# what is percentage change in distance if the force of attraction between two point charges increases to 4 times keeping of charges constant?

Narendran
34 Points
5 years ago
As,
$F\alpha \frac{1}{r^{2}}$.

$r\alpha \frac{1}{\sqrt{F}}$.

there for  reduces to half as F goes upto 4 times.Hence the percentage change is -50% (reduces).