Two point charges q1 = 2 μC and q2 = 3 μC are separated by d = 0.5 m. Using k = 9 x 10^9 N m^2/C^2, what is the magnitude of the electrostatic force between them?

• A. 0.108 N
• B. 0.432 N
• C. 0.216 N
• D. 2.16 N

Answer: (C)

Coulomb’s law gives the force as F = k |q1 q2| / r^2. Convert the charges to coulombs: q1 = 2 μC = 2×10^-6 C and q2 = 3 μC = 3×10^-6 C, so the product is 6×10^-12 C^2. The separation is d = 0.5 m, so r^2 = 0.25 m^2. Plug in: F = (9×10^9) × (6×10^-12) / 0.25. The numerator is 9×6×10^-3 = 0.054, then 0.054 / 0.25 = 0.216 N. So the magnitude of the force is 0.216 N. Both charges are positive, so the force is repulsive along the line joining them.