I Ionization energy of an electric field

1. Dec 20, 2016

Bhope69199

Hi,

If I had a charged parallel plate capacitor which was not connected to a circuit (so the charges stay on the plates) creating an electric field strong enough to cause ionization, and I then fired a neutral atom between the charged plates (the charge starts outside of the electric field) perpendicular to the electric field. I assume the electric field will impart energy onto the atom and cause ionization.

If these two products (the electron and positive ion) still had enough energy to escape the electric field (from the initial K.E from being fired, and don't have a chance to effect the charges on the plates) would this result in a decrease in the energy of the electric field, as the energy that the electric field has imparted to cause ionization has now left the electric field within the positive ion?

2. Dec 20, 2016

phinds

I think if the charge is great enough to cause ionization, you'd get discharge via a spark across the gap before you could fire a neutral atom between the plates

3. Dec 20, 2016

Staff: Mentor

I dunno, this seems like a pretty interesting question. I can see how work is done and energy is lost in the trajectory of the molecule, but I'm not sure what happens to the charges on the plates. Maybe nothing -- all the extra energy lost is in the decease in the KE of the particle?

4. Dec 20, 2016

phinds

So would ionization not cause a spark discharge the way I thought it would?

5. Dec 20, 2016

Staff: Mentor

Good question. But there is a time of travel issue for the arc or electron/ion physical travel, no? I've marked this thread for Homework Helpers assistance...

6. Dec 20, 2016

Staff: Mentor

Wait, even if the electron and ion don't go directly to the ionizing plates, they will probably eventually make it to them outside the capacitor structure. That may be the reason that the capacitor energy decreases...

7. Dec 20, 2016

TSny

Fundamentally, is the question essentially the same as the simpler case of firing a positive point charge between the plates?

The particle exists the plates with an increase in KE, even though there has not been any change in the energy of the electric field of the plates.

There is no violation of conservation of energy.

This easily extends (pardon the pun) to two oppositely charged particles connected by a blue rubber band that snaps just before exit. ("Ionization")

It wouldn't surprise me if I'm missing the point of the original question.

8. Dec 20, 2016

kuruman

What if ionization occurs when the atom is far from the exit point? Where does the ionization energy come from? I think it is misleading to draw two "disembodied" capacitor plates with a high potential difference between them and pretend that that's all there is. If that were the case, the plates would attract and smash into each other. So let's suppose that the plates are kept in place by perfectly insulating stretched springs. These springs are stretched just enough to balance the attractive electrical force. Now when ionization occurs, the ionization energy comes from the potential energy stored in the electric field. For example, if the atom is hydrogen, the potential energy is reduced by 13.6 eV, which means that the voltage across the plates is reduced by 13.6 V. This results in an additional stretching by a bit of the springs and a decrease, also by a bit, of the plate separation. The plate separation decreases because at constant charge, when the voltage decreases the capacitance must increase. This makes sense because the springs stretch more as a positive charge moves closer to the negative plate and at the same time a negative charge moves closer to the positive plate. Additional spring stretching and hence voltage reduction occurs as the components of the dissociated atom gain kinetic energy until they exit the plate region.

Am I stretching the truth? (Also pardon the pun.)

9. Dec 21, 2016

Staff: Mentor

The electric field strength of the capacitor goes down. To avoid discharges we can use a lower electric field and a Rydberg atom (which is very weakly bound), but those details don't matter. Let's shoot the particle into the capacitor in the middle: The positive charge will end up in the half with the negative plate, and the negative charge will end up in the half with the positive plate. Both charges go to regions of lower potential energy. As a result, the energy stored in the electric field goes down. This reduction in energy provides the energy for ionization and acceleration. The particles end up in a potential equal to the middle of the capacitor plates, with a slighly slower speed compared to their initial speed as the ionization process took some energy.

Edited after the first version was wrong.

Last edited: Dec 21, 2016
10. Dec 21, 2016

vanhees71

I thought the charges of the disintegrated atom both leave the capacitor? Then in the asymptotic state (both particles gone infinitely far away) there shouldn't be any change of the field within the capacitor and thus the em. fieldenergy should stay the same (density $\vec{E}^2/2$).

11. Dec 21, 2016

Bhope69199

TSny why would there be no violation of conservation of energy in this case? Where has the energy come from to increase the K.E, if not from the electric field?

kuruman Ok, so the electric field loses energy? The voltage reduces but in order to keep the same capacitance (due to no loss of charge) the plates essential move together. So if the plates were fixed and couldn't close in would there still be a reduction in energy of the field, or is there energy coming from where the plates are fixed?

If the electric field energy reduces and there is no way for the plates to move, how could the same amount of charge (still applying the same force I assume) attract the plates together?

vanhees71 yes, the charges do leave the capacitor. So the electric field doesn't lose energy? Where does the energy come from to ionize the particles, if not from the electric field?

12. Dec 21, 2016

vanhees71

Hm, obviously I was too quick in answering. Energy should of course be conserved here...

13. Dec 21, 2016

Staff: Mentor

Hmm, thought a bit more about it. The field far away should be equal to the field centrally between the plates, of course, so the particles lose the speed they gain between the capacitor plates. They will even end up a bit slower as the ionization process needed energy.

