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Answer to Question #100169 in Physics for Angela
Question #100169
Two point charges are separated by a distance r. If the separation n is reduced by a factor of 1.5, by what factor does the electric force between them change?
Expert's answer
"F\\sim r^{-2}"
Thus,
"\\frac{F'}{F}=\\frac{r^2}{r'^2}=1.5^2=2.25"
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Comments
Dear Sandra, electrostatic force is inversely proportional to the squared distance (F~r^-2) which gives us factor F’/F = r^2/r’^2
What does F~r^-2 and F’/F = r^2/r’^2 mean?
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