Talk:Displacement current/Archive 3

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Contents 1 Experiment to measure displacement current in a parallel plate capacitor, or a transmission line equivalent 2 Brave challenger 3 Experimental setup 4 Difficulties in measurement of 'displacement current' 5 Results 6 Comments on Mak17f's experimental work 7 Experimental difficulty

Experiment to measure displacement current in a parallel plate capacitor, or a transmission line equivalent

I would be interested to know if any one has any ideas on how to measure displacement current in an air spaced capacitor. My only stipulation is that the procedure must not use the measuremrent of any magnetic field to infer the displacement current.

This is because Im trying to prove that there is no current, not that there is no magnetic field. The presence of magnetic field will not be taken as evidence of displacement current. Suggestions welcomed here!--Light current 02:19, 25 December 2005 (UTC)

Just an additional note about vacuum capacitors. Obviously, any so called 'displacement current' will be equal in magnitude to the conduction current in the leads, so it should be easy to measure one would think! If its not easy to measure or it can only be measured as changing electric filed, magnetic filed, flux density or some other derived quantity, then perhaps disp current is an erroneous concept and real curernt does not actaully pass from one plate to another. But then you get the problem of how the external conduction current gets from one plate to the other!!--Light current 17:42, 12 January 2006 (UTC)

Brave challenger

Can I measure the displacement current? Sure. I’ll put a CVR in series with the capacitor. (I could also be redundant and put a current transformer around the leads) Is there real actual bona fide electrons flowing through the leads to your capacitor? Yes. Is there an electron or ion beam between the plates of your capacitor? No. If you want me to, I’ll send current traces, photographs, and even go the extra step of performing this in an ultra high vacuum. Mak17f 03:39, 7 March 2006 (UTC)

Sorry Im not sure what a CVR is (unless you mean current viewing resistor), but all you'd be measuring is the conduction current in the leads. We all know that exists. I need proof that something flows between the plates.--Light current 08:53, 7 March 2006 (UTC)

Yes, current viewing resistor. Also, yes, in the leads it is conduction current. However, the act of the plates of the capacitor charging and discharging is displacement current. Because the charge flows in our leads does not remove from the fact that the current is do to a time-varying electric field across the capacitor. Charge flowing through the medium between the plates is irrelevant. In that case, it would also be conduction current (as in a lossy capacitor). Mak17f 13:30, 7 March 2006 (UTC)

It is my contention that nothing flows from one plate to the other.(Unless someone can show otherwise by experiment- can you?)--Light current 14:57, 7 March 2006 (UTC)

OK then. Since no one has come up with even a suggestion in over two months on how to measure DC in a vacuum (or air) capacitor, I will offer my suggestion.

It involves placing a very low value CVR or the same shape as the parallel plates between the plates. Each side of the CVR is then taken to a differential amplifier, then a scope/voltmeter etc. Hopefully this will disturb the electric field distribution minimally esp if the capacitor is driven differentially and the CVR is nominally at ground potential. THe few mV or uV across the CVR would prove the existence of DC.

Has anyone the resources to do this expt? It will probably need large area plates to get sufficient capacitance in order to measure the DC (which should be the same as the conduction current in the leads)

Can anyone see any problems with this experimental procedure in terms of accuracy/reliability?.

Unfortunately no one has won the prize!! --Light current 15:31, 7 March 2006 (UTC)

Experimental setup

--Light current 03:59, 8 March 2006 (UTC)

Hey don't give up on me!!! :-) I just saw the page. It is my contention that charge flows from one plate to another (not through the bulk of the capacitor, but through the generator). I have the means to do the experiment. I'll also apply the voltage at a few kV so there is little question as to the current. If you want me to do it in high vacuum rather than air, I'll oblige, however, the answer will be ~ the same. I'll give redundant current measurements if desired (pearson coil and cvr). Any other requirements?Mak17f 15:51, 7 March 2006 (UTC)

Yes I have a few other tips which may make the experiment more valid. However I have to go out now (work). I will reply in approx 6 hrs time.--Light current 17:20, 7 March 2006 (UTC)

Just so we are clear as to the question. I have provided a simplified circuit. I am measuring the current two ways (CVR and pearson current transformer)and am using a high voltage probe to measure the applied voltage. I am stating that when a time varying voltage is applied across the parallel plates, a flow of charge goes from one plate to another. This charge is supplied by the generator and does not flow through the dielectric (air or vacuum in our case). In a 1-D circuit schematic it would be identical to stating the charge flowed from one plate to another. Afterwords, I'll show the calculations as to exactly how much charge was displaced and related this to the geometry. For extra credit, I can put another dielectric in and state how its polarization effected the amount of charge displaced from one plate to the other. Mak17f 17:56, 7 March 2006 (UTC)

I am not familiar with the Pearson coil. Is it similar to a CT?

