Question

Two charges \( +4 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are placed 2 m apart. What is the magnitude of the electric force between them? (Take \(k = 9 \times 10^9 \, \text{N.m}^2/\text{C}^2\).

• A. 0.036 N
• B. 0.072 N
• C. 0.108 N
• D. 0.144 N

Hint:

For Coulomb’s law, use \( F = k \frac{|q_1 q_2|}{r^2} \), ensuring charges are in coulombs and distance in meters.

Answer 1

The Correct Option is A

The electric force between two point charges is given by Coulomb’s law: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \(k = 9 \times 10^9 \, \text{N.m}^2/\text{C}^2\), \( q_1 = +4 \times 10^{-6} \, \text{C} \), \( q_2 = -4 \times 10^{-6} \, \text{C} \), and \( r = 2 \, \text{m} \). The magnitude of the force is: \[ F = 9 \times 10^9 \cdot \frac{(4 \times 10^{-6}) \cdot (4 \times 10^{-6})}{2^2} \] \[ = 9 \times 10^9 \cdot \frac{16 \times 10^{-12}}{4} = 9 \times 10^9 \cdot 4 \times 10^{-12} = 36 \times 10^{-3} = 0.036 \, \text{N} \] This matches option (A). \[ {0.072} \]