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We have a capacitor which is connected to a voltage source. The top plate of the capacitor is connected to the positive terminal of the source. Therefore, electrons go from the top plate to the positive terminal. What happens to the electrons when they arrive to the voltage source? I think when electrons arrive at the positive terminal the source obtains a net negative charge.

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When electrons leave the top plate it then has a net positive charge which will tract electron to the bottom plate which thus becomes negatively charged.
So there is a movement of electrons from the top plate (which becomes positive) through the voltage source and onto the bottom plate (which becomes negative).

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    $\begingroup$ But it's not the very same electrons that move from the + side to the - side of the cap. It's a shift in charge density profile. $\endgroup$ – Bill N Aug 12 at 12:26
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    $\begingroup$ @BillN I am sorry of my answer gives that impression as it was never my intention to do so. $\endgroup$ – Farcher Aug 12 at 12:38
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The voltage source does work on the charges to put them onto the bottom plate until the amount of charge $Q$ on the bottom plate is given by $Q=CV$, where $C$ is the capacitance and $V$ is the potential difference between the two sides of the battery. The power source doesn't store charges like you propose.

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    $\begingroup$ in voltage source direction of electric field is from top to bottom and electrons must move from bottom to top. $\endgroup$ – user3728644 Aug 12 at 10:43
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    $\begingroup$ I think positive and negative poles of voltage source reduced, but voltage is constant. how? $\endgroup$ – user3728644 Aug 12 at 10:45
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    $\begingroup$ @user3728644 Batteries do work against the usual forces caused by the field. That's the point of batteries. To do work on the charge. There is some process in the battery that does this. What that is depends on the type of battery. $\endgroup$ – Aaron Stevens Aug 12 at 10:46
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    $\begingroup$ Ok. that is mean when electrons arrive to the positive terminal of voltage source, voltage source push them to the negative terminal and negative terminal push them to bottom plate of capacitor? $\endgroup$ – user3728644 Aug 12 at 10:53
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    $\begingroup$ @user3728644 Kind of, yes. Once charges leave the battery then the battery isn't influencing them. The battery just gives the charges energy so that there can be a charge separation between the two plates of the capacitor. $\endgroup$ – Aaron Stevens Aug 12 at 10:55
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I will rephrase your question as I interpret it. Perhaps this is not what you're getting at, but here goes regardless.

Even a small flashlight battery can push a coulomb of charge through a circuit. If you start with an even mix of protons and electrons everywhere (so there is no net accumulation of charge anywhere), and then you remove a coulomb of electrons from the negative terminal and add a coulomb of electrons to the positive terminal (which is what you're doing if you drain the battery through the circuit), now you should have +1 C and -1 C accumulated at the terminals, an inch or two apart. Thus there should be a ginormous force between the terminals, $10^{12}$ newtons or something, far more than the battery could possibly withstand structurally. How can this be?

The answer is that the change in charge at the terminals due to the departure and arrival of electrons is compensated by the movement of ions within the electrolyte. There is a chemical reaction at the anode that donates electrons to the negative terminal and creates positive ions in the electrolyte, and another one at the cathode in which the positive terminal gives electrons to the positive ions, and the flow of ions within the electrolyte keeps charge from building up anywhere. This may be a bit of an oversimplification of the electrochemistry involved, but I think it captures the general concept.

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