Question

Given two positively charged particles, of equal magnitude, separated by a distance “d”. What will happen to the force field between the two particles when d is halved

Answer 1

The force between the two particles will quadruple

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{d^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

d is the separation between the two charges

In this problem, let's call F the initial force between the two charges, when they are at a distance of d.

Later, the distance between the two particles is halved, so the new distance is:

$d'=\frac{d}{2}$

This means that the new electrostatic force will be:

$F'=k\frac{q_1 q_2}{(d/2)^2}=4(k\frac{q_1 q_2}{d^2})=4F$

Therefore, the force between the two particles will quadruple.

Learn more about electrostatic force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly