What's the electric potential of the Earth?

[jaumzaum]

I was wondering, we constantly assume the reference of zero potential is the surface of the Earth. But if we consider the reference to be the infinity, what would be the electric potential of the Earth?

As Faraday says, the Earth is charged with a -580 kC of negative charge. If we consider the radius of the Earth to be 6400 km, and consider the charge to be radially uniform, the electric potential of Earth would be -800 MV.

I know that it is electric current that kill us, but a potential of -800 MV in Earth (and in us) would imply a lot of excess of electrons in our body. Also, the calculations I did for the excess charge in our body does not seem to make sense. All pages in the interent says that human body has a capacitance of about 100 pF. That would imply an excess charge of -0,08C if we consider we are in voltage equilibrium with the Earth (considering we are here for hundreds of millions of years, I would assume this is true)

But if we multiply 7 billion people with 0,08C we find 1000 times more than the -580C. It seems that I made some assumption wrong. Can anyone help me to identify what I am missing?

 

[Delta2]

Well you have two contradicting lines of reasoning.

One line of reasoning considers our bodies as part of the Earth spherical conductor and the other line of reasoning considers our bodies as separate conductors that are in electrostatic equilibrium with the Earth conductor. That's what goes wrong with your post in my opinion.

 

[jaumzaum]

Well you have two contradicting lines of reasoning.

One line of reasoning considers our bodies as part of the Earth spherical conductor and the other line of reasoning considers our bodies as separate conductors that are in electrostatic equilibrium with the Earth conductor. That's what goes wrong with your post in my opinion.

Hello Delta! Thanks for the answer!

Do you think the parallel assossiation of our bodies to the Earth would change Earth ‘s capacitance significatively?

 

[Delta2]

Hello Delta! Thanks for the answer!

Do you think the parallel assossiation of our bodies to the Earth would change Earth ‘s capacitance significatively?

Hmm, it depends what is your exact model (for example you model Earth as a spherical conductor, WITHOUT the 7 billion human bodies and having total charge -540kCb?) and then you want to find how this charge is split to 7 billion conductors of capacitance 100pF .., I think after some thought that the electrostatic equilibrium will not be at a potential of -800MV but at a much lower potential...

 

[Delta2]

So yes in a way yes the total capacitance of the system of earth+7 billion conductors of 100pF is significantly different than the capacitance of Earth alone.

 

[Vanadium 50]

People do not have one foot on the Earth and the other at "infinity" as your 0.08C calculation assumes.

 

[Baluncore]

The ionosphere is conductive and is between infinity and the Earth's surface. You must analyse the system between the ionosphere and the Earth, or the ionosphere and infinity.
Currents flow between Earth and the ionosphere, and between the ionosphere and space.

 

[jaumzaum]

People do not have one foot on the Earth and the other at "infinity" as your 0.08C calculation assumes.

Thanks for the answer Vanadium!
In my head I was not assuming this. Can you explain to me why do you think I made that assumption?

My reasoning: If we have 2 spheres, a big one (Earth) with capacitance C1 and a small of (human) with capacitance C2. When we put both in contact, as C2 is much bigger than C1, the voltage of C1 in the end would be the voltage of C2 initially, that is -800MV compared to infinity. If a sphere has -800MV of voltage compared to infinity and a capacitance C1=100pF, its charge would be
Q=C1V=0,08C

Also, do you think the -800MV is right?

 

[Vanadium 50]

It's the potential difference across a capacitor that matters. 800 MV is the difference between the Earth and "infinity", not between one foot and the other.

 

[hutchphd]

As Faraday says, the Earth is charged with a -580 kC of negative charge.

Please provide reference for this number. This may be the surface charge but as mentioned by @Baluncore most of the charge separation is relative to the ionosphere. Interesting physics but your numbers are not at all correct I think.

 

[hutchphd]

If a sphere has -800MV of voltage compared to infinity and a capacitance C1=100pF,

That is for an isolated sphere. As soon as you are near Earth it changes and if you attach it to the Earth it is just a bump on a log.

 

[jaumzaum]

It's the potential difference across a capacitor that matters. 800 MV is the difference between the Earth and "infinity", not between one foot and the other.

I know that. If we had a potential difference of -800 MV between both of our feet, this would imply a large electric current between our feet. Considering we have 100 kOhm - 1 MOhm of resistance, and 100mA is enough to kill us, we would be imediately dead.

But as I said, it's not with the current that I am concerned, I am concerned with the excess charge. That is, the amount of electrons that we have more than protons. And, by what I understand, we calculate the electric charge of a solid capacitor using the potential relative to infinity.

Sure this is the potential difference between our bodies and infinity, and sure this will not make any electric current, but this would create a lot of excess negative charge in our bodies, and that is what I am concerned.

It seems a little absurd to say that we have 0,08C of excess negative electric charge, More so because if we multiply this with the amount of people, we get more than the charge of earth. And we didn't even account for all the others things on Earth, like animals, building, trees...

 

[Delta2]

Even if we ignore the problem mentioned at post #6, your value of 0.08C is still not correct. The Earth capacitance is $$C_{Earth}=\frac{Q}{\frac{KQ}{R_{Earth}}}=\frac{R_{Earth}}{K}=\frac{6.4\cdot 10^6}{8.9\cdot 10^9}=0.719mF$$. 7 billion conductors of 100pF have a total capacitance $$C_{human}=7\cdot 10^9\cdot 10^2\cdot 10^{-12}F=7\cdot10^{-1}F=700mF$$. So its like connecting a conductor of 0.719mF to a conductor of 700mF. Most of the charge will go to the conductor of 700mF, so that $$V_{equilibrium}=\frac{Q}{0.719}=\frac{Q'}{700},Q+Q'=580kC$$. The common potential $V_{equilibrium}$ of the equilibrium won't be 800MV but much much lower.

Your model has the problems, that Vanadium, Baluncore and hutchphd mentions, but even if we ignore these problems, your value of 0.08C per human is still not correct, and that's what I am trying to explain here.

If you solve the above system of equations you ll find that $Q'=579,4kC$ and that $V_{equilibrium}=\frac{Q'}{700mF}=827.7kV$ and also that each human body has a charge of $q=\frac{Q'}{7\cdot10^9}=8.2\cdot 10^-5C$

 

[hutchphd]

Again this is totally wrong.

1. Our bodies are not isolated
2. The capacitor is the surface of the Earth relative to the ionosphere. If you work it out the capacitance is about 1F so the potential difference is 500kV. In fact there is an electric field of about 100v/m at the surface because of this.

It would be a good Idea to read the answers already supplied and try to understand them.

.edit: I am referring here to #12 by @jaumzaum. I was too slow!