If I double the electric field, that should double the acceleration of electrons inside the conductor in the general direction of the electric field. But why does that double the drift velocity, and in turn, double the current? Normally when we have a fixed path length the velocity ends up getting multiplied by $\sqrt{2}$ when we double the acceleration. Thanks!

  • $\begingroup$ Related physics.stackexchange.com/q/323959/104696 $\endgroup$ – Farcher Aug 13 '18 at 22:36
  • $\begingroup$ There is no "fixed path length" that an electron travels through between collisions. In the Drude model, there's a fixed amount of TIME in between an electron's collisions with atoms. And if the acceleration of the electrons gets doubled, but the time between collisions doesn't change, that doubles the average velocity of the electrons as they move between one atom the other, thus doubling the drift velocity and the current overall. $\endgroup$ – joshuaronis Mar 3 '19 at 19:09
  • $\begingroup$ There is not a fixed time between collisions in the Drude model. There is however an average time between collisions. In my answer I have called this average time $\tau$. It is all to do with averages. $\endgroup$ – Farcher Mar 4 '19 at 12:50
  • $\begingroup$ Thanks, @Farcher just saw and upvoted your answer - for some reason I hadn't seen it before. My comment was more a response to my past self than a response to you!! $\endgroup$ – joshuaronis Mar 4 '19 at 13:04

Recall that Ohm’s law relates the three physical objects namely the electric field $\vec{E}$, the conductivity $\sigma$ and the current density $\vec{J}$:

$ \vec{J} = \sigma \vec{E} $.

Assuming that the electrons will flow in a direction perpendicular to some surface in the direction of $ \vec{J}$ we can find the current using the following equation:

$I = J A$ , where $J = |\vec{J}|$ and $A$ the surface area.

Another useful thing to note is that the current of some electrons with electron density $n$ and some velocity which we define to be the drift velocity $v_d$ then we can also write:

$I = n A v_d e$ where $e$ is the charge of an electron.

This equation could best be derived using for example a cylinder with cross section surface area $A$ and electrons moving with a speed $v_d$ a time $t$ in the direction perpendicular to the surface. Then by definition of the current we have that $I = Q/t = \frac{A n v_d t e}{t} = A n v_d e$.

Now we take these equations together.

$ I = n A e v_d = J A = \sigma E A $ and solving for the drift velocity we get:

$ v_d = \sigma E / (n e) $.

This shows that the drift velocity is indeed proportional to the applied electric field. How to check this? Do an experiment, I just did a basic derivation of the drift velocity that consist of a lot of assumptions. As a final note mark that a drift velocity is the average velocity of the electrons. Please also note that since the acceleration is constant we have that $v_d = a \tau$ with $a$ the acceleration and $\tau$ the mean free time. This is always the case, could you give an example where the velocity changes quadratically with $a$ ? Hope it helps, I am not an expert in this subject and I also had to look it up.

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  • $\begingroup$ Hey Dani. Thats all good and it works out mathematically, but the reason that the current density is proportional to the electric field is that the drift velocity is proportional to the electric field in the first place (as you said yourself, I=nAvde, which means J = nvde). But from basic concepts, why is that in the first place? Thank you!! $\endgroup$ – joshuaronis Aug 14 '18 at 0:57
  • $\begingroup$ The reason that the current density is proportional to the electric field is due to Ohm’s law. Ohm’s law is very general and does hold for other situations. Now since we are dealing with a flow of electrons we can define a drift velocity and talk about a electron density. Then you use a definition of the current $I$ and try to relate it to the current density $J$ and therefore the electric field $E$. The two $\textbf{definitions}$ of classical electromagnetism that I used are: $J=\sigma E, I=Q/t $ and I applied it to this situation, that’s it. $\endgroup$ – Mathphys meister Aug 14 '18 at 6:55
  • $\begingroup$ I do not know whether you have difficulties with these definitions (which could be measured in lab) or the notion of drift velocity. For the latter I cannot help you further, since my knowledge is not good enough on that subject. $\endgroup$ – Mathphys meister Aug 14 '18 at 6:58
  • $\begingroup$ $I$ is just how many electrons pass per second. Therefore you could look at how much charge is transported from one place to another, this is $Q=nAtv_d e=nAle=nVe=Ne$ where $V$ is the volume and $N$ is the total number of electrons contained in that volume and $l$ is the path length. And this you relate to the two definitions given in the first comment. $\endgroup$ – Mathphys meister Aug 14 '18 at 7:02
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    $\begingroup$ I understand, but the reason Ohm's law is true is because the drift velocity is proportional to the electric field. Imagine if it wasn't. Then doubling the electric field for a certain distance (aka doubling the voltage) would not mean doubling the current. $\endgroup$ – joshuaronis Aug 14 '18 at 9:26

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