Question

What is the ratio of electrostatic force and gravitational force between a proton and electron at 1 nm distance?
Given $m_p=1.67\times {10}^{-27}kg$, $m_e=9.1\times {10}^{-31}kg$, $q_p=1.6\times {10}^{-19}C$

Answer 1

$\frac{|\overrightarrow{F_e}|}{|\overrightarrow{F_g}|}=\frac{\left(\frac{K{q}_1{q}_2}{r^2}\right)}{\left(\frac{G{m}_1{m}_2}{r^2}\right)}=\frac{9\times {10}^9\times 1.6\times {10}^{-19}\times 1.6\times {10}^{-19}}{6.67\times {10}^{-11}\times 9.1\times {10}^{-31}\times 1.67\times {10}^{-27}}=2.2\times {10}^{39}$