Force of Repulsion between two people with 1% more electrons than protons

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SUMMARY

The discussion centers on calculating the force of repulsion between two individuals, each possessing 1% more electrons than protons, as proposed by Richard Feynman. Utilizing the formula F(elec) = k*q^2 / r^2, participants derived that the electric force could reach approximately 7.46 * 10^24 N, which is 2 * 10^9 times greater than the weight of the Earth (4 * 10^15 N). This calculation confirms that the repulsive force could indeed lift a weight equivalent to that of the entire Earth, given reasonable assumptions about the number of protons and electrons in a human body.

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  • Understanding of Coulomb's Law and electric force calculations
  • Basic knowledge of atomic structure, specifically protons and electrons
  • Familiarity with mass and weight concepts, including gravitational force
  • Ability to perform order-of-magnitude estimations and unit conversions
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  • Research the mass of protons and neutrons, and their role in atomic structure
  • Study the relationship between mass and weight, particularly in gravitational contexts
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Students in physics, educators teaching electrostatics, and anyone interested in the fundamental forces of nature and their implications in real-world scenarios.

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Homework Statement


Richard Feynman once said that if two persons stood at arm's length from each other and each person had 1% more electrons than protons, the force of repulsion between the two people 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.


Homework Equations


F(elec) = k*q^2 / r^2
mass of Earth = 5.9742 × 10^24 kilograms

The Attempt at a Solution


I am stuck with the beginning; how do I know how many electrons that is? And how do I find the charge given the number of electrons?

Thanks a lot!
 
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There is no specific number of protons. You have to figure out a reasonable number using reasonable assumptions. It is not 3 and it is not 1040, but something in-between. You know the mass of an average person, so how many protons is that? Like I said, make some assumptions and use the mass of a proton, which that you can look up. You can also look the charge of an electron which the same in magnitude as the charge of a proton.
 
Ok. Amazing. Thanks for the quick reply. Here's what I've come up with:

protons:electrons ~= 1:1

average human mass ~= 6.00x10^4 g

mass of one proton is 1 amu = 1.6607x10^-24 g

mass / amu mass = #protons + #neutrons
-> 6.00x10^4g / 1.6607x10^-24g = 3.61x10^28

divide by 2 to get the # of protons
-> 3.61x10^28 / 2 = 1.80x10^28 protons

--> there are approximately 1.80x10^28 electrons in a human body

1% of that is 1.80x10^26

1 Coulomb = −6.24151 × 10^18 electrons

-> a human body with 1% more electrons than protons has a charge of...
1% of electrons in body / # electrons in Coulomb = -2.88x10^7 C

Say arms length is 1m => r^2 = 1m^2

WOW I GET IT!

F(elec) = k * q^2 / r^2 = (9 * 10^9 N*m^2/C^2) * (-2.88x10^7 C)^2 / 1 m^2
= 7.46 * 10^24 N

weight of Earth = mg = 4*10^15 N

=> the electric force is 2*10^9 times larger than the weight of the Earth (according to this approximation), so it is probably true that the repulsion of two people, each with 1% more electrons than protons, can have a force of repulsion enough to lift a weight equal to that of the entire earth.

How is that?
 
Looks fine except for the weight of the Earth. The mass of the Earth is 6x1024 kg.
 
Please advise me how to cal the weight of the Earth?? Is it related to the mass of the Earth about 6 x 10^24kg?? Thanks!
 

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