Is the Earth really electrically neutral (0.00000000V), or do we just say it is? - Page 1
Forum section: electronics / beginners

author: JoeN
author_rank: Frequent Contributor
timestamp: June 18, 2015, 02:31:09 am
content: For any normal circuit it wouldn't matter. And how would we know anyway? Voltages are all relative, right? Might the Earth not be truly neutral - it has some net charge one way or the other? Are all heavenly bodies neutral - or might there be a potential of difference between the Earth and the Moon or Venus?Inquiring minds want to know.

author: hamster_nz
author_rank: Super Contributor
timestamp: June 18, 2015, 02:50:19 am
content: See...http://vixra.org/pdf/1008.0071v1.pdfQuote

author: Skimask
author_rank: Super Contributor
timestamp: June 18, 2015, 02:50:46 am
content: Monster will makes some test leads for that...

author: JoeN
author_rank: Frequent Contributor
timestamp: June 18, 2015, 03:06:43 am
content: Quote from: Skimask on June 18, 2015, 02:50:46 amI just bought a 1000' spool tonight. How much more do I need?

author: Skimask
author_rank: Super Contributor
timestamp: June 18, 2015, 04:13:23 am
content: Quote from: JoeN on June 18, 2015, 03:06:43 amBest wait for the next indiegogo campaign.That PDF linked above made for an interesting read though.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 18, 2015, 06:37:23 am
content: Potentials in the positive MV to GV range occur (or are presumed to) up in space. This corresponds to solar wind knocking away electrons preferentially. Likewise, the Sun itself has a pretty high voltage (in terrestrial terms). Neither matters much for overall matter or charge balance, because such a field pretty quickly overcomes the energy pushing it (i.e., MeV of particles in solar wind) and the vast majority of matter remains neutral.An ion thruster works by ionizing a propellant gas (usually xenon) and accelerating it with as much voltage as possible (usually in the 100k to several meg range). This would leave the craft negative, however, and the charged ions would eventually circle around and return (after travelling a pretty good distance I suspect). Therefore, they also provide an electron gun, of similar voltage level and equal current flow, to shoot electrons into the ion wake and maintain neutrality. Whether the gas ever deionizes, who knows, but as long as it remains neutral (plasma or gas), it's fine.There is such a concept as potential at infinity, which corresponds to, taking the average over the rest of the universe. This would be "ground" in the most fundamental sense. In almost every other situation, however, "ground" is whatever is a convenient reference. Sometimes it's the Earth's surface, sometimes it's a nearby metal plane or enclosure. Sometimes it's the inside of an enclosure, which externally is at some massive potential against something else!Voltage is only ever meaningful as a difference.Tim

author: jeroen79
author_rank: Frequent Contributor
timestamp: June 18, 2015, 10:34:50 am
content: Isn't earth considered 'neutral' because it is a really big ball of matter that you can freely draw current from or into it without having a significant impact on it's charge or potential?

author: John Coloccia
author_rank: Super Contributor
timestamp: June 18, 2015, 10:54:11 am
content: It's the other way around. Earth isn't considered neutral. In US power distribution (not sure how it is elsewhere), the "Neutral" wire is simply the center tap on a transformer. That's a crappy name for it...it's not "neutral" in any sense of the word. When the wire gets to your house, it's attached to the Earth. That references the secondary of the transformer to Earth, as opposed to simply floating.So it's "neutral" that is considered to be at Earth potential, not the other way aroundEarth is convenient to use because:1) we're all attached to it, and ideally we're all sitting at Earth potential, whatever that is, so it makes sense to use it as our reference2) It's really big and can happily supply and absorb however many electrons you wish, so it's a good place to dump fault currents

author: Shock
author_rank: Super Contributor
timestamp: June 18, 2015, 11:20:10 am
content: Dr Martin Uman (University of Florida)"so there is constantly current flowing into the earth from thunder storms and flowing out where there is not thunder storms, and in that process an electric field is generated in the atmosphere. We walk around the voltage between your head and toes is 200 or 300 volts and noone notices that because you grew up in it."To the best of my knowledge..Neutral is actually the return path of current flow in a circuit. In many countries the neutral side of mains is earth referenced to prevent dangerous DC charges from building up on floating mains lines from the effects generated in atmosphere and storms.So the earth is earth (not neutral), confusing terms regarding neutral/ground/earth/chassis are often interchanged which makes it initially a little harder to grasp.Negatively charged would be a better term to describe earth.

author: Kjelt
author_rank: Super Contributor
timestamp: June 18, 2015, 11:24:52 am
content: Quote from: Shock on June 18, 2015, 11:20:10 amMy Fluke doesn't notice it either when I put it on ACV and hold the probes on the floor and the other one floating 2m in the air, what is he talking about?

author: EEVblog
author_rank: Unknown
timestamp: June 18, 2015, 11:27:51 am
content: Quote from: Kjelt on June 18, 2015, 11:24:52 amYou've got to stick your tongue on it.

