a +0.0129 c charge feels a 4110 n force from a -0.00707 c charge. how far apart are they? ? m

Explanation:

Step1: Recall Coulomb's law

$F = k\frac{q_1q_2}{r^{2}}$, where $F$ is the force between two charges, $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. We need to solve for $r$.

Step2: Rearrange the formula for $r$

$r^{2}=k\frac{q_1q_2}{F}$, so $r=\sqrt{k\frac{q_1q_2}{F}}$.

Step3: Substitute the given values

$q_1 = 0.0129\ C$, $q_2= 0.00707\ C$ (taking the magnitudes as the sign just indicates attraction/repulsion), $F = 4110\ N$, and $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$. $r=\sqrt{\frac{9\times 10^{9}\times0.0129\times0.00707}{4110}}$ $r=\sqrt{\frac{9\times 10^{9}\times9.1203\times10^{-5}}{4110}}$ $r=\sqrt{\frac{820827}{4110}}$ $r=\sqrt{200}$ $r = 14.14\ m$

Answer:

$14.14$