Would you please tell me how the charge would spread itself on the capacitor plates (asuming it's a parallel plate capacitor) and the wire? Are there any electrons accumulate at the terminal labeled V1? (please see the diagram attached below)

If the capacitor is NOT charged, why would the voltage at node V1 have a value of 12V? (node V1 is open-circuited)

Thank you!

 

What are your thoughts about these? Post those and then we can discuss.

 

What are your thoughts about these? Post those and then we can discuss.

I think that at the instant the battery's positive side is connected to one side of the capacitor, leaving the other side of the capacitor non-connected, the positive charge accumulated on the positive side of the battery will attract electrons that were originally on the capacitor plate, causing that plate charged positively and the other plate charged negatively, and the wire labeled V1 has positive charge, so this distribution can explain why node V1 has a voltage 12V.

But someone told me that because the circuit path is not completed, there is no current, and the capacitor is not charged.

So I'm comfused, why is that there is a voltage of 12v at node V1?

ps, Would you please tell me how to place a circuit diagram directly on my post?

 

I think that at the instant the battery's positive side is connected to one side of the capacitor, leaving the other side of the capacitor non-connected, the positive charge accumulated on the positive side of the battery will attract electrons that were originally on the capacitor plate, causing that plate charged positively and the other plate charged negatively, and the wire labeled V1 has positive charge, so this distribution can explain why node V1 has a voltage 12V.

But someone told me that because the circuit path is not completed, there is no current, and the capacitor is not charged.

So I'm comfused, why is that there is a voltage of 12v at node V1?

ps, Would you please tell me how to place a circuit diagram directly on my post?

It's important to keep in mind that voltage has to do with electric fields and potential energy. So the battery produces a voltage of 12V relative to the bottom wire. Now, any electron that can move from it's present location to the positive terminal of the battery without having to be pushed or held back is at this same voltage. This applies to all of the electrons on the wire that is connected to the positive terminal (even as it is connected), so the plate of the capacitor that is connected to that wire is also at this same voltage and no electrons have to move. If the capacitor is uncharged then there is no electric field between the plates and hence an electron could be moved from one plate to the other without gaining or losing any potential energy. Therefore, the other plate is at the same potential and, again, without any charges moving anywhere.

 

WBahn, Thank you very much for your detailed explanation.

There are still a few points I don't understand.

First of all, if a chemical battery is a device that seperates charge, causing positive ions stay on its positive terminal and negative ions on its negative terminal, thus creating electric field around the battery. Is this concept correct?

This applies to all of the electrons on the wire that is connected to the positive terminal (even as it is connected), so the plate of the capacitor that is connected to that wire is also at this same voltage and no electrons have to move.

If now we connect a piece of wire to the positive terminal of the battery, since there's an electric field near that terminal built up by the positive and negative ion piles, why don't the electrons on the wire be attracted and move, at the very beginning the wire is connected to the positive terminal of the battery?

 

The actual charge separation in the battery is not very much. The chemisty only occurs until enough charge is built up to stop the chemistry. As charge is drawn off, the voltage drops and the chemisty starts up again until the voltage is re-established. Yes, there is an electric field around the battery, but when you bring the wire to the terminal the electric field is rearranged so that now the equipotential surface that includes the battery terminal extends around the entire wire as well.

Now, at a very, very fine level of detail, the wire has capacitance to the negative terminal (as well as everything else in the universe) and so there is a very slight charge flow in order to charge this tiny capacitance. Perhaps that is the key to getting your mind wrapped around this.

 

Thank you, WBahn.

After reading your explanation and some other materials over, I would like you to check if I've got the correct concept. (But English is not my native language, I'll try to express it clearly.)

1. When we connect a capacitor and a battery as in the figure attached below, an approximate surface-charge is then created as shown. On the bottom wire and the negative electrode of the chemical battery, there are some electrons coming from the chemical reaction in the battery, but not many. An equal amount of positive charge lies mostly on the positive electrode and partly on the surface of the wire and the left plate, making the right plate and the wire connected to it have a charge separation as shown in the figure.

2. The whole charge distribution doesn't creat electric field in the wire or between the plates.

3. Once the circuit path is completed with, for example, a light bulb, the surface-charge will be forced to make a rapid rearrangement and each electron at the instant when the circuit is completed can do an amount of work that is equal to the amount of work the battery gave every electron.

4. The electrodes of a chemical battery don't have IONs accumulated. The action of pushing and attracting electrons occurs only when the chemical reaction begins.

 

After reading your explanation and some other materials over, I would like you to check if I've got the correct concept.

I was thinking that maybe I was saying something wrong, inappropriately, that might lead to misunderstanding. Sorry! Actually I was trying to say "after I have read your explanation and some other materials over...".