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I have been studying Capacitors for the past year now and the one thing I don't understand is how a charge is stored on the capacitor. Essentially, a circuit with a capacitor is an incomplete circuit right?

So why do the electrons start to gather up on one of the plates. Its like they are tricked into thinking its a complete circuit and get trapped once they realise it isn't! What is this sorcery!

What is happening to make the electrons do that and if it simply is electrons moving and then turning back or getting stuck, would a switch have the same effect? (Assuming the switch is open).

Hopefully someone can finally fill the major gap in my knowledge because everything else I read doesn't explain this.

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    $\begingroup$ Actually a good question, though worded in a strange manner.. I thought it was simple, but one can go quite deep into this. "Sorcery" made me lol.. $\endgroup$
    – Manishearth
    Commented Mar 26, 2012 at 9:16
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    $\begingroup$ elecromotive force. $\endgroup$
    – Jasen
    Commented Jan 21, 2016 at 20:19

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Imagine taking a disc of some conductor and putting it in an electric field (with the plane of the disc at right angles to the field lines). The free electrons in the conductor will move in response to the field, so the face at the positive side gets a net negative charge and the other face gets a net positive charge. You get a transient current through the body of the disk when you turn on the electric field and a transient current in the other direction when you turn it off.

Now split the disk through the middle, i.e. turn it into the two plates of a capacitor, and connect a wire between the two outside faces, then switch on the external electric field again. Once more the disk gets polarised, but this time you get a transient current flowing from one side of the (now split) disk to the other through the wire. You end up with the same charge separation as before, but the electrons flowed through the piece of wire instead of through the body of the disk.

Finally, connect a battery in the middle of the piece of wire. This produces an electric field between the two sides of our split disk (i.e. capacitor), and just like the externally applied field it will polarise the disk and electrons will flow through the wire from one face of the disk to the other.

That's why when you connect a battery to a capacitor you get a transient current as the capacitor polarises.

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  • $\begingroup$ By the way, how does a battery create an electric field similar to the one described in your previous example? $\endgroup$
    – Vinayak Pathak
    Commented Mar 26, 2012 at 11:46
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    $\begingroup$ There is a potential difference between the anode and the cathode of the battery (I assume you're not asking how a battery works). When you connect a conductor of any form to a terminal of the battery the whole of that conductor rises to the same potential of the battery - there can't be a potential difference in the conductor otherwise free elctrons would simply flow until they had removed the potential difference. Hence the two plates of the capacitor are at the same potential as the battery anode and cathode. Therefore there is a potential difference between the plates of the capacitor ... $\endgroup$
    – John Rennie
    Commented Mar 26, 2012 at 11:57
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    $\begingroup$ ... and therefore there is an electric field between the plates of the capacitor. $\endgroup$
    – John Rennie
    Commented Mar 26, 2012 at 11:57
  • $\begingroup$ I GET IT! The split disks are raised to the same Potential difference as the battery, this means an electric field would be produced in the gap. This electric field would attract the electrons to one side and as a result keep the other side positive. $\endgroup$
    – Pejman Poh
    Commented Mar 26, 2012 at 12:15
  • $\begingroup$ @Pejman Poh, what exactly did you get? :) isn't potential difference raised by electrons accumulating on one plate and moving away from another? So what causes them to move, before potential difference is actually raised? $\endgroup$
    – jayarjo
    Commented Nov 24, 2015 at 8:33
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It's perhaps better to look at it the other way around: "why do electrons not enter a single piece of metal (at least not so many)?"

The reason is obviously that the electrons repel each other, so if you try to squeeze in much more than there are protons in the nuclei, you'll have to work against great forces. In the same way, of course, you have to exert a lot of force to remove electrons from a piece of metal.

But if you have two capacitor plates, it looks a bit different: if you push electrons into one of the plates, it still requires some force, but once they're in they also repel the electrons in the other plate. So if you then remove electrons there, it's easier than it would be without the first plate. Once you have removed electrons there, this plate attracts the electrons in the first plate, so it's now again easier to add even more. And so on.

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The capacitor is connected to an outside source of voltage (battery, generator ...), this charges the capacitor until the voltage between the plates is the same as the one applied from outside.

You can see the capacitor as a space where charges can sit.

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  • $\begingroup$ Actually, I think the OP knows that, he's asking why the charges want to approach the capacitor in the first place. Normal charge flow happens because of a train-like effect, with a source and sink of electrons. The OP doesn't see how this works with capacitors, AFAICT. $\endgroup$
    – Manishearth
    Commented Mar 26, 2012 at 9:21
  • $\begingroup$ Yes exactly Manishearth, from birth I was taught that if the circuit is incomplete, charge won't even try to flow. With a capacitor, the circuit is still incomplete and yet charge flows. $\endgroup$
    – Pejman Poh
    Commented Mar 26, 2012 at 9:36
  • $\begingroup$ @PejmanPoh: you were just taught wrong. Electrons flow in an open circuit, they just build up, and stop flowing after a little bit of time. $\endgroup$
    – Ron Maimon
    Commented Mar 27, 2012 at 1:25
  • $\begingroup$ @Ron Maimon, so you mean that if we connect two separate conductors with the lightdiods in them to the battery ends and leave the circuit open - lightdiods will still light up? :) $\endgroup$
    – jayarjo
    Commented Nov 24, 2015 at 8:39

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