I can get 2 Mb/s download with my internet package. How many electrons are needed to communicate that data rate to my computer?
It’s probably measureable but not meaningful. You could derive the amount of energy in the signal of whatever medium you are using but there is no direct correlation from electrons to bits. This is true even inside a computer. A simple circuit called a flip flop can store one bit. One made with a vacuum tube is going to suck down some serious juice and of course discreet transistors would take less and modern chip based memory less and less per bit as memory density gets higher.
Perfect for looking at porn!
(It’s ‘discrete’.)
I will assume it is not useful, but if it is measureable it is going to be meaningful (at least to me)!
You NEED one electron.
There’s no single simple conversion number from bits (or bits/s) to electrons. Even assuming you’re using electrical signals and not fiber optics :), the answer depends on the specifics of your setup–specifically, on what the input stage of your connection looks like (FET or BJT design?) and on the signal and encoding. FETs are designed to have low quiescent current (the input stage acts like a capacitor) so only changing the input voltage level requires current; input FETs tend to have capacitances in the few-pF range. So we can calculate the charge required for a given voltage swing; a ~1.5V swing requires ~1.5pJ, or about 10 million electrons. But different encodings and modulation schemes make different voltage transitions (some examples); it is possible (as some of those examples show) that a string of 0s requires a different number of electrons than a string of 1s or an alternating string 010101… .
Also consider that since you are most likely communicating on a twisted pairs, as many electrons leave your computer as enter it…the data is conveyed by noting the direction of flow, or more often (thinking of modems here) the phase (which is to say time) at which the flow occured.
Is there a simple way to find a typical current rating for, say, an ADSL connection? After all, amps = quantities of electrons.
Only for a one, for a bit reading zero, you don’t need any.
This is actuall the reasons programmers try to write thier code using mostly zero’s, it saves valuable electrons when the code is run.
It doesn’t work that way. You are confusing sending power with sending information. Power is needed to overcome signal losses over distance but has no direct bearing on how many bits can fit in a given bandwith. I’m reminded of a B-movie cliche where more power is routed to a spaceship’s computer until it blows up. I can’t even think of a good analogy other than trying to make a tinny mono AM radio sound more like the orignal CD by turning the volume up. More power != more information.
There is a correlation between analog and digital bandwidth but it’s been so long since I was in that field I can’t remember what the theoretical maximum bits per second can be transmitted over a 6MHz TV channel.
While I was not whooshed by the joke but for non-programmers the reason that doesn’t work is that you have to be able to tell a zero from a null otherwise zeroes become meaningless. All zeroes is just as much information as all ones or any mixture of them. Of course all zeroes or all ones will compress more than a mixture but that is a different matter.
I’m asking for one possible real-world answer to the OP (given that the theoretical answers have been given).
[nitpick]amps = quantities of electrons/second.[/nitpick] Or maybe not so nit.
An electron has a charge of 1.602177 X 10[sup]-19[/sup] Coulombs. And an ampere is 1 Coulomb/sec. So an electric charge which is the equivalent of 6.142 X 10[sup]18[/sup] electrons/sec is going past a point if the current is 1 ampere.
I believe the international ampere is defined in terms of the force exerted between two long (infinitely long ideally), parallel conductors of negligible diameter that are 1 meter apart. I could look up what the force is but I’m too lazy. It used to be defined in terms of the amount of silver deposited out by a current through a standard solution in a certain amount of time. However I think that method lost out to the force method in the last election by not carrying enough of the swing states.
Well hell. If a one in the system requires 1 volt across 300 Ohms and a bit lasts for 2 nanoseconds the number of electrons is the number in 1/300 amp for one second times 2*10[/sup]-9[/sup] seconds.
I get about 4*10[sup]7[/sup].
If a zero is zero volts across the same 300 ohms then the number of electrons is 0.
Disregarding noise which over the long term should balance out.
It’s a meaningless question. Information is transmitted with electromagnetic radiation, not electrons. Electromagnetic radiation can travel through a vacuum.
It depends on the signal-to-noise ratio (Shannon’s Law).
C = B log2 (1 + S/N)
C = channel capacity in bits per second.
B = channel bandwidth in Hertz.
S = signal power.
N = noise power.
Shannon, C.E., “Communication in the Presence of Noise,” Proceedings of the Institute of Radio Engineers, Volume 37, No. 1, January 1949, pgs 10 to 21.
Not if it’s sent over a fiber optic link. 
If I remember my physics classes, it doesn’t take much-in fact it hardly takes any. Even metals where the outer ring electrons are relatively free to move, current flow doesn’t involve the actual movement of electrons over significant distances (microns and such). If I remember correctly, current flow involves “waves” passing through the electron cloud. Each electron moves a very small amount pushing the next electron a very small amount and so on. The electrons don’t leave their locations, they just influence the next electron in the wire. So looking at the very tip of the connector in the computer, very few if any electrons actually move into the computer, put their influence travels as far as necessary.