Question

the force between two point charges is 100 N in air.Calculate the force if the distance between them is increased by 50%.

Answer 1

F=q1q2/r*
100= q1q2/r*_____(1)
when the r is increased by 50% then r2= 3/2r
let the force calculated be x then
x= q1q2/(3/2r)*______(2)
By solving eq 1 and 2 , we get:
F= 44.4N

Answer 2

Answer:∵ Force is inversely proportional to distance between two charges .

Let initial distance between two charges q₁ and q₂ is r

And force acts between them , 100N = Kq²/r²

Now, when distance between charges = r + 50% of r

= 3r/2

Then, force acts between them , F' = Kq²/(3r/2)² = 4Kq²/9

Now, dividing both the equations ,

100/F' = {Kq²/r²}/{4Kq²/9} = 9/4

F' = 400/9 = 44.44N

Hence, answer is 44.44N