Question

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

Answer 1

Answer:

We know that:

Force is inversely proportional to the distance between the two charges.

Now:

Initial distance between two charges q₁ and q₂ is r.

And:

Force acts between them,

$\implies$ $\boxed{\sf{100\:N=\frac{Kq</p><p>^{2}}{r^{2}}}}$

Now,

Distance between charges:

$\implies$ r + 50% of r

$\implies$ 3r/2

Then,

Force acts between them,

$\implies$ F' = Kq²/(3r/2)²

$\implies$ F' = 4Kq²/9

Now,

Dividing both the equations,

$\implies$ 100/F' = {Kq²/r²}/{4Kq²/9} = 9/4

$\implies$ F' = 100 × 4 / 9

$\implies$ F' = 400 / 9

$\implies$ F' = 44.44 N

Hence,

Answer is 44.44 N.

___________________

Answered by: Niki Swar, Goa❤️

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

Explanation:

answered by anjali