Talk:Capacitor/Archive 3

Archive 1

Archive 2

Archive 3

Archive 4

How to read the numbers on a capacitor

A useful section would be an explanation of how to read the markings on a capacator. THanks.

There is no single source that I can find that explains what the labels on capacitors mean, for example my air conditioner capacitor states: 30/7.5mdf 370VA, I am left guessing. 71.127.24.125

Different types

Please include a dicussion on the different types of caps (electrolitic, ceramic, film, tantlum, etc) and their various strength and weaknesses. Dept of Alchemy 09:10, 12 January 2007 (UTC)

Have you seen Capacitor (component)? I admit that you might expect capacitors to be described under Capacitor, but that's not the way it has panned out, unfortunately. --Heron 18:25, 26 May 2007 (UTC)

I don't understand!

You need to talk about the physics FIRST so it's easier to understand the beginning of it!!!

Yes, well it is debatable as to whether to have the history before the physics or the physics before the history. Some people like it one way, some like it another. Why dont you try changing it round to see if it looks any better your way?;-)--Light current 02:35, 28 December 2005 (UTC)

Pictures

I suggest moving the pictures to the capacitor (component) page as they are more appropriate there--Light current 15:00, 3 December 2005 (UTC)

TODO

I would suggest doing the following:

- Move capacitor non-idealities before the section describing types of capacitors. The former is needed to understand the latter.

- Remove the bit about replacing old electrolytics in audio equipment. Or possibly move this into the section on capacitor types

- Are the ESR numbers good? I've seen caps with much higher ESR. Someone should verify this.

- I'd clean up/organize the section on non-idealities. Parts of it are a bit rambling. I'd have one paragraph on each type of non-ideality, I'd also move the parts on accuracy/precision from the top into this section.

At some point, it'd be very nice to have a table of type of capacitor vs. accuracy, tempco, ESR, ESL, resonant frequency, leakage, etc. This would be a valuable resource.

I don't think it's bad to have more information in a Wikipedia article (as someone mentioned about keeping it basic/for beginners, rather than an engineering reference). But I do agree that the additional information should come in seperate sections at the end. A beginner should be able to read the first bit. An expert should be able to use the rest for reference. The reference should be in a reference format. We could add a section 'engineering practice' at the end for this. Organize it as:

Overview

Physics

History

Non-idealities

Types of capacitors

Applications/Engineering practics

Parrallel Plate Capacitor vs Other Geometries

I have the gut feeling that the equation C=Q/v works for point charges and infinitely long parallel surfaces - but that it isn't true for spherical or - more realistically - cylinderical capacitors. Am I right? Note of this should be on this page if I am. Fresheneesz 18:17, 20 November 2005 (UTC)

Fresheneesz: C=Q/V works for any capacitor. C is a geometrical factor that depends on the shape and size of your capacitor. Of course, you're right that infinitely close parallel plates and point charges are just an approximation. These don't affect the relationship C=Q/V, just the constant C itself. Jwyrwas 17:02, 31 January 2006 (UTC)

However, I would like to add something to the Physics: Capacitance section, that $C\approx {\frac {2\pi \epsilon L}{ln(b/a)}}$ , where a is the inner radius and b is the outer radius for a cylindrical capacitor, and $C\approx {\frac {4\pi \epsilon ab}{b-a}}$ for a spherical capacitor. Jwyrwas 17:11, 31 January 2006 (UTC)

Synchronous Capacitor

One of the very important theories in Electrical engineering is regarding the use of motors in control systems. throughout the article, there is no mentioning about the servo motor exhibiting capacitance effect. note the point. --Davy Jones 08:15, 16 October 2005 (UTC)

Does a servo motor have a capacitor effect or are you thinking about angular momentum?--Light current 09:10, 16 October 2005 (UTC)

Synchronous Motor as a Capacitor

1. As load on the motor increases, I_a increases regardless of excitation.
2. For under and over excited motors, the power factor (p.f.) tends to approach unity with increase in load.
3. Both with under and over excitation, change in p.f.is greater than in I_a with increase in load.
4. With normal excitation, when load is increased, change in I_a is greater than in p.f. which tends to become increasingly lagging.

Important Points

1. The magnitude of armature current varies with excitation. The current has large value both for low and high values of excitation (though it is lagging for low excitation and leading for higher excitation). In between, it has minimum value corresponding to a certain excitation. The variations of I with excitation are known as V curves because of their shape.
2. For the same input, the armature current varies over a wide range and so causes the power factor also to vary accordingly. When over-excited, motor runs with leading power factor and with lagging power factor when under-excited. In between, the power factor is unity. The curve for power factor looks like inverted V curve. Also, the minimum armature current corresponds to unity power factor.

