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

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

Solution: Using Coulomb's law, F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * |(2 x 10^-6 C) * (-3 x 10^-6 C)| / (0.5 m)^2 = -1.08 N.

Steps: 8

Step 1: Identify the values of the charges. Charge 1 (q1) is +2 µC, which is equal to 2 x 10^-6 C. Charge 2 (q2) is -3 µC, which is equal to -3 x 10^-6 C.
Step 2: Identify the distance between the charges. The distance (r) is 0.5 m.
Step 3: Use Coulomb's law formula: F = k * |q1 * q2| / r^2, where k is Coulomb's constant (8.99 x 10^9 N m²/C²).
Step 4: Calculate the product of the charges: |q1 * q2| = |(2 x 10^-6 C) * (-3 x 10^-6 C)| = 6 x 10^-12 C².
Step 5: Calculate r squared: r^2 = (0.5 m)² = 0.25 m².
Step 6: Substitute the values into the formula: F = (8.99 x 10^9 N m²/C²) * (6 x 10^-12 C²) / (0.25 m²).
Step 7: Calculate the force: F = (8.99 x 10^9 * 6 x 10^-12) / 0.25 = -1.08 N.
Step 8: Note that the force is negative, indicating that the force is attractive due to the opposite charges.