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

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Homework Help Overview

The discussion revolves around calculating the force of repulsion between two individuals who each possess 1% more electrons than protons. The context involves concepts from electrostatics and mass estimation, particularly focusing on the implications of this charge imbalance.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to estimate the number of protons in an average human body and the corresponding number of electrons. There are attempts to calculate the charge based on these estimates, and questions arise regarding the relationship between mass and the number of protons.

Discussion Status

The conversation is active, with participants providing calculations and seeking clarification on the mass of the Earth and its relation to weight. Some guidance has been offered regarding assumptions and calculations, but no consensus has been reached on the final values.

Contextual Notes

Participants are working under the constraints of making reasonable assumptions about human mass and the number of protons, as well as the need to clarify the relationship between mass and weight in the context of gravitational force.

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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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