Question

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

Answer 1

Answer:

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

Answer 2

Hi Mate Here is your Answer.........

If F1 =100 then the required force is 4/9(F1)=4/9(100)4=400/9=44.4N

Hope it helps