Question

How far part the two electrons be, if the force between them equals the weight of an electrons?

Answer 1

$m_e=9.1\times {10}^{-31}kg$
$e=1.6\times {10}^{-19}C$
$F=mg$ and
$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$9.1\times {10}^{-31}\times 10=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{r^2}$
$r^2=\frac{9\times {10}^9\times 1.6\times {10}^{-19}\times 1.6\times {10}^{-19}\times 1.6\times {10}^{-19}}{9.1\times {10}^{-30}}$
$r^2=\frac{23.04\times 10}{9.1}$
$r^3=\frac{230.4}{9.1}=25.319$
$r=\sqrt{25.39}=5.032m$.

$r\approx 5m$