Challenge Questions

highest dielectric strength..

Just to add to the above answer, if glass is the highest dielectric, then the distance between the capacitor plates is less. Therefore if the distance between them is less, then the resistance is less, giving a larger spark when the glass is removed.

Regards JD.

This the operation of the equation q=1/2 C*V*V

The charge is there. You decrease the capacitance, thus the voltage must increase to balance the equation.

Does the act of removing the glass create a static charge on the surface enabling elelctrons to travel across the charged plate?

Glass is capable of holding a higher static charge than the others. Assuming an equal charge across all of the test pieces, when you slide it out the glass, the friction generates some static charge, which adds to the spark energy.

Because glass has a high dielectric constant, ε, which determines the capacitance. Removing the glass lowers the capacitance. Since the charge is constant, the voltage increases. V=Q/C. This increase is more with glass than with a dielectric with a lower dielectric constant.

There is a difference between dielectric strength: the voltage gradient at which a material breaks down:- From http://www.allaboutcircuits.com/vol_5/chpt_3/6.html

and the permittivity which is related to the dielectric constant: the amount of electrostatic energy which can be stored in a unit of a material:

From http://www.csgnetwork.com/dieconstantstable.html

But this observation doesn't seem to lead to an answer the question as asked.

The question says we partially discharge first so it is unclear how much charge remains on the plates and wether there is a circuit still attached that can drain charge as we remove the dielectric. I think the author probably was thinking of charging to maximum capacity then removing it. The dielectric breakdown strength would then be what it relevant and the highest ones are for Si-O compounds. Fused silica is about 670 kV/mm.

Having NO Electrical training and just throwing in my two cents worth I feel that it has something to do with the ionization of the gap immediately prior to the spark.

Glass has high dielectric constant with very low loss tangent. Which allows higher potential to be sustained between the plates until fully with drawn.

I am not sure what loss tangent is so maybe I am just restating your point here. As you pull the glass out, the high dielectric tends to draw the charges with it creating a larger field where the glass is at the expense the potential in the rest of the gap. Once the fringe fields get too large there is breakdown across the surface of the glass plate that remains in the gap.

Other materials' dielectric constants may be higher than the value for glass.

So they will produce bigger sparks. But the spark produced by glass looks bigger provided it's transparent glass...

The spark is the capacitor discharging across the plates; as the glass is removed the dielectric constant changes from 2000 to 20 the charge stored on the plates has a potential difference great enough to ionize air and provide a path for the current, the capacitor discharges until a potential difference is sustainable by air. The biggest step change between the initial stored charge and the final stored charge is given by the biggest change in dielectric strength, the bigger the step change the bigger the spark. (Was that too many words?)

Apparently, glass simply has a higher dielectric constant than the other materials that were tried, but there is a new polystyrene dielectric that promises to make supercapacitors a realistic replacement for Lithium ion cells.

The potential that the capacitor is charged to should matter only to the extent that the potential is above the breakdown potential for the remaining dielectric (in this case I assume that you are replacing the glass with air.

The question is unclear. Are we discharging the plates or the dielectric? Am I to assume that if I pull the glass out of a charged capacitor, that the glass will have no charge when it comes out?

Perhaps my answer is sophomoric, but I always "assumed" the charge was primarily stored on the plates of the capacitor, and the role of the dielectric was to allow more charge to be stored at a given voltage. Removing the dielectric (glass), and "allowing" air or a vacuum (depending on the experimental design) to be in place as the glass is removed, will force the charge to exist at a higher potential, hence the spark.

When the glass is removed, the index of refraction between the plates suddenly changes.

Here's where the guess-work starts... Sudden refractive index change causes a sudden increase in the speed of light (a photon) between the plates, which in turn creates a path for the charged particles to follow that has far less resistance than the discharge circuit???

I wonder if the question might have been poorly worded. Glass has such a high breakdown strength that, in a partial discharge, you end up lowering the voltage to a value where there is not much leakage at all (Old tv tubes would leak down from 20 kV to 3 kV very quickly, but might take months then to lose the 3 kV). Materials with a much lower breakdown strength don't show this problem since they continue to leak.

So, is there a time between the partial discharge and the sparks?

If the charged unit is discharged by a shortcut of the unit then the complete charge is discharged in one time.

If the charged unit is first discharged by waiting for an hour (or something like that) the mark of discharge depends on the dielecticity of the Dielectric media and the remaining charge is the difference between total charge and discharged - so higher the dielectricity so higher is the remaining charge and the spark(plug).

