Electric field misconception

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So I know I have a misconception in one of these ideas and I would appreciate it if someone could correct me and thoroughly explain my error.

Say I have two parallel plates connected to a battery, now there will be an electric field with magnitude, V/d between these plates. Now if I place a negative ion into the field it will move away from the cathodes. I used to think that is because the electrons travel from the cathode to the anode and so make one side , in this case the anode, less negative than the cathode; however; I also recently found out that in order for the electrons to actually escape the metal it usually needs to be heated.
Another thought I had was that the electrons simply build up and then simply repulse the ion but that doesn't seem mathematically correct since by that definition the strength would keep increasing over time.

So in summary I would greatly appreciate it if someone could thoroughly explain to me what happens at the atomic level between the two plates connected by a battery.
 

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  • #2
davenn
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Now if I place a negative ion into the field it will move away from the cathodes. I used to think that is because the electrons travel from the cathode to the anode and so make one side , in this case the anode, less negative than the cathode;
well it's a dual action
1) the electron placed in the field is repelled by the like charge on the negative plate of the " capacitor " <--- which is what you are basically describing
2) and they will also be attracted to the positively charged plate

I also recently found out that in order for the electrons to actually escape the metal it usually needs to be heated.
yes well that is a different process to what you initially described as now you are heating the negative plate rather than just placing a negative charge ( an electron between the plates) ... a whole different ball game

when you heat the negatively charged plate either directly or indirectly, you produce thermionic emission
https://en.wikipedia.org/wiki/Thermionic_emission
and now you have a diode ... see the drawing in that link around half way down on the right


Dave
 
  • #3
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I am still a bit confused. So when there is just two parallel plates does this mean the electrons do travel from the cathode to the anode? If so wouldn't that mean that when calculating the energy transferred to this negatively charged ion, you would have to take into account electrons slamming into it? Also why is the anode positively charged, I always thought it was merely less negative, I mean there are no protons involved as far as I know.
 
  • #4
nasu
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1. You are switching between electron and negative ion. They are not the same thing. A negative ion is an atom with one or more extra electrons.
2. Rather than using "cathode" it will be more clear to just use the polarity of the plates. The sign of what is called cathode depends on the device and what the device does. A negatively charged particle (ion or electron) placed between the plates will move away from the negative plate and towards the positive plate.
3. For a parallel plate capacitor (as described in OP), the "positive" plate has a deficit of electrons, as compared with the neutral state of the same plate. There is no motion of protons in this setup.
4. How is the emission of electrons related to your first question (about motion of that negative particle)?
 
  • #5
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1. You are switching between electron and negative ion. They are not the same thing. A negative ion is an atom with one or more extra electrons.
2. Rather than using "cathode" it will be more clear to just use the polarity of the plates. The sign of what is called cathode depends on the device and what the device does. A negatively charged particle (ion or electron) placed between the plates will move away from the negative plate and towards the positive plate.
3. For a parallel plate capacitor (as described in OP), the "positive" plate has a deficit of electrons, as compared with the neutral state of the same plate. There is no motion of protons in this setup.
4. How is the emission of electrons related to your first question (about motion of that negative particle)?
Apparently I was not specific enough, I apologize. I am placing a negatively charged ion in between the two parallel plates. Now it will move towards the positive plate, obviously.
My questions are,
1. Why is the positive plate positive? I am not understanding how it gains that polarity.
2. If there are just two parallel plates connected to a battery, without any heat source or such, will electrons travel from the negatively charged plate to the positive plate?
3. If yes, to number 2, how do you take into account that the electrons when traveling from the negative plate to the positive plate, don't collide with the ion.

Look, I am obviously completely clueless on how the electric field between the two plates is produced by the battery at a molecular level. If you could simply describe the process in great detail it might clear up my questions.
 
  • #6
Khashishi
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In an ideal capacitor, there is no current flowing across the gap. If you connect a battery to a parallel plate capacitor, one plate (attached to the -) will gain a negative charge, and one plate will gain a positive charge (assuming it is neutral overall). The amount of charge (q) depends on the voltage of the battery and the capacitance of the plates.
##q = C/V##
Now if you put a charge between the plates, it will be attracted to one plate and repelled from the other, depending on the sign of the charge.

If you heat up the capacitor enough that you have significant thermionic emission, then it is no longer a capacitor.
 
  • #7
davenn
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Apparently I was not specific enough, I apologize. I am placing a negatively charged ion in between the two parallel plates. Now it will move towards the positive plate, obviously.
My questions are,
1. Why is the positive plate positive? I am not understanding how it gains that polarity.
because you stated you connected one plate to the battery positive and the other to the battery negative
Doing so you have effectively connected a capacitor across the battery ... as I stated in my initial post

When you do that that, there will be an initial brief current flow .... electrons will flow into the plate connected to the battery negative making that plate negatively charged
This will create the electric field between the plates and will force electrons off the other plate and back towards the battery positive terminal resulting in the plate connected to the battery positive becoming positively charged.
Once the capacitor reaches equilibrium = the full battery voltage is seen across the capacitor, then the current flow stops, this happens very quickly

2. If there are just two parallel plates connected to a battery, without any heat source or such, will electrons travel from the negatively charged plate to the positive plate?
no ... electrons need a conductive path when there is no force to propel them as in an electron tube
as I linked to in my first post, describing thermionic emission
caveat .... electrons will only flow across the gap when the electric field reaches the dielectric breakdown voltage then there will be a discharge across the gap

think... lightning strike from cloud to ground, spark gap in a car sparkplug etc



Dave
 
  • #8
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because you stated you connected one plate to the battery positive and the other to the battery negative
Doing so you have effectively connected a capacitor across the battery ... as I stated in my initial post

When you do that that, there will be an initial brief current flow .... electrons will flow into the plate connected to the battery negative making that plate negatively charged
This will create the electric field between the plates and will force electrons off the other plate and back towards the battery positive terminal resulting in the plate connected to the battery positive becoming positively charged.
Once the capacitor reaches equilibrium = the full battery voltage is seen across the capacitor, then the current flow stops, this happens very quickly



no ... electrons need a conductive path when there is no force to propel them as in an electron tube
as I linked to in my first post, describing thermionic emission
caveat .... electrons will only flow across the gap when the electric field reaches the dielectric breakdown voltage then there will be a discharge across the gap

think... lightning strike from cloud to ground, spark gap in a car sparkplug etc



Dave
Okay, but the second plate, the positive one, does not actually have a positive charge such as +1 Coloumb or such. Right? It simply is less negative than the first plate.
Otherwise, thanks that was precisely what I was looking for.
 
  • #9
nasu
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Yes, it does have a charge of +1 C. Which means that the positive charge on that plate is 1 C more than the negative charge on the same plate.
When you remove electrons from an object the object is called "positively charged". The net charge is positive. The total positive charge from protons is larger than the negative charge from electrons.
 
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  • #10
Khashishi
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If the circuit is neutral overall, then the positive plate will have a positive charge (+q) and the negative plate will have a negative charge (-q). The difference in charge between the plates is 2q. It's possible for the entire circuit to possess some static charge, in which case the charge on the positive plate and negative plate will be different, but the difference between them is still 2q. This is a detail that I'm only bringing up in response to your specific question in post #8. Voltages on circuits are all relative, so they don't fully specify the charges.
 

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