Theory

As per the first point, an over-excited motor can be run with leading power factor. This property renders it extremely useful for phase advancing (and so power factor correction) purposes in the case of industrial loads driven by induction motors and lighting and heating rods supplied through transformers.

Both transformers and induction motors draw lagging currents from the line. Especially on light loads, the power drawn by them has has a large reactive component and the power factor has a very low value. This reactive component, though essential for operating the electrical mechinery, entails appreciable losses in many ways. By using synchronous motors in conjunction with induction motors and transformers, the lagging reactive power required by the latter is supplied locally by the leading reactive component taken by the former, thereby relieving the line and generators of much of the reactive component. Hence, they now supply only the active component of the load current.

When used in this way, a cynchronous motor is called as synchronous capacitor, because it draws, like a capacitor, leading current from the line. Most synchronous capacitors are rated between 20 MVAR and 200 MVAR and many are hydrogen cooled.

I'll try to get the characteristics graphs too. But now, after all this BLAH, what is the definition of capacitor? Is it a charge storing device or a device with leading power factor? Also, is it an electronic component or an electrical Device?--Davy Jones 01:33, 23 October 2005 (UTC)

Effect of Excitation on Armature current and power factor.

Well I think a motor is a motor and not a capacitor even tho it may have a leading PF. It should be included in the page on motors though! and definitely on the page about power factor correction. BTW Its an electrical device!--Light current 01:46, 23 October 2005 (UTC)

But there are a lot of well, standard books that refer them as servo capacitors'. We must consult some industrialist as they deal with larger capacitances. what do u think? --Davy Jones 01:56, 23 October 2005 (UTC)

I think the capacitor page is too big alrady! Put it in PFC power factor correction or motors or something :-)--Light current 01:59, 23 October 2005 (UTC)

I wouldn't call it a capacitor, per se. (i forgot about this thing) It will supply a leading reactance load as long as it is always synchronized to the lagging (inductive) load. If they become unsynchronized, it's not functionling like a capacitor. A capacitor is a capacitor is a capacitor all the time. Snafflekid 18:46, 23 October 2005 (UTC)

The effect is exhibitted no only on varying load but also on varying excitation (refer the first point). How can we comapre normal V/s synchronous capacitors? --Davy Jones 02:16, 26 October 2005 (UTC)

Category?

Why does Inductor belong to [[Category: Electromagnetic components]] while Capacitor belongs to [[Category: Electronics]]?--Astor Piazzolla 11:55, 14 December 2005 (UTC)

I dont really know. But I would suggest that an inductor relies on a electromagnetic field more than a capacitor does for its operation.--Light current 02:46, 23 December 2005 (UTC)

Missing definition

In the discussion of impedance Z you don't give a definition for j. If it's supposed to be obvious from the mention of "imaginary component" right below -- it's not.

Maybe we should add this to that section to make it really obvious what j is:

"where $j={\sqrt {-1}}$ and is the imaginary unit."

Well, whatdayaknow - it already IS in the article. Wonder how you missed that? Alfred Centauri 13:14, 9 April 2006 (UTC)

Cleanup

This article is not clearly defined:

• As I understand, all the component-related stuff goes into Capacitor (component). But the physics is already present in capacitance, and shouldn't be duplicated here.
• The topic on safety hazards of capacitors used with tesla coils has, in my opinion, nothing to do in an article that is supposed to be the root of the Category:Capacitor (i.e the one with the broadest scope).
• The topic about tolerance values is available in much more details in Capacitor (component).
• The Application section should be categorized: Signal decoupling and tuned circuits are a subtopic of signal processing, as well as filtering the output of a rectifier is a subtopic of energy storage.

This is why I added the cleanup tag. What is your opinion about this? -- CyrilB 23:13, 8 April 2006 (UTC)

I dont think the tag is really appropriate to this situation. Why dont you just try to edit the page as you see fit? ;-)

Well, I think it is appropriate, because I'd like to get some attention on that issue. As I can see, many people have worked on this article, and I don't want to start a revert war... I'll start some light modifications, but I really think this article must be clarified, as any person looking for information about capacitors on wikipedia will arrive on this page. In my opinion, this page must give an oversight of the Category:capacitor structure. And this requires heavy changes that I'd prefer to discuss first. -- CyrilB 09:57, 9 April 2006 (UTC)

CyrilB, the points you make above are valid. LC and I have had similar discussions before. The reality is that after you refocus this article, the information you remove (because you believe it to be superfluous) will be added back over time by other contributors. Certain articles seem to attract contributions and the Capacitor article appears to be an example.