14. Dec 21, 2016

TSny

What matters is the field energy of the net electric field of all charges. For the single charge $q$ passing between the plates $\vec{E} = \vec{E}_{pl}+ \vec{E}_q$, where $E_{pl}$ is the field due to the charge on the plates and $E_q$ is the field due to the charged particle. So $E^2 = E_{pl}^2+ E_q^2 + 2\vec{E}_{pl}\cdot\vec{E}_q$.

If the charge on the plates remains fixed, then any change in field energy of the system is due to a change in $\int \vec{E}_{pl}\cdot\vec{E}_q \; dV$ over all of space.

The value of this integral decreases as the particle passes between the plates which accounts for the increase in KE of the particle.

15. Dec 21, 2016

Bhope69199

mfb, so the energy that the field imparts for ionization to occur should be accounted for in the reduction of velocity of the particles, caused by the fringe fields de-accelerating the particles? This results in the electric field not losing energy, so the sum of the two charged particles' energies is equal to the initial energy of the neutral atom?

Last edited: Dec 21, 2016
16. Dec 22, 2016

Staff: Mentor

I wouldn't call that the cause. They appear in the overall energy sum but they are not relevant at the same place and time during the process.
Minus the ionization energy.

17. Dec 22, 2016

Bhope69199

Is this thinking right:

As the neutral atom enters the electric field it becomes ionised from energy imparted by the electric field. The total velocity of the two newly formed ions is less than the velocity of the neutral atom. (The velocity is slowed to a value where the energy is equivalent to the energy required for ionisation) As the two ions leave the electric field their total energy is the sum of their energies with a lower velocity (which is smaller than the initial energy of the atom) plus the ionisation energy which is equal to the initial energy of the atom.

The electric field is imparting energy onto the atom to cause ionization and the ions are imparting energy to the electric field by their velocities changing.

This way no energy is lost from the electric field, there is no reduction in charges and energy is conserved.

18. Dec 22, 2016

TSny

Suppose the atom is a hydrogen atom. There is electric potential energy $U_{\rm atom}$ associated with the electrical attraction between the electron and the proton. However, there is additional potential energy $U_{\rm e \, plates}$ associated with the interaction of the electron with the charges on both parallel plates. Likewise, the proton has a potential energy $U_{\rm p \, plates}$ associated with its interaction with the charges on both parallel plates. Finally, there is a potential energy $U_{\rm plates}$ associated with the interaction of different elements of charge of the plates with each other. So, the total electric potential energy of the system is $U_{\rm total} = U_{\rm atom} + U_{\rm e \, plates} + U_{\rm p \, plates} +U_{\rm plates}$.

As the electron and proton are pulled away from each other, $U_{\rm atom}$ increases while $U_{\rm e \, plates}$ and $U_{\rm p \, plates}$ decrease. $U_{\rm plates}$ remains constant. The energy required to pull the electron and proton away from each other comes from the decrease in $U_{\rm e \, plates}$ and $U_{\rm p \, plates}$.

The forward speed of the electron and proton does not decrease. These particles do not feel any force component parallel to the initial velocity of the atom. (We assume the ideal case where the E field of the plates is uniform and perpendicular to the plates.)

The total KE of the proton and electron as they leave the plates will be at least as large as the KE of the atom originally. (I am not going to worry about the orbital KE of the electron while it was in the atom. I'm still thinking in terms of the rubber band model of post #7.) If the ionization occurred somewhere before the exit point, then the total KE of the proton and electron will be greater than the original KE of the atom due to the fact that the particles will pick up components of velocity perpendicular to the plates. This increase in KE of the particles compared to the initial atom is due to additional decreases in $U_{\rm e \, plates}$ and $U_{\rm p \, plates}$. But, again, $U_{\rm plates}$ does not change. Overall, the energy required to ionize the atom and to also increase the speeds of the electron and proton comes from the decrease in potential energy of interaction of the electron and the proton with the charge on the two plates.

The total potential energy of the system can be expressed in the form $U_{\rm total} = \frac{1}{2} \varepsilon_0 \int {E^2 dV}$ where $E$ is the magnitude of the net electric field due to all the charges in the system and the integration is over all of space. So, the potential energy can be thought of as "stored" in the total electric field of the system.

$\vec{E} = \vec{E}_{\rm electron} + \vec{E}_{\rm proton} + \vec{E}_{\rm plates}$. So,

$E^2 = E_{\rm electron}^2 + E_{\rm proton}^2 + E_{\rm plates}^2 + 2\left( \vec{E}_{\rm electron} \cdot \vec{E}_{\rm proton} + \vec{E}_{\rm electron} \cdot \vec{E}_{\rm plates} + \vec{E}_{\rm proton} \cdot \vec{E}_{\rm plates} \right)$

The last three terms are the terms corresponding to the energy associated with the interaction of the electron and proton with each other and with the plates. The volume integrals of these terms change as the atom is ionized and the separated particles move away from each other. The volume integral of each of the first three terms remains constant. In particular, $\frac{1}{2} \varepsilon_0 \int {E_{\rm plates}^2 dV}$ represents the energy of the electric field of the plates. This energy does not change during the process of ionization.

Last edited: Dec 23, 2016