Final answer:
Using Coulomb's Law, two point charges of 75.0 nC each must be approximately 0.50 meters apart to have a force of 1.00 N between them. None of the given options a, b, c, or d match this result.
Step-by-step explanation:
To determine how far apart two point charges must be to have a force of 1.00 N between them, we use Coulomb's Law, which states that the force (F) between two charges (q1 and q2) separated by a distance (r) in a vacuum is given by:
F = k * |q1 * q2| / r^2,
where k is Coulomb's constant (8.99 x 10^9 Nm^2/C^2), and F is the electrostatic force between the charges.
Given that both charges are 75.0 nC, we must first convert nanocoulombs to coulombs (1 nC = 1 x 10^-9 C), so the charge in coulombs is 75.0 x 10^-9 C. We then plug the values into Coulomb's Law and solve for r:
1.00 N = (8.99 x 10^9 Nm^2/C^2) * (75.0 x 10^-9 C)^2 / r^2
After rearranging and solving for r:
r = √((8.99 x 10^9 Nm^2/C^2) * (75.0 x 10^-9 C)^2 / 1.00 N)
r ≈ 0.50 meters
Therefore, none of the options a, b, c, or d are correct, and the two point charges must be approximately 0.50 meters apart to have a force of 1.00 N between them.