Two charges in vacuum are 2.0 m apart. The strengths of the charges are 0.00030 C and 0.00080 C. What is the electric force between the charges? (k = 8.99 × 10^9 N m^2/C^2)

Two charges in vacuum are 2.0 m apart. The strengths of the charges are 0.00030 C and 0.00080 C. What is the electric force between the charges? (k = 8.99 × 10^9 N m^2/C^2)

• A. 540 N
• B. 27 N
• C. 1080 N
• D. 2.7 N

Answer: (A)

The main idea is Coulomb’s law: the electric force between two point charges in vacuum is F = k q1 q2 / r^2. Use k = 8.99 × 10^9 N·m^2/C^2, q1 = 3.0 × 10^-4 C, q2 = 8.0 × 10^-4 C, and r = 2.0 m. Compute the product q1 q2 = (3.0 × 10^-4)(8.0 × 10^-4) = 2.4 × 10^-7 C^2. The distance squared is r^2 = 4.0 m^2. Then F = (8.99 × 10^9) × (2.4 × 10^-7) / 4 = 539.4 N, about 5.4 × 10^2 N. Because both charges are positive, the force on each charge points away from the other—it's a repulsive force.

The main idea is Coulomb’s law: the electric force between two point charges in vacuum is F = k q1 q2 / r^2. Use k = 8.99 × 10^9 N·m^2/C^2, q1 = 3.0 × 10^-4 C, q2 = 8.0 × 10^-4 C, and r = 2.0 m.

Compute the product q1 q2 = (3.0 × 10^-4)(8.0 × 10^-4) = 2.4 × 10^-7 C^2. The distance squared is r^2 = 4.0 m^2. Then F = (8.99 × 10^9) × (2.4 × 10^-7) / 4 = 539.4 N, about 5.4 × 10^2 N.

Because both charges are positive, the force on each charge points away from the other—it's a repulsive force.