How Many Electrons Are Needed to Levitate a Person?

AI Thread Summary
To levitate a person weighing 60 kg at a height of 2.0 m, the number of electrons needed is calculated using Coulomb's law, which requires considering both the person's charge and the plate's charge. The initial calculation mistakenly treated the person's charge as equal to the electron charge without squaring the number of electrons. The correct approach recognizes that the positive charge of the person must equal the negative charge of the electrons moved, adhering to the conservation of charge. The textbook answer of 3.2 x 10^15 electrons reflects this proper application of the principles involved. Understanding the interaction between the charges is crucial for achieving equilibrium in this scenario.
dekoi
A small metal plate is bolted to the ceiling, and an "electron pump" is connected between the metal plate and yourself (mass of 60 kg). The pump starts pumping electrons from the metal plate to you.
How many electrons must be moved from the metal plate to you in order for you to hang suspended in the air 2.0 m below the ceiling? HINT: Assume that both you and the plate can be modeled as point charges.
My diagram is as such:
____
- - -

:smile:
_|__

The electrons move downward, toward me. However, I am confused as to where the charge is directed. Is it directed toward me or toward the metal plate? (Since charges move toward the negative charge.

So we have:
m = 60 kg
r = 2.0 m
q_e = 1.6 * 10^{-19} C
k = 9 * 10^{9} \frac{Nm^2}{C^2}
n = number of electrons

The net force must be zero since I am suspended in the air at this point and am not moving, therefore:
F_{net}=F_q - F_g
F_q = F_g
\frac{kqn}{r^2} = mg
n = \frac{mgr^2}{kq} = 1.6 * 10^{12} electrons
However, this is the wrong answer.
The answer in the textbook is 3.2 * 10^{15}. What is wrong with my method?

Is there an electric force coming from myself also? If so, what is it?ALSO: If you happen to know anything about sound interference/light waves, please post in the other two threads started by my boyfriend, as he is in desperate need right now of answers but no one wants to reply to his threads for some reason. His posts are under the name "dekoi" also. Thank you.
 
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Anyone? .
 
dekoi said:
The net force must be zero since I am suspended in the air at this point and am not moving, therefore:
F_{net}=F_q - F_g
F_q = F_g
\frac{kqn}{r^2} = mg
Your error is in applying Coulomb's law, which is:
F = \frac{k q_1 q_2}{r^2}
(You only included one of the charges.)
 
I know that's where I am confused.
I don't know what the charge on myself is. At first I thought it was 0, since a human does not have a charge.
But then I thought that if the electrons are holding me up, then I must have a positive charge.
A positive charge equal to the electron charge?

That's what I don't know.
 
Nevermind

I got it.

My error was when I made the positive charge equal to the electron charge, I forgot to square the number of electrons, since it applies to both charges.

And the positive charge HAS to be equal to the electron charge or else I would not be in equilibrium. Right?
 
dekoi said:
My error was when I made the positive charge equal to the electron charge, I forgot to square the number of electrons, since it applies to both charges.
Right.
And the positive charge HAS to be equal to the electron charge or else I would not be in equilibrium. Right?
The reason the positive charge must equal the negative charge is due to the conservation of charge. For every electron moved, you gain -e of charge, but the plate (which loses the electron) gets +e of charge.
 
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