Charges q₁ and q₂ exert repulsive forces of 10 N on each other.What is the repulsive force when their separation is decreased so that their final separation is 80% of their initial separation? A)16 N B)12 N C)10 N D)8.0 N E)6.4 N

Question

Quizplus · Accepted Answer

The repulsive force between two charges is given by Coulomb's law, which states that the force is inversely proportional to the square of the distance between the charges. If the final separation is 80% of the initial separation, the distance between the charges is reduced to 0.8 times the original distance. Since the force is inversely proportional to the square of the distance, when the distance is reduced to 0.8 times the original, the force becomes (1/0.8)^2 = (1/0.64) times stronger, which is approximately 1.5625 times the original force. Therefore, the new force is 10 N * 1.5625 = 15.625 N, which rounds to 16 N, making option A correct.

Charges Q1 and Q2 Exert Repulsive Forces of 10 N

Charges Q₁ and Q₂ Exert Repulsive Forces of 10 N