Question

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

Answer 1

The new force becomes 4 times the initial force.

Step-by-step explanation:

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

`F=k(q_1q_2)/(d^2)`

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 :

`F'=k(q_1q_2)/(d'^2)\\\\=k(q_1q_2)/(((d)/(2))^2)\\\\=k(q_1q_2)/((d^2)/(4))\\\\=4* (kq_1q_2)/(d^2)\\\\F'=4F`

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