author: Mechatrommer
author_rank: Super Contributor
timestamp: June 18, 2015, 11:33:35 am
content: Quote from: Kjelt on June 18, 2015, 11:24:52 amthere's your problem. it should be DCV. anyway i disagree there is 200V differential between head to toe.

author: opty
author_rank: Regular Contributor
timestamp: June 18, 2015, 11:36:31 am
content: Quote from: Kjelt on June 18, 2015, 11:24:52 amPossibilities:1) 10Meg input Fluke might still be to low to measure the voltage?2) He might be referring to situation when you are passing under power lines?3) ...btw. I can clearly see around 200VAC 50Hz on my body when I check on oscilloscope...

author: tszaboo
author_rank: Super Contributor
timestamp: June 18, 2015, 11:50:16 am
content: Voltage by definition is charge difference. Talking about the charge difference of one point is pointless. Please don't confuse people with US mains, it is an abomination by any standards.

author: SL4P
author_rank: Super Contributor
timestamp: June 18, 2015, 01:35:01 pm
content: Quote from: Shock on June 18, 2015, 11:20:10 amThat's a cool reference...I remember seeing a documentary back in 70s or 80s explaining the problem for those tall transmission towers and skyscrapers - and lightning charged clouds...Basically what I recall is that proportional to the density of the charge in the cloud mass, there is a 'shadow' of opposite charge that tracks across the ground below the clouds... when it meets a tall structure (or fool standing out in the open), the charge takes that opportunity to equalise.Accompanied by a large flash, and burnt things!So in that model, there is no universal level of charge that floats the whole planet at one time. Just a lot of very large capacitors !

author: Gyro
author_rank: Super Contributor
timestamp: June 18, 2015, 01:36:55 pm
content: QuoteThere _is_ an atmospheric potential gradient, and it _is_ in the ballpark of 200V/m depending on the weather. The reason you can't see it - It's very high impedance of course. You can't just stick a meter probe of negligible contact area up in the air and expect to measure it with a typical DVM. It's static, and it's DC not AC.To measure it you need a well insulated plate or wire grid of significant area and measure it with an electrometer. There have been several project published over the year. Meteorologists studying atmospheric electricity do it all the time, one experiment (I think, in the French Alps) has a wire strung across a valley between two peaks.It's common for radio Hams to have a shorting switch and/or discharge tubes to ground on their antennasIn thundery weather the potential gradient rises to some major levels until...Now if you want AC then look up Schumann resonance, an ELF resonance excited by lightning discharges around the globe.

author: TimFox
author_rank: Super Contributor
timestamp: June 18, 2015, 02:10:40 pm
content: When I was in high school, I visited the local university physics department and the professor had an experiment in progress to measure the electrostatic field in the atmosphere at different locations. The apparatus was a pair of electrodes with one rotating rapidly. If I remember correctly, the top electrode was three 60-degree segments with 60-degree gaps, and the lower electrode had six 60-degree segments with short gaps. In order to get a huge input impedance on the amplifier for the AC voltage produced by this sensor, he used an acorn triode (955) with the input connected between the plate and cathode and the output from the grid-cathode circuit. He showed me how to clean the glass carefully to avoid parasitic resistance.

author: rolycat
author_rank: Super Contributor
timestamp: June 18, 2015, 03:46:44 pm
content: There is a (naturally) brilliant discussion of atmospheric electricity inchapter 9of the second volume of The Feynman Lectures on Physics.These are rather wonderfully nowavailablefor everyone to read free of charge, courtesy of Caltech.

author: Gyro
author_rank: Super Contributor
timestamp: June 18, 2015, 04:37:51 pm
content: QuoteWow, good find, thanks!Chapter 9 covers it all with the clarity that we mere mortals could only dream ofP.S. Just realized I've actually got the book he references sitting on the bookshelf (Chalmers, 1957, Pergamon Press 63/- net). What are the chances

author: AG6QR
author_rank: Frequent Contributor
timestamp: June 18, 2015, 07:18:23 pm
content: Quote from: NANDBlog on June 18, 2015, 11:50:16 amThat's one common practical definition, and from that, it seems there would be no absolute reference for zero volts. Voltage would always be relative. You'd only be able to measure voltages between two points. All voltmeters would need two leads.But an electroscope shows that, in fact, there IS such a thing as an absolute reference for zero volts, even though it may be hard to measure precisely. A gold leaf electroscope is a crude sort of voltmeter, with only one terminal.https://en.wikipedia.org/wiki/ElectroscopeThe two leaves of a gold leaf electroscope will separate if there is a positive voltage on them, and they'll also separate if there is a negative voltage on them. Since they're conductive, they'll each have the same voltage, and since like charges repel, if there is a non-zero net charge on them, they will repel. The two leaves will have no repulsive force between them only if they've got zero volts on them.Zero volts would be when the body as a whole has no net charge, that is, the same number of protons as electrons.In practice, an electroscope can't measure small voltages. It requires voltages in the range of thousands of volts to produce a visible indication. It's much easier to precisely measure relative voltages than absolute voltages, and for almost all electronics purposes, the only thing that matters is relative voltage.But sometimes it's worth remembering that there's a bit more to the story.

