Question

If the distance between two point charges is increased by $3$, then calculate the decrease in the force between them.

Answer 1

Initial case when distance between charges is $r$ the force can be determined as, ${\overrightarrow{F}}_1=\frac{K{q}_1{q}_2}{r^2}$

But when the radius between the charges increases that is when $r\to 3r$, New distance b/w charges, $r^1=3r$, then force can be defined as,

${\overrightarrow{F}}^1=\frac{K{q}_1{q}_2}{{\left(r^1\right)}^2}=\frac{K{q}_1{q}_2}{{\left(3r\right)}^2}=\frac{K{q}_1{q}_2}{{9r}^2}$

${\overrightarrow{F}}^1=\frac{\overrightarrow{F}}{9}$

As the distance increases by 3 times the force reduces by 9 times.