why can the capacitor be charged by a battery (DC)

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You will never see it on the finish line if it is a capacitor as an obstacle. There is an insulator that prevents it from doing so.
We both agree.
Haven't you ever heard of the term "free electrons" in metal?
I've heard of them. Never seen them, though.
Are you trying to twist what I said?
It's not my intent, but it seems we have a different view on what "passing" means. Passing is transmitting.
No, that concept is completely valid. The current referred to here is the charging and discharging current.
Yes, current. And since the current coming from one side is reflected integrally on the other side, I call it "passing". And so did my tutors.
No, you obviously misunderstood something I said, which is not impossible because English is not my native language.
Neither mine.
Does this current flow through capacitor capsule, from one to other side?
Of course, since it is on both sides of the capsule (in absolute value).
 
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The heat is mostly generated by the current flowing thru metal plates. The plates are in practice made of a thin material wound into rollers or such to obtain a large effective surface of the plate. This material is heated by charging and discharging the capacitor.

I do not mean to drag this on but I think it would be good to clarify things from learning point of view.

According to your argument current does not flow through the capacitor. So, surely no current flows through the plates.

The heat is generated by the ESR, hence current flow through it, not the charge/discharge action. We al know the power loss in capacitors. If there is power loss then there is current flow.

Please see attached. This is from my own library. Note underscored bits in the second page. Here he is not talking about shunt capacitor, hence what Abbey has been trying to say.

I am also in line with Matador's input. Maxwell rocks.

I completely understand where you are coming from but you have to revise your view on this. Perhaps think outside solid conductor.
 

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I do not mean to drag this on but I think it would be good to clarify things from learning point of view.
No problem.
According to your argument current does not flow through the capacitor. So, surely no current flows through the plates.

The heat is generated by the ESR, hence current flow through it, not the charge/discharge action. We al know the power loss in capacitors. If there is power loss then there is current flow.
You should read about dissipation loss or factor which in simple terms speaks of losses and heating within dielectrics but not because of current flow but high frequencies. It doesn't happen in vacuum capacitors used in microwave spectrum.

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I am also in line with Matador's input. Maxwell rocks.
Then you should watch together with Matador the video he inserted, this part in particular.

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Perhaps we could make an interesting experience.
The circuit on the right is the one where, according to our esteemed member, current doesn't pass through the capacitor. However, practice would show that some current passes in the circuit.
The circuit on the left, where the capacitor has been removed, has no current circulating.
Doesn't that mean that current in the right circuit circulates via the capacitor, or by the grace of the capacitor, or thanks to the presence of the capacitor, if the word "through" is unacceptable.
I believe my take is better suited to the original question, at least on a macroscopic scale.
 

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You should read about dissipation loss or factor which in simple terms speaks of losses and heating within dielectrics but not because of current flow but high frequencies. It doesn't happen in vacuum capacitors used in microwave spectrum.

With huge respect again.

Yes I do and I did. That's why I posted this article but it seems that, you are not only ignoring what the guy says in the first page (power losses on ESR), but you also do not seem to make up your mind.

You stated that the heating was caused by current flow through the metal plates, but now you are saying high frequency. High frequency of what? Without a current flow what is the meaning of high frequency?

I have read many application notes in the past and also some since last night, and they all amplify "ripple current through ESR, ripple current through ESR " yet you still maintain that the current does not flow through a capacitor. I am not questioning your professional integrity here, but what makes your claim to be true than so many other authors (engineers) who are saying otherwise and you still failed to answer that.

My way of verifying a statement is to check to make sure that it reconciles with other electrical principals.

Is there a power loss, and is dissipated as heat? Yes? in which case there must be a current flow somewhere..

But I'll agree to disagree and bow out.
 
Doesn't that mean that current in the right circuit circulates via the capacitor, or by the grace of the capacitor, or thanks to the presence of the capacitor, if the word "through" is unacceptable.
The charge on the capacitor is determined by the voltage across its plates.
Q=CV
If the voltage changes, so must the charge on the capacitor plates.

Simply put, your capacitor charges and discharges at a frequency of 1kHz. You can say that the signal passes through the capacitor, I have absolutely nothing against it, but you have to be aware that in reality that current does not pass through the capacitor. The current only charges or discharges it.
 
This reminds me of something that happened when I was working at a large brokerage in NYC. Someone sent an email notice about something useless but they accidentally sent it to a large global email group that was just an alias for multiple other large email groups that collectively added up to literally tens of thousands of people. Someone hit "Reply-All" to ask "I don't think I was supposed to get this. Can you please remove me from your list?". That prompted several other intellectually challenged individuals to "Reply-All" to say "Take me off this list too. Thanks.". Of course each email went out to everyone on the list and so each email prompted more people to reply some of whom start to explain that it was just a mistake and that everyone should just stop. But at that point there was no stopping it. The thread lasted days. It was aggravating for most people but others chimed in with jokes. Disciplinary action was required. They literally had to shut down the mail servers and run a script to seek-and-destroy every single message so that there would be nothing left for people to reply to.
 
