Problem — Ranking Coulomb forces (pairs A-D)

Problem

Ranking. Four pairs in vacuum, given $|q_1 q_2|$ and distance $r$: (A) $|q_1 q_2|=2\ (\mu\text{C})^2$, $r=2\ \text{cm}$. (B) $|q_1 q_2|=8\ (\mu\text{C})^2$, $r=2\ \text{cm}$. (C) $|q_1 q_2|=2\ (\mu\text{C})^2$, $r=1\ \text{cm}$. (D) $|q_1 q_2|=1\ (\mu\text{C})^2$, $r=2\ \text{cm}$. Order the magnitudes from smallest to largest.

Solution

There is no need to calculate absolute values: Coulomb’s law gives $F = k\,|q_1 q_2|/r^2$, so it suffices to compare the index $|q_1 q_2|/r^2$ (in units of $(\mu\text{C})^2/\text{cm}^2$).

$F_A \propto \frac{2}{2^2} = 0{,}5;\quad F_B \propto \frac{8}{2^2} = 2;\quad F_C \propto \frac{2}{1^2} = 2;\quad F_D \propto \frac{1}{2^2} = 0{,}25$

Ordering the values $0{,}25 < 0{,}5 < 2 = 2$:

$\ev{D<A<C=B}$

Links

Topics: Elettrostatica Concepts: Legge di Coulomb Skills: Ragionamento di ranking