Question about the movement of electrons in a conductor

[a7madfmj]

Hello,
I know that the movement of a charged particle in space or a fluid caused by an electric field is accelerated(a = F / m) by the Coulomb force(Fc = k q1 q2 / d^2). And the Kinetic Energy of the particle is calculated through this equation ΔKE = q ΔV. But does an electron accelerate in a conductor because of a potential difference between the conductor's two ends ?. If it does, then does it affect the current's voltage or Intensity ?

 

[BiGyElLoWhAt]

I prefer to think of current as an exchange of elementary particles i.e. w & z bosons, which essentially transfers energy through the wire. That's all you need to make stuff work.

Whether or not this is "correct", I don't know, but free electrons don't actually shoot through the wire; the whole "sea of valence electrons" is just a model we use.

 

[dauto]

I prefer to think of current as an exchange of elementary particles i.e. w & z bosons, which essentially transfers energy through the wire. That's all you need to make stuff work.

Whether or not this is "correct", I don't know, but free electrons don't actually shoot through the wire; the whole "sea of valence electrons" is just a model we use.

That makes no sense. there is no exchange of W and Z bosons in an ordinary wire with a current coming through. The sea of valence electrons on the other hand is quire real.

 

[dauto]

Hello,
I know that the movement of a charged particle in space or a fluid caused by an electric field is accelerated(a = F / m) by the Coulomb force(Fc = k q1 q2 / d^2). And the Kinetic Energy of the particle is calculated through this equation ΔKE = q ΔV. But does an electron accelerate in a conductor because of a potential difference between the conductor's two ends ?. If it does, then does it affect the current's voltage or Intensity ?

Yes. the electrons will accelerate but than they collide with some obstacle in the wire losing its energy. After the collision it starts accelerating again and so on.You end up with an average velocity (called the drift velocity) which is proportional to the force. So, instead of F=ma the equation F= const * velocity should be used. So we have Potential V ~ Force F ~ Velocity v ~ Current I. In other words the potential V is proportional to the current I: V+RI where I is a constant. That just turns out to be Ohm's law.

 

[BiGyElLoWhAt]

That makes no sense. there is no exchange of W and Z bosons in an ordinary wire with a current coming through. The see of valence electrons on the other hand is quire real.

Hmmm... upon looking it up, I see that I was mixing up exchange particles.

That aside, yes there "is" a SEA of valence electrons, but it doesn't behave like water flowing through a pipe like so many people seem to think it does. You don't have some certain number of electrons racing around the copper wire, causing dq/dt. In fact, it's plausible (at least in my mind) that no single electron makes it around a loop of wire within a reasonably finite amount of time.

 

[a7madfmj]

Yes. the electrons will accelerate but than they collide with some obstacle in the wire losing its energy. After the collision it starts accelerating again and so on.You end up with an average velocity (called the drift velocity) which is proportional to the force. So, instead of F=ma the equation F= const * velocity should be used. So we have Potential V ~ Force F ~ Velocity v ~ Current I. In other words the potential V is proportional to the current I: V+RI where I is a constant. That just turns out to be Ohm's law.

Thank you.