Question

When you double the distance between a pair of charged particles what happens to the force between them?

Answer 1

Force between the two charges becomes one fourth of the initial force.

Step-by-step explanation:

The electrostatic force acting between any two charges is given as,

`F = k(q_(1)q_(2))/(r^(2))`

Here,

F = force

k = Coulomb's constant

`q_(1)` = magnitude of charge of the first particle

`q_(2)` = magnitude of charge of the second particle

`r` = separation between the two charges

From the above relation,

`F \propto (1)/(r^(2))`

Thus,

`(F_(1))/(F_(2)) = \left ( (r_(2))/(r_(1)) \right )^(2)`

`(F_(1))/(F_(2)) = \left ( (2r)/(r) \right )^(2)`

`\Rightarrow \ F_(2) = (1)/(4)F_(4).`