author: Marco
author_rank: Super Contributor
timestamp: June 18, 2015, 08:05:51 pm
content: Quote from: JoeN on June 18, 2015, 02:31:09 amMeasuring the force between two earthed metal plates in vacuum?

author: Gyro
author_rank: Super Contributor
timestamp: June 18, 2015, 08:23:28 pm
content: QuoteThat's a tricky one. I agree 100% with the premise, they will only deflect when there is a net charge, positive or negative, but surely it must be relative to something - such as the outer case, assuming there is one? or it's surroundings.To take an example - take a discharged electroscope, put it in a faraday cage (mesh so you can see through it) at the same potential and observe that it still shows no deflection. Now raise the Faraday cage to several kV (enough to ensure that the electroscope can reliably detect). i suspect that the electroscope will remain flat despite it's absolute potential (or potential relative to Earth) having changed.Not sure how that works.

author: Sigmoid
author_rank: Frequent Contributor
timestamp: June 18, 2015, 08:47:13 pm
content: So... if I understand correctly, if I set up an iron pole that reaches from ground level to 4 meters above, then what I have is a voltage source of 800V connected across the pole, with an extremely high internal resistance?

author: John Coloccia
author_rank: Super Contributor
timestamp: June 18, 2015, 09:02:20 pm
content: Think of it like this...if you put 1000V on one end of a 4m piece of glass, and ground the other end, and then start probing around, what might you expect to measure? Air is a much worse conductor than glass. There's a voltage difference for sure, but there aren't any charge carriers available to do anything useful.

author: rolycat
author_rank: Super Contributor
timestamp: June 18, 2015, 09:04:49 pm
content: Quote from: Sigmoid on June 18, 2015, 08:47:13 pmNo. Why would an iron pole have an extremely high internal resistance?Take a look at the Feynman reference given earlier for a lucid explanation of potential distribution.

author: AG6QR
author_rank: Frequent Contributor
timestamp: June 18, 2015, 09:08:07 pm
content: Quote from: Gyro on June 18, 2015, 08:23:28 pmLike charges repel. And that's what the electroscope detects.In a vacuum, two electrons will repel each other. Two protons (hydrogen nuclei) will repel each other. Two hydrogen atoms, since each one has a balanced charge of exactly zero, will have no net electrostatic attraction or repulsion. Likewise, the leaves of a gold leaf electroscope will repel each other if there are more electrons than protons, or if there are more protons than electrons, but they won't repel each other if the number of electrons and protons are equal.In your experiment, if the ball at the top of the electroscope is connected to the faraday cage, then yes, it will indicate the charge on the faraday cage. If the ball is isolated from the faraday cage, and there's no conductive path, then the voltage on the faraday cage won't affect the gold leaf. In practice, there's probably at least a weak conductive path via ions in the air, so given enough time, the charge from the faraday cage might eventually bleed over into the gold leaf.

author: AG6QR
author_rank: Frequent Contributor
timestamp: June 18, 2015, 09:15:52 pm
content: Quote from: rolycat on June 18, 2015, 09:04:49 pmMy reading was that the voltage source is what would have the extremely high internal resistance.Of course, a voltage source, by definition, normally has a very low internal resistance. If it has a high internal resistance, it's not a very good voltage source. And a bar of conductive metal pretty accurately fits the definition of a very stiff voltage source of precisely zero volts.If you consider the atmosphere to be a voltage source with very high resistance, then the experiment describes the act of shorting it out with the iron bar. Since the air has such an incredibly high resistance, effectively no current will flow.

author: Gyro
author_rank: Super Contributor
timestamp: June 18, 2015, 09:26:01 pm
content: Again, no argument with your explanation of why they deflect.I just don't think that the electroscope inside the Faraday cage would deflect (connected or unconnected) as the potential of the cage is raised. Wasn't there an electrostatics experiment that shows that charge only exists on the outside surface of a conductor.It's a bit like saying that you can only make an electroscope that doesn't deflect if you're here on earth. Just can't rationalize it. Where's Prof. Feynman when you need him!

