The separation between them is [tex]\frac{r}{\sqrt{20} }[/tex]
Concept :
If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.
Given,
Two point charges are brought closer together, increasing the force between them by a factor of 20.
Original force is
F = [tex]\frac{kq_{1} q_{2} }{r^{2} }[/tex] -------- ( 1 )
Here, [tex]q_{1} , q_{2}[/tex] are charges and r is the distance between them
New force F' = [tex]\frac{kq_{1q_{2} } }{r'^{2} }[/tex] ----------- (2 )
Divide ( 1 ) and ( 2 )
[tex]\frac{F'}{F}[/tex] = [tex]\frac{\frac{kq_{1}q_{2} }{r'^{2} } }{\frac{kq_{1}_{2} }{r^{2} } }[/tex]
20 = [tex]\frac{r^{2} }{r'^{2} }[/tex]
r' = [tex]\frac{r}{\sqrt{20} }[/tex]
Given that force between them are increasing and therefore distance between them decrease by [tex]\frac{r}{\sqrt{20} }[/tex]
Learn more about two point charges here : https://brainly.com/question/24206363
#SPJ4
