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4hv.org :: Forums :: General Science and Electronics
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Capacitor stored charge

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Mattski
Thu Mar 31 2011, 03:42PM
Mattski Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527
Ash Small wrote ...

Dr. Slack wrote ...

All of the charge is stored on the plates, that's where the eclectrons are, they don't jump off the plates into the dielectric. Each electron carries a charge of 1.6e-whatever (19?) Coulombs, and you can (in theory) count them on and off the plates as the capacitor charges.

.

With all due respect (and I'm more than happy to be proved wrong), how is this measured? If you are simply extracting electrons from the system you can't be certain if they are 'on the surface' or 'in a cloud'.
Ultimately with all science we have come up with "models" which describe the world, and these models are compelling because they agree with a vast body of experimental evidence. In the case of capacitors, the Leyden jars are a special case where the dielectric is not an insulator as it has a conductive layer of water, allowing charge to build up on the surface of the glass. If you were to take any number of other dielectrics (as mentioned in my Wikipedia link) and repeat the disassembly experiment you would find that the findings are consistent with charge stored mainly on the plates instead of in the dielectric. This is also consistent with the idea that a good dielectric must be an insulator. Dielectrics which are poor insulators have more high frequency loss (although it's not the only source of that loss).

wrote ...
Take a 'cold cathode', for example (basically a capacitor plate). Elactrons are 'vibrating' on/near the plate. If they gain sufficient energy they 'escape'. A similar mechanism occurs in a liquid electrolyte in 'wet chemistry'.

I would argue that 'electrons are stored on the plate' is an oversimplification, as is a lot of 'schoolboy physics'.
In a cold cathode the voltage is high enough to cause field emission, consistent with the work function of the material as measured by the photoelectric effect. A cathode can also be heated above room temperature to give each electron more energy, thus reduce the work function. You could design a capacitor to do this: it's called breakdown, and you would soon find your capacitor is not working smile

So there may be a tiny bit of simplification because some dielectrics are known to trap some charges, but for the most part the classical view of the capacitor describes them extremely well.
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Ash Small
Thu Mar 31 2011, 04:27PM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
How does this then explain 'corona discharge' from a 'corner' of a charged plate?

EDIT: I've a few other points to 'bring up', but I'd prefer to cover them one at a time for the sake of simplicity.
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big5824
Thu Mar 31 2011, 05:03PM
big5824 Registered Member #1687 Joined: Tue Sept 09 2008, 08:47PM
Location: UK, Darlington
Posts: 240
Corona is due to the increased E field on corners and points caused by the charge on the plate. This E field is then able to ionise the air surrounding it, creating the corona.
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Ash Small
Thu Mar 31 2011, 06:08PM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
big5824 wrote ...

Corona is due to the increased E field on corners and points caused by the charge on the plate. This E field is then able to ionise the air surrounding it, creating the corona.

If the E field was simply ionising the air it would produce a plasma with an 'overall' neutral charge. Corona discharge involves the emission of electrons.

I'd argue that the 'electron cloud' surrounding the plate extends further from the plate at the 'corners', therefore these electrons are the first to aquire (preferentially aquire) sufficient energy to escape. (as in a cold cathode or thermionic emission as described above)
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Steve Conner
Thu Mar 31 2011, 06:24PM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
I'd argue that this is true but completely irrelevant, hence off topic.

Reason: Capacitance is a linear effect that "works" the same at all voltages. Corona only appears above some threshold voltage. Electrons in the conductor are sufficient to cause capacitance, but corona only happens when the E field gets strong enough to eject them from the conductor into the air. (or rip them off nearby oxygen and nitrogen atoms which is much easier: true field emission from metals requires a much stronger field, so can only happen in a vacuum)

However, corona adds capacitance as any Tesla coil builder soon learns. The cloud of plasma acts as an extension of the conductor, increasing its surface area. It is a very lossy capacitance, as the ions and electrons are always recombining and releasing the energy as light and heat.
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Ash Small
Thu Mar 31 2011, 07:01PM
Ash Small Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Steve McConner wrote ...

I'd argue that this is true but completely irrelevant, hence off topic.

I'll accept your point and will not pursue this line further, after this post.

Steve McConner wrote ...



Reason: Capacitance is a linear effect that "works" the same at all voltages. Corona only appears above some threshold voltage. Electrons in the conductor are sufficient to cause capacitance, but corona only happens when the E field gets strong enough to eject them from the conductor into the air. (or rip them off nearby oxygen and nitrogen atoms which is much easier: true field emission from metals requires a much stronger field, so can only happen in a vacuum)

However, corona adds capacitance as any Tesla coil builder soon learns. The cloud of plasma acts as an extension of the conductor, increasing its surface area. It is a very lossy capacitance, as the ions and electrons are always recombining and releasing the energy as light and heat.

