How many volts are generated by the difference of 1 electron?

The question is meaningless.

An electron has CHARGE not voltage.

Graham

It depends on how far you move it and the geometry of the conductors, Basically as many as you like. Getting one electron moved would be rather tricky though and if there was a significant P.D. afterwards, it might not stay there . . .

On the plates of a 160 zF capacitor, 1 V.

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John

Depends on what you dump it onto.

If you hang a medium-sized coin from a string, it might have a picofarad of capacitance to the world. Since

Q = C * V

and Qe, the charge on an electron, is -1.6e-19 colombs, then if you whack the coin with one electron, the change in voltage is negative about 160 nanovolts. Numbers like that aren't impossible to measure, so it's feasible that some science-project level of effort could measure single electron charges using Mouser-type parts.

I repeat: you should take an elementary physics course and get straight on basic stuff like this.

John

Radium has never asked a meaningful question. He just searches for something stupid, and starts asking. Then if he gets an answer he does not grok, he changes the rules and the question. All to get trolling glory.

Oh, just tried it. A US quarter is just about 0.4 pF to the universe, less than I'd expected.

John

One electron is -1.6e-19 coulomb. So if the coin is hit with one electron, wouldn't there be a negative voltage of 1.6e-19 volt?

Your head is all screwed up, Radium. Why don't you try reading a book on the subject? Why don't you take a class in basic physics?

Thanks.

Bob

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== NOTE: I automatically delete all Google Group posts due to uncontrolled 
SPAM ==

An electron is a particle. Its *charge* is -1.6e-19 colombs.

So if the coin is hit with one

A colomb is not a volt, just like a quart is not a kilowatt. That's why people gave them different names, so most of us wouldn't mix them up.

Geez, do you enjoy not understanding things?

John

It would be 1.6e-19 volt divided by the coin's capacitance (to ground) in farads.

Charge in a capacitor is voltage times capacitance. Voltage across a capacitor is charge divided by capacitance.

If you want some good reading, I totally recommend my favorite physics textbook, "University Physics" by Sears, Zemansky and Young. When I was young, I often read that one when I had time to read, before I was in highschool physics. Any edition is good. Libraries sometimes have it.

- Don Klipstein ( snipped-for-privacy@misty.com)

We've known this for ages.

He dislikes educating himself. Instead he prefers to ask stupid questions, ad nauseam in fact.

Graham

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That depends on the perihelion of the ampliphon.

JF

In the context of 'electronics basics' this raises a discussion of capacitance.

I prefer, however, the universal context, and the most common items in the universe are hydrogen and stupidity.

So, a difference of one electron to a hydrogen atom is my answer, the familiar 'ionization potential', 13.527 volts.

I can't think, right offhand, of a stupidity-based answer.

a

it.

Good answer but not complete. Voltage is a measure of potential energy and an electron can have as much energy as you wish if it is moving. Your answer considers only the stationary charge component.

my 2c

Cheers

10e-Graham's # of volts would be nice. Now that is an extremely small voltage.

Think about it. 10-the-power-NEGATIVE-Graham's-number.

That is a decimal followed by a Graham's-number amount of zeros followed by one 1.

's_number

Just where can such a low voltage exist?

Voltage is not a measure of energy. And the charge caried by an electron does not depend on its velocity.

People who insist on tangling units can never calculate, and probably never understand, things.

John

I seem to recall an experiment by somebody named Millikan balancing oil drops on a potential difference of several thousand volts. So, of course, Wikipedia has an article on it, with pictures:

(why remember anything these days?)

His experiment was fundamental in determining the charge on a single electron, which is now thought to be

1.602176487(40) x 10?19

In order to generate a microvolt, that would need to be put on 1/6 of a picofarad. So, your coin (which you measured later to be .4 picofarad) would be high by a factor of 3. Maybe a dime would work?

BTW, how did you measure the capacitance of the quarter? It must be fun to have all those cool toys laying around.

Regards, Bob Monsen

Which, on a pure units basis, implies that the capacitance of a proton must be around 1e-20 farads.

Which would make its radius around 1e-10 meters.

None of which are true, of course.

John

Everybody should have one of these:

Oh, I repeated the measurement more sensibly and got 0.72 pF, very close to the theoretical value.

John