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Someone invents a machine that creates electrons. It takes in some electrically neutral matter, and produces electrons of equal mass.1

Let's consider two quantities of negative charge:

$q$ is the amount of negative charge that a typical individual might reasonably want to create and use, not counting in lab experiments.

$Q$ is the amount of negative charge that if added to the Earth, would soon destroy human society as we know it. (It wouldn't necessarily kill everyone, but it would trigger a huge leap backwards.)

I'd like to keep $Q\over q$ within a few orders of magnitude of $10^{10}$. This way I can keep $q$ in the budget range of an upper-class individual, while making $Q$ too expensive to create outside a major government attempt.

What are reasonable values for $q$ and $Q$?

1 Of course this violates the principle of conservation of charge. Some skeptics claim that the machine actually sucks the electrons from another part of the universe, but it doesn't really matter.

abcde
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  • Why would the typical individual want to use the machine in the first place? – Frostfyre Nov 02 '15 at 04:33
  • @Frostfyre That's part of the question. – abcde Nov 02 '15 at 04:34
  • If that's part of the question, my hypothetical people will naturally try to use it to break science. Conservation of energy would be my first check. If the region the machine outputs into doesn't want to become more negatively charged, such as if its the negative side of a charged capacitor, do I have to put energy into it to account for the increased potential energy being stored in the capacitor? – Cort Ammon Nov 02 '15 at 05:12
  • may I know why and how your dooms day device can destroy the world? – user6760 Nov 02 '15 at 05:38
  • yr machine sounds like electron gun I keep thinking of quasar, but surely you must meant something else hence I like to clarify. – user6760 Nov 02 '15 at 05:49
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    Well, if you want an upper bound for complete and utter annihilation: http://what-if.xkcd.com/140/ – Brenn_ Nov 02 '15 at 06:04
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    Electrons repel each other and are not attracted to the Earth. Neither do the repel neutral atoms. As such the only effect would be lots of electrons flying into space? I guess that if you did it carelessly, it might create a very spectacular explosion. – Ville Niemi Nov 02 '15 at 07:33
  • @VilleNiemi Electrons have mass, this makes them Catholic (and attracted to Earth). – Samuel Nov 02 '15 at 20:29
  • Try reading this first: https://what-if.xkcd.com/140/ – Thucydides Nov 03 '15 at 01:36
  • @Samuel Nice. The mass is low enough for the gravity not really matter. The explosion caused by the explosion of electrons will dwarf any gravitic attraction easily. – Ville Niemi Nov 03 '15 at 08:28
  • Indeed. Full fledged helium atoms escape from earth all the time. And that atom contains 2 electrons and 4 barryons, so about 4000x electron mass. Add enough electrons and we will start loosing them to space, so in the long run we can sustain this if we do not add too many too fast. – Hennes Nov 05 '15 at 00:09
  • Photon with a wavelength (λ) of 6e^-52 meter. (h∙c)/λ h: 6.62607015e^-34 J∙s c: 2.99792458e^8 m/s h∙c = 1.9865e-25 J / 6e^-52 m = 3.26e^26 J – slimer Nov 01 '19 at 01:26

1 Answers1

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Electrons are very light. If we added about 20 billion extra electrons to the surface of the earth, the charge would repel electrons with about the same strength that gravity attracts them. Add more than that, and any extra electrons would simply float away.

Twenty billion is a minuscule number of electrons, especially spread across the entire planet.

Update with more clear explanation and more accurate numbers.

The ratio of attractive force of gravity of two protons to the repulsive force of electricity is about $10^{-38}$.

One gram of Hydrogen is one mole of Hydrogen, contains $6.02 \times 10^{23}$ Hydrogen atoms. One gram of any other substance has essentially the same combined number of protons & neutrons as one gram of Hydrogen.

The Earth weighs $6 \times 10^{27}$ grams, which corresponds to the weight of $3.6 \times 10^{51}$ protons. Thus, the gravity of the Earth exerts about as much attractive force on a proton as the electrical-repulsive force of $3.6 \times 10^{13}$ protons.

Electrons weigh about 1800 times less than protons, so the gravitational pull on an electron by the Earth is that same ratio smaller. Thus, it would only take about $2 \times 10^{10}$ electrons to repel a single electron with the same electrical force that the Earth attracts the electron with, gravitationally.

Yes, 20 billion is a truly miniscule amount, essentially undetectable. Spread out evenly over the surface of the Earth, it would be about 1 extra electron per 6 acres. Yet, this extra electric charge would be enough to cancel the gravitational pull electrons feel toward the Earth. If we added any more than that, then electrons liberated in the ionosphere would accelerate away from the Earth and be lost. Balance would quickly be restored.

Note that the gravitational (or electrical) attraction toward a point mass is the same as towards a sphere centered on that point, from anywhere outside the sphere. Likewise the attraction toward a spherical shell. Thus, electrons scattered evenly over the surface of the Earth would attract/repel anything above the surface the same as if they were at the center of the Earth.

user3294068
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    These numbers don't make sense. The ratio of forces is 4.1E42 for electrons (1.2E36 for protons), you look like you're actually trying to multiply by Avagadro's number (but are off by an order of magnitude) which would only give the number of atoms for a given substance, not protons. A billion electrons is nothing, there are 2E13 times that many in a AA battery. – Samuel Nov 02 '15 at 18:49
  • Updated to take into account your comments. – user3294068 Nov 02 '15 at 20:15
  • You have a small section discussing the surface area now which is good, but it's not accounted for in your calculation. The gravitational force is towards the Earth, the repulsive force of the charge is away from other electrons. The charge density you've calculated for is a point of mass and charge. The vertical component of the force (the force opposing gravity) of an electron sitting in a 6 acre area is not enough to eject another electron into space. – Samuel Nov 02 '15 at 20:27
  • Added another update. – user3294068 Nov 02 '15 at 20:48
  • Your update is incorrect. It's only correct for a single point to point interaction; a 2D world. However, you're trying to calculate the forces between three points: the gravitational center of the Earth, a point charge on its surface, and a second point charge which is approaching the surface (or being ejected). You can not assume this entire interaction is bound to a single line connecting all three points. – Samuel Nov 02 '15 at 21:02
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    Mass spread evenly on a spherical shell provides the exact same gravitational attraction outside the sphere as the same amount of mass at the center of the sphere. The calculus is complicated, but it works out. – user3294068 Nov 02 '15 at 21:11
  • I'm not disagreeing with that. I'm disagreeing that you can use that simplification to solve this problem. The fact that you didn't need to use any trigonometry to solve this problem should be a give-away to you that you did it wrong. – Samuel Nov 02 '15 at 21:21
  • Let us continue this discussion in chat. – user3294068 Nov 03 '15 at 15:10
  • I think, there is a mistake in the calculation: the ratio of attractive force of gravity of two protons to the repulsive force of electricity should be $10^{-36}$, thus yielding $2\times 10^{12}$ electrons. – Vadim Alekseev Dec 06 '15 at 16:41