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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 atom 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?

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No. Kinetic energy of the products should be less than of the incoming neutral atom. There are always fringe fields that affect the outgoing charges.

In practice, the experiment could only be done with Rydberg atoms, where the binding energy is tiny.

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  • $\begingroup$ Why is the kinetic energy less? If the product particles are attracted to their respective plates would they not get accelerated in that direction, increasing their velocity, hence increasing their kinetic energy? $\endgroup$ – bhope69199 Dec 21 '16 at 16:15
  • $\begingroup$ Yes, between the plates, but that would only be a short time. At the edges and beyond, the fringe fields would decrease the velocity of the charges. $\endgroup$ – Pieter Dec 21 '16 at 17:10
  • $\begingroup$ OK. So passed the plates the they would feel de-acceleration (essentially the fringe fields would be pulling them back to the plates). So the energy that it has gained from the plates to cause ionisation is imparted back to the field by the particles de-accelerating and having a reduction in K.E? Is this the case even if the particles where able to get sufficiently far away from the plates? $\endgroup$ – bhope69199 Dec 21 '16 at 17:19
  • $\begingroup$ Calculations of the work will be difficult with integrals out to infinity (also between the charges of products). My reasoning is just based on consideration of the energies, and then I can only see edge effects as a possibility to make up the balance. $\endgroup$ – Pieter Dec 21 '16 at 17:33
  • $\begingroup$ Thanks. If edge effects didn't make up the balance and the electric field lost energy, you could theoretically remove the energy and still have charges on the plates? Which would make no sense. $\endgroup$ – bhope69199 Dec 21 '16 at 17:56

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