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# Coulomb's Law

Nobel laureate Richard Feynman once said that if two persons stood at arm&#039;s

length from each other and each personhad 1% more electrons than protons

,the force of repulsion between them would be enough to lift a &quot;weight&quot; equal

to that of the entire Earth. Carry out an order-of-magnitude calculation to

substantiate this assertion.

### 1 Answer

- 六呎將軍Lv 71 decade agoFavorite Answer
Taking the following data:

Mass of proton = 1.66 x 10-27 kg

Electronic charge = 1.6 x 10-19 C

Permittivity of free space = 8.85 x 10-12 F/m

Now, suppose that each person weighs 70 kg, standing at a distance of 0.5 m apart, we have:

Mass of each person contributed by protons = 35 kg (assume that 1 proton weighs equally as 1 neutron and the mass of electrons is negligible)

No. of protons of each person = 35/(1.66 x 10-27) = 2.108 x 1028

Proton charge = 2.108 x 1028 x 1.6 x 10-19 = 3.37 x 109 C

Then, excessive elctronic charge = 3.37 x 109 x 0.01 = 3.37 x 107 C

So the repulsive force between the persons is:

(3.37 x 107)2/(4π x 8.85 x 10-12 x 0.52) = 4.09 x 1025 N

which is about 4.17 x 1024 kg, of the order of the mass of the earth (5.98 x 1024 kg)

Source(s): Myself