Question

If the distance between two charges is doubled, what will happen the force between the charges?

Answer 1

the force between 2 charges, by coulomb's law is given by,

`F = (q_1q_2)/(4piepsilon_0r^2)` where, `q_1` and `q_2` are the magnitude of the charges. and `r` is the distance between them.

So, according to the question, if the distance between two charges is doubled, that is `color(red)(r') = 2r` what happens to the force between them.?

the new force, `F' = (q_1q_2)/(4piepsilon_0color(red)(r')^2)`

`F' = (q_1q_2)/(4piepsilon_0color(red)(2r)^2)`

`F' = (q_1q_2)/(4piepsilon_0 (4r^2)`

`F' = (q_1q_2)/(4(4piepsilon_0 r^2)`

`F' =F/4`

therefore the new force if 4 times the old one :)