Answer :
Distance between them is decrease by an element of 0.21
Evaluating :
Two point charges are brought closer together, increasing the force by an element of 22
Let the first force be
F = [tex]\frac{kq_{1} q_{2} }{r^{2} } ...........1[/tex]
where q₁ q₂ are charges and r is that the distance between new force F' = [tex]\frac{Kq_{1} q_{2} }{r^{2} } ................2[/tex]
divide equation 1 & 2
[tex]\frac{F}{F'}[/tex] = [tex]\frac{r^{2} }{r'^{2} }[/tex]
[tex]22= \frac{r^{2} }{r'^{2} }[/tex]
r' = [tex]\frac{r}{\sqrt{22} }[/tex] ≈0.213r
F₁ represents the first electric force between the two point charges.
F₂ represents the new electric force between the 2 point charges.
r₁ represents the first separation distance between the two point charge.
r₂ represents the new separation distance between the 2 point charge.
Distance between them is decrease by an element of 0.21
Electric Force:
The term 'electric force' exists when two or more electric charges are situated at a specific distance. The variation of electrical force with point charge magnitude is a linear one, whereas the variation of electrical force with the separation distance between the charges is an inversely square one.
Learn more about electric force :
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