Question

Two charged objects have an attractive force of 10 n. if the charge of both objects increases by a factor of 5, and the distance separating the objects decreases by 1/2, then what is the new force?

Answer 1

The new force between two charged objects, with each charge increasing by a factor of 5 and the separation distance halved, would be 1000 N.

Step-by-step explanation:

The subject of this question is Physics, specifically dealing with Coulomb's Law and the electrostatic force between two charged objects. Given that the initial force is 10 N, and if both charges increase by a factor of 5, the force due to the charges would increase by a factor of 5 squared, according to Coulomb's Law, because the force depends on the product of the charges.

If the distance between the charges is then halved, the force would increase by the square of the factor by which the distance is decreased, which is 4 (since (½)² = ¼, and 1/¼ = 4). Therefore, the new force would be the initial force times the product of those two factors. Since 5 squared is 25, and the second factor is 4, the final force would be 10 N × 25 × 4, which equals 1000 N.