If two charges of +3μC and +5μC are placed 0.3m apart, what is the magnitude of the force between them?

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Q: If two charges of +3μC and +5μC are placed 0.3m apart, what is the magnitude of the force between them?

Solution: Using Coulomb's law, F = k * |q1 * q2| / r² = (9 × 10^9 N m²/C²) * |(3 × 10^-6 C) * (5 × 10^-6 C)| / (0.3 m)² = 0.45 N.

Steps: 9

Step 1: Identify the charges. We have two charges: q1 = +3μC and q2 = +5μC.
Step 2: Convert the charges from microcoulombs (μC) to coulombs (C). 3μC = 3 × 10^-6 C and 5μC = 5 × 10^-6 C.
Step 3: Identify the distance between the charges. The distance r = 0.3 m.
Step 4: Use Coulomb's law formula: F = k * |q1 * q2| / r², where k = 9 × 10^9 N m²/C².
Step 5: Substitute the values into the formula: F = (9 × 10^9) * |(3 × 10^-6) * (5 × 10^-6)| / (0.3)².
Step 6: Calculate the product of the charges: (3 × 10^-6) * (5 × 10^-6) = 15 × 10^-12 C².
Step 7: Calculate the square of the distance: (0.3)² = 0.09 m².
Step 8: Substitute these values back into the formula: F = (9 × 10^9) * (15 × 10^-12) / 0.09.
Step 9: Calculate the force: F = (9 × 15 / 0.09) × 10^9 × 10^-12 = 0.45 N.