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Feynman’s electric force analogy

April 22, 2012

From Feynman Lectures II 1-1:

If you were standing at arm’s length from somenone and each of you had one percent more electrons than protons, the repelling force would be incredible. How great? Enough to lift the Empire State Building? No! To lift Mount Everest? No! The repulsion would be enough to lift a “weight” equal to that of the entire Earth!

Really? Where is that back of the envelope? Alright here we go:

Coulomb’s law: F=\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2},

where \frac{1}{4\pi\epsilon_0}\approx 10^4, q_1q_2\approx10^{-38} for two electrons and r^2 = 1, since the distance is “arm’s length”. So we have F\approx 10^{-34}N if the two persons had one excess electron each. A single electron in person A’s body feels the force of each of the M individual excess electron in person B’s body, that is M\times 10^{-34}N. The total force on person A is the sum of forces of person B on each of its M excess electrons, that is M^2\times 10^{-34}N.

I model a 70kg person as a 70kg piece of sugar (C_6H_{12}O_6). A single molecule has 96 electrons, one percent excess is about one electron per molecule. Each such molecule weighs about 180 atomic mass units which is about 10^{-25}kg. The person is thus made up of about 10^{27} sugar molecules, and one percent excess electrons amounts to M=10^{27}. So we have F\approx 10^{54} \times 10^{-34} = 10^{20}N.

The Earth is an Iron sphere with radius 6\times 10^6m and density 10^4kg/m^3. So it weighs about 10^{25}kg which is the equivalent of a gravitational force of 10^{26}N which is more than the repelling force between the two charged sugar cubes.

We are off by 6 orders of magnitude. We might get them right by modeling the 70kg person as a wet sugar cube or something. Anyway, the analogy is nice, even though we might only be able to lift a fraction of the Moon after all.

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