\begin{problem*}{3}
Nobel laureate Richard Feynman once said that if two persons stood at arm's length from each other and each person had $p=1$\% more electrons than protons, the force of repulsion between them would be enough to lift a ``weight'' equal to that of the entire Earth.
Carry out an order of magnitude calculation to substantiate this assertion. 
\end{problem*}

\begin{solution}
Let $m = 70\U{kg}$ be the mass of one person, and $q_e$ be the charge of one electron.
Assume that there are approximately equal numbers of protons, electrons, and neutrons in a person.
Electrons have much less mass than protons or neutrons, so we ignore their mass contribution.
Protons and neutrons have very similar masses, so $N = (m/2)/m_p$ is the number of protons, and
 $N_q = N \cdot p$ is the number of extra electrons in each person.
Assume they are seperated by $r = 1\U{m}$.
The force of repulsion $F$ is given by
\begin{equation}
 F = k_e \frac{q^2}{r^2}
   = k_e \left(\frac{m p q_e}{2 m_p r}\right)^2
   = 9.0\E{9}\U{N$\cdot$m$^2$/C$^2$} \left(\frac{70\U{kg} \cdot 0.01 \cdot 1.6\E{-19}\U{C}}{2 \cdot 1.7\E{-27}\U{kg} \cdot 1\U{m}}\right)^2
   \approx 1\E{10} \left( \frac{0.3\E{-19}}{1\E{-27}} \right)^2 \U{N}
   = \ans{1\E{25}\U{N}}
\end{equation}
And a ``weight'' the mass of the earth would be $F_g = Mg \approx 6\E{24}\U{kg}\cdot 9.8\U{m/s$^2$} \approx 6\E{25}\U{N} \sim F$.
\end{solution}
