Suppose each person has mass $70kg.$ In terms of elementary charges, each person consists of precisely equal numbers of protons and electrons and a needy equal number of neutrons. The electrons comprise very little of the mass, so for each person we find the total number of protons and neutrons, taken together:
$$(70kg)({u\over{1.66×10^{−27}kg}})=4×10^{28}u$$
of these, nearly on half, $2×10^{28}$,are protons, and 1% of this is $2×10^{26}$, constituting a charge of $$(2×10^{26})(1.60×10^{−19}C)=3×10^7C.$$
Thus, Feyman's force has magnitude
$$F={{keq1q1\over {r2}}={{(8.99×109N.m^2/C^2)(3×10^7C)^2}\over {(0.5m)2}}}∼1026N$$
where we have used a half-meter arm's length. According to the particle in a gravitational field model, if the Earth were in an externally-produced uniform gravitational field of magnitude 9.80m/s2, it would weigh $$Fg=mg=(6×10^{24}kg)(10m/s^2)∼10^{26}N.$$
Thus, the forces are of the same order of magnitude.
Suppose each person has mass$70kg.$ In terms of elementary charges, each person consists of precisely equal numbers of protons and electrons and a needy equal number of neutrons. The electrons comprise very little of the mass, so for each person we find the total number of protons and neutrons, taken together:
$$(70kg)({u\over{1.66×10^{−27}kg}})=4×10^{28}u$$ $2×10^{28}$ ,are protons, and 1% of this is $2×10^{26}$ , constituting a charge of $$(2×10^{26})(1.60×10^{−19}C)=3×10^7C.$$
$$F={{keq1q1\over {r2}}={{(8.99×109N.m^2/C^2)(3×10^7C)^2}\over {(0.5m)2}}}∼1026N$$ $$Fg=mg=(6×10^{24}kg)(10m/s^2)∼10^{26}N.$$
of these, nearly on half,
Thus, Feyman's force has magnitude
where we have used a half-meter arm's length. According to the particle in a gravitational field model, if the Earth were in an externally-produced uniform gravitational field of magnitude 9.80m/s2, it would weigh
Thus, the forces are of the same order of magnitude.