Upon rereading, I think my previous answer might be wrong. Because glass capacitors have such a high dilectric breakdown, they are often charged to a higher voltage and experience a surface charge from corona. This occurs very little or not at all in other dielectrics. But, in this interpretation, the glass must have been charged to a higher voltage. Was it?

Rixter qualitatively answers the question. Here's an analysis...

Consider the capacitor filled with dielectric.
By applying a charge Q to one of the capacitor's plates, the resulting potential difference between the plates will be

V₁ = Qd/(ε₀ε_rA)

using the definition of capacitance C=Q/V and the parallel plate formula C=ε₀ε_rA/d

where d is the plate separation,
ε₀ is the vacuum permittivity
ε_r is the relative permittivity or dielectric constant of the medium between the plates

As specified, the dielectric is removed. Being a physicist I assume this is carried out swiftly, with no free charge removed by the dielectric.

As the charge applied to the plate of the capacitor is a conserved quantity the resulting voltage between the plates is now

V₂ = Qd/(ε₀A)
= V1*ε_r

So the resulting inter-plate voltage has now increased by a factor ε_r.
Glass, typically, has a high relative permittivity compared to other common dielectric materials, so the final voltage difference will be high compared to the original voltage. If the resulting voltage gradient across the capacitor (V2/d) exceeds the dielectric strength of air, you'll get a spark and it's more likely to happen with a high permittivity material such as glass

Look at the question again: "...ceramic, glass,...".

Some ceramics have ε_r one thousand times bigger than glass.

brgds

Snel

You can't pull it out so fast that the electrons don't rearrange on the surface for intermediate cases.

I believe this challenge question is not well enough specified for an accurate answer. Depending on the geometries, a variety of results may occur.

I hope that this in not too tangential, but here is my story:

In a college physics class, the instructor assembled a Leyden Jar capacitor, consisting of an outer metal cup sitting on a grounded plate, a taller glass cup fitting inside of the grounded metal cup, and an inner metal cup with a protruding ball/electrode that extended above the top of the glass cup. He charged the capacitor system by applying multiple big sparks from a static electrical generator to the exposed ball/electrode. He reached up towards the ball/electrode with a grounded shorting rod and produced an exciting spark to show that the Leyden Jar could store a charge. He charged the system again, and then he grabbed the electrode/ball with insulated tongs and slowly removed the inner cup, setting it on an insulated surface. Then he grabbed the glass cup with the insulated tongs, removing it from the outer grounded cup, setting it on the grounded plate. He grabbed the inner cup with the tongs, and started moving the inner cup towards the outer grounded cup, asking the class what was going to happen. We were surprised when the 2 cups touched and there was no spark.

Using the tongs, he reassembled the Leyden Jar, then he reached over with the shorting bar drawing a big loud spark from the ball/electrode, surprising the majority of the class. His timing was excellent . . . the bell rang and he said "Come back tomorrow and tell me what happened."

Our conclusion: The energy was stored in the dielectric. Contrary to this discussion's problem, the geometries were configured so that as the capacitor was being disassembled, the capacitance was dropping, the voltage was rising (Q=CV), but no spark was produced. The glass between the metal plates stored the energy, with the energy density increasing as there was less and less overlap of the metal plates.

Yes. The air gap between electrode and dielectric rapidly breaks down and charge is stored on the surface. Certain geometries like spheres allow one to avoid fringing fields that are not going to drive the breakdown if you are far enough from the maximum field for the configuration. Our configuration of parallel plates creates a lot of fringing so I think the spark travels down the edge of the glass.

You're right, challenge question not well enough specified. Yours is also a nice story with the Leyden jar. But we may assume our challenge question is about a flat construction (or a cylindrical one - but w/o a bottom). In a Leyden jar you cannot take away the dielectric and leave the electrodes.

brgds

Snel

I guess my main question/complaint about the stated problem is it implies that "glass" dielectric makes the biggest spark. I'm not sure if that is even true, or under what conditions it might be true. There are so many different dielectric characteristics, that I can't believe that an unspecified type of glass could cause the most impressive effect.

Later,

Carl (the skeptic)

Now here's an added effect that could also be a factor. Maybe the glass that was used is transparent. The other dielectrics used in this demonstration were opaque. So with the glass dielectric more of the light from the spark was able to reach light detectors (eyeballs) than when light absorbing dielectrics were used.

Hi redfred,

please check my comment (#14).

brgds

Snel

What can I say... Great minds think alike.

Dielectric relaxation...

Clue is in the question, charged, then partially discharged....

Does the glass, being a silicate crystalline structure, exhibit a Piezoelectric affect?

My two cents. I followed all the rest, but it seems like Occam might be hovering in the answer to this one.

