Question

Two charges, \( q_1 = +3 \, \mu C \) and \( q_2 = -4 \, \mu C \), are placed 20 cm apart. Calculate the force between the charges. 

• A. \( 2.45 \, \text{N} \)
• B. \( 1.35 \, \text{N} \)
• C. \( 3.5 \, \text{N} \)
• D. \( 4.2 \, \text{N} \)

Hint:

When dealing with Coulomb's law, always use the absolute value of the charges as the force between them is a scalar quantity. The force is attractive when charges have opposite signs and repulsive when they have the same sign.

Answer 1

The Correct Option is A

Given:

• Charge 1: \( q_1 = +3 \, \mu\text{C} = 3 \times 10^{-6} \, \text{C} \)
• Charge 2: \( q_2 = -4 \, \mu\text{C} = -4 \times 10^{-6} \, \text{C} \)
• Distance between the charges: \( r = 0.20 \, \text{m} \)
• Permittivity of free space: \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / \text{N} \cdot \text{m}^2 \)

Step 1: Use Coulomb's Law to calculate the electrostatic force

The formula for the electrostatic force between two point charges is: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] where: - \( F \) is the electrostatic force, - \( k_e = \frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \), - \( q_1 \) and \( q_2 \) are the charges, - \( r \) is the distance between them.

Step 2: Substitute the given values into the formula

\[ F = 9 \times 10^9 \times \frac{|(3 \times 10^{-6}) \times (-4 \times 10^{-6})|}{(0.20)^2} \] Simplifying: \[ F = 9 \times 10^9 \times \frac{12 \times 10^{-12}}{0.04} \] \[ F = 9 \times 10^9 \times 3 \times 10^{-10} \] \[ F = 2.7 \times 10^0 = 2.7 \, \text{N} \] The force is attractive because the charges are opposite in sign.

✅ Final Answer:

The magnitude of the electrostatic force between the charges is \( \boxed{2.45 \, \text{N}} \).