author: edy
author_rank: Super Contributor
timestamp: June 18, 2015, 11:07:49 pm
content: There are a few ways to interpret the original post question:If you short 2 probes of your voltmeter you will get 0 V difference, correct? If you now place your 2 probes in the ground, depending on what you are touching (rock, sand, soil, etc.) on each end of the your probe, and how far apart your probes are.... I am sure they will no longer be showing 0 volts. Now space them 1, 2, 5 meters apart.... 10 meters.... 1 kilometer. Will it still show 0 volts? So is the Earth electrically neutral... well, compared to what? Compared even to Earth at different places it is not. It may be close... but unlikely to be 0.0000000000 V.Taken to the extreme, if you stuck one probe in China and the other in the USA and could measure the potential difference (never mind all the other problems with this analogy... just bear with me a second).... again I am sure they would not show 0 volts difference.Why? There must be influence from electron distributions through different soil, rocks, effect of atmospheric discharges and polarization/capacitive charges due to cloud cover, and convective currents through the mantle and interior to the Earth's crust due to flow of magma. Essentially, the Earth itself has some unequal distribution of charges within it due to various influences and it takes time for them to flow and neutralize to balance out. Sometimes it may actually not be able to easily flow due to properties of different minerals and dielectric materials in the ground from region to region. And even if it did, the effect of atmospheric static electricity causing charge movement within the crust will also affect what you consider "neutral" compared to another part of Earth some distance away.So another way to interpret the question is if you ADD ALL POSITIVE AND NEGATIVE CHARGES of the entire planet including the atmosphere and the surrounding ionosphere up to a distance of 50 km into space, do they all balance out? Unlikely. Again you constantly have a solar wind hitting the magnetosphere, with particles deflecting all over the place. Who knows what the overall balance of charge really is, and whether it gets skewed one way or another.How about the entire solar system then? Add up all particles in the entire solar system out to furthest reaches of barren inter-solar-system space... Does the total balance out? Same number of electrons as protons? And what about other charged particles? Good question. You would think conservation of charge would need to be observed so if it started equal, it would have to remain equal (barring any influence from external particles coming in with charge). But we don't know.For all we know there may be a net imbalance but we'd never notice unless somehow on Earth we had it as well and could take extremely careful measurements and see that indeed.... there are more electrons than protons, but for whatever reason they seem to be sticking around (like a huge static charge on the Earth) because there is not enough of an imbalance to cause a sufficient mutual repelling of charges to drive them out of the region of Earth.

author: Marco
author_rank: Super Contributor
timestamp: June 18, 2015, 11:27:31 pm
content: Quote from: Gyro on June 18, 2015, 08:23:28 pmCoulomb's law is an r squared law, so the local repulsion of excess electrons/protons in the leafs overrides whatever effect the surrounding has on them (the leafs are close, the surroundings are far ... or in the case of air have negligible density of charge carriers).I'm pretty sure Faraday cages don't block electrostatic effects, unless they are earthed, it's just that the inside of it is still equipotential so it doesn't form EM waves inside.

author: BennVenn
author_rank: Regular Contributor
timestamp: June 19, 2015, 02:38:06 am
content: A 4m evaucated glass tube with an electron source at one end. 200v/m constant acceleration across the tube. That could be detected, yes?

author: helius
author_rank: Super Contributor
timestamp: June 19, 2015, 03:00:20 am
content: If the ball of the electroscope is connected to the inside of the Faraday cage, there will be no net charge on the device for it to detect. All imbalanced charge moves to the outside of a Faraday cage, that's the whole point.

author: slateraptor
author_rank: Frequent Contributor
timestamp: June 19, 2015, 03:02:13 am
content: Quote from: Kjelt on June 18, 2015, 11:24:52 amWhatever he wants...Prof. Uman is director of one of only a handful of lightning research labs in the US.

author: Kjelt
author_rank: Super Contributor
timestamp: June 19, 2015, 06:47:02 am
content: Yeah well bear with me for this simple experiment:I have two fully isolated copper wires one side with a banana plug in the powersupply plug other end fully isolated.There is one black wire on the 0V and a red one on the 300VDC.If I use my probes and stick the pins through the isolation I measure the 300VDC. If I attach the probes to the outside isolator I measure nothing.The question is what is the potential difference between two isolated wires?

author: hamster_nz
author_rank: Super Contributor
timestamp: June 19, 2015, 07:41:51 am
content: Quote from: Kjelt on June 19, 2015, 06:47:02 amWhat current would flow through a 1 megaohm resistor touching the insulation on both wires? If it is so close to zero that you cant measure it then the potential difference must be almost zero.

author: Gyro
author_rank: Super Contributor
timestamp: June 19, 2015, 09:45:40 am
content: QuoteThat's the bit I can't get my head around - I suspect the same would actually happen whether the ball is connected or not (someone needs to try this). But either way the potential of the electroscope has changed relative to Earth by charging the Faraday cage and the Electroscope hasn't deflected. I know there is a logical answer but I can't think of it.My other confusion is that if you ground (Earth) yourself and touch the ball of a charged electroscope (we're not in the Faraday cage any more), you expect its leaves to collapse - implying that earth is at (give or take the sensitivity of the electroscope) at 0V. However my guess is that exactly the same would happen if you did the same on another planet even though there will probably be MV or GV or relative potential between it and Earth. I was going to say on the ISSI but of course that's a Faraday cage.I think what I'm asking is... Is absolute voltage always relative to the biggest thing you have to hand, be it the earth, the Solar system or the inside of a Faraday cage? I just can't reconcile it with the obviously equal balance of positive and negative charges on a colapsed Electroscope - which should be absolute.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 19, 2015, 11:21:20 am
content: Voltage is relative. That's all there is to say!Perhaps it would be more illuminating to think in terms of electric field, which produces the force on the electroscope. Electric field is the spacial derivative (gradient) of voltage, so it has no absolute reference, the constant term disappears. Likewise, voltage can be measured as the integral (along a given path) of electric field; the absolute voltage is the"plus a constant"of integration -- a free variable that doesn't matter to the problem, internally.Tim

