IDNLearn.com: Your one-stop destination for finding reliable answers. Ask your questions and get detailed, reliable answers from our community of knowledgeable experts.

Two point charges are brought closer together, increasing the force between them by a factor of 20. By what factor was their separation changed?

Sagot :

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