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I was trying to find out why the drift velocity is proportional to the electric field applied in a circuit. I came across this website which provided a derivation. However I am stuck on one point.

In the derivation, it is stated that

Average random velocity of free electrons, u=0. If ‘v’ is the velocity just before the start of next collision the n,

$Drift \, \, Velocity \, \, v_d = \dfrac{u + v}{2} = \dfrac{0 + v}{2} = \dfrac{v}{2}$

My question is: why is it the case (or if it is not exactly true, why can we assume) that the electron looses ALL of its kinetic energy in a single collision and therefore the starting speed (just after a collision) is zero?

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I had to use this link http://www.sciencehq.com/physics/current-electricity.html to view the derivation. The u being zero isn't in reference to the actual starting speed of the electron. Its the average velocity of the electrons in the system in the equilibrium state, and the v is the velocity which is superimposed ontop of the equilibrium state.

See the top of the derivation:

"When no potential difference is applied across a conductor, the free electrons are in thermal equilibrium with the rest of the conductor and are in random motion. That is the average velocity vector of free electrons is zero and consequently this motion does not constitute a net transport of charge across any section of the conductor and hence there is no current in the conductor."

I hope I didn't misunderstand your questions

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  • $\begingroup$ Agreed. Just to add a detail, its easier to think of the ensemble of electrons. when an ensemble of electrons scatter, they won't loose all there energy, but the averaging over all directions means the starting velocity of all the electrons u is 0. Since the field acts on all of them the same way, the new added kinetic energy for each electron is v, which averages to v for the ensemble too. $\endgroup$
    – daFireman
    Oct 3, 2017 at 4:26

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