Charges q₁ and q₂ exert repulsive forces of 15 N on each other.What is the repulsive force when their separation is increased so that their final separation is 160% of their initial separation? A)1.3 N B)5.9 N C)9.8 N D)8.7 N E)6.4 N

Question

Quizplus · Accepted Answer

The 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 160% of the initial separation, the distance between the charges increases by a factor of 1.6. Therefore, the new force is the original force divided by (1.6)^2. Calculating this gives (15 N) / (1.6)^2 = 5.859375 N, which rounds to 5.9 N, making option B the correct answer.

Charges Q1 and Q2 Exert Repulsive Forces of 15 N

Charges Q₁ and Q₂ Exert Repulsive Forces of 15 N