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MiffedMouse t1_iznvgqe wrote

It will be close to 0. Charge imbalances create very strong forces to rebalance the charge. Space also isn’t completely empty, but it is filled with charged particles that will quickly move to rebalance any charge. This is especially true around earth, where we sit in the solar wind (the stream of charged particles coming from the sun).

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smnms t1_izo4h64 wrote

Replying to valdocs_user (#900,567)

Not on the long run. If solar wind carried away from the sun more positive then negative charges, the sun would build up a positive charge, and this would pull back all the negatively charged particles.

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PatrickKieliszek t1_izo5c6g wrote

Replying to valdocs_user (#900,567)

The solar wind is full of both positively charged hydrogen and free electrons.

The Earth's magnetic field captures more electrons than protons because the protons are more massive.

Eventually the net negative charge of the earth is enough to balance out the effect of the magnetic field and the rate of capture becomes even.

This causes the earth as a whole to have a slight negative charge.

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KingoPants t1_izo990t wrote

Replying to MiffedMouse (#900,065)

Putting numbers to this:

If we approximate earth as a conductive sphere with a particular radius in a vaccuum. It's capacitance to infinity is equal to  C = 4πϵ_0R.

Plugging in the radius of the earth into that you get a capacitence of Q=710 uF.

By definition of capacitence Q=CV. V=Q/C gives you the voltage.

Now for a mass (m) to gravitationationally escape earth then you need E=GMm/R energy where M is earth's mass and R is earth's radius.

Now if you set that energy to E>V*q where q is the charge of that charged mass it is energetic favourable for that mass to escape earth.

Put together you get the final big equation out:

Q = GM4πϵ_0*(m/q). This is how much charge you need on the surface of earth to start to fling stuff off earth.

If you want to fling out electrons then it turns out you need a net negative charge of 252 nano culombs.

To fling out a hydrogen ion / a proton you would need 463 uC. To fling out an alpha particle would take 920 uC.

It's not a ton really. Considering how big the earth is.

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TheTalkingMeowth t1_izojcf2 wrote

Replying to [deleted] (#900,153)

The article you link appears to be an attempt to explain gravity as the result of electrical interactions. It's riddled with misunderstandings. For example, it seems to think that you need to supply a constant stream of energy to keep the Earth spinning.

Basically, the author's a practicing electrical engineer who's not touched the rest of his physics education in far too long.

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Koffeeboy t1_izoms1a wrote

Probably the closest analogy to this would be holding a charged capacitor in a vacuum and asking what the potential between the capacitor and vacuum was. The thing is, (voltage/potential) is measured between areas of high and low electron charge. The vacuum is a chargeless or close to chargeless media, and it has no real way of retaining a charge. Vacuum capacitors are actually a thing used in high voltage scenarios because of how good of an insulator a vacuum is. Any charge you would measure would be between the positive and negative plates of the capacitor. If you pumped enough energy into the capacitor to start some sort of run off, well im pretty sure that would be equivalent to a cathode ray tube where instead of the vacuum retaining a charge, the electrons would shoot out until they hit something.

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FolkSong t1_izomyo9 wrote

Replying to [deleted] (#901,029)

>In addition, voltage decreases exponentially the farther the 2 reference points are from each other. If there were a voltage, all the particles in space would be pulled toward the earth.

Voltage isn't affected by distance, it's the electric field that drops exponentially as objects get further away.

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Pink_Slyvie t1_izovpfj wrote

Wasn't there an experiment that involved tethering something to the space shuttle, and lowering back towards earth, and it generated electric due to differences in the atmosphere? My brain is super foggy this morning, but I vaguely remember learning about it.

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TheSmartestBanana t1_izxzhap wrote

Replying to FolkSong (#901,531)

What I'm referring to is Coulomb's law, so technically I was wrong. It is the electrostatic force between 2 charged bodies that decreases with the square of the distance.

I may be wrong about this, but I would argue that if you could measure it with a voltmeter, the electrostatic force would be the same as the voltage in a practical sense. Granted, Coulomb's law isn't the best model for this situation since earth is not a single point, and you can't just stick a voltmeter in it. But I think my logic still applies.

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FolkSong t1_izyap8v wrote

Replying to TheSmartestBanana (#914,122)

I'm with you on Coulomb's law but I don't see how you're relating it to voltage.

A voltage difference can be defined as the work done to move a charged particle between two points. Work is force times distance. If you move the points further apart the work to move the particle stays the same - the average force is lower but the distance is higher.

For example you can imagine a negatively charged Earth and a positively charged spaceship, and imagine moving an electron from the Earth to the ship. Near the Earth the main force on the electron is repulsion from the Earth. Near the ship the main force is attraction to the ship. In between it's a combination of both, but the force is relatively low. You need to integrate along the path to calculate the total work. So you can roughly break this down into 3 sections - near the Earth, near the ship, and the middle section. If the objects move further away, this stretches out the middle section, making the force lower but the distance longer. The main contribution at each end stays about the same. This is hand-wavey but I think it helps give an intuitive sense that the integral will work out the same as distance changes.

You can also consider the case of two objects with the same charge (equally positive or negative). The voltage difference between them is obviously zero, regardless of distance. But from Coulomb's law you know there is a force between them (repulsive) that depends on distance.

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