Oddball thought experiment with electrons

[Jdo300]

Hello All,

I was thinking about how electrostatic fields are used to accelerate electrons in different applications (like particle research) and I thought of a question I might run past you all. Let's say that we have an electron that is at on side of a set of charged capacitor plates. Assuming that the electron is at the negative end of the plate, it will accelerate towards the positive end, and if the positive side happens to have a hole or something in it, the charge would presumably fly out somewhere about it's merry way.

Now, what if we consider the same situation only instead of the electron sitting in open air/vacuum/whatever, it is inside of an electrical conductor (like a piece of copper wire), and this wire extends from one plate to the other, going through the hole in the positively charged plate (see my attached picture). I am wondering if the electron in the wire will still be accelerated towards the positive plate like it was in open air (i'm assuming a lot slower if at all). If this doesn't work, could someone explain to me why?

Thanks,
Jason O

P.S. if having the hole in the positive plate messes things up, then that doesn't have to be there then :-).

 

[ZapperZ]

P.S. if having the hole in the positive plate messes things up, then that doesn't have to be there then :-).

Then how is this any different than connecting one end of a wire to one terminal of a battery and then connecting the middle of that wire to the other end of that battery? We all know what happens here, don't we?

Zz.

 

[Jdo300]

Then how is this any different than connecting one end of a wire to one terminal of a battery and then connecting the middle of that wire to the other end of that battery? We all know what happens here, don't we?

Zz.

Well in this case, there isn't any direct physicsl contact between the plates producing E-field, and the wires. So, are you saying that the electron would still move through the wire as if the voltage potential difference were applied directly across the ends of the wire?

Thank you,
Jason O

 

[ZapperZ]

Well in this case, there isn't any direct physicsl contact between the plates producing E-field, and the wires. So, are you saying that the electron would still move through the wire as if the voltage potential difference were applied directly across the ends of the wire?

Thank you,
Jason O

No. Connect one end of the wire to one terminal of the battery, then connect the other terminal to somehere in the MIDDLE of the wire. There's physical contact at two points on the wire. You still have one end of the wire dangling. Isn't this similar to what you want?

Zz.

 

[Jdo300]

Ok, I do know how the voltage potential affects charges in a wire when the potential is applied via physical contact to the wire itself. What I am essentually asking though is if the wire is placed inside a uniform external electric field, will the free charges in the wire be affected by that outside field?

 

[actionintegral]

Yes, free charges in a conductor are affected by an external field. They move in such a way to cancel the field inside the conductor.

 

[Jdo300]

Yes, free charges in a conductor are affected by an external field. They move in such a way to cancel the field inside the conductor.

Ahhh ok, that clears that up. Thanks :-). I have another related question but I'll post it in a separate thread.

- Jason O

 

[Jdo300]

Relationship of internal and external E-fields

Hello,

I am still trying to gain a good understanding of how e-fields cause current flow in wires. I attached a diagram showing two different scenarios where a charge buildup occurs in a wire. *I think*. In the top case, a segment of copper wire is sitting between the plates of a capacitor with a constant E-field. In the second diagram is just a piece of copper wire hooked directly to a voltage source. I am interested in understanding how the voltage potential in the wire is different in either case (if there is a difference). If I assume that the voltage source is the same in both cases, could I assume that the E-field that is inside the wire will be the same in both cases? In the bottom picture, I have made the assumption that there is an electric field produced in the wire from the voltage source rather than an external field from the plates.

In case 1, I know that current will not flow continuously through the wire because it is an open circuit. But if we just concentrate on the voltage aspect, would the same voltage potential difference appear on the wire in both cases? If not, then why?

Thanks,
Jason O

 

[Jdo300]

Charged plate thought experiment

Hello All,

I was thinking about some various scenarios involving a piece of copper wire immersed in an E field between two plates which made me come up with some questions I thought I'd run past you all here. I attached a couple of images to illustrate my point.

We know from basic physics that if you take two charged plates (like a parallel plate capacitor) and you place an electron up against the negative plate, it will accelerate through the E field between the two plates from the negative plate to the positive one.

Likewise, as pictured in my first diagram, we would expect that if this electron happens to be inside a segment of copper wire, that it would also experience an acceleration from the negative plate to the positive one. Of course, it would be retarded by the resistance of the wire and ultimately the e-field that would be created from the charge separation inside the wire would inevitably stop the charge(s) from moving further.

My question is, if you compare the setup in scenario1 with the setup in scenario2. Will the charges inside the copper wire still collect on the far ends in both cases? (Assuming that the metal plates have no physical contact with the wire).

Thank you,
Jason O

 

[ZapperZ]

You seem to be obsessed with what appears to be a question about "conductors" in an electrostatic field. This is nothing more than a similar thread that you had created before, so I'm merging it to that thread.

Please don't create another thread that basically covers the same issue.

Zz.