What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original magnitude?

F' = 64 F

Explanation:

The electric force between charges is given by :

Where

q₁ and q₂ are charges

r is the distance between charges

When  each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

Apply new values,

So, the new force becomes 64 times the initial force.

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow