Question

The electric force between a negatively charged paint droplet and a positively charged automobile body is increased by a factor of two, but the charges on each remain constant. how has the distance between the two changed? (assume that the charge on the automobile is located at a single point.)

Answer 1

The electric force between the two objects is given by

`F= k_e (q_1 q_2)/(r^2)`
where

`k_e` is the Coulomb's constant

`q_1` and
`q_2` are the charges of the two objects

`r` is the distance between the two objects

in the problem, the charges q1 and q2 remain constant, so the only quantity that changes is r, the distance. We can see that the force is proportional to
`(1)/(r^2)`: this means that the distance r is proportional to

`r \sim (1)/(√(F))`
So, if the force has increased by a factor 2:
`F' = 2F`, the new distance is

`r'= (1)/(√(2F)) = (r)/(√(2) )`

so, the distance has decreased by a factor
`√(2)`.