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This is the phrasing that I commonly hear when talking about voltage's relationship to charge velocity...something along the lines of:

In a resistor circuit, raising the voltage will increase the current. Current is the flow of charge and therefore, for a given conductor cross-section, increasing the current must increase the charge velocity.

And this certainly makes sense! However, couldn't the following phenomenon also produce greater current without raising the velocity of the individual charges?:

From what I understand about batteries, mobile electrons moving through the circuit can be derived from either the atoms of wire itself (i.e. external to the battery) or derived from the anode of the battery due to chemical reaction (i.e. internal to the battery). How do I know that as the voltage of a battery increases ...

A) more "external-to-the-battery" electrons are being ripped away from the atoms that comprise the wire and/or

B) more "internal-to-the-battery" electrons are being supplied to the wire?

In either case, you would certainly have a "greater total" of electrons in the circuit as compared to lower voltages...and therefore, assuming these extra electrons are mobile and moving at the same velocity as they were for lower voltages, the current would be higher as well.

So why do we know that this other possible phenomenon is not occurring.

Thanks! Cheers~


Edit: After reading the comments below, I have realized that I can phrase my question more precisely to better capture the essence of my inquiry.

At the moment a circuit is closed, does the wire pick up a net negative charge? It has been brought to my attention that the number of electrons in the wire does not change with time (i.e. whatever enters the wire at the anode exits the wire at the cathode...sort of like conservation of mass for the water-pipe analogies)...but what about at the very instant the circuit is closed?

For example, if at some voltage x, the wire immediately became net negatively charged, and retained this exact net negativity over time (as long as the circuit was closed), it would technically not be changing its number of electrons as a function of time. So does this occur?

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From what I understand about batteries, mobile electrons moving through the circuit can be derived from either the atoms of wire itself (i.e. external to the battery) or derived from the anode of the battery due to chemical reaction (i.e. internal to the battery).

That seems OK but be sure that you understand that the mobile charges are moving rather like a bicycle chain. When you stand on the pedal the effect is felt on the rear wheel sprocket "at the speed of light" while the chain itself moves "at the drift velocity of the electrons". (Electron drift velocity will be in the order of mm/s.)

How do I know that as the voltage of a battery increases ...

A) more "external-to-the-battery" electrons are being ripped away from the atoms that comprise the wire and/or

They're not being ripped. A long chain of electrons is being nudged along by the battery.

B) more "internal-to-the-battery" electrons are being supplied to the wire?

The same number of mobile charges that leave on one terminal are returned on the other.

In either case, you would certainly have a "greater total" of electrons in the circuit as compared to lower voltages ...

No, there is no change to the number of electrons in the circuit.

... and therefore, assuming these extra electrons are mobile and moving at the same velocity as they were for lower voltages, the current would be higher as well.

You've lost me here. I don't understand what you are proposing.

So why do we know that this other possible phenomenon is not occurring.

By measuring the current?

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  • \$\begingroup\$ Thank you for the thoughtful response. I'll respond to several of your comments: "They're not being ripped. [...]". My understanding of mobile external-to-the-battery electrons is that they are derived from the valence shells of the atoms comprising the wire. Therefore, isn't it true that they are being "ripped off"? \$\endgroup\$
    – S.Cramer
    Mar 10 '20 at 22:47
  • \$\begingroup\$ "The same number of mobile charges that leave on one terminal are returned on the other". Sure, this is certainly true. But this just states that there is no charge accumulation in the wire as a function of time. \$\endgroup\$
    – S.Cramer
    Mar 10 '20 at 22:48
  • \$\begingroup\$ I guess the real question I should ask is, "Does the wire ever possess a net negative charge once a circuit is completed?" For instance, when the circuit is first completed, does the wire acquire a net negative charge and then retain this net negative charge (which would satisfy the 'no charge accumulation as a function of time' property)? Or is there, at no point in time, a change in the total number of electrons "housed" by the wire REGARDLESS of circuit completion? \$\endgroup\$
    – S.Cramer
    Mar 10 '20 at 22:52
  • \$\begingroup\$ I edited my question, which may be a little more accessible. Cheers~ \$\endgroup\$
    – S.Cramer
    Mar 10 '20 at 23:16
  • \$\begingroup\$ There is almost no change in the totally number of electrons; this is slightly complicated by the parasitic capacitance of the wire to ground. \$\endgroup\$
    – pjc50
    Mar 11 '20 at 1:09
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Current is basically the product of current density and drift velocity. Increasing the voltage increases both density and velocity, with the exact characteristic dependent on the material carrying the current.

Electron drift velocity (what we’re discussing here) is in the mm/s category, that is, very slow. Charge propagation is much faster - a fraction of the speed of light.

More here: https://opentextbc.ca/physicstestbook2/chapter/current/

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  • \$\begingroup\$ In your first sentence do you mean charge density instead of current density? \$\endgroup\$
    – The Photon
    Mar 10 '20 at 22:28
  • \$\begingroup\$ Here we get to strangle Ben Franklin. Anyway, Big picture: current (ampere) is defined as flow of charge, in coulomb/s. It’s even an SI base unit. So current density would be expressed as (coulomb/s)/area, or more simply, ampere/area. \$\endgroup\$ Mar 10 '20 at 22:38
  • \$\begingroup\$ Yes, so \$I=JA\$, not \$I=Jv\$. Alternatelly, \$J=qv\$ where \$q\$ is charge density. \$\endgroup\$
    – The Photon
    Mar 10 '20 at 22:40
  • \$\begingroup\$ @hacktastical i edited my question, which may be a little more accessible. \$\endgroup\$
    – S.Cramer
    Mar 10 '20 at 23:16

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