2 balloons and a doubled distance (physics)
posted by help! .
EARLIER SOMEONE POSTED:
"Two like-charged balloons, placed at a distance of .5 meters, experience a repulsive force of .32 N. What is the force if the distance between the two balloons is doubled?
If you use Coulomb's Law, the equation used to find .32 N was Force = the constant of K multiplied by the product of the charges divided by the square of the distance. Therefore, I think to solve this, you need to double r, which would mean that force is one-fourth of what it originally was. Following that, I think you divide .32 N by 4 and the answer would be .08 N =F."
I WAS WONDERING WHY IT WOULDN'T BE 4 multiplied by the original force of .32 N SO THAT THE ANSWER WOULD BE 1.28 N.
Can someone PLEASE explain why this is?