Question

Two point charges \( +4 \, \mu\text{C} \) and \( -2 \, \mu\text{C} \) are separated by a distance of 0.3 m in air. What is the magnitude of the electrostatic force between them?

• A. \( 24 \, \text{N} \)
• B. \( 16 \, \text{N} \)
• C. \( 8 \, \text{N} \)
• D. \( 32 \, \text{N} \)

Hint:

When applying Coulomb’s law, use the absolute values of the charges to find the magnitude of the force. Ensure all units are in SI (Coulombs for charge, meters for distance) to avoid errors in calculation.

Answer 1

The Correct Option is C

Given:

• Charge 1: \( q_1 = +4 \, \mu\text{C} = 4 \times 10^{-6} \, \text{C} \)
• Charge 2: \( q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C} \)
• Distance between charges: \( r = 0.3 \, \text{m} \)
• Permittivity of free space: \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{N} \cdot \text{m}^2 \)

Step 1: Calculate the electrostatic force using Coulomb's Law

Coulomb's Law for the electrostatic force between two point charges is given by: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] where: - \( F \) is the electrostatic force, - \( k_e = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \) is Coulomb's constant, - \( q_1 \) and \( q_2 \) are the magnitudes of the charges, - \( r \) is the separation distance between the charges.

Step 2: Substitute the provided values into Coulomb's Law

\[ F = 9 \times 10^9 \times \frac{|(4 \times 10^{-6}) \times (-2 \times 10^{-6})|}{(0.3)^2} \] After simplification: \[ F = 9 \times 10^9 \times \frac{8 \times 10^{-12}}{0.09} \] \[ F = 9 \times 10^9 \times 8.89 \times 10^{-11} \] \[ F = 8 \, \text{N} \]

✅ Final Answer:

The magnitude of the electrostatic force between the charges is \( \boxed{8 \, \text{N}} \).