Question

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

Answer 1

The separation between them is $\frac{r}{\sqrt{20} }$

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 = $\frac{kq_{1} q_{2} }{r^{2} }$ -------- ( 1 )

Here, $q_{1} , q_{2}$ are charges and r is the distance between them

New force F' = $\frac{kq_{1q_{2} } }{r'^{2} }$ ----------- (2 )

Divide ( 1 ) and ( 2 )

$\frac{F'}{F}$ = $\frac{\frac{kq_{1}q_{2} }{r'^{2} } }{\frac{kq_{1}_{2} }{r^{2} } }$

20 = $\frac{r^{2} }{r'^{2} }$

r' = $\frac{r}{\sqrt{20} }$

Given that force between them are increasing and therefore distance between them decrease by $\frac{r}{\sqrt{20} }$

Learn more about two point charges here : brainly.com/question/24206363

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