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lilywaterlily · Jan 10, 2012, 06:46 AM

Consider two point charges that are separated by a distance,2r.if this distance between them is increased to 5r, the force between the charges is what?

Aurora2000 · Jan 12, 2012, 06:47 AM

Use Coloumb's formula: the force between point charges $q_1,q_2$ separated by a distance
$r$ is

$\frac{1}{4\pi\vep_0} \frac{q_1q_2}{r^2}$

Thus changing the distance from 2r to 5r will change the force by

$\frac{\frac{1}{4\pi\vep_0} \frac{q_1q_2}{(2r)^2}}{\frac{1}{4\pi\vep_0} \frac{q_1q_2}{(5r)^2}} <br /> =25/4$
thus if you call F the force between your charges with 2r between them, when the distance is 5r the force between them is $\frac{4}{25}F$ .

Aurora2000 · Jan 12, 2012, 06:48 AM

$\vep_0$ is a constant, not influent for your question.