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According to this video, and multiple other sources, we can calculate the energy stored by a capacitor using the idea integrating the work done by moving a charge from one plate to the other at instantaneous charge levels $dq$. But I do not understand how we can talk about the work done moving each charge from one plate to the other, when in reality, charges are not jumping from a plate to the next. They have to go through the battery, where they are given a potential difference $\epsilon$ and then move from that higher voltage to the instantaneous lower voltage of the capacitor, $V$. So I see how there's an electric field caused by the battery having a higher voltage than the capacitor, but I don't understand how the voltage difference between the plates has to do with what's happening if considering moving one charge to the other plate.

I'll give you an example of what I'm thinking to maybe clear it up. At the beginning, when both plates are neutral in charge, and a $10V$ battery is connected, (in conventional current), protons will flow through the battery and gain a potential of 10V, so the work done by the battery is $10 * dq$, and that should happen for every charge, as each charge is given a potential of $10V$ as it moves through the battery, no matter where you are in the capacitance charging process. But it is said that for the first charge no work is done to move charge from one plate to the other. But the battery had to do work to move the charge through the battery! So where have I gone wrong in thought?

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Where you have gone wrong is in the model you assume on how the currents are built up at the level of atoms.

To start with one needs a conductor to have a current.

Then one imposes a potential difference which acts on the electrons in the conduction band , so it is the electrons that are mobile in the conductors, not protons.

Then, because there are innumerable atoms in the conductor, the electrons have a drift velocity, no single electron starts from one side of the circuit and ends on the other. So the capacitor faces are charged, by electrons depleted from one plate and electrons appearing on the other, but the real electrons just "push" each other along ,

drift velocity

What the battery does, it is like a chemical electron sink on the + side and an electron source on its - : it attracts the electrons which drift off from capacitor face A and start the drift velocity in the circuit, electrons pushing each other along , through the battery,to face B. Thus a potential difference is established between the two plates.

By energy conservation, measuring the potential between two plates by a test charge is a correct method. One need not go through the long process of motion through the circuit.

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  • $\begingroup$ Thanks for your answer. So then how do electrons start flowing right when you connect them to the battery? They want to flow from low to high voltage but the negative end of the battery is at 0 volts and so are both plates of the capacitor, so what gives them the field to move towards the battery? $\endgroup$ – rb612 Mar 8 '17 at 7:09
  • $\begingroup$ the electrons next to the electrodes go to the battery, the space they leave generates a positive charge attracting the next electron in the row and so on. $\endgroup$ – anna v Mar 8 '17 at 8:41
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But it is said that for the first charge no work is done to move charge from one plate to the other. But the battery had to do work to move the charge through the battery! So where have I gone wrong in thought?

I believe you've gone wrong by thinking that the work done by the battery must be the work done on the capacitor. Once you recognize that this isn't true, it should be clear why no work is done (on the capacitor) 'for the first charge'.

Assuming the capacitor is initially uncharged, there is zero volts across the capacitor. Then there must be some resistance (for example, an actual resistor or the internal resistance of the battery) between the capacitor and the $10\,\mathrm{V}$ battery in your example.

Initially, the entire battery voltage is dropped across this resistance and so, while there is work done by the battery in 'moving that first charge', the work is done on the resistance rather than the capacitor. Put another way, the initial instantaneous power delivered by the battery goes to heat the resistance and not to charge the capacitor.

As the capacitor charges, the voltage across the capacitor increases while the voltage across the resistance decreases. This means that less of the power delivered by the battery goes to heat the resistance and more of the power goes to increase the energy stored in the capacitor.

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