BTW, I wasn't aware that there was a Capacitor (component) article. This seems awkward. The capacitor article should, IMHO, have three sections - fundamental theory (the physics of the capacitor, not capacitance!), circuit theory (the ideal capacitor circuit element with applications), and practical (non-ideal real world components).

Also, I changed the indent of your last comment. LC is a stickler for proper talk page indentation! Alfred Centauri 13:42, 9 April 2006 (UTC)

(Sorry if I don't indent correctly, I thought I had to indent one step more than the paragraph I reply to) Actually, it took me some time to find the Capacitor (component) page, although there is a link on top of Capacitor. I was about to propose a merge, but it might be simpler to remove some stuff from this article first. I agree with the sections you propose, although I would reduce the circuit theory as much as possible, as there is already a page about parallel and series connections. Non-idealities are currently in the Capacitor (component) article, but I agree they should be moved here. This would result in this page being about capacitor theory and phenomena, and Capacitor (component) about technology. CyrilB 14:37, 9 April 2006 (UTC)

Well normally indentation increases by one : for each post, but we've found that this quickly leads to tabbing off the page and using a lot more vertical space than needed. I tend to prefer therefore that everyone keeps his own level of indentation under a heading. For instance , you started this thread, so your indentation is zero,I chipped in and Ill keep on 1 :. AC replied second so his is 2:. This is by no means a WP rule, but I think youll find its sensible esp in longish exchanges!--Light current 01:52, 10 April 2006 (UTC)

Removing tag. Its clean -not perfect -but clean enough for now--Light current 14:24, 17 April 2006 (UTC)

Photo captions

While the pictures at the top of the page are nice it would be good to say whther the rulers are graduated in inches or centimetres. I think they're both in inches but I'm not sure. Lisiate 01:56, 10 April 2006 (UTC)

Both are in cm and mm but its not obvious I agree!--Light current 18:07, 9 April 2006 (UTC)

Added this info--Light current 14:25, 17 April 2006 (UTC)

Variable Capacitor Photos?

They're seen less and less in modern radio electronics, but some photos of variable capacitors common in radio receivers, transceivers, and antenna matching devices would be nice to see...

Yes they would.. have you got any? I got a nice 365pF airpaced tuning cap somewhere. Do you want to see it?--Light current 00:24, 23 April 2006 (UTC)

There are lots of variable capacitors in the Elecraft K1 I built. I put a photo on wikimedia which anyone is welcome to crop/resize/use. The variable ('trimmer') capacitors are on the upper board, the small blue components with screw-like tops. If you'd like a closer picture / different angle (or more pictures from the construction, of which I have many), feel free to contact me.

Variable Caps, here's one for you! Fit this in your QRP Rig OM!

I've got loads more ranging from matchbook sized to this 18" monster I'll work on getting some snaps to upload.

--Dp67 | QSO | Sandbox | UBX's 19:59, 27 September 2007 (UTC)

im confused!!!

im an aspiring electronics student shortly to join the course (within in 2 months probably) so im juss a beginner and i see that im confused with the usage of capacitors in parallel and vith capacitors in series...can anyone please help me out?! —The preceding unsigned comment was added by 59.144.67.119 (talk • contribs) .

Think of it this way:

• When the capacitors are placed in parallel, it's as if the their metal plates were placed side-by-side (with each capacitor contributing its own area of metal plate) so there's more total metal area so more total capacitance. C_t = C₁ + C₂ + ... + C_n

• When the capacitors are placed in series, it's as if the dielectrics of the two capacitors are stacked-up so the metal plates become farther apart, reducing the total capacitance. C_t = 1 / ( 1/C₁ + 1/C₂ + ... + 1/C_n ) (The capacitance goes down by the reciprocal of the increase of the thickness of the dielectric.)

Atlant 14:08, 10 May 2006 (UTC)

Hmmm. Thats a good way of thinking about it Atlant, esp the series config. However, have you ver wondered about actually making a capacitor in this way by actually interspersing floating screens within a solid dielectric?--Light current 17:16, 10 May 2006 (UTC)

Actually, I think thats how capacitor bushings are made!--Light current 23:49, 10 May 2006 (UTC)

Within any electric field, you can envision "equipotential planes" (a possibly-curved plane where the voltage stress is the same, similar to the way isobars on a weather map show areas where the barometric pressure is the same). Aside from the fact that the conductive floating screens would "force" certain contours into the equipotential planes, infinitesimally-thin floating screens would have no effect. And you're right, this is quite similar to the situation with series capacitors, except that the plate+wire+plate isn't exactly infinitessimally thin.

(Credit where credit is due: I learned about equipotential planes from an excellent volume once published by Tektronix that explained in great detail how CRTs used equipotential planes to focus and deflect electron beams.)

