Question

At what distance would the repulsive force between two electrons have a magnitude of 4.0 N?

Answer 1

Answer:

$7.6\cdot 10^{-15} m$

Explanation:

The electrostatic force between two electrons is given by:

$F=k\frac{e^2}{r^2}$

where

$k=8.99\cdot 10^9 Nm^2C^{-2}$ is the Coulomb's constant

$e=-1.6\cdot 10^{-19}C$ is the electron charge

r is the distance between the two electrons

In this problem, we know F=4.0 N, so we can re-arrange the equation to calculate r:

$r=\sqrt{\frac{ke^2}{F}}=\sqrt{\frac{(8.99\cdot 10^9)(-1.6\cdot 10^{-19})^2}{4.0 N}}=7.6\cdot 10^{-15} m$