When the distance between two charges is halved, the electrical force between them?

If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:

where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:

the magnitude of the force changes as follows:

so, the force increases by a factor 4.

U okay bro