Atlant 12:21, 11 May 2006 (UTC)

Well I cant see how thin floating screens would have no effect. As I said, capacitor bushings are made in this way and the foils are floating. THese create a string of capacitors in series although their main purpose is to manage the electric stress field. Hence the conundrum: ie in what way do the floating screens differ from the interconnected plates of two series capcitors?(these are also floating) --Light current 13:28, 11 May 2006 (UTC)

I think you've hit on the exact point. The floating screens equalize the electric stress fields, even against a slightly-non-uniform dielectric. I think this prevents the problem that variations in the dielectric could lead to local build-ups in stress which, in turn, could lead to localised punch-through and subsequent failure of the overall insulator. But we're now pretty far outside my area of expertise.

Atlant 13:53, 11 May 2006 (UTC)

Yes of course. Any electric field strength that exceeds the withstand of the dielectric will naturally cuase breadown. But partial discharge activity is likely to start at the most stressed parts of the insulation well before this. You are correct in assumimg the foils will form equipotential planes, but I think these planes should be carefully shaped to create a very smooth transistion in the electric field strenth all the way from the HV condutor right down to ground with no great concentration of equipotential lines. Im not sure if 'shaped' foils are actually used tho' in HV bushings! If they were, I think it would need to be done at the ends of the foils.

--Light current 14:49, 11 May 2006 (UTC)

So, If you have a high capacitance substrate that you cant thicken to reudce the capacitance between top surface conductors and the ground plane say, you could always put a number of conducting (floating) foils inside the substrate to achieve the same effect! Yes? 8-)--Light current 14:53, 11 May 2006 (UTC)

V = E·d ?

The picture text in section "Capacitance in a capacitor", says that V = E·d. Shouldn't this be V = E/d ? Since the voltage becomes smaller with an increased distance.

Use dimensional analysis to see that V = E/d cannot be correct. E has units of volts/meter. To get units of volts. E must be multiplied by a distance in meters. Also, it is a well known result that increasing the distance between the plates of a charged capacitor increases the voltage across the capacitor. This makes sense because work must be done to increase the distance. The charged plates of a capacitor are equal and opposite so the plates are attracted together. Alfred Centauri 13:36, 22 May 2006 (UTC)

More confusion

In parallel, the total charge stored is the sum of the charge in each capacitor. While in series, the charge on each capacitor is the same.

This sentence is very confusing as in series the total charge is also the sum of the charge in each capacitor.

It was confusing, so I changed it to relate to the overall capacitance through the effective capacitor created (as per "Guide to Electronic Components", William Segallis) which uses that argument. The series and parrallel section could do with a bit more work as its a bit disjointed. Softgrow 11:54, 22 June 2006 (UTC)

There's some talk up above us here that discusses this as well; perhaps it will spark some ideas for you?

Atlant 12:18, 22 June 2006 (UTC)

Half the work done...

Say you energise/charge an uncharged, ideal capacitor using a battery. The work done by the battery to move the charge from one plate to the other will be Q*V. Yet the energy stored in the capacitor is only half of it (½*Q*V). Where does the remaining half go? 59.94.138.200 13:09, 26 June 2006 (UTC)

The other half is lost and goes into Pandoras box, which you are on the point of opening! 8-|--Light current 15:08, 26 June 2006 (UTC)

Regarding the inquiry by 59.94.138.200: The problem you pose is ill defined. On the one hand, you say the voltage across the battery is a constant V in order to get the result that w = Q * V. On the other hand, you say the voltage across the capacitor is zero initially and charges to V over time but this capacitor is in parallel with a constant voltage V so you have a contradiction unless V = 0. To properly define the problem, add an arbitrarily small series resistance to see that the missing energy is converted to heat in the resistance. Of course, if the resistor is small enough, you must take into account EM radiation from the rapidly changing current. This radiation resistance carries away the missing energy in the limit as the series resistor goes to zero. Alfred Centauri 16:48, 26 June 2006 (UTC)

Alfred, I dont think there is necessarily a contradiction, unless the writer is thinking of an ideal (unrealisable) 'circuit theory' type capacitor. As we all know, all practical capacitors will have some non zero value of Zo.

If the voltage source is mismatched with the Zo of the capacitor, then there will be multiple reflections that will cause the excess energy to be progressively dissipated in any real resistance connecting the source to the capacitor. The question of how much (if any) energy is radiated away is a tricky one and would depend upon the radiation resistance of the antenna formed by the wiring in relation to the impedance of freespace etc.

