The force of attraction that a -40.0 μC point charge exerts on a +108 μC point charge has magnitude 4.00 N.How far apart are these two charges? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A)2.10 m B)3.67 m C)3.12 m D)2.49 m E)1.13 m

Question

Quizplus · Accepted Answer

The force between two point charges can be calculated using Coulomb's law,
F = k(q1q2/r^2)
where k is Coulomb's constant, q1 and q2 are the magnitudes of the two charges, and r is the distance between them.

In this problem, we know the magnitudes of both charges and the force between them. We can rearrange Coulomb's law to solve for r,
r = √(k(q1q2)/F)

Substituting the given values into this equation, we get
r = √((8.99×10^9 N·m^2/C^2)(40×10^-6 C)(108×10^-6 C)/(4 N))
r = 3.12 m

Therefore, the two charges are separated by a distance of 3.12 meters. The answer is (C).

The Force of Attraction That a -40