I have a few comments on your proposed setup:

1. The places you have shown as current measurement points will IMO only record the conduction current thro' the wires. In order to measure the true displacement current in the capacitor, you must put your current probe between the plates of the capacitor. Now, remembering that a magnetic field is produced by a changing (strictly an accelerating) electric field, the current sensor should not be sensitive to magnetic fields otherwise a false result could occur.
2. The signal frequency should be high enough such that any effect in the wires due to movement of charge carriers is eliminated.(a few MHz?)
3. If you can do the expt at high vacuum, this would be good as it would eliminate most effects due to free or bound charges in the air dielectric. What I suggest is that the expt is done before pumping the air out and then when the desired low pressure has been reached. Im not sure of the effect with air as a dielectric altho' I expect the current to be minimal.
4. If you are going to use a few kV, please be careful not to get corona discharge as this would completely foul up the results!(ie radius your sharp edges)
5. The capacitor plates should preferably be rectangular and form an end fed transmission line structure of Z0 =50R or be properly matched to the characteristic impedance of the generator at the test frequency by matching sections.
6. The drive to the capacitor should preferably be differential about ground and the CVR or other current probe (non magnetic) should be coupled to a differential oscilloscope plug in to avoid any possible common mode pickup.
7. Polarisation due to real dielectrics is not required to prove the point, but I suppose it may be interesting to see what a piece of polethylene (say) does in the gap.

What I suggest for the current viewing resistor is a stiffened piece of resistive Teledeltos paper with an output lead glued onto each side by conductive epoxy or silver paint etc. This CVR is then placed midway between the capacitor plates to detect any current flowing. Please respond with any questions, comments on the above notes before proceeding!. Thanks.--88.109.116.24 23:42, 7 March 2006 (UTC)

We are going to get hung up on your first point. The conduction current in the leads is due to displacement current in the capacitor. There will not be any charges at any time in the gap between the plates (unless there are effects such as field emission which is aside from the concept of displacement current). There will however be a charge flow from one plate to the other (through the generator). Consequently, when we are solving 1-D circuit equations, it is perfectly acceptable to say current, in some cases exclusively displacement current, is through the capacitor… even when there is never at any point a charge carrier between the plates. When there is a lossy capacitor, capacitor conduction current is in parallel with the displacement current.Mak17f 00:40, 8 March 2006 (UTC)

Well of course I dont agree. Otherwise the problem is trivial. The conduction current in the leads (if there is any) is certainly not equal to the displacement current (if there is any). The displacement current (from one plate to another) is zero!.

But hey, why are we arguing about it when it can all be proven (or disproven) by the experiment. Te experimantal results will surprise either:

• You, or
• Me, or possibly
• Both of us

What I would like to know is whether you have any serious objections to the form of the experiment and perhaps we could sort these out first. Also, lets not get side tracked by lossy capacitors-- here's where the vacuum part comes into it (no loss that I know of). BTW do you know what Teledeltos paper is and can you get hold of any> --Light current 01:45, 8 March 2006 (UTC)

I just wanted to make sure we defined our hypothesis. Maybe we are too scientific for Wikipedia :-)))) I am stating that the conduction current in the leads equals the displacement current in the capacitor. I’ll go as far as to calculate the capacitance of the capacitor (using approx. geometric equations) and check it with an LCR meter. Then, I’ll relate the current through the capacitor to the charge on the plates. This charge density on the plates will be equal to the magnitude of the displacement current.

As far as your other points, things seem reasonable. I will most likely be in the ~500kHz range for the frequency. However, exact parameters will depend on what I have in the lab. Off hand, I think I should have some appropriate sized (~4x4inch) stainless squares to use as the electrodes. Its not a problem to work in a vacuum, except photographs won’t be as easy as in air. Also, my generator circuit won’t be visible. What I have been calling a generator is store-bought RF amplifier in series with a home-made impedance matching transformer. From earlier, yes a Pearson coil is a current transformer. It will be measuring the current in the leads. If redundancy is insisted upon, I will also use a CVR, but it just makes things more difficult. The voltage probe is a Tektronix p6015a. I want to be as accommodating to your requests for the test as possible, but I hope you understand there may be some limitations with the setup. One of which will be going to much higher frequencies; this is because of the shortening of the wavelength and transmission line effects start to come into play. I’m not sure of the paper he is talking about for the CVR.Mak17f 02:32, 8 March 2006 (UTC)

No its not too scientific. Other experiments have been conducted over WP to try to solve differences of opinion. I think its quite legitimate as long as we dont publish our actual original research on the article page. But were not going to do that. We are just trying to understand what displacement current is! Anyway there dont seem to be many reports of actually measuring DC so you may be able to publish elesewhere if you find any thing interesting!. More later--Light current 02:42, 8 March 2006 (UTC)

Yes the 6015A probe @ 75MHz bw should be OK for monitoring the applied voltage. Im not insisting on redundancy in measuring the current, its just that I think if you measure it in the supply leads you are only going to get the conduction current. Im pasting an earlier discussion to try to give you the idea of what Im looking for.

