The force between two charges of -3.0 x 10-9 C that are 5.0 x 10-2 m apart is

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The force between two charges of -3.0 x 10-9 C that are 5.0 x 10-2 m apart is

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We start by identifying the given quantities and the formula to use. The two charges are q1 = q2 = -3.0 × 10^(-9) C, the separation is r = 5.0 × 10^(-2) m, and Coulomb's law gives F = k |q1 q2| / r^2 with k ≈ 8.99 × 10^9 N·m^2/C^2.

Option 1: 1.8 × 10^(-16) N. To reach such a tiny force, one would need either much smaller charges or a much larger distance. Here, the charges are relatively small but not tiny enough, and the distance is modest; plugging in the numbers does not yield that order of magnitude, so this option is far too small.

Option 2: 3.6 × 10^(-15) N. This value is also several orders of magnitude off. Even with the given charges and distance, the calculation produces a force on the order of 10^(-5) N, not 10^(-15)–10^(-14) N, so this is incorrect.

Option 3: 1.6 × 10^(-6) N. This is closer in magnitude to the correct result but still too small by about three orders of magnitude. If you mis-handle the square of the distance or the constants, you might land in this range, but the actual computation shows a larger force.

Option 4: 3.2 × 10^(-5) N. This aligns with the standard Coulomb calculation: F = (8.99 × 10^9) × |(-3.0 × 10^(-9)) × (-3.0 × 10^(-9))| / (5.0 × 10^(-2))^2. The product |q1 q2| = 9.0 × 10^(-18) C^2, r^2 = 2.5 × 10^(-3) m^2, so F ≈ (8.99 × 10^9) × (9.0 × 10^(-18)) / (2.5 × 10^(-3)) ≈ 32.4 × 10^(-6) N ≈ 3.2 × 10^(-5) N. Therefore, this option matches the correct magnitude.

Additionally, note the physical meaning: since the charges are both negative, the force is repulsive, not attractive, so the direction would push them apart along the line joining them. The calculation gives a positive magnitude, consistent with that repulsive interaction.

APPHYSC142-031 with labs Exam Review - Final

The force between two charges of -3.0 x 10-9 C that are 5.0 x 10-2 m apart is

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