One (more important) question is: if there is no real resistance (such as could be arranged by having an ideal coaxial TL connecting the two elements) and of course no radiation from the coax , what happens to the xs energy there? 8-)--Light current 21:02, 26 June 2006 (UTC)

LC, I did assume that our anonymous puzzler was referring to ideal circuit elements for otherwise, there is no paradox. However, your point is well taken. Next time, I will ask first if the context is ideal circuit theory. Alfred Centauri 00:03, 29 June 2006 (UTC)

Aah, you are (becoming) 2 Ys 4 me Alfred 8-) Now try the next poser below!--Light current 00:07, 29 June 2006 (UTC)

The other half is not lost and there is no need to use resistors. the work done is the integral of qV(t) as you add charges. The first charges you add very little electric field to work against. As you add more charges, the voltage gets bigger and each charge requires more work than the last. you are basically calculating the area of a triangle - 1/2 base * height. Here it is 1/2 * Qtotal * Vfinal = 1/2 CVfinal * VfinalTjmozdzen 08:19, 20 September 2007 (UTC)

Charging capacitors without energy loss.

Is it possible? 9-)--Light current 15:20, 26 June 2006 (UTC)

I think I know of (at least) one way! 8-)--Light current 00:36, 29 June 2006 (UTC)

The two piece TL problem

The above post reminds me of a question I asked many moons ago on another page:

=== #1 The two piece TL problem ===

Imagine two identical ideal transmission lines (say pieces of coax). Their central conductors are separated by a single pole ideal lossless switch (say an ideal in line reed switch). Their outers are are connected together at the switch location. Both TLs are open circuit at the ends not connected to the switch. Initially the left hand TL is charged to a voltage V and the right hand TL is discharged. At time t=0, the switch is closed and remains closed. If the capacitance of each piece of coax is C, application of conservation of charge before and after switch closure shows that half of the original energy (0.5CV^2) has been lost. If this energy cannot radiate away and there is no skin loss or dielectric loss in the TLs, where has this energy gone?--Light current 15:00, 8 October 2005 (UTC)

Has the energy really gone somewhere else? --Light current 15:31, 26 June 2006 (UTC)

I hate to say this again but the problem you have presented is ill posed. Consider the statement "If the energy cannot radiate away...". That statment severely constrains the problem statement. Do you see what I'm getting at? Alfred Centauri 00:10, 29 June 2006 (UTC)

Well no, because I have said energy can't radiate away. I have proposed a coaxial line. How can energy radiate from a completely enclosed system?--Light current 00:13, 29 June 2006 (UTC)

Consider the switch. Alfred Centauri 02:21, 29 June 2006 (UTC)

The ideal 'in line' reed switch is totally enclosed within the continuous coaxial screen as below. 8-)

                    -------------------------------
                    _______________/ ______________

                    -------------------------------

--Light current 08:33, 29 June 2006 (UTC)

Charge stored on the dielectric or the plates? (Yes, this again...)

older discussion

This sentence was added to both Capacitor and Leyden jar:

Benjamin Franklin investigated the Leyden jar, and proved that the charge was stored on the glass, not in the water as others had assumed.

Yet our article on Leyden jar says that the charge is stored on the plates, and that the "dissectable Leyden jar" demonstration is a myth caused by high-voltage conductors "spraying" charge onto the dielectric when it is disassembled. Which is true? Where did the sentence about Ben originate? It needs a reference, at least. — Omegatron 14:39, 8 July 2006 (UTC)

Follow the PDF link on this page to some university notes by Robert Morse. The PDF contains long quotes from a letter by Franklin about this very experiment, so we can use that letter as our source. Franklin found that the charge in a dismantled Leyden jar remained on the glass. He didn't realise that, while the jar was intact, the charge would have been within the conductors. --Heron 16:24, 8 July 2006 (UTC)

Interesting. It would be good to have another reference besides Bill. I trust him, but I don't know if others would. — Omegatron 17:13, 8 July 2006 (UTC)

These notes by Donald Simanek at lhup.edu mention (in the "Footnote" about half way down) the electrons jumping the gap during disassembly. Unfortunately, though, Simanek doesn't mention the case where the field is too weak for this to happen. All the other Google hits I can find consist of arguments involving Bill, which never reach a conclusion because nobody has done the necessary experiments. --Heron 18:40, 8 July 2006 (UTC)

Don't trust me, I could be wrong! But here's even more thought-experiments: in an air-dielectric capacitor, does the charge reside on the metal plates, or on the air? (And if we blew some wind between the plates, would this sweep away the charges?) In the dissectable capacitor, what if there is an air-gap between the metal plates and the slab of plastic? Do the charges in the dissectable Leyden Jar leap across the air (and only do this above a certain high voltage?)