Good idea, but that would apply only if the glass was flexed in some way (and since its glass it is brittle so it can't really be flexed) and the question doesn't mention physical stress of that type.

My relaxation comment was to prompt people...if the glass is charged up and discharged (partially) then left for a few seconds it will relax to a higher level of charge than other dielectrics that don't relax.

In other words I am adding a little extra charge (and therefore volts at breakdown) to the glass than the other dielectrics would have stored on them, as it has relaxed back more.

So that is the "brighter" spark

If the capacitor is disconnected and we pull out the dielectric, I think it should be noted that the potential difference between the plates will go up, even though the charge is fixed. The breakdown will occur at the corner when the dielectric meets the plates.

Yes of course it will, but so will any other dielectric being moved out, same work done...so the question remains why glass?

This boils down to:-

1. Either glass has a higher dielectric constant than the others to hold a larger charge -- but its not the largest mentioned in the suggested types in the question
2. Maybe glass has some electrical property we are over looking (surface conductivity higher/lower) which makes a bigger spark (I see none...so this is a stretch to me)
   
3. It holds some special property the other dielectrics don't

• its some form of piezo effect (but no stress was added, and its fragile)
• it has a greater relaxation than the others (one of my suggestions above)
• 'cause its transparent we only see a brighter/bigger spark but really its the same (suggested above too)
• there's still something we haven't understood or noted.

To me the question is loosely worded (I suspect on purpose) so I opt for relaxation...but we could all have missed another property!

I was trying to allude to something about the fringe fields at the corner. I can't quite tell if the breakdown at this corner is somehow important. In the idealized solution (e.g. Jackson problem 4.10 but for parallel plates) the fringing field is exactly cancelled and so there is a huge charge build up just behind the corner of the material.

Breakdown is a very nonuniform process that often involves a tree-like branching of plasma arcs. If it were just V_breakdown that mattered in bulk then I think some polymers might do better than glass. If it is the field at this narrow corner that matters then some microscale breakdown limit might matter more. Glass would be more homogeneous on smaller scales so have fewer effective "defects" (inhomogeneities) to start the cascade.

Isn't the relaxation time for glass under a static field very short (picoseconds)? Such a relaxation time often implies some hysteresis which, for glass, I think would mean irreversible degradation before breakdown. Does this happen?

CCHAF, I was thinking of dielectric relaxation that restores a partially discharged charge - this type of relaxation doesn't dissipate in small time frames (unless the dielectric is conductive and that's not a good dielectric).

Here's what I understand happens;Dielectrics enhance capacitance by their molecules storing stress/strain under the charge storage conditions, because they suffer a force moment from the applied e-field. Normally on discharge they return to their previous positions in the lattice. But some materials have a relaxation of that stress/strain and they actually recharge again after discharge from a pure mechanical effect of their molecule's positions springing back.

I'm not familiar with relaxation of a type that only exists for pico-seconds (and that means I'm ignorant of it not that its not true!)

That to me is a disappointing answer...all dielectrics polarize, that's how they work. If glass is materially different (pun intended) than educate us and tell us how its different...then it would be a good answer.

Sorry to be critical but for me the value of a puzzle like this is the education - I guessed at relaxation as the quality of glass that is different from other "normal" dielectrics but that's not the answer. So I misunderstood the puzzle.

However I'm no further along understanding this answer either...

Also your answer would imply that a conductor placed between the plates would also spark more intensely...a conductor would give 100% polarization as all the charges would gather under the opposite ones on the outer plates since in a conductor the charge has 100% mobility. How does the glass polarize more intensely? It would have to allow the charges to move away from local areas of E-stress (the very action of a dielectric) to the surface and glass is not conductive at this scale...?

I see that glass is unusually large as a dielectric but there are much stronger ones (like water and Barium Titanate). I think they meant for materials commonly used for this purpose. Still, saying other dielectrics polarize negligibly is overstating things. Conjugated polymers and metals give huge values but if they conduct the charge away they won't work. A slab of metal with insulation would be the best. The part I don't get is where the spark occurs. It we pull out the dielectric to the point that we achieve breakdown in the cavity it vacates and the breakdown is there, this is alway the same voltage (up to humidity changes). Is there more current? It seems like I will always generate the same delta V so the sparks would be the same.

I agree that this "answer" is lacking. None of my concerns in the setup have been clarified at all. We don't know if the same voltage, charge or energy have been stored in the bigger spark comparison. We don't even know which type of glass (polarized, safety, Crown, Flint, Leaded, Cobalt blue) is being used or if any of these make a difference at all.