author: TimFox
author_rank: Super Contributor
timestamp: June 19, 2015, 02:01:54 pm
content: Quote from: Gyro on June 18, 2015, 08:23:28 pmA traditional gold-leaf electroscope measures charge, not voltage. Voltage is always a difference between two points, but the net charge on an object is an absolute value.

author: Gyro
author_rank: Super Contributor
timestamp: June 19, 2015, 02:13:23 pm
content: Of course guys, that's the difference isn't it! Voltage has to be relative, charge isn't. I was tying myself in knots. Thanks.I was beginning to feel a bit like my son in the back of the car repeating "but why..?"

author: Chaos_Klaus
author_rank: Contributor
timestamp: June 19, 2015, 03:44:52 pm
content: There is one thing that is still a little off.Voltage is not difference in absolute charge. It is difference in potential and that again is about electric fields.You could have two electron reservoires, one is very large but not densly populated with electrons. The other one is small but densly packed with electrons. Electrons will want to flow into the reservoir with less charge density.In electronics the word "ground" is often mixed up with "neutral".A single atom will be neutral in electric charge when it has the same amount of electrons as it has protons. However, a single atom will not be a particularly good ground, because as soon as I sink some current into it, it changes it's potential significantly.Gound ideally is supposed to be able to absorb any amount of charge (and current is just moving charges). We use "earth" as a ground reference, because we know that our close environment is at that same potential and therefore we won't run into large voltages that could cause problems. Also "earth" is a large chunk of matter that will not change it's potential.I think the actual thing here is that in electronics people think in electronic terms. They are used to talking about voltages and drawing currents arrows in the "wrong" directions. In physics terms, one would not ask the question about a neutral voltage, because voltage is just defined as a difference in potential.

author: IanB
author_rank: Super Contributor
timestamp: June 19, 2015, 04:34:24 pm
content: Quote from: Shock on June 18, 2015, 11:20:10 amThis is not really correct. Current flows in loops, conceptually like a circle, so there is no "return path". At any point you choose in a current loop, the current is simply flowing to the next part of the loop. If you take any two separated points in the loop there is a potential difference you can measure.The neutral conductor in power distribution circuits isdefinedby connecting it to the earth. This makes the neutral conductor have the same potential as the earth, thus "neutralizing" its potential. If all conductors are floating and none are grounded (as happens in some power distribution systems), then there is no neutral.

author: helius
author_rank: Super Contributor
timestamp: June 19, 2015, 05:55:49 pm
content: Continuing that thought, in an IT-System, there is no hot conductor, either (because there is no neutral there can be no hot). You simply have two "line" conductors that are supposed to be floating, and indistinguishable from each other.If the wires are twisted a certain way, does this scheme have advantages in terms of EMI emissions?

author: AG6QR
author_rank: Frequent Contributor
timestamp: June 19, 2015, 10:17:19 pm
content: Quote from: TimFox on June 19, 2015, 02:01:54 pmBut charge and voltage are directly related to one another by capacitance. Capacitance is defined as the ratio of a change in charge to the corresponding change in potential.The capacitance of Earth is about 710 microfaradsThat's another way of stating that, if the Earth's net charge changes by 1 coulomb, its potential will change by 1400V.Since there is an absolute value for charge, there can be an absolute value for voltage.(edit to fix math error -- I mixed up milli and micro)

author: IanB
author_rank: Super Contributor
timestamp: June 20, 2015, 04:17:36 am
content: Quote from: AG6QR on June 19, 2015, 10:17:19 pmBut a change in charge is arelativemeasure, as is a change in potential.QuoteAs before, these are relative changes in charge and voltage, not absolute values.QuoteI don't think your conclusion follows from the argument.

author: IanB
author_rank: Super Contributor
timestamp: June 20, 2015, 04:35:43 am
content: Quote from: TimFox on June 19, 2015, 02:01:54 pmAn electroscope will respond to electric fields. For the leaves to deflect there has to be a non-uniform electric field surrounding them. To measure charge the electroscope exploits the fact that a charge differential between the instrument and the surroundings produces a field gradient that leads to a deflection of the leaves.If we place the electroscope inside a Faraday cage and raise the whole system to a potential of a million volts relative to the outside world, the electroscope will still show no deflection. The electric field inside the Faraday cage will be uniform in all directions with no gradients and therefore the leaves will not experience any forces other than gravity.It is clear that this must be the case if we consider that the surface of the Earth is at a huge potential relative to the upper atmosphere (hence lightning). In spite of this enormous charged potential on the ground the electroscope is quite unmoved.

