Electric field along an electric current

[ran13]

Hello fellow PFers,

I'm having some trouble visualizing electric fields along an electric current. Assuming that charge flows from high to low potential, and since electric field lines point from high to low potential, does that mean the electric field is always in the direction of the current? Also, I understand that a wire, if closed, will eventually become equipotential because of the fact that it is closed, but how does that effect the field and current?

Thank you.

 

[Antiphon]

It does not.

It's like balls on a hill, a gravitational analogy. If you grab the charge and move it against the field ("uphill") you are "charging" something. After all this is how capacitors get charged.

If you "release" the charge it will want to move with the field possibly releasing energy as it moves (that is, it speeds up and the charge gains kinetic energy from the field).

In a resistor, this is like a rock tumbling down a hill at a steady speed. The rock speeds up as it freefalls but every time it bounces against the hill it slows down a bit heating the spot where it hit the hill.

 

[cragar]

I saw a demo that walter lewin did with a neon flash tube and a van-degraff generator.
The Van-degraff creates an E field because charge builds up on it and he holds the neon tube radially away from it and it lights up because the E field causes current to flow because one end is in a stronger E field than the other. But when you rotate it 90 degrees it won't light up cause there is no potential difference everywhere along the tube the E field is the same. Does this help?

 

[ran13]

It does not.

It's like balls on a hill, a gravitational analogy. If you grab the charge and move it against the field ("uphill") you are "charging" something. After all this is how capacitors get charged.

If you "release" the charge it will want to move with the field possibly releasing energy as it moves (that is, it speeds up and the charge gains kinetic energy from the field).

In a resistor, this is like a rock tumbling down a hill at a steady speed. The rock speeds up as it freefalls but every time it bounces against the hill it slows down a bit heating the spot where it hit the hill.

Sorry, but when you said, 'It does not', which question were you referring to? Regarding your analogy, what would be the counterpart for a charge at the top of the hill, that is, right before it rolls down the hill? Would the resistor be fully charged and equipotential, or is this after the charged gains KE and comes to a steady motion, like with no net force? Thank you for your help

 

[ran13]

I saw a demo that walter lewin did with a neon flash tube and a van-degraff generator.
The Van-degraff creates an E field because charge builds up on it and he holds the neon tube radially away from it and it lights up because the E field causes current to flow because one end is in a stronger E field than the other. But when you rotate it 90 degrees it won't light up cause there is no potential difference everywhere along the tube the E field is the same. Does this help?

I actually saw that, so let me see if I get what you're saying... The E field is uniform on the surface, it's initially radial because it's not interacting with anything (these are ideals but I'm assuming it's okay to assume this for the sake of simplicity?) or maybe doesn't have a definite shape since it isn't interacting with anything. Once a conductor tube is placed near the sphere, the field lines interact/accelerate the electrons in the tube, causing the light?

So when you say that holding the tube parallel to the Van de Graff will not light it up, does that mean the field is actually not radial? Or is it just because an appreciable amount of the field is interacting with particles all along the tube, making it roughly equipotential? Does that mean equipotential surfaces have no net field, and if so, why is that? Do the electric forces acting on the particles somehow cancel? Sorry for the onslaught of questions and thanks for your help.

 

[kame]

If there is an equipotential there will be no current and no e-field.