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?

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.

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.