And what about vacuum capacitors?  :)

Here's an argument based on a practical issue which analog engineers should recognize. See:

Bob Pease on "capacitor soakage

We know that the charge which moves to the dielectric is almost insignificant when compared to the charge on the plates because in real-world capacitors the dielectric isn't perfectly insulating and instead has a large resistance. In a "charged" capacitor, some charge slowly leaves the surface of the capacitor plates and migrates into the dielectric. This effect is called "dielectric absorbtion." When dielectric absorbtion occurs, the charge which has moved into the dielectric becomes trapped and unavailable until it migrates out again. So for example, if we short out a charged capacitor (discharge it,) there will be an enormous discharge current, followed by an extremely small continuing current as the trapped charge leaks from the dielectric and back into the plates. So... how do we answer the original question? Easy: in normal capacitors the charge involved in charge/discharge is enormous when compared to the charge trapped in the dielectric, so we usually assume that no charge moves into the dielectric at all. Also, dielectrics are chosen to give a minimum dielectric absorbtion effect ...so if most charge was stored in the dielectric, that would be a weird and flakey capacitor, and nobody would buy it. On the other hand, if a commercial high-volt capacitor with bad (cheap) dielectric is left charged for awhile, significant charge will leave the plates and move to the dielectric. During a later discharge, the net charge transferred will be unexpectedly smaller because a significant portion is stored in the dielectric surface. And if this discharged capacitor is left alone with no shorting bar, it will "charge back up" as the trapped charges slowly leak back to the metal plates. A similar problem can plague any low-volt precision capacitor circuit operated at very low frequencies, such as op-amp integrators. Note that this explanation assumes that the metal plates are bonded to the dielectric over a large area, while the air gap of a Leyden jar would introduce a higher resistance which goes nonlinear above a certain voltage.

--Wjbeaty 05:26, 9 July 2006 (UTC)

Yeah, but you always make sense. Like you just did.  :-)

Capacitor (component)#Dielectric absorption (soakage) — Omegatron 14:32, 9 July 2006 (UTC)

I don't understand! Part Two

Does Light current own this article? If so, I'll just stay away. No point in getting frustrated for nothing.

Interesting comment ! .. I had a similar experience with him some months back when he appointed himself god over some pages I had worked with earlier. He seemed to be remarkably ignorant of the subject matter at hand given his alternately imho high handed,hypocritical,condescending,meglomaniac attitude, which got me very frustrated before I simply gave up, he seemed to have more time to waste that I had available.Sadly the main victim was the page.tubenutdave 09:24, 27 December 2006 (UTC)

I have a problem with the intro definition: "charges ... have been placed." laughable. Imagine a little gnome plucking a +charge out of a little vial and placing it onto a plate, then plucking out a -charge and putting it on the opposite plate. The thinking of physicist, not an engineer.

This an article about a device, there is no section on how physicists use capacitors, ergo this is an EE article. Physics takes a subservient position.

The article provides minimal material to help the inexperienced engineer. It's virtually useless to the novice- say a 14 year old. It's not written well, no sense of top/down or high level stuff first/detail later. For example, see section on Capacitors with DC circuits. There nothing to speak of about circuits, it starts off talking about the dielectric, there isn't even a circuit diagram to help the reader.

If it is supposed to do any real teaching, things have to be improved, especially when it comes to organization. As for the comment below about length, you make as long as you need to and no more. (See Amadeus, "too many notes")

In its present state, I can not recommend this article as a useful source of information about capacitors. blackcloak 07:35, 27 July 2006 (UTC)

No I dont own this article. Why did you say so? Any comments from people like yourself are most welcome! We will try to act on them. 8-)--Light current 08:34, 27 July 2006 (UTC)

Because you dismissively removed edits I made to the overview. Usually when I remove something, even if it is a small change, I indicate why. It's a courtesy. Beyond that, when I edit another person's edits, I am careful to find what is valuable in their contribution and incorporate it wherever possible. If it is long and poorly thought out, I say so, and may suggest that the material belongs elsewhere, perhaps in its own section, after a rewrite.

Yeah sorry buddy but the edits were crap! (no offence)--Light current 00:26, 28 July 2006 (UTC)

While I won't claim the edits I made are perfect (I leave it to others to make the final improvements, and sometimes they do that really well), but I do claim that I filled in many of the details the would help the reader to see the subject matter more clearly, and without some giant leaps. I think that those who have contributed to this article so far know their subject matter too well, and they have little sense for what it takes to communicate with a novice.

And what is this we stuff. Is this article closed to those outside the we group? This is the first time I've run across such possessiveness in wikipedia. I've been working under the assumption that articles are improved my many minds looking at the same material and, seeing the need for improvement, do so when so motivated. I'm seeing something more heavy-handed here and I don't like it.

