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

1 question

1 item

Q: What is the force between two point charges of +2 µC and -3 µC separated by a distance of 0.5 m 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 = -24 N.

Steps: 9

Step 1: Identify the charges. We have two point charges: q1 = +2 µC and q2 = -3 µC.
Step 2: Convert the charges from microcoulombs (µC) to coulombs (C). +2 µC = 2 × 10^-6 C and -3 µC = -3 × 10^-6 C.
Step 3: Identify the distance between the charges. The distance r = 0.5 m.
Step 4: Use Coulomb's law formula: F = k * |q1 * q2| / r^2, where k = 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 = -24 N.
Step 9: Interpret the result. The negative sign indicates that the force is attractive since one charge is positive and the other is negative.