A negative charge, q1, of 6 µC is 0.002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the electrical force, Fe, applied by q1 on q2?

Use Coulomb’s law: F = k|q₁q₂|/r². Substituting the values gives F = (9 × 10⁹)(6 × 10⁻⁶)(3 × 10⁻⁶)/(0.002)² ≈ 4 × 10⁴ N. Because the charges have opposite signs, they attract; q₁ is north of q₂, so the force on q₂ points north. The correct choice is “magnitude: 4 × 10⁴ N, direction: north.”

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