Question

You are given two positively charged particles, of equal magnitude, separated by a distance, "d". What will happen to the force field between the two particles when "d" is doubled? A) remain constant B) decrease by a factor of two C) decrease by a factor of four D) nothing; no force field between particles

Answer 1

force goes as 1/d^2 ... (2d)^2 => 4d^2 ...

C) decrease by a factor of four

Answer 2

Explanation :

There exists a force between charged particle given by Coulomb's law.

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

where,

k is the electrostatic constant

`q_1,q_2` are charged particles

d is the distance between the charges.

When d is doubled (d' = 2d) , let F' be the force.

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

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

`F'=(1)/(4)F`

So, the force decrease by a factor of four.

Hence, the correct option is (C) "decrease by a factor of four".