Question

the force between 2 point charges is 100N in air. calculate the force if the distance between the charges is increased by 50%.​

Answer 1

Explanation:

∵ 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

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