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. \( 1.35 \, \text{N} \)
• B. \( 2.45 \, \text{N} \)
• 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

The force between two charges is given by Coulomb's Law:

F = k × |q₁ q₂| / r²

Where:

• k = 9 × 10⁹ N·m²/C² is Coulomb's constant,
• q₁ = +3 μC = 3 × 10^-6 C is the first charge,
• q₂ = -4 μC = -4 × 10^-6 C is the second charge,
• r = 20 cm = 0.2 m is the distance between the charges.

Substitute the values into Coulomb's law:

F = 9 × 10⁹ × |(3 × 10^-6) × (-4 × 10^-6)| / (0.2)²

Now, simplify the equation:

F = 9 × 10⁹ × (12 × 10^-12) / 0.04

F = 9 × 10⁹ × 3 × 10^-10 = 1.35 N

Thus, the force between the charges is 1.35 N.