So in the interest of clarification, who do you see as the audience of this article? Is it first year physics/engineering students and above? Do you see any value in presenting material a 14 year old might find useful as he/she tries to figure out what electronics is all about? Should separate sections be created to provide the more basic material? Maybe there should be a "capacitors for dummies" section. I wrote "An elementary explanation ..." section in the article on "Ohm's law." You might want to take a look at it because it will give you an idea what I think should be done to help young/non-technical readers through some fairly difficult (though elementary by an absolute standard) material. blackcloak 16:56, 27 July 2006 (UTC)

Herons comments

I don't think Lightcurrent should have reverted three consecutive edits by different contributors without an explanation. That was rude. On the other hand, Blackcloak, there were some problems with your edit. You used some unconventional terms, like 'interconnection conductor', and your style was more verbose than is usual for a technical article. If you're writing for 14-year-olds then you need to use simpler syntax. However, I'm sure that a compromise can be reached. You just need to proceed in smaller steps, and then you won't feel so bad when other editors change your words. Also bear in mind that the explanations given to young science students aren't always correct, and Wikipedians tend to rank correctness above educational conventions. That said, I agree with most of your criticisms of the article: it is badly organised and unfocused.

As for the intended audience, don't assume that Wikipedia is mainly a teaching resource. It is an encyclopedia. You wouldn't expect 14-year-olds to pick up a major encyclopedia and understand its articles about electronics without some adult guidance, would you? --Heron (self-appointed arbiter) 20:33, 27 July 2006 (UTC):

Sorry Heron, can you explain why Im not allowed to revert obviously bad edits without apologising for it on the talk page. Did I not fill in edit summaries? Also can you show me the 3 consecutive reverts I made? I can only find one in my contributions and that was to a version by Omegatron 8-( --Light current 00:31, 28 July 2006 (UTC)

I agree that that was was rude. I'm surprised to still see this after a year of editing.

O, Get off your high horse 'O' otherwise you might fall off!--Light current 00:31, 28 July 2006 (UTC)

I agree that these articles are poorly organized. We need to decide on a layout for the article and stick to it. We have Capacitor, Capacitor (component), and Capacitance. Significant amounts of information are duplicated in more than one article. Other information should be split into more appropriately-named articles.

I disagree somewhat about the intended audience. The introduction should be accessible even to a 14-year old ("A capacitor is a device for storing electrical energy" is perfectly accessible) and the article should get more complicated and detailed after that. Make technical articles accessible. — Omegatron 21:36, 27 July 2006 (UTC)

Thanks for your thoughtful response, Heron. While I don't want to get mired down in a philosophical discussion of style and intent, I do want to make a couple of points, mostly because I have thought about them as I've made other edits. Unfortunately there does not appear to be an easy way for authors to guide other editors through the thinking process used when making a series of edits.

1) I use the term "interconnection conductor" because I could not come up with any other term that would cover a wire, a terminal post, a clip lead, surface mount solder pad, a probe, etc. If you've got a standard one, I'll be the first to use it. I had hoped that others would make such edits when terms unknown to me are commonly used. I have not searched for it to see it is already there, but perhaps the solution is to provide a dictionary within wikipedia where special terms can be defined (the entry's apearance being clearly distinguishable from articles).

2) My style is verbose. You need losts of words when you are trying to be clear and simultaneously show how the various components are interrelated. Having written a lot of material to be incorporated into patents, I've thought through what it takes to communicate. Not only do you need to show how things are interrelated/interconnected, but sometimes you should also explain how they can not be interconnected. Reading is easy when things flow and are logically developed. Terse prose is like poetry, you have to think about it. Now thinking about it is good, but providing so little detail that you become ambiguous is not good (a problem poetry does not have because the authors are trying to create mental images, and likely images that the author had no intention of creating). I think you have to look at the total average time it takes for a reader to grasp the content.

3) I understand that wikipedia is an encyclopedia. Encyclopedias are not technical reference books. I don't remember ever using an encyclopedia as a source of technical information after I started college. I think wikipedia should be geared to the 12 to 18 year range. Students who go on to college and do not take any technical (engineering/physics) courses need a source of information that will help them do their work (say a reporter, or a business analysist, or even a doctor who wants to understand what went wrong with his chart recorder). Those who do go on to a technical degree and career will probably accummulate mountains of reference material and books. Wikipedia will never compete with those sources. And yes, I do believe that the articles can be written in such a manner that a 14 year old (ok, a motivated one) can pick up most of the content of an article about something like a capacitor, especially when links are liberally incorporated. blackcloak 22:01, 27 July 2006 (UTC)

LCs reply to Blackcloak

Well, to my knowledge, the only edit I made that could have caused this furore was one where I reverted to a stable version by Omegatron. This included reverting 3 bad edits in one go. This is not unacceptable. Also it seems that Blackcloak is getting paranoid. The 'we' I refer to of course includes all the editors. I was making a concilaitory statement; I hoped Blackcloak would see it that way. I am surprised that Omegatron sees this action as rude. Perhaps he could explian why in detail? Do we need to give excuses when reverting obvious rubbish? Did I not fill in an edit summary? If U 2/3are so perfect, tell me what I did wrong! 8-(--Light current 00:17, 28 July 2006 (UTC)

Oh. I don't really think I'm getting paranoid, I just want to understand the editing rules you're following. I do appreciate Light current's candor and I understand better why we are where we are.

