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There are examples that describe electrical potential as two oppositely charged plates facing each other to make an electric field. They place a positive test charge close the positive charged plate which would lead for the test charge to have gained electric potential. It is never however concluded, what would happen after the positive test charge is dropped to let it collide with the negative plate? Would the kinetic energy, the current, (caused by the attraction to the negative plate) be transformed into heat or would there be an electric discharge?

My confusion derives from wanting to connect a similar scenario (a battery in a circuit connected with a light bulb) to the waterfall analogy where the kinetic energy can be used to produce electricity.

Forgive my misunderstanding please.

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  • $\begingroup$ If you want an example where a kind of electric discharge happens (although that's not the usual term we use for it), look at photomultiplier tubes. $\endgroup$
    – The Photon
    Jul 25, 2022 at 20:46

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Yes, just like any collision there would be a transfer of momentum and energy - each separately conserved. The kinetic energy of the positive charge would eventually become heat in the molecules of the plate - no different than any other object of the same size hitting the plate. It would effectively neutralize one of the free negative charges in the plate

That's more or less what's going on inside a light bulb - kinetic energy of electrons being converted to heat and light. An important distinction is that in addition to "molecular heat" (which travels by conduction & convection) you have "radiative heat" (no different from visible light except in wavelength), because when charged particles decelerate in collisions they emit radiation.

Light bulbs also benefit from "thermionic emission", which is when a material gets hot enough that the average energy of bound electrons becomes high enough to overcome that binding energy and become free electrons (leading to more collisions/radiation/etc)

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