Question

the force of repulsion between two point charges kept at distance "d" apart in vaccum is "f". what should be the seperation between the two charges to make the force F/3 ?​

Answer 1

In the 1st case , the force of repulsion between 2 point charges is F for a separation of d .

As per Coulomb's Law :

$F = k \bigg (\dfrac{ {q}^{2} }{ {d}^{2} } \bigg)$

Now let the separation in the 2nd case be D

New force shall be F/3 . The same point charges have been taken in the same medium (i.e vacuum)

Again As per Coulomb's Law :

$\dfrac{F}{3} = k \bigg (\dfrac{ {q}^{2} }{ {D}^{2} } \bigg)$

Dividing the 2 equations, we get :

$\therefore \: \dfrac{ {D}^{2} }{{d}^{2} } = 3$

$= > \: {D}^{2} = 3 {d}^{2}$

$= > \: D = d \sqrt{3}$

So new separation will be d√3