You stated that the heating was caused by current flow through the metal plates, but now you are saying high frequency. High frequency of what? Without a current flow what is the meaning of high frequency?
Yes.
I have read many application notes in the past and also some since last night, and they all amplify "ripple current through ESR, ripple current through ESR " yet you still maintain that the current does not flow through a capacitor. I am not questioning your professional integrity here, but what makes your claim to be true than so many other authors (engineers) who are saying otherwise and you still failed to answer that.
There are several reasons why a capacitor can be heated. One of the reasons is high charging and discharging currents, which happens for example in power supplies. Another reason is dissipation in the dielectric, especially at high frequencies. The article you posted on the first page talks about it. The mechanism of why this occurs is described in this box.
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Is there a power loss, and is dissipated as heat? Yes? in which case there must be a current flow somewhere..
How the microwave heats the water?
 
Doesn't that mean that current in the right circuit circulates via the capacitor, or by the grace of the capacitor, or thanks to the presence of the capacitor, if the word "through" is unacceptable.
I certainly hope so, because otherwise superposition doesn't work. :ROFLMAO:

If we dispose of the notion of a capacitor, and just consider it a black box with two terminals, what behavior would we see in abbey's diagram if we subjected the input to a step response? We would see a current "through" the black box (as in "entering" one set of terminals, and leaving via the other), and that is what causes the terminal voltage of load resistor to change (because there's nothing else that could!).

This is why current isn't defined as the number of electrons passing a point: it is the net charge that "changes" through that point. It's an electricma flux. Perhaps as abbey hint's "through" is the real fundamental argument point, because electrons don't fly across a capacitor. But Maxwell and Gauss have very easy to understand mechanism (at least mathematically) to define what "through" means, which are the magnetic or electric fields lines that penetrate the surface and then terminate. An ideal capacitor subjected to a perfect step input response is electrically equivalent to a zero ohm wire. As in they are exactly the same thing.

Since current is defined as a changing charge, this is why "displacement current", although not the flow of electrons, is exactly equivalent to it, because it takes a change in charge to establish an electric field. In other words, changing charge can be the result of establishing an electrical field (or a magnetic field), in exactly the same way as movement of electrons. They are exactly the same thing.
 
Okay, we can conclude that you think the current is passing through the capacitor, I don't think it's passing. Well let everyone if they’re interested do a little research and they’ll find out what’s right.
Let's move on.
I think you are both right. Electrons move onto and off the capacitor plates but none flow from one plate to the other (except for any leakeage). However, when an ac waveform is applied, there is what is known as a displacement current that flows through the capacitor. Check out Mr Maxwell's excellent work:

Displacement current - Wikipedia

Cheers

Ian
 
There are several reasons why a capacitor can be heated. One of the reasons is high charging and discharging currents, which happens for example in power supplies. Another reason is dissipation in the dielectric, especially at high frequencies. The article you posted on the first page talks about it. The mechanism of why this occurs is described in this box.

I am getting frustrated by you, but this time I am definitely off. I have better things to spend my time on.

You keep throwing at me the article that I posted. I know all that.

At the same time you keep conveniently avoiding the ESR. In case you miss it.

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I am getting frustrated by you, but this time I am definitely off. I have better things to spend my time on.

I am very sorry about that and I apologize.
You have warned of an ESR that has a value of 0.01 to 2 ohms. In this case, it is mostly the resistance of equivalent metal terminals (plates) inside the capacitor at low frequencies and losses in the dielectric at high frequencies. As I said before, the effective size of the plates inside the actual capacitor is very large, and has a resistance that cannot be ignored. The problem also increases because this effective area changes with frequency in some types of capacitors. This resistance is by no means the resistance to the passage of current from one plate to another.

For example:
"Plates of electrodes are made of high purity, very thin aluminium foil (0.05 to 0.1 mm thickness). To get the maximum capacitance for a given electrode surface area, an electrochemical process called “etching” is used to dissolve metal and increase the surface area of the foil in the form of a dense network of microscopic channels. The increase in surface area is referred to as foil gain and can be increased as much as 100 times for foil being used in low voltage capacitor applications and 20 to 25 times for higher voltage applications."
 
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I see this has dragged on despite some clear explanations.
Briefly, in physics terms current is a net flow of electrons in a circuit.
Just to throw in alternative terminology - it can be a net flow of holes rather than electrons.
 
Briefly, in physics terms current is a net flow of electrons in a circuit.
There is a microscopic definition, which justifies one's position of saying that current doesn't pass through because there's an insulating material.
But on a macroscopic level, which IMO is what most members are concerned with, current passes via a capacitor, so it can be modelled as a dipole with an impedance.
I believe that denying that current passes in, via, through, whatever, a capacitor because the electrons are stopped at one armature (plate) is a somewhat ayatollesque position.
 
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