Question

Two point charges \( +2 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are placed 10 cm apart in vacuum. What is the electrostatic force between them?

• A. 4.5 N
• B. 9 N
• C. 18 N
• D. 2.25 N

Hint:

Remember: Coulomb's law is used to calculate the electrostatic force between two point charges: \( F = k \frac{|q_1 q_2|}{r^2} \).

Answer 1

The Correct Option is B

Given:  \( q_1 = +2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C} \) 
 \( q_2 = -3 \, \mu\text{C} = -3 \times 10^{-6} \, \text{C} \) 
 Distance between charges, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \) 
Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2 / \text{C}^2 \)

Step 1: Formula for Electrostatic Force The electrostatic force between two point charges is given by Coulomb's law: \[ F = k \frac{|q_1 q_2|}{r^2} \] 

Step 2: Substitute the given values Substitute the given values into the formula: \[ F = (9 \times 10^9) \frac{|(2 \times 10^{-6})(-3 \times 10^{-6})|}{(0.1)^2} \] \[ F = (9 \times 10^9) \times \frac{6 \times 10^{-12}}{0.01} \] \[ F = 9 \times 10^9 \times 6 \times 10^{-10} = 9 \, \text{N} \] 

Answer: The correct answer is option (b): 9 N.