Just come across another problem quoted by Nigel Cook. When a capacitor is in series with a load, how exactly does the energy get from one plate to the other. Does this need displacement current?. If so, how is this didsplaement current generated? --Light current 20:37, 19 December 2005 (UTC)

The energy is transmitted when the field goes across the places. And, the energy goes through the wires on either side of the plate. Pfalstad 20:43, 19 December 2005 (UTC)

Ahh but there can be no energy flow across (perpendicular to) the plates beacuse the Poynting vector points in a direction parallel to the plates-- no?. --Light current 20:52, 19 December 2005 (UTC)

Right you are! Interesting. Well, then the energy goes down the wires. It doesn't go across the plate. Pfalstad 20:58, 19 December 2005 (UTC)

What, and by passes the capacitors as if they werent there?--Light current 21:17, 19 December 2005 (UTC)

No, I meant it stops at the capacitors as if they were an open circuit. Pfalstad 21:53, 19 December 2005 (UTC)

If thats the case, how does energy get to the load? capacitor | | -----------------| |----------------| | | |--| ac source | | Load (resistor) | | |--| ------------------------------------| --Light current 21:58, 19 December 2005 (UTC) The energy gets to the load by the bottom wire. And to anticipate your next question, if you put another cap at the bottom, then I don't know what happens. There must be some energy flow across the cap... I can't simulate the fields with the resistor present, so I don't know what the poynting vector looks like in that case. Pfalstad 22:12, 19 December 2005 (UTC) Ok, consider the case where there is a capacitor on either side of the load. Straighten everything out so the wire, caps, and resistor are in a line. Consider the time when current is maximum (voltages across each cap = 0). There is a magnetic field around the wire/caps/resistor. The electric field lines lead from the top wire to the bottom wire; near the resistor, they are parallel to the wire. If you work out the poynting vector, you'll find that it's pointing to the resistor. Energy is flowing from the wires into the resistor. It's not flowing across the caps, though; it just goes around them. Weird. Pfalstad 22:24, 19 December 2005 (UTC)

Interesting! Im going to think about that for a while. Needless to say, I dont know the answer.--Light current 22:30, 19 December 2005 (UTC)

Do you mean arrange cct as below: ?

----------------------------| |___ ____ C (Zo is very low) | Energy flow |--| -----> | | | | R Load ac source(Zout=R) | | |--| |___ ____ C (Zo is very low) | ----------------------------| yes. I've drawn the Cs as little transmission lines to help us in thinking about this cct. This arrangement of components would indicate that the energy gets shared between the 3 components initially but we know that the Cs do not charge up and can therefore hold no energy. If they do not hold energy, they must reflect it. Ah ha! Maybe the energy stolen by the capacitors in the first few ns (2 way transit time of TL) is given back to the load or source after reflection. Probably given back to the source actually (assuming matched source and load resistances). (ie the capacitors introduce a small but positive mismatch to the source) Any reaction to my thoughts?--Light current 23:07, 19 December 2005 (UTC)

Well the capacitors definitely charge, and they hold energy when charged. I'm not sure I understand what you are saying. Pfalstad 23:16, 19 December 2005 (UTC)

No. The capacitors do not charge up because you have an ac source! (the voltage on each side of each capacitor is the same--roughly). Charging means holding separated charges on each plate and having a steady voltage difference between the plates (Q=CV)-- we dont have that here. In normal circuit theory, large capacitors act like a low impedance to the ac and so dont drop any voltage. In the em field representation, they dont drop any voltage because their characteristic impedance is so low compared with the resistance of the load. Understand so far?--Light current 23:43, 19 December 2005 (UTC)

There you go again, assuming the high-frequency limit without saying it. Ok, in that case the caps charge very little. I don't think I understand enough about your cap-as-TL model to say whether the energy is reflected or what happens to it. Pfalstad 23:52, 19 December 2005 (UTC)

Sorry. Yes I am assuming the capacitors heve low impedance compared to the resistor. But thats what you expect for coupling capacitors isnt it? OOPs looks tho I didnt mention that. But the argument still holds for any size capacitor. Anyway so far so good. What Im trying to show here is that energy can get to the resistor without the need for displacement current going from one plate to the other in the capacitors. EM energy can flow up and down the transmission line capacitors which are o/c at the far end of course. My propostion is that it all must happen by EM fields.

