{C}\)
- Distance \(r = 0.1 \, \text{m}\)
- Coulomb's constant \(k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\)
Step 1: Calculate the magnitude of the electrostatic force using Coulomb's Law
\[
F = \frac{k |q_1 q_2|}{r^2}
\]
\[
F = \frac{(9 \times 10^9) \cdot |(2 \times 10^{-6}) \cdot (-2 \times 10^{-6})|}{(0.1)^2}
\]
\[
F = \frac{9 \times 10^9 \cdot (4 \times 10^{-12})}{0.01} = \frac{36 \times 10^{-3}}{0.01} = 3.6 \, \text{N}
\]
Step 2: Determine the direction of the force
The charges have opposite signs, indicating an attractive force.
Answer: The electrostatic force between the charges is \(3.6 \, \text{N}\). This corresponds to option (1).