Question

if distance between two charged particle decreased by 25percent then find the percentage change in force.​

Answer 1

Answer:77.78 percent

Explanation:

Let the charges be q1 and q2. They are separated by a distance of r. New distance =3/4r. Initial force=Kq1q2/r2. New force=16/9F Therefore , percentage change=(16/9F-F)/F*100=77.78 per cent