What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

So, the new force becomes 4 times the initial force.