I don't think anyone disputes that the transmission of energy across the capacitor all happens by EM fields. I certainly don't. I don't understand how the TL representation helps. Of course I mentally picture a TL as a chain of caps and inductors. I suppose you have lots of experience with TL's, so perhaps they are more comfortable for you to work with. User:Pfalstad

Now in one limit, if we make the capacitors smaller and smaller the energy transferred to the resistor will tend to zero. In this case all the energy must be reflected back to the source by the capacitors. In the other limit, where C-->oo, none on the energy is reflected back to source. So the only effect the capacitors have is to decide how much of the incident radiation is reflected! Current does NOT need to pass from one plate to another. Energy enters the resistor sideways. Energy enters the capacitors sideways and is reflected (to a greater or lesser degree). Nigel Cook is WRONG!! HA HA!! (See : Talk:displacement current to see what Nigel said).--Light current 00:19, 20 December 2005 (UTC)

--Light current 03:59, 8 March 2006 (UTC)

I’ll put together some waveforms/pics/schematics in the next couple days or so. Mak17f 04:56, 8 March 2006 (UTC) Excellent. I look forward to seeing your results.--Light current 05:37, 8 March 2006 (UTC)

---

If you could get hold of a thin sheet of graphite and attach a wire to each side using conductive epoxy, this could be used as the current probe between the parallel plates of the capacitor and would show what current if any was passing.--Light current 17:36, 8 March 2006 (UTC)

Difficulties in measurement of 'displacement current'

Seen on the web:(Possible **crank** alert!) A Problem With Ammeters and Measurement of dØ/dt

Note that an ammeter cannot differentiate between displacement current dØ/dt and normal current dq/dt. In the ammeter, the sample dØ/dt will couple to free electrons, producing a normal dq/dt inside the ammeter. The driving of this dq/dt through a precision resistance, e.g., is measured and the instrument is calibrated to show the dq/dt amperes flowing. One of the major needs of free energy researchers is the development of a good current meter that will differentiate between dØ/dt and dq/dt, and measure each one. Short of using a mass spectrometer to differentiate the mass current dm/dt, and comparison of those mass current measurements with an ammeter's measurements of the "current" dq/dt, and calculating the dØ/dt from that, I presently know of no way to precisely and simply measure and separate the two current components. I have been thinking of utilizing a multi-channel sampling meter set arrangement, where one channel uses a dm/dt blocking device such as the Fogal semiconductor in this respect, but have not yet developed the complete concept. Tom Bearden??

--Light current 19:30, 11 January 2006 (UTC)

Results

Rather than write a book about what I did, I’ll answer specific questions as they arise. I’ll give the numbers and some calculations here.

I built the air gap capacitor out of two 24mm x 75mm stainless steel plates. I separated them with a 0.8mm thick polyethylene washer and bolted the plates together with Teflon nuts and threaded rods.

For the experiment, I drove the capacitor with an RF amplifier in series with an impedance matching transformer. The output of the transformer is connected directly across the capacitor. The current transformer is placed around the ground lead to the capacitor; the voltage probe is touching the opposite lead.

Assuming a parallel plate capacitor, the capacitance should be 8.85e-12*1.8e-3/0.8e-3=20 pF. This is confirmed by putting the device in an LCR meter.

I drove the capacitor at 500 kHz. The calculated impedance magnitude is 15.9 kOhm. The current and voltage traces are shown first. The traces come directly from the oscope. Yellow is voltage and blue is current (1V=1A).

Numerically integrating the current waveform gives the total charge transferred. The charge trace is the magnitude of charge on the bottom plate plus the top plate.

First, the impedance of the capacitor is measured. One approximation is to take the peak magnitude of the voltage and divide by the peak magnitude of the current. As shown in the waveforms, my amplitude is not constant over time, so this might introduce some error. Vpeak/Ipeak=1060V/.074A=14.3 kOhms.

At the peak voltage, we can look at the charge on the capacitor plate. The max total charge is 50.87 nC, therefore the charge on one plate is 25.44nC. The charge density on one plate is equal to the average electric flux density between the paltes. Q/A=14.13 microC/m^2. Dividing this by the permittivity of free space gives the average electric field, 1.59 MV/m. Multiplying this by the gap distance gives the expected voltage across the plates, 1.28kV.

