Why is the Coulomb 6.25 times 10 to the 18th electrons?

I suppose it has something to do with basic units of mass, length and time The coulomb would be a more standard unit of charge if it was say 10 times

10 to the 18th instead if some weird number of 6.25 times 10 to the 18th.. One volt is defined as one joule of work per coulomb of charge. So, what is the charge of one electron and how do we measure that? Why are the numbers so screwed up? .
Reply to
Bill Bowden
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The Coulomb was defined - in 1881 - before we knew that the electron existed. It was the charged transferred by an ampere of current flowing for one second.

Back then there was an idea that a discrete negative charge carrier existed, and in 1881 it was given the name "electrolion" which didn't stick.

The Coulomb is the charge carried by a "mole" of negative ions, so depends on Avegadro's number.

Ernst Mach, died in 1916 didn't believe in discrete atoms, though he was willing to believe that they were a handy mathematical devices, comparable with the infinite subdivision imagined in calculus. Einstein, in 1904, should have set him straight ...

There's nothing screwed up about the number, but it took us a while to find out what it was.

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Bill Sloman, Sydney
Reply to
bill.sloman

That's the Faraday; the coulomb is a smaller unit of charge, derived from magnetic attraction of wires through the 'Ampere' unit.

To put it tersely % units faraday coulomb * 96485.309 / 1.0364272e-05

Reply to
whit3rd

the whole set:

% units faraday e * 6.0220827e+23 / 1.6605551e-24 % units coulomb e * 6.2414504e+18 / 1.6021917e-19

Reply to
whit3rd

The important point is that the coulomb is defined in terms of the ampere so no constant is needed in electrical calculations, it's just 1. The Faraday was defined in chemistry terms so a constant is needed to convert from amperes, but since it is a mole of electrons, no conversion is needed to use it in chemical calculations.

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Reply to
rickman

Sorry about that. I was trying to emphasise that the idea that charge was quantised post-dated the establishment of the Coulomb as a unit.

The process of determining precisely how big a Coulomb was might have been determined that way when it was defined, but conceptually it has always been a unit of electric charge, and counting charge carriers is the natural way to define it.

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Bill Sloman, Sydney
Reply to
bill.sloman

Of course, units like the Ampere were devised before the detailed knowledge of electrons and the charge per electron was known, and the weird factor was calculated and standardized later.

It is like the definition of a meter as the distance light travels in 1/299.792.458 second. Of course it was not deliberately specified that way, it originally was originally defined as 1/10.000.000 the length of the meridian across Paris (from north pole to equator). That is a nice factor but a difficult to standardize base. So the definition was changed a couple of times.

The definition of the Ampere is similar:

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in vacuum, would produce between

Reply to
Rob

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It is now. Those interesting people at the various national standards labor atories are now into counting single electrons (in a integrated circuit sun k in liquid helium, in the version I sat through a talk about).

Watch this space.

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Bill Sloman, Sydney
Reply to
bill.sloman

The actual values are completely arbitrary and historical, the only thing that matters is the relationships between the fundamental constants.

See:

Reply to
bitrex

how was that figured out? How do you measure the energy of 1 electron? I hear there are photo detectors that can count individual photons, but I have no idea how to measure the energy of a single electron, I guess the first step is to believe electrons exist?

Reply to
Bill Bowden

So,

Not the energy, the attractive force of a field, onto a microscopic oil droplet. The mass of the droplet was gained from measuring its diamet er, its speed was observed in a known electric field.

Then, you do the same thing for a thousand other droplets. Hiire students to observe... That was the Millikan experiment.

Knowing the viscosity of air, the drift velocity and charge and mass all go into a formula. Solve for charge, and do a histogram; there's peaks at integer multiples of a tiny amount of charge, and THAT'S the charge on a single electron.

A coulomb-volt is exactly one watt-second, and an electron-volt is a lot less. 1.6 e-19 factor less.

Reply to
whit3rd

The coulomb is derived from the ampere and the second.

The ampere from a measurement of electromagnetic force between parallel conductors.

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Reply to
Jasen Betts

For the same reason there are 4.5 E26 atoms of C and H in a simple gallon of gasoline.

Reply to
bloggs.fredbloggs.fred

10^-19 I think 1.6E-19

believe first, (eg: Millikan's experiment)

then measure the charge of one electron a watt is one amp.volt

a coulomb is one amp.second

a joule is one watt.second or one coulomb.volt

an electron is just a smaller measure of charge than a coulomb.

So the ration of coulombs to electrons is the same he ratio of joules to electron-volts (except the sign is opposite)

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Reply to
Jasen Betts

The Millikan experiment entailed observing tiny electrically charged droplets of oil located between two parallel metal surfaces, forming the plates of a capacitor. The plates were horizontally orientated with one plate above the other. A mist of atomized oil drops was introduced through a small hole in the top plate. First, with zero applied electric field, the velocity of a falling droplet was measured. At terminal velocity, the drag force equals the gravitational force. As both forces depend on the radius in different ways, the radius of the droplet, and therefore the mass and gravitational force, could be determined (using the known density of the oil). Then, a voltage, inducing an electric field, was applied between the plates and adjusted until the drops were suspended in mechanical equilibrium, indicating that the electrical force and the gravitational force were in balance. Using the known electric field, Millikan and Fletcher could determine the charge on the oil droplet. By repeating the experiment for many droplets, they confirmed that the charges were all small integer

about 0.6% difference from the currently accepted value of

of the negative charge of a single electron.

But I don't see how they knew how many droplets of oil were involved or the mass of any one droplet? Maybe it doesn't matter since the total mass was known? So, how do you figure out how many electrons are involved? Seems like the electron could be twice the size with twice the charge and only half as many per Coulomb?

Reply to
Bill Bowden

Not if they found drioplets with 3e charge (using the accepted value of eletron charge)

neither would it be likely to be be half the accepted value, 10 repetitions with no odd charges has a probabiltiy of about 1/1000.

20 repetitions is a million to one. The probability of a missing (arithmetic) factor being common to all measurements becomes vanisingly small after measuring "many" droplets. (many as described above)
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Reply to
Jasen Betts

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You can measure the charge of the electron by looking at shot noise. About a 1% measurement.... depending on how well yo measure the bandwidth.

George H.

Reply to
George Herold

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That was essentially Einstein's 1904 insight with Brownian motion (though t hat gave him a route to Avogadro's number, rather than the charge on the el ectron, but the central idea is the same).

--
Bill Sloman, Sydney
Reply to
bill.sloman

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Hmm OK, I thought the big deal with Brownian motion, was that it lead the way to the fluctuation dissipation theorem.

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Which leads right into Johnson noise.

George H.

Reply to
George Herold

No they are not. The actual definitions are updated to reflect the most accurate method to measure the physicial entity. E.g. now that we can count individual atoms Avogrado's constant is defined as a simple integer number.

Suppose the meter was still *defined* as one 1/10000 of the distance between the equator and the northpole. How would you do accurate measurements based on that? The situation with the ever shrinking platinum standard meter in Paris is not better, in principle.

Groetjes Albert

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Reply to
Albert van der Horst

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