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Drift velocity is proportional to the applied electric field. If there is an electric field, why doesn’t drift velocity increase as you move along the wire? For example, consider electrons that have speed v1 at point A. When these electrons reach point B some distance d further down the wire, why don’t they have a new drift velocity v2, where v2>v1? What is counteracting the acceleration from the electric field?

The electrons have decreased in voltage by E*d from A to B. What is the electric potential energy being converted to? I assumed it was kinetic energy, but since the drift velocity is constant, I am not sure if this is true.

Akash
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  • The electrons are constantly bouncing off atoms, defects, as each other, essentially randomizing their velocity each time. Their mean free path is quite short. – Jon Custer Feb 25 '22 at 03:36
  • You are contardicting yourself, The drift velocity is indeed proprtional with the electric field. The electric field does not increase or decerase so why would the drift velocity do so? Do you understand why is proportional with the electric field? – nasu Feb 25 '22 at 04:21
  • electrons are quantum mechanical particles, the current is a statistical macroscopic variable. this link may help http://hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html – anna v Feb 25 '22 at 05:46

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The electrons interact with the nuclei in the metal, transferring energy to them. This transfer of energy to the nuclei then manifests as thermal energy, and the wire heats up as it conducts electricity.

Of course, thermal energy is kinetic energy (and potential energy), just "randomized" among many degrees of freedom rather than moving in one coherent direction. So in that sense, your idea that the energy ends up as kinetic energy is not too far from the mark.

Michael Seifert
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