Two point charges, q1 = +4.0 microCoulombs and q2 = -3.0 microCoulombs, are placed 0.30 meters apart in a vacuum. What is the magnitude of the electrostatic force between them? (Coulomb's constant k = 8.99 10^9 Nm^2/C^2)

Answer and explanation

Correct answer: B. 1.20 N

  1. Step 1: Apply Coulomb's Law: F = (k |q1 q2|) / r^2.
  2. Step 2: Convert charges: q1 = 4.0 10^-6 C, q2 = 3.0 10^-6 C. Product |q1 q2| = 12.0 10^-12 C^2.
  3. Step 3: Compute r^2: (0.30 m)^2 = 0.090 m^2.
  4. Step 4: Calculate: F = (8.99 10^9 12.0 * 10^-12) / 0.090 = (0.10788) / 0.090 approx 1.199 N approx 1.20 N (attractive).
Answer
1.20 N.
Why not the others: '3.60 N' divides by r = 0.30 instead of r^2 = 0.09; '0.40 N' divides by 0.30^3; '12.0 N' makes an exponent error in the microCoulomb conversion.

Practise more Physics: Mechanics, Electricity, and Modern Physics.

Free questions with an explanation for every one, and full timed General Secondary Certificate (Tawjihiya Kuwait) mock tests.

All Physics: Mechanics, Electricity, and Modern Physics questionsTake a mock test

More Physics: Mechanics, Electricity, and Modern Physics questions