What is the force between two point charges of +2μC and -3μC separated by a distance of 0.5m in vacuum?

1 question

1 item

Q: What is the force between two point charges of +2μC and -3μC separated by a distance of 0.5m in vacuum?

Solution: Using Coulomb's law, F = k * |q1 * q2| / r^2 = (9 × 10^9) * |2 × 10^-6 * -3 × 10^-6| / (0.5)^2 = -1.08 N.

Steps: 8

Step 1: Identify the values of the charges. Charge 1 (q1) is +2μC and Charge 2 (q2) is -3μC.
Step 2: Convert the charges from microcoulombs (μC) to coulombs (C). +2μC = 2 × 10^-6 C and -3μC = -3 × 10^-6 C.
Step 3: Identify the distance (r) between the charges, which is 0.5m.
Step 4: Use Coulomb's law formula: F = k * |q1 * q2| / r^2, where k is Coulomb's constant (approximately 9 × 10^9 N m²/C²).
Step 5: Calculate the product of the charges: |q1 * q2| = |(2 × 10^-6) * (-3 × 10^-6)| = 6 × 10^-12 C².
Step 6: Calculate r squared: r^2 = (0.5)^2 = 0.25 m².
Step 7: Substitute the values into the formula: F = (9 × 10^9) * (6 × 10^-12) / 0.25.
Step 8: Calculate the force: F = (9 × 10^9 * 6 × 10^-12) / 0.25 = -1.08 N.