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Capacitor stored charge

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ZakWolf

Thu Mar 31 2011, 03:46AM

Registered Member #3114 Joined: Sat Aug 14 2010, 08:33AM
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Posts: 608

Oh! OK so the glass doesn't hold the charge its the surface air around the jar that holds the charge and is redistributed to the metal when its put back together.

Mattski

Thu Mar 31 2011, 05:44AM

Registered Member #1792 Joined: Fri Oct 31 2008, 08:12PM
Location: University of California
Posts: 527

wrote ...
When not properly explained, this demonstration promotes the myth that capacitors store their charge inside their dielectric. This erroneous theory, due to Franklin, was taught throughout the 1800s, and is still sometimes encountered. However this phenomenon is a special effect caused by the high voltage on the Leyden jar.[7] In the dissectible Leyden jar, charge is transferred to the surface of the glass cup by corona discharge when the jar is disassembled; this is the source of the residual charge after the jar is reassembled. Handling the cup while disassembled does not provide enough contact to remove all the surface charge. Soda glass is hygroscopic and forms a partially conductive coating on its surface, which holds the charge.[7] Addenbrook (1922) found that in a dissectible jar made of paraffin, or glass baked to remove moisture, the charge remained on the metal plates.[8] Zeleny (1944) confirmed these results and observed the corona charge transfer.[9] In capacitors generally, the charge is not stored in the dielectric, but on the inside surfaces of the plates, as can be seen from the fact that capacitors can function with a vacuum between their plates.[10]

Apparently this effect is due to the hygroscopic nature of the glass forming a conductive layer of water. In an air/vacuum capacitor the charge is definitely stored only on the plates, until voltages get extremely high that is and you have air breakdown or vacuum emission :)

When you add the dielectric all that you want it to do is allow you to store more charge at the same voltage by polarizing the dielectric's molecules which cancels out and reduces local field intensity from what it would otherwise be. But not all dielectrics are perfect and small (or large) amounts of charge trapping can occur even without the conductive layer of water.

Myke

Thu Mar 31 2011, 05:49AM

Registered Member #540 Joined: Mon Feb 19 2007, 07:49PM
Location: MIT
Posts: 969

Aren't charges in a dielectric stored via the alignment of the polar molecules in the structure? Charges are stored on the plates... just not much in comparison to the capacitor with a dielectric.
Think about walking across a carpet on a dry day. Your body, a conductor, is one of the plates that is stripping electrons from the carpet. If you stand still, you still hold the charge (for a short time due to losses to particles in the air around you such as water, dust, and others). Touching a neutral, insulated sphere transfers the build up electrons from you into the sphere until both you and the sphere are at equal potentials in relation to the ground.
Charges are stored on the plates in a vacuum. There are many physics books that talk about this subject...

EDIT: Ninjaed... Look at how the electric field ties in with capacitance in parallel plate capacitors.

Dr. Slack

Thu Mar 31 2011, 10:22AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Vaccuum is a dielectric, with er=1.

Where you have a capacitor with a physical diecltric, the polarisation of the material allows more charge to be stored, so er>1.

As to *what* is storing stuff, and what *is* a vaccuum, then you're into very hairy stuff. The purest description seems to be photons and quantum mechanics, which we know to be incomplete anyway, perhaps there's strings underneath that? But my brain hurts when I think around this so I just leave the worry to the theorists, and use the energy storage to make bright sparks and loud bangs.

Steve Conner

Thu Mar 31 2011, 10:42AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

I would say that 1 worth of the charge is stored on the plates, and the remaining (er-1) worth in the body of the dielectric.

So in a vacuum capacitor, because the dielectric can't be polarised (has er=1) then er-1 = 0, and the charge is all stored on the plates.

Ash Small

Thu Mar 31 2011, 12:19PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

I'm going to confuse things here. I agree with the above posts that say vacuum caps have a dielectric constsnt of 1, and other dielectrics have a higher er, but what about the 'field'?

A charged plate will have an associated field, will it not?

If the charge in a vacuum cap is stored in the field around the plate, and the polar nature of dielectrics allow more charge to be stored in the field (at a given voltage), this would explain the phenomenon quite simply.

This also seems to fit in with the fact that the charge on a cathode consists of an 'electron cloud' surrounding the cathode, the 'extra electrons' are not within the lattice structure of the cathode. (this applies to cathode in an electrolyte as well as cathodes in vacuum tubes, etc.)

The above is an oversimplification, but I'm sure you get the gist of what I'm on about.

(EDIT: We know energy is stored in fields associated with inductors, so why should it be any different with capacitors?)

Dr. Slack

Thu Mar 31 2011, 12:25PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

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.

All of the energy is stored in the space between them.

The dielectric stores the extra energy, the er-1's worth, and in doing so, allows the plates to store more charge at the same voltage, in a way that's ameanable to high school physics.

As the dielectric polarises, its shifting but bound charges reduce the electric field between the plates, allowing more electrons onto the -ve plate for any given potential differnce. The shift of the bound charges against the spring force that holds them bound stores 0.5*kx^2 of energy, where k is trhe effective srping consdtant and x is how much they move. If they aren't bound, then you have a conductor, and not a capacitor.

In the absence of a dielectric, what stores the er=1's worth of energy? That's the bit that needs QM and photons, an ice-pack for the forehead and a PhD, or a refusal to worry about it.

Ash Small

Thu Mar 31 2011, 12:50PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

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'.

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'.

Steve Conner

Thu Mar 31 2011, 12:57PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

By that same argument, you can show that electric power does not flow in wires but in the spaces around and between them: Poynting vectors and so on.

If you draw a control volume inside the capacitor, including the space between the plates but not the plates themselves, can you visualize how much Poynting vector enters this volume as the capacitor is charged?

Equivalent problem: does displacement current have an H-field? Maxwell's equations say yes. Therefore E x H can be non-zero over the surface of the above control volume during charge and discharge.

Therefore, since Poynting vector enters the control volume during charge, and leaves it during discharge, the energy must be stored in there somehow. But how? This is basically the same question that motivated classical physicists to search for an "aether"- an invisible elastic medium in which electromagnetic energy could be stored as some kind of stress.

Pinky's Brain

Thu Mar 31 2011, 03:23PM

Registered Member #2901 Joined: Thu Jun 03 2010, 01:25PM
Location:
Posts: 837

Coronafix wrote ...

If a capacitors charge is stored in the dielectric, how does this work with a vacuum?
What "holds" the charge?

The electrons and holes are at the outside of the conductors, not in the dielectric.

The dielectric can be said to store the energy, but not the charge.

PS. it might be schoolboy physics ... but advanced physics doesn't replace that, it just teaches you when schoolboy physics stops being a good approximation. I don't think that applies here.

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