Discussion
The electrostatic force between two point charges at a distance of separation d is F. If one of the charge is moved away by a distance d/2 then the force between them is
Answer: C
The new force is \(\frac{4}{9}F\).
Moving one charge away by \(d/2\) increases the separation from \(d\) to \(d + d/2 = 3d/2\). Coulomb's force varies as the inverse square of the separation:
\[ \begin{aligned} \frac{F'}{F} &= \left(\frac{d}{3d/2}\right)^{2} = \left(\frac{2}{3}\right)^{2} \\ F' &= \frac{4}{9}F \end{aligned} \]
\(\frac{2}{3}F\) forgets to square the ratio, and \(\frac{9}{4}F\) and \(\frac{3}{2}F\) invert it, as if the charges had moved closer. The force must decrease, because the charges are farther apart.
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