Question

Two point charges \(+2 \, \mu\text{C}\) and \(-2 \, \mu\text{C}\) are placed 0.1 m apart in air. What is the electrostatic force between them?

• A. \(3.6 \, \text{N}\)
• B. \(1.8 \, \text{N}\)
• C. \(4.2 \, \text{N}\)
• D. \(0.72 \, \text{N}\)

Hint:

Always convert microcoulombs (\( \mu\text{C} \)) to coulombs (\( \text{C} \)) before using Coulomb’s Law.

Answer 1

The Correct Option is A

We are given: - \(q_1 = +2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C}\) - \(q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C}\) - Distance \(r = 0.1 \, \text{m}\) - Coulomb’s constant \(k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\) Step 1: Use Coulomb’s Law \[ F = \frac{k |q_1 q_2|}{r^2} \] \[ F = \frac{9 \times 10^9 \cdot (2 \times 10^{-6})^2}{(0.1)^2} \] \[ F = \frac{9 \times 10^9 \cdot 4 \times 10^{-12}}{0.01} = \frac{36 \times 10^{-3}}{0.01} = 3.6 \, \text{N} \] Step 2: Direction of force Since the charges are opposite in sign, the force is attractive. Answer: The electrostatic force between the charges is \(3.6 \, \text{N}\). So, the correct answer is option (1).