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It’s my understanding this device can be used to measure voltage. How is this accomplished? What electromagnetic equations are applied and how is it calibrated?

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    \$\begingroup\$ I've never attempted the idea, nor thought about it -- using an electroscope to measure voltage. But I think it could be calibrated for it. Though I think we'd be talking about rather higher voltages than we often use. Here's a chart I was able to find on the web, for example. One could also determine the voltage-variable capacitance using other procedures. \$\endgroup\$
    – jonk
    May 27 at 18:00

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The gold-leaf electroscope can't be calibrated very accurately but there are voltmeters based on the the same electrostatic effects that cause the gold leaves to repel more at higher voltage.

They are of historic interest only or for educational purposes. Electronic instruments can now be implemented that are more sensitive and accurate.

In the case of the electrostatic voltmeter the voltage to be measured is applied to the static plates marked 'Q' in the diagram. The moving plate 'N' that can swivel on a jeweled bearing is connected to ground.

When a voltage is applied there is an attractive force between the two plates that causes the moving plate to rotate more as the voltage increases.

They are not useful for voltages less than a few hundred volts as the forces are too small but they have the advantage that they consume essentially zero current.

Electrostatic Voltmeter Wikipedia

Electrostatic Voltmeter

Modern instruments that can measure electrostatic potentials, also with essentially zero current, often use the concept of what is referred to as a Field Mill where the electric field from the point being measured is periodically interrupted with a mechanical vane. This can induce an AC signal in a nearby electrode that can be amplified and measured and represents the voltage or electric field.

Electric Field Mill

Image credit Wikipedia

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  • \$\begingroup\$ They consume zero DC current after an initial charging surge. For AC, the current consumption might be considerable, especially at high frequencies. \$\endgroup\$ Jun 8 at 23:09
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This doesn't really measure voltage, it measures charge. The gold leaf and fixed plate (or two pieces of gold leaf), which are either in a vacuum or at least shielded very well from air currents, are charged to the same charge, and the degree by which they repel each other is proportional to that charge. The gold leaf is extremely lightweight, so the repulsion is enough to counteract the force of gravity. It's exactly the same effect that makes your hair stand up if you touch the top load of a van de Graaf generator.

The only one in your pictures that has any sort of scale on it at all appears to just be measuring angle (notice how the gradations are 0 to 90, and it looks to be about a 90° angle); I don't think these are calibrated at all. Potentially useful for demonstrating that a van de Graaf generator or a Wimshurst machine is doing anything at all, but I would be very surprised if any quantitative measurements can be taken with these. These instruments are from the very early days of electrical experimentation.

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    \$\begingroup\$ In another post comments someone claims this device can measure voltage \$\endgroup\$
    – Nick
    May 27 at 18:39
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    \$\begingroup\$ It's a little complicated; the old electrometer does respond to the amount of transferred charge but as long as the reservoir is a lot bigger than the electrometer that charge is proportional to voltage because the electrometer is a physically large but very low capacitance capacitor. \$\endgroup\$
    – uhoh
    May 28 at 4:06
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    \$\begingroup\$ Hmm... why is that the reason you make a distinction? All voltmeters measure potential difference, all electrometers and ammeters measure absolute charge and current. I'm probably missing something obvious. \$\endgroup\$
    – uhoh
    May 28 at 14:49
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    \$\begingroup\$ @Hearth No it really isn't impossible. When you use an electroscope to measure a voltage you use a return path, so it can also be used as a two terminal device. If you know the capacitance, then use V = Q/C. If you don't think so, maybe I'll ask that as a new question. \$\endgroup\$
    – uhoh
    May 28 at 14:55
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    \$\begingroup\$ @Hearth - there is a second terminal to the electroscope - it is the rest of the universe, usually dominated by close by conductors. If you enclose the entire electroscope in a conducting container and connect that container to the electroscope terminal so there are no electric fields to its view of the universe; the leaves will go to the vertical position. It is a fairly meaningless distinction to distinguish between voltage and charge. \$\endgroup\$ May 28 at 22:08
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Electroscopes don't measure voltage directly, they measure charge.

They use the movement of objects due to the repulsion of like charges. Every object has a charge on it which causes a force between it and other objects. Most of the time other forces, such as gravity and air pressure, exert much greater forces so you don't notice the electrostatic force. Some times that you will is when something clings from static, such as rubbing a balloon on your sweater and having it stick to the wall, or your hair standing up after combing it on a dry day.

In a gold leaf electroscope very thin sheets of gold leaf are kept in a vacuum, and connected to an electrode that receives the charge to be measured. Since the pieces of gold leaf are electrically connected they receive the same charge, and since like charges repel each other, the leaves move apart.

