Question

The force of repulsion that two like charges exert on each other is 3.4 N. What will the force be if the distance between the charges is increased to 4 times its original value?

Answer 1

Let `q_1` and `q_2` be the charges, `K` be the Coulomb's constant and `r` be the distance between the charges when the force is `3.4N` and let `F` denote this force.

`F=(kq_1q_2)/r^2`

`implies 3.4=(kq_1q_2)/r^2`

Now, every other condition remains same except the distance between the charges. The distance between the charges has been increased `4` times its original value. Let `r'` be then new distance between the charges, then
`r'=4r`
Let the new force between the charges be denoted by `F'`.
`F'=(kq_1q_2)/r_'^2`

`implies F'=(kq_1q_2)/(4r)^2`

`implies F'=(kq_1q_2)/(16r^2)`

`implies F'=(1/16)((kq_1q_2)/r^2)`

Since `3.4=(kq_1q_2)/r^2`

`implies F'=(1/16)(3.4)`

`implies F'=0.2125N`

Hence the force between the two charges becomes `0.2125N`