I agree about your point regarding 'threshold voltage'. I'd argue that when the threshold voltage is reached in a capacitor, the dielectric breaks down and the capacitor fails. I'd also argue that the dielectric 'allows' the electron cloud to be greater/more dense (not sure which) before breakdown occurs (at which point it becomes conductive/fails to be a uniform dielectric), and that the 'lossy corona' of a tesla coil demonstrates this basic principle. (a cold cathode is just a capacitor plate in a vacuum that has reached 'breakdown potential' and becomes 'lossy'.)

I won't go into comparisons with 'wet chemistry' in this thread as it would be 'off topic')

EDIT: I'm not saying "I'm correct and others are wrong", I'm just suggesting an alternative explanation which seems to have 'some' plausibility.

EDIT: I apologise to the OP if my comments are too far 'off topic'.
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Steve Conner
Thu Mar 31 2011, 07:31PM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Well, I think the wet chemistry angle is interesting. I read somewhere that a car battery has an electrostatic capacitance of several farads, over and above its electrochemical capacity. Can't remember the source, and I have no idea how you tell the two capacities apart.

In the light of Dr. Slack's earlier post, I think the best explanation is that a polar dielectric builds up an internal charge that mirrors the charge on the capacitor plates, partly cancelling it and allowing more charge for a given P.D. between the plates. This polarisation of the dielectric is not caused by a bunch of loose electrons all rushing to one side: if that were the case, it would be a conductor. It's caused by molecules with electrical asymmetry, that rotate under the influence of the E field so that their positive ends face the negative electrode and so on. The more asymmetrical they ar and the easier they rotate, the greater the permittivity of the dielectric. But if they ionise and move around, it's not much of a dielectric any more.

Water is an extreme example of this. Being very polar it has a dielectric constant of about 80. But it is only useful for a few nanoseconds before it breaks down and conducts. It is used as a dielectric in ultra-low impedance transmission lines for Z-pinches.

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Bored Chemist
Thu Mar 31 2011, 08:06PM
Bored Chemist Registered Member #193 Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
I read somewhere that a car battery has an electrostatic capacitance of several farads, over and above its electrochemical capacity.
I doubt that, but I may be wrong. There's no dielectric.

"Can't remember the source, and I have no idea how you tell the two capacities apart."
Frequency dependence of the impedance.


". It's caused by molecules with electrical asymmetry, that rotate under the influence of the E field"

Spherically symmetrical molecules like methane and carbon tetrachloride have dielectric constants > 1.

Solids, in which the molecules cannot rotate, have dielectric constants>1

The electron clouds round the molecules are distorted by the applied field. When they "spring" back that energy can be recovered. That's why the capacitance is bigger.
To get back to the original question.

Since capacitors with a vacuum dielectric work perfectly well, the charge cannot (always) be stored in the dielectric- it simply isn't there.

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Steve Conner
Thu Mar 31 2011, 08:26PM
Steve Conner Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
BC, yes, that is the whole topic of this thread. Where is the energy stored in a vacuum capacitor? I think we agree the charge is on the plates. A 19th century physicist would have said the energy was stored in distortion of the aether or something, but that explanation went out in favour of all the virtual photon stuff.

I forgot about the non-polar dielectrics. Polythene, polypropylene, carbon tet, oil and so on, the polarisation comes from distortion of the electron clouds. But these never have a really impressive dielectric constant. They are however almost lossless. Polar dielectrics on the other hand, like water, Mylar, aluminium oxide in electrolytic caps, can have a whopping dielectric constant, but are much lossier at high frequencies. I understood that the polarisation of these comes from some more gross distortion of an asymmetrical molecule.
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klugesmith
Thu Mar 31 2011, 10:35PM
klugesmith Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
I agree with what Dr. Slack said: the charge in capacitors is stored on the plates.
Here is another view of the same thing.

Consider a parallel-plate capacitor in vacuum, with net charge of +/- Q in its plates.
The charge produces a certain electric field in the space between the plates.
We can integrate the E-field intensity to get the voltage difference between the plates,
then figure C = Q/V.

(If we move the plates closer by a factor of 2 and don't change Q,
the electric field strength will be the same but voltage will be half, so C will be double.
But that's not what this thread is about.)

Now instead of moving the plates, let's insert a slab of solid dielectric material.
To keep things simple, we won't let it actually touch either plate.
The plates' charge is unchanged, as is the E-field in the space between plates and dielectric.
No charge has been transferred to or from the dielectric material.
But within that material, the E-field is partly cancelled by the polarization of molecules.
So plate-to-plate voltage is reduced, i.e. capacitance is increased.

Steve: [edit] Would be interesting to calculate the saturation polarization of water, if the molecular dipoles were all aligned in the same direction.
I bet that with the strongest E-fields used in water-dielectric HV capacitors, the molecular orientations are still dominated by thermal randomness, Van der Waals forces, etc., with a slight bias toward aligning with the E-field.

BC: Materials don't have to be insulators to exhibit dielectric polarization.

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