- #1

Strafespar

- 47

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Strafespar
- Start date

- #1

Strafespar

- 47

- 0

- #2

Khashishi

Science Advisor

- 2,815

- 493

Yes.

- #3

Crazymechanic

- 831

- 12

What drives the electrons then if the electric field is zero?

And is it then that the higher the resistance of a conductor , the higher the electric field ?

- #4

sophiecentaur

Science Advisor

Gold Member

- 27,545

- 6,182

When you have a finite resistance then, of course, there is a PD across it. As for the field, that (the Volts per Metre) will depend upon the length and the path of the wire. When considering what goes on in circuits, the fields (in the large scale) are not really relevant and don't usually come into it.

- #5

Strafespar

- 47

- 0

You initially have two wires connected to the same battery; but they are not connected (incomplete circuit). When the Circuit is connected we assume that the surface charges move in some manner (creates a distribution) which creates zero electric field within the non resistive wires and a large electric field in the resistor. Because there is no field in the wire they are initially stationary in wire. However the electrons in the resistor are not. They experience a force and move in a direction. The velocity of these electrons is constant because the resistance forces act similar to air resistance. When the electrons exit the resistor they encounter the "stationary" electrons in the wires. The electrons exiting the resistor transfer their momentum to the stationary ones. Because all electrons have the same mass, in a 1 dimensional case, all their momentum is lost and is transfered to the stationary ones. Think of this like an elastic collsion-a subatomic newton's cradle.

In the wire behind the resistor, the electrons in the resistor in front are moving away. This exposes positive charges in the resistor that attract nearby electrons in the wire. The electrons then move into the resistor. This may appear as a force in the wire, but as soon as electrons begin to move in this manner the net force becomes zero when they all move with the same velocity.

- #6

Crazymechanic

- 831

- 12

so zero resistance , zero electric field , now if we have some conductor with some finite value of resistivity then measuring the potential difference at the beginning of the wore and at the end we would get different results , lower voltage the further we go down the length of the conductor , now since we get lower voltage that tells me there also has to be a lower/decreasing electrical field strength the further we go in a conductor with non zero resistance.?

Now in a conductor with finite resistance but infinite length we would get zero field and zero volts at the end right?

Now then what do we get with an infinitely short conductor with an infinitely large resistance?

Since the resistance is infinite we should get also zero volts after it but since it is infinitely small the distance from the " high" side where there is electric field of certain strength and voltage is very small , now how that affects the outcome? Assuming there is no electric discharge or short circuit across the conductor ofcourse.

- #7

mikeph

- 1,235

- 18

I'm struggling to understand what "an infinitely short conductor with infinitely large resistance" could mean. If it has an infinite resistance then it is in no way a conductor.

- #8

Crazymechanic

- 831

- 12

The question is how do these two differ since the electric field falls of with distance obeying the inverse square law ? The electric field you say is constant all the way to have a linear voltage drop as we see in a resistor, but at the end of the resistor you get a lowered voltage , is that because the field has got weaker due to the drop in voltage ? I guess not , only why not?

So if it does imply a constant field then I have a question , put a metal sphere (similar to a point charge) at both ends of the resistor , now as much as I know you should get a stringer charge on the one which is before the resistor than on the one which is after it, yet the field is constant all the way through the resistor , this kinda amuses me.

- #9

mikeph

- 1,235

- 18

Resistance is a quantity you assign to an object equal to V/I. That's all it is. Your objects can be made of silver, aluminium, wood, air, whatever, if the resistances are the same then the same current will flow due to the same potential difference.

I have no idea what the last paragraph is, you're not just analysing a resistor in isolation are you? You're presumably connecting it to a circuit, so where do these spheres go? How is a metal sphere similar to a point charge?

I think you should find a book on EM and read it from the beginning, these hypothetical questions seem an exhausting way to learn a subject.

- #10

Strafespar

- 47

- 0

so if we have a very long wire with some resistance, the electric field will be small. But b/c there is a long distance E(small)D(long) = change in voltage = 4

Now with a resistor of the same resistance with shorter length. The electric field will be greater to arrive at the same change in voltage:

E(large)D(small) = change in voltage = 4.

Mathematically an infintely long conductor will need an infinitely small electric field and an infintely short conductor will need an infinitely large electric field...but this is not a physical possibility.

- #11

K^2

Science Advisor

- 2,470

- 29

No, you really shouldn't think of electricity like that. It's ok to think of electricity as a fluid under pressure for simple circuits, just to help you visualize things. But the moment you start thinking of electrons as bouncing balls, you've gone off the safe path. They really, really don't work like that.They way I think of it is like this:

[...]

When the electrons exit the resistor they encounter the "stationary" electrons in the wires. The electrons exiting the resistor transfer their momentum to the stationary ones. Because all electrons have the same mass, in a 1 dimensional case, all their momentum is lost and is transfered to the stationary ones.

In a conductor with no voltage or thermal gradient applied, the Fermi surface is a sphere centered about zero momentum. So the ground state of the conductor has zero average current. And that means that any net current will quickly die out. This is why an ordinary conductor with zero resistance simply isn't going to happen.

On the other hand, we have superconductors. But these are completely different beasts with completely different ground state, and any classic picture needs to be thrown out of the window. To be honest, I'm having very hard time picturing what happens at a junction of a superconductor and a conductor myself. But within the superconductor, the magnetic and electric fields are both zero.

- #12

Crazymechanic

- 831

- 12

So the bit with the colliding electrons may be a bit off here.

- #13

Strafespar

- 47

- 0

No, you really shouldn't think of electricity like that. It's ok to think of electricity as a fluid under pressure for simple circuits, just to help you visualize things. But the moment you start thinking of electrons as bouncing balls, you've gone off the safe path. They really, really don't work like that.

In a conductor with no voltage or thermal gradient applied, the Fermi surface is a sphere centered about zero momentum. So the ground state of the conductor has zero average current. And that means that any net current will quickly die out. This is why an ordinary conductor with zero resistance simply isn't going to happen.

On the other hand, we have superconductors. But these are completely different beasts with completely different ground state, and any classic picture needs to be thrown out of the window. To be honest, I'm having very hard time picturing what happens at a junction of a superconductor and a conductor myself. But within the superconductor, the magnetic and electric fields are both zero.

Yes, but classically I thought we could treat it like this. The classical derivations of the V/I = R relationship work on principles of elastic collisions and average time between collisions; this is how most intro level classes derive it.

Share:

- Last Post

- Replies
- 2

- Views
- 3K

- Replies
- 2

- Views
- 3K

- Replies
- 19

- Views
- 3K

- Last Post

- Replies
- 11

- Views
- 10K

- Replies
- 7

- Views
- 3K

- Replies
- 10

- Views
- 1K

- Replies
- 44

- Views
- 613

- Replies
- 5

- Views
- 55K

- Replies
- 3

- Views
- 2K

- Replies
- 12

- Views
- 3K