0

https://physics.stackexchange.com/a/18806

Apparently he said 'at the same time', so when the electrons are 'pushed' by the electric field it is 'blocked' by the resistance simultaneously, if he said at the 'same time', and electrons don't accelerate, they move at a constant velocity, this means both forces (Electric Field and Resistance) are equal (F=M. A)

Using Newton's first law, how could the electron even move at the very first time if both equal forces act on the electron AT THE SAME TIME? Or shouldn't he used the word 'at the same time'?

2 Answers2

1

TL;DR: yes the forces Apply "at the same time", but no, they are not equal and opposite if you are not in the stationnary state described in the answer

The answer you refer to is talking about a quasi-stationnary picture, i.e. when the "friction" from the medium and the driving force induced by the field have equilibrated ($eE = kv$). It shows you why the electron eventually reaches a constant speed.

However, if you ask the question of how it gets to that speed, you are in a non-stationnary picture. If you imagine that your electron starts from perfect rest ($v=0$), and that you Apply an electric field, you can easily see that the dragging force is initially 0, and so the electron will start Moving as if it was under the influence of the field only. As it picks up speed, the dragging force itensifies, so that the acceleration of the electron is reduced. This goes on until both reach equilibrium, as explained in the answer you mention. Only then are the forces equal and opposite

The key here is the linear dependance of the dragging force with speed ($F_{drag} = kv$). In this very simplified model, the problem is that same as describing any object pushed through a viscous fluid (e.g. the air). So the result should not appear any more counter-intuitive than the one you get when studying an object falling in the air for example

Barbaud Julien
  • 983
  • 5
  • 15
0

$$m_e\frac{dv}{dt} = eE-k\cdot v$$

This is exactly $F = m \cdot a$: on the left hand side, there is $m$ and $\frac{dv}{dt}$ which is just fancy notation for $a$ in this case.

On the right hand side, there are two forces that are opposing each other (hence the difference is used as total force):

$eE$ is the force on the electron due to the electric field $E$. $kv$ is the force that is assumed to be proportionally increasing with the velocity $v$, the "friction" of the electrons in the conductor.

The $eE$ part "always" pushes the electron, the $kv$ part pushes "when the electrons collide". But they do it all the time, so it does not make sense to differentiate between both processes. Imagine a Galton board: Drawing of a Galton board

There is the constant downward force due to gravity ($mg$, or $eE$ for electrons) and the non-constant force due to collisions ($kv$). It makes sense to look at the average speed of the balls falling through, even if we know that the balls don't always move at this speed.

Jasper
  • 1,778