Question

The force of attraction that a -40.0 X10-6 C point charge exerts on a +108 X10-6C point charge has magnitude 4.00 N. How far apart are these two charges? (k = 9.9  109 N • m2/C2)

Answer 1

Answer:

The answer to your question is 3.27 m

Explanation:

Data

q₁ = -40 x 10⁻⁶ C

q₂ = 108 x 10⁻⁶ C

F = 4 N

d = ?

K = 9.9 x 10⁹ N m²/C²

- Use Coulomb's law to solve this problem.

              F = Kq₁q₂ / d²

- Solve for d²

            d² = Kq₁q₂ / F

- Substitution

            d² = 9.9 x 10⁹ (40 x 10⁻⁶)(108 x 10⁻⁶) / 4

- Simplification

            d² = 42.768 / 4

            d² = 10.692

- Result

            d = √10.692

            d = 3.27 m