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    \$\begingroup\$ yes they measure charge but that's how they measure voltage. It's not an either-or. The old current meters measured current by putting a volt meter across a fixed, low value resistor. You could say they measure current but they really measure voltage. \$\endgroup\$
    – uhoh
    May 28 at 4:08
  • \$\begingroup\$ They measure charge, when you connect the electroscope to a voltage source, see the leafs move to a new position, disconnect the voltage sources, the deflection remains because the charge is still there without the voltage. \$\endgroup\$
    – Uwe
    May 29 at 15:57
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Well, voltage is a potential difference between two terminals, so if you want to measure it simply, it takes two electroscopes. Just connect one to each terminal and add enough additional charge to the system so that neither electroscope reads 0.

The difference in charge between the two electroscopes will be proportional the the voltage difference between the terminals, multiplied by the self-capacitance of the electroscopes. It will take a large voltage to get a meaningful reading, and of course this only works if the whole system is isolated so that you can apply the bias charge.

If you want to measure voltage with one electroscope, you can connect one terminal, and a separate wire, to a big metal ball. Then observe the difference in charge that results from connecting it to either the wire or the other terminal in turn.

Doing this with the kind of voltages that would produce a reading is incredibly dangerous if those voltage sources can produce any meaningful current.

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The movement of the electroscope leaf depends on its internal electric field. And electric field is proportional to VOLTAGE difference between the input terminal and the instrument frame. Charge, on the other hand, is the capacity (size) of the object being measured times the voltage.

Note that for very small objects with low capacity, the capacity of the electroscope can load the combined system so as to give lower readings than the original voltage of the object under test, just as any other voltmeter might do in a high impedance circuit.

An interesting idea is energy storage: some energy is used to raise the center of gravity of the gold leaf. Presumably this potential energy is recovered when the system discharges through a load. Perhaps this can be used to increase energy storage of capacitors beyond the standard 1/2 C V^2 formula.

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The physical mechanism which determines the position of the gold leaf/leaves will be a combination of electrostatic repulsion, and whatever forces oppose that repulsion. Opposing forces are chiefly gravity pulling the leaf/leaves downwards, and in electroscopes where the leaves are not hinged on friction-less bearings, the restoring force due to elasticity of the gold under deformation.

The gold leaf will settle in a position where all these effects produce zero net force on the leaf, and leaf is in a state of equilibrium.

If the leaf bends in any way, its centre of mass will be displaced somewhat, and any expression you can derive for the lateral component of force due to gravity will likely be an integration of smaller components over the entire length of the leaf. At best it will be an approximation of that.

Calculation of the restoring spring force suffers in the same way, being an application of Hook's law integrated over the leaf's length, a complicated affair, or, if one exists, a simpler approximation.

The force separating the leaf from its support (or two leaves) is the Coulomb force, due to charges of like polarity accumulating on both surfaces, and repelling each other. Each such individual charge in one surface will contribute to a repulsive force on all charges on the other surface, in accordance with Coulomb's law:

$$ \frac{1}{4\pi \epsilon _0} \frac{q_1 q_2}{r^2}$$

The distribution of charges within the conductors will not be uniform, and will change as the geometry of the system changes. Consequently any attempt to predict the final geometry mathematically will probably be nightmarish, and, in the end, be some compromise between precision and alegbraic managability.

All this means that it's really hard to find any practical formulaic predictor of the electroscope's behaviour, and the only sensible means of calibrating the thing would be by experiment. That is, apply a known potential difference and then mark the resulting leaf position on a scale next to it.

The frame surrounding the actual leaf mechanism is an integral part of the device, and while much documentation exists that claim to describe the operation of the gold leaf electroscope, it seems that this enclosure is mostly overlooked.

The enclosure needs to be metallic, or at least partly metallic, because the operation of the device depends on the capacitance formed between the gold leaf (or leaves) and its support, and a nearby metallic surface. Essentially, the enclosure forms one plate of a capacitor, while the leaf and its support form the other.

In normal operation both elements must be connected in a closed circuit, since by Kirchhoff's Current Law, current entering the capacitor on one side must equal current leaving the other. In other words, if you wish for charge to accumulate on the leaf/support, you must allow an equal amount of charge to vacate the metal of the enclosure. That requires a circuit, and that's why you must apply a potential difference between the case and the leaf for it to work.

As you can imagine, determining the actual capacitance between the leaf/support and the enclosure is not trivial at all, but again, it may be determined experimentally. Assuming you are able to find that capacitance \$C\$, then you can apply a well established principle of capacitance that helps explain why the electroscope measures voltage, and not some other quantity.

The voltage \$V\$ across a capacitor is proportional to the charge \$Q\$ that it stores, and the constant relating the two is capacitance \$C\$:

$$ Q = C \times V $$

Since charge is proportional to voltage, and since at least some of that charge will accumulate in the leaf and its support, the resulting repulsive Coulomb force that sperates leaf and support will have some (probably complicated) relationship with the potential difference that moved the charges around in the first place. Thus, all other things remaining unchanged, the dominant cause of the leaf's deflection is the potential difference across the entire device.

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