At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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