Is the Earth really electrically neutral (0.00000000V), or do we just say it is?

[AG6QR]

The 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.

That'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.

Like 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.

[AG6QR]

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?

No. 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.

My 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.

[Gyro]

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!

[edy]

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. 

[Marco]

That'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

Coulomb'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.

[BennVenn]

A 4m evaucated glass tube with an electron source at one end. 200v/m constant acceleration across the tube. That could be detected, yes?

[helius]

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.

[slateraptor]

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."

My 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?

Whatever he wants...Prof. Uman is director of one of only a handful of lightning research labs in the US.

[Kjelt]

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?   

[hamster_nz]

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? 

What 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.

[Gyro]

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.

That'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.

[T3sl4co1l]

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

[TimFox]

The 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.

That'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.

A 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.

[Gyro]

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..?"   

[Chaos_Klaus]

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.

[IanB]

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.

This 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 is defined by 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.

[helius]

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?

[AG6QR]

A 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.

But 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 microfarads

That'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)

[IanB]

But 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.

But a change in charge is a relative measure, as is a change in potential.

The capacitance of Earth is about 710 microfarads

That's another way of stating that, if the Earth's net charge changes by 1 coulomb, its potential will change by 1400V.

As before, these are relative changes in charge and voltage, not absolute values.

Since there is an absolute value for charge, there can be an absolute value for voltage.

I don't think your conclusion follows from the argument.

[IanB]

The 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.

That'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.

A 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.

An 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.

[jpb]

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.)

[TimFox]

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.

[IanB]

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.

I 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.

[TimFox]

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.

[onlooker]

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.