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I have been scratching my brain for a long time to figure out where the energy comes from when charges on two plates are switched on and there is already a charged particle between the plates?

I’ve only seen examples where the charge was placed there after the field is present. I’m also trying to figure out how the charges goes from plus to minus, given that the circuit is connected to a battery source?

We have two charged plates with a very electric force permeable wall between the plates and a positively charged particle. As the plates are switched on, the particle will slam in to the wall M and create thermal energy. The question is where did the energy come from? And where is the charge decrease?. I put a wall there to ask how the discharge occurs? Without the particle touching the other plate. In my mind, just a change of electric field must also somehow discharge the plates? Discharge, is how the battery goes to a lower energy state. Otherwise energy is created from thin air enter image description here

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Such an arrangement of two parallel plates is known as a capacitor. The energy of a capacitor arises from the separation of charges on the two plates. The charges on the two surfaces exert an electrostatic field between the plates of magnitude $\frac {q}{\epsilon_0}$. The charge on the two plates depends on the capacitance which is given by $C=\frac {A{\epsilon_0}}{d}$ where $A$ is the area of the plates and $d$ is the distance between them. Discharge cannot occur unless there is a medium for the charges to go from from the negative plate to the positive plate.

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  • $\begingroup$ I think this doesn’t quite answer my question. But also if the charge is not discharged somehow we can practically create energy from thin air? A practical same question, where does the energy comes from if you switch on and off a capacitor multiple times with a certain frequency? If it never discharges we can switch on and off forever? $\endgroup$
    – einstein
    Dec 31, 2019 at 15:21
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    $\begingroup$ Energy never appears from thin air, the energy of the capacitor comes from the power source. When you disconnect the source, the energy stays in the capacitor. $\endgroup$
    – Sam
    Dec 31, 2019 at 15:38
  • $\begingroup$ If you disconnect the capacitor the energy gained by the particle is lost from the electric field of the capacitor (that is also the source of the tiny amount of energy radiated away). $\endgroup$ Dec 31, 2019 at 18:34

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