author: jpb
author_rank: Super Contributor
timestamp: June 20, 2015, 09:38:04 am
content: One of our modules in A-level Physics (the exam you take at the end of school at age 18) was on thunderstorms and the earth's field.I don't remember much (it was decades ago) other than thunder clouds have a charge distribution, thunder storms act to replenish the earths charge and that you can get an electric shock from an isolated aerial for example. (Most other things conduct well enough to remain close to the earths potential.)

author: TimFox
author_rank: Super Contributor
timestamp: June 22, 2015, 03:03:59 pm
content: In a traditional gold-leaf electroscope, the charge on the electrode splits between the two foils, which then repel each other (like charges repel) with a force given by the usual electrostatic formula (Coulomb's law).With the electrode disconnected, when you move the jar around the capacitance changes, but the charge remains constant. Therefore, the potential difference between the electrode and your cold-water ground system will change.

author: IanB
author_rank: Super Contributor
timestamp: June 22, 2015, 04:41:54 pm
content: Quote from: TimFox on June 22, 2015, 03:03:59 pmI understand what you are saying here, but the application of Coulomb's law requires the assumption that the system is unaffected by its surroundings (for example with the point charges in free space). If the surroundings are sufficiently distant they can usually be neglected due to the inverse square law. However, the forces between charges relate to the electric fields and potential gradients generated by those charges. If the surroundings introduce other significant electric fields then the electroscope deflection will vary and simple application of Coulombs' law will not apply. Another way of looking at this is to say that the charge on an electroscope is a relative difference to its near surroundings and not an absolute quantity.To clarify what I was saying about the Faraday cage: if you place the electroscope inside a Faraday cage and connect the electrode to the cage (local "ground"), then the electroscope will have no charge on it relative to its local environment and will have no potential difference relative to its surroundings. As such the two foils will not repel, even if the outside of the Faraday cage is charged up to a high potential.

author: TimFox
author_rank: Super Contributor
timestamp: June 22, 2015, 05:40:04 pm
content: For the two-leaf electroscope, if there is an external electrostatic field, uniform across the spacing between the foils, it will have the same force on both foils, since they split the charge. With the device open-circuited, the field will not induce a charge on the foils.

author: onlooker
author_rank: Frequent Contributor
timestamp: June 23, 2015, 11:25:57 am
content: If there is only an E field acrossing the two leaves, a "dipole" would be induced in each leaf that makes the two attract each other. The split experiment is just the effect of a vertical field. When an angled field is present, attract or repel is more complicated to calcalate. Uniform or not is really secondary.

author: TimFox
author_rank: Super Contributor
timestamp: June 23, 2015, 12:43:35 pm
content: From Wikipedia:"The electroscope can also be charged without touching it to a charged object, by electrostatic induction. If a charged object is brought near the electroscope terminal, the leaves also diverge, because the electric field of the object causes the charges in the electroscope rod to separate. Charges of the opposite polarity to the charged object are attracted to the terminal, while charges with the same polarity are repelled to the leaves, causing them to spread. If the electroscope terminal is grounded while the charged object is nearby, by touching it momentarily with a finger, the same polarity charges in the leaves drain away to ground, leaving the electroscope with a net charge of opposite polarity to the object. The leaves close because the charge is all concentrated at the terminal end. When the charged object is moved away, the charge at the terminal spreads into the leaves, causing them to spread apart again."Again, the external E field, applied to the entire device, induces a charge (monopole) on the metal guts, including the leaves, which then deflect by Coulomb force. The excerpt above includes a specific procedure for discharge, which includes connection (finger) to the terminal.

author: houkensjtu
author_rank: Contributor
timestamp: June 23, 2015, 12:49:42 pm
content: OT: I think the problem related with "earth" is really not simple for newbies, hoping Dave could make a "fundamental Friday" on this topic!

author: Marco
author_rank: Super Contributor
timestamp: June 23, 2015, 01:53:29 pm
content: Quote from: IanB on June 22, 2015, 04:41:54 pmOops, you're right of coarse ... the excess charge is on the outside of the cage.