Using the parallel plate approximation for the capacitor, the measured charge gives a voltage that is within <20% of the measured voltage across the capacitor. My goal was to show that the conduction current measured in the leads is equal to the displacement current in the capacitor. This was done by showing charge transfer and its relation to the E and D fields. Any additional info I'll be happy to add. Mak17f 19:16, 8 March 2006 (UTC)

Comments on Mak17f's experimental work

Very speedy work well done. I just have a few points to mention:

1. I envisgaed both parallel plates as being fed from the same end, although at 500kHz this may not be important.
2. If you had the system properly matched with the 50R coax from the 50R generator going thro a balun that was matched to the Zo of the // plates, then a distance varying standing wave would have been set up across the plates and the capacitor would act like an open ended TL. THese are the ideal test conditions as you can predict what should be happening at every point in the cct.
3. You have not achieved any permanent charge separation in the capacitor (ie when you turn off the generator the capcitor is not charged is it?). You have only assumed this from the voltage waveform. Can charges round your circuit actually travel that fast? Or is it (as I suggest) the electric field doing the job?
4. THe calcualtions you have made do indeed seem to have shown consitency with the accepted theory (Kirchoffs Laws), but this does not in any way prove that actual charge carriers of other form of electric current actually passes from one plate to the other. To show that, you would need to put a current sensor in betweeen the plates, as I mentioned earlier then see what you get. My bet, if the setup is correct, is that no current will be measureable. If it is, then I am wrong and you have discovered something significant!--Light current 20:00, 8 March 2006 (UTC)

With regard to the first two points. The wavelength (for a 500 kHz driving frequency) I am working at is 600m. This is much greater than any dimension in my circuit. Therefore, transmission line effects do not need to be taken into account. If I were working at a higher frequency, these issues would come into play. For the third point, because the charge separation is not permanent does not mean it does not occur. If, at the peak voltage, I opened the circuit between my source and the capacitor, I would have the charge I measured on the plate of the capacitor (they would “permanently” stay there). Because of my small (wrt driving wavelength) dimensions, there is no issue with “speed” of electrons. Your forth point I believe gets to the crux of the issue. I was, and still am, adamant that there are no charge carriers between the two plates at any point in time. If there were, they would result in conduction current. This is a separate issue completely from displacement current. This does not mean there is not current from one plate to the other; it is through the generator.

You mentioned the word Kirchoff, and I think that this may be the point of confusion. Our basic circuits classes teach us that current into a node equals current out of a node. This is undeniably correct. For my experiment, during one cycle of voltage, there is current into one plate (none coming out!). This is the effect of charging the capacitor. On the opposite cycle, the current direction is opposite and the plate begins to discharge. Because there is net flow of charge from one plate to the other, it is correct (in our circuit diagram) to say there is current through the capacitor.Mak17f 22:32, 8 March 2006 (UTC)

Charge separation thro' the extenal circuit cannot occur at this frequency and Im sorry but I cannot accept your assertion as true.--Light current 02:29, 9 March 2006 (UTC)

Any reason why not? Mak17f 02:46, 9 March 2006 (UTC)

Because charge carriers in the metal travel very slowly and cannot cause actual charge separtaion in your cct in the times involed (~2us).So any apparent charge (if there is any) on the plates must be due to some other phenomenon.--Light current 14:48, 9 March 2006 (UTC)

You have a fundamental misunderstanding as to how charge is transferred in a conductor. The fact that an electron may move “slowly” in mediums has no bearing as to how charge is transferred from one plate to another. The specific electron that is on one plate during one half cycle will not be on the other plate on the other cycle. Two microseconds is much longer than is needed for charge transfer. There is no dependence at all on electron mobility in a conductor. Mak17f 15:16, 9 March 2006 (UTC)

you say:The fact that an electron may move “slowly” in mediums has no bearing as to how charge is transferred from one plate to another.

In that case, how is the charge transferred from one plate to the other?--Light current 15:25, 9 March 2006 (UTC)

A decent explanation: http://answers.yahoo.com/question/index?qid=1006020502146 There are devices which charge capacitors to charges on the order of 100’s of microC in less than 100ns (one is in our lab). However, this is not a huge feat; charge mobility is not the limiting factor in this.Mak17f 15:45, 9 March 2006 (UTC)

Experimental difficulty

I have just realised that the insertion of a resistive sheet between the plates will intercept some of the em wave and act like any other series resistor in the cct. There will therefore probably be some pd measured across it. So another way of determining the presence of any 'interplate' current needs to be found. I dont know if there is a reliable method that doesnt give spurious answers. Anyone have any suggestions? --Light current 03:11, 11 March 2006 (UTC)