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FolkSong t1_izomyo9 wrote

Reply to comment by [deleted] in What is the average voltage between Earth and outer space? by dvorahtheexplorer

>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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TheSmartestBanana t1_izxzhap wrote

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

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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