author: niconiconi
author_rank: Frequent Contributor
timestamp: June 20, 2024, 05:30:31 pm
content: Quote from: T3sl4co1l on June 19, 2015, 11:21:20 amSorry for bumping an old thread. I came across this thread while trying to find the answer to the same problem on the Web. After getting confused by it for several years, I think I have finally found the answer.In engineering textbooks, voltage is always defined with respect to a reference, so engineering students believe all voltages are relative. Meanwhile, many physics textbooks on electrostatics say that an object's absolute electric potential is defined with respect to point at infinity for several reasons. So physics students believe an absolute voltage exists in some forms, at least in theory. In fact, it can be really difficult to analyze non-circuit problems in physics if you don't accept the existence of an absolute potential:1. Imagine a single electron in a vacuum, the absolute electric potential at point A is the work required (with a negative sign) to overcome the electric force to bring a test charge from point A to a point at infinity without any electric field.2. All objects are formed by subatomic particles, some particles like electrons and protons, carries an elementary charge, and it can be both polarities. Theoretically one knows the exact charge distribution of an object, it's possible to calculate its electric potential and electric field, with respect to point at infinity.3. Like charges always repel, opposite charges always attract, and both charges always attract neutral objects, it's just Coulomb's law. Even two objects that have never met each other before, still apply electrostatic forces to each other - circuit theory doesn't allow us to analyze this situation but electrostatics can, with respect to point at infinity. Also, if you add charges to all objects to increase their potentials, the outcome of electrostatic experiments may change. This suggest one can detect charges absolutely in theory.Even in engineering, absolute electric potential is sometimes used when there's no well-defined reference. For example the Human-Body Model of ESD said the human body's capacitance is a "free-space capacitance". In other words, the "self-capacitance" in electrostatics - it's the extra charges needed to increase the absolute electric potential of an object by 1 V. So accepting absolute potential gives you a wildcard when your reference is not well-defined - with respect to what? Just everything else, whatever it may be...So is voltage absolute or relative, for real? It turns out that, after reading more physics... Ultimately, it's still relative. The trick is to consider what happens when you increase the electric potential of the vacuum itself together with all objects inside it, once you do that, the outcome of all electrostatic experiments remain unchanged. Coulomb's law works just fine. The electric field remains unchanged since the (static) electric field is the gradient of electric potential, and only E causes physical effects, so only the potential difference matters. Furthermore, it turns out that the entire theory of classical electromagnetism is known as gauge-invariant in theoretical physics. Adding a gauge transformation to the electric potential V or the magnetic vector potential A, does not change the observable electric field E and magnetic field B.So TL;DR: Absolute electric potential exists, in a sense, using a far vacuum as the reference. Sometimes the concept is even useful too. But this definition assumes the electric potential of the "background" vacuum itself is 0 V - which is only a convention. One is free to choose other values.To put it in another way, one can imagine the universe as a Faraday cage with Perfect Electric Conductor as its boundary condition. It's natural to define the cage itself to be at 0 V for simplicity, but in fact any voltage would do, since the potential of everything inside the cage increases by the same amount so it has no detectable physical effect - observers outside that cage will have different ideas about its electric potential as well.Now here's the question I don't know yet... Is it possible to approximately measure the "absolute electric potential" defined by electrostatics experimentally? If not a point at infinity, at least a point at somewhere else in the solar system...

author: bdunham7
author_rank: Super Contributor
timestamp: June 20, 2024, 05:40:14 pm
content: QuoteThese statements reflect a common understanding of how we usually think of voltage in practical circuits, but they are, in fact, incorrect. Voltage and charge are inherently absolute values.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: June 20, 2024, 06:09:11 pm
content: Quote from: bdunham7 on June 20, 2024, 05:40:14 pmStill confusion exists. Charge is an absolute value? Yes the fundamental unit of charge is fixed, and charge of any defined region or object is just the count of those fundamental units contained in that defining box. So absolute once your box is defined. I don't see voltage as absolute in any sense. Take a single electron in an otherwise empty universe. Voltage can be thought of in many ways, but one obvious one is the difference in e field strength between two points. So there is a unique voltage between the electron and the end of that empty universe. But a different one between it and a point a few centimeters away.Thinking numerically would clarify some of the issues. I haven't actually done the math, but I am sure that the ratio of positive to negative charges associated with the earth (defined as the region of space around us to some radius well beyond the moon) is one to a great many decimal points. I am equally sure that there is a net surplus (or deficit) of electrons of a huge number of coulombs. Depending on what you are doing one or the other of those facts will be more important. The same two facts apply (with different numerical values) apply to the atmosphere. And the existence of lightning indicates that the second fact is important even though the balance of positive and negative charges is nearly perfect.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 20, 2024, 06:39:57 pm
content: Note that voltage at infinity is indeterminate, and any choice is arbitrary. Zero is perhaps the least arbitrary real number, so it's a common choice -- but it is still just as arbitraryTim

