# If we consider the electric field to act upon charges with a force, how does it stay in line with Newton's laws?

This should be a relatively simple question.

Let's say we have a constant electric field $\textbf{E}$ in a conductor. The electric current density would then be $$\textbf{j}=\rho \textbf{v}$$ where both charge density $\rho$ and the velocity of charges $\textbf{v}$ should be constant.

The movement is obviously created by a constant force acting upon the each charge $q$: $$\textbf{F}=q\textbf{E}$$

But constant force should give rise to acceleration and not constant movement. (Newton's Laws). Now I know there is an explanation here, that satisfies both ends, I was just unable to think of it.

• I think charges get accelerated , collides, decelerated, repeats – InQusitive Jan 2 '16 at 15:33
• In a conductor with zero resistance, all the energy acquired by the charges would remain in the charges. But you can think of resistance as a sort of friction that is stripping kinetic energy away from charges and turning it into heat. The drift velocity is just the equilibrium regime of this scenario. – Phoenix87 Jan 2 '16 at 15:45

Electrons does accelerate, increasing its drift speed until it collides with a positive ion of the metal lattice. It loses its drift speed after collision but starts to accelerate and again gains drift speed only to suffer a collision again and so on. On the average only, does the electron acquire drift speed and doesn't accelerate.

• Electrons collide with electrons as well I think, am I wrong? – InQusitive Jan 2 '16 at 15:43
• of course, they do. – user36790 Jan 2 '16 at 15:44
• Thank you, I had this explanation in my mind, wasn't sure it was the correct approach! :) – Henrikas Jan 2 '16 at 15:46

I know there's already an answer, but I wanted to comment with a little more vocabulary--the concept people are bringing up here is called the Drude model. The key difference between the scenario you've outlined and this one is that we include some damping, conceptualized as a series of collisions. The change in the current now has some increase (from the electric field) and some "drag" coefficient (from the collisions--more current means more collisions), and just like a falling body under air resistance will reach an equilibrium value. In most electrical situations, this response is very, very fast.

As you'll see on Wikipedia, this classical model doesn't actually characterize the microscopic nature of conductance because a quantum picture is needed.

Inside of a conductor, like in the air, there are many obstacles for electrons to move. These obstacles give rise to a friction-like force that is proportional to a velocity of an electron. When this velocity becomes such that friction force is equal to electric force, situation of no total force is present. Forces that act like friction forces are vibrations of a lattice and collisions of electrons with them selves. NOT collisions with ions of a lattice, might I add, although this question does not ask for this remark. If a lattice would be perfectly still electrons would not scatter-off of it. That is because electrons in a etal are quantum particles in an already existing periodic potential which has defined its wave function. So, electron knows all about the lattice and where it is. But if this lattice should start to vibrate, this would effect electrons wave functions and this would be called scattering. So electrons scatter on vibrations of a lattice not lattice it self.

So you have something like: $qE = A . v$, where v is the speed of electrons, $A . v$ is a friction speed dependent force. In some instant, like the one I just wrote, $qE$ and Av are equal, and we get in the simplest form, $v= \frac {qE}{A}= \frac {q}{A} . E$. This is terminal speed, like the one you get when you are falling down through a medium, speed at which no total force is acting any more.