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With my current knowledge of electric circuits and electrostatic fields, I would expect that the electric field generated by the potential difference of, say, a cell should exert a constant force upon the charge carriers in the wire.

From Newton’s Second Law, which states $F = ma$, I would therefore expect the charge carriers to constantly accelerate.

However, from the equation $I = nAqv$, we can see that the drift velocity stays constant so long as the current and charge of the carriers does.

So my question is: how come the charge carriers don’t accelerate constantly? Is there some kind of balancing force that keeps them at a constant drift velocity?

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  • $\begingroup$ The carriers aren't free. You're correct in considering energy conservation, but that energy is lost in heat due to the carrier interactions with the conductive medium. $\endgroup$
    – user253088
    Commented Nov 21, 2023 at 12:24
  • $\begingroup$ I'll also add the "drift velocity" is an average and there is a wide range of velocities among the individual carriers as well as potentially other degrees of freedom. It is a misunderstanding to believe that a constant drift velocity implies no work has been done. $\endgroup$
    – user253088
    Commented Nov 21, 2023 at 12:30
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    $\begingroup$ It sounds like your questions would be answered by reading about the Drude model. It's about as good of an explanation as you can get without going to a full quantum-mechanical description. $\endgroup$ Commented Nov 21, 2023 at 12:37
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    $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html $\endgroup$
    – anna v
    Commented Nov 21, 2023 at 13:26

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So my question is: how come the charge carriers don’t accelerate constantly? Is there some kind of balancing force that keeps them at a constant drift velocity?

The charge carriers alternatively acquire kinetic energy from the electric field and give up kinetic energy due to collisions with atoms, other electrons, or impurities in the conductor that cause resistance.

Essentially, the positive work done by the electric field on the charge carriers equals the negative work done on the charge carriers by the resistance, for a net work of zero and no change in kinetic energy.

Hope this helps.

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