author: TimFox
author_rank: Super Contributor
timestamp: June 20, 2024, 08:26:23 pm
content: Another way of stating the point above is that if you were to add, say, +100 V to the voltage at every point in the universe, it would make no physical difference, since the electrostatic field (E) that induces forces on test charges would remain the same.However, you cannot arbitrarily add 1 Coulomb of charge to everything: any given physical object contains a fixed quantity of charge carriers (electrons, protons, etc.) that add up to the charge on that object.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 21, 2024, 02:16:09 am
content: The way I heard it explained, there is apparently a shortage of electrons in the universe, and as a result of that "stuff" will be positively charged. But I don't know how big this charge is. Apparently it's not very big, or those "goldleaf electrometers" (Mentioned 9 years ago in this thread) would always push their leaves apart.And you have to consider the differenced between electrical charge and voltage.Electrical charge is simply the difference between electrons and protons in some object, while voltages are always differentials, and thus you need two points to measure the voltage (differental) between those points.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 21, 2024, 03:16:12 am
content: A volumetric charge imbalance in the universe wouldn't need to be very large for the energy level around any given (astronomical) volume to be so high as to rip balance charges out of the very fabric of space-time; although there's also no known mechanism I think for C parity violation in and of itself (i.e. creation of charge out of absolutely nothing)? But by the same token, there's no reason the universe shouldn't be neutral since creation, nor remain that way.Other than that, local regions can have charge, but they'll very quickly draw a balance toward them, and the inter{planetary|stellar|galactic} medium is, however tenuous, also filled with plasma and so is conductive over long enough time scales. I would guess most low-density astronomical objects have on the order of GV on them, tops, and the more dense ones (neutron stars, black holes) can manage... somewhat more than that? I don't think I've seen it discussed specifically but there'll be a charge loss mechanism by the same argument as Hawking radiation so even a black hole is not a perfect insulator.Tim

author: Nominal Animal
author_rank: Super Contributor
timestamp: June 21, 2024, 04:44:49 am
content: Quote from: bdunham7 on June 20, 2024, 05:40:14 pmNo, that's not correct either. Electric potential is an absolute value, yes; and so is charge. But voltage is a purely relative measurement: the difference between two electric potentials.It is easy to confuse electric potential (also called electrostatic potential) with voltage, because both use "volt" as the SI unit.To put it crudely, if you point to some single thing, you can say it hasQcoulombs of charge, andVvolts of electrostatic potential; but you cannot say it has a voltage ofVvolts.The original post from a decade ago is a good example of this confusion.The short answer to that is"No, the Earth has quite a large electric potential", but a proper answer would also add"but since everything in the human scale tends to that same potential, it makes sense to choose that as the reference potential forvoltage measurements:to compare any other electric potential against."

author: iMo
author_rank: Super Contributor
timestamp: June 21, 2024, 07:05:46 am
content: Quote from: Doctorandus_P on June 21, 2024, 02:16:09 amLet us wait on the ali and ebay sellers offering the electrons..

author: RJSV
author_rank: Super Contributor
timestamp: June 22, 2024, 05:36:14 am
content: I would agree with that field strength, 200 V per meterWhat about capacitance ? Regardless of accumulated charges, from solar wind, etc. the huge earth itself must have large capacitance, and accessable through a single connection.I learned that from watching the various ElectroBoom videos, on Tesla coils. I learned, you CAN get a high voltage shock, having only the one wire or other conducting path.Actually, thinking about that, a lightning strike is a form of 'single connection' (rather than a completed circuit loop.)

author: radiolistener
author_rank: Super Contributor
timestamp: June 22, 2024, 06:21:23 am
content: Quote from: JoeN on June 18, 2015, 02:31:09 amYes, it may have some charge relative to other sky object. Depends on a reference, Earth charge may be positive or negative, or zero.If we get Earth charge as a reference, then Earth charge will be zero because Earth charge - Earth charge = 0.If we get average charge of entire universe (all objects) as a reference, Earth charge probably will be close to zero, but if there is balance between positive and negative charges in universe. But it is possible that there is some imbalance, and in that case Earth charge will not be zero relative to average universe charge.Quote from: JoeN on June 18, 2015, 02:31:09 amit depends on the object which you select for comparison

author: jmelson
author_rank: Super Contributor
timestamp: June 22, 2024, 02:40:07 pm
content: Quote from: TimFox on June 18, 2015, 02:10:40 pmThis is called a field mill, and the rotating shutters convert the static field into AC, so a high impedance op-amp can measure it. There was an "amateur scientist" article on building one in Scientific American, but I can't find it now. I did find this article:https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://www.vaisala.com/sites/default/files/documents/Recent%2520Work%2520on%2520the%2520Georgia%2520Tech%2520High%2520School%2520Field%2520Mill%2520Project.pdf&ved=2ahUKEwjyrJqCt--GAxVAkYkEHQRLBfEQFnoECCsQAQ&usg=AOvVaw1xQt9JcPe1TUR-fsJWcJFsJon

author: ejeffrey
author_rank: Super Contributor
timestamp: June 22, 2024, 07:49:13 pm
content: Quote from: AG6QR on June 18, 2015, 07:18:23 pm

author: Marco
author_rank: Super Contributor
timestamp: June 22, 2024, 08:26:54 pm
content: So lets say the earth has excess electric charge relative to the rest of the local universe, would most of the excess charge be in the upper atmosphere? (Similar to any charged conductor.)And if the universe as a whole had a significant net charge it wouldn't be measurable, because solids would not exist?