But rather than pick up my marbles and play elsewhere, I rewrote the intro to conform to what I think needs to be said (and by omission, not said) in the intro. (I like it when the glaze-over visage doesn't occur right up front.) There's plenty of time to go into the details, and the further down in the piece that we place the mathematical constructs (like imaginary numbers) the better. Now I'll admit, as I did above, that the intro is not perfect- the sentence structure is a little complicated and perhaps should be broken up. I like the style I used but can certainly understand why others won't. The way I read these things when they become too complicated is to reform the sentence without the clarifying phrases separated by commas. Using this approach (easily taught if someone does not already use it intuitively), properly written long sentences can be easily understood. blackcloak 05:34, 28 July 2006 (UTC)

Patents

Ah, Blackcloack, you mentioned patents. Now I see where you're coming from. I have had the misfortune to read many electrical patents, and I believe that patentese is the very opposite of good writing. It is to communication what an atom bomb is to diplomacy. It gets results, but does so without regard for the pain it causes the reader. I'm happy to work with you to build a better article, but I think you need to accept that the best Wikipedia articles don't read like legal documents. --Heron 17:16, 28 July 2006 (UTC)

See the talk page for reed switch for what turned out to be a futile attempt to keep "patentese" out of Wikipedia. Patent language is meant for very special purposes almost perpendicular to the purposes of an encyclopedia article. --Wtshymanski 17:50, 28 July 2006 (UTC)

Good, I'm glad it's not just me that hates patentese. And in response to your earlier question, Lightcurrent, I think the answer is at WP:REVERT under 'Explain reverts'. --Heron 17:57, 28 July 2006 (UTC)

Reverting 'bad' edits

This extract from WP:REVERT

Being reverted can feel a bit like a slap in the face — "I worked hard on those edits, and someone just rolled it all back". However, sometimes a revert is the best response to a less-than-great edit, so we can't just stop reverting. What's important is to let people know why you reverted. This helps the reverted person because they can remake their edit, but fixing whatever problem it is that you've identified.

Explaining reverts also helps other people. For example, it lets people know whether they need to even view the reverted version (in the case of e.g. "rv page blanking"). Because of the lack of non-verbal communication online, if you don't explain things clearly people will probably assume all kinds of nasty things, and that's one of the possible causes for edit wars.

If your reasons for reverting are too complex to explain in the edit summary, drop a note on the Talk page. A nice thing to do is to drop the note on the Talk page first, and then revert, rather than the other way round. Sometimes the other person will agree with you and revert for you before you have a chance. Conversely, if someone reverts your change without apparent explanation, you may wish to wait a few minutes to see if they explain their actions on the article's talk page or your user talk page.

is not very clear on the question of reverting to a much earlier version. Perhaps I should have said on the talk page that I was going to do it but these were all anon edits (I believe) at the time and the article was getting out of control. This is the first time (I think) where Ive reverted more than one less-than-great edit in one go! Perhaps an apology is due to someone, but to whom? 8-(--Light current 18:23, 28 July 2006 (UTC)

If you're going to revert edits that are not obviously vandalism, you need to explain why; either on the talk page or in the edit summary. This is not difficult to understand. — Omegatron 18:18, 29 July 2006 (UTC)

Yes I should have made it clear in the edit summary why I was reverting back to your version. Sorry! 8-(--Light current 20:43, 29 July 2006 (UTC)

Heron and others: As a technical person, one of the reasons I got involved in writing material that would be incorporated into patents was to create clear text for the technical reader. In the environment where I worked, the patents were either written by a lawyer or by an outside contractor. They could get pretty bad. But the reason I mentioned my patent experience was to indicate the thinking process around writing claims in particular. This is a highly structured form of writing that has a required dicipline. By that I mean, all the relevant (and only the relevant) components must by described, and all the relevant interconnections among the various components must also be described. For example, in the intro, the final phrase of the first paragraph explicity states that the terminals are connected to the plates. This is what must be done in writing claims. Now we could assume that readers can figure this out, but do we want to leave any doubt?

Heron, thanks for your improvements to the intro. At this point I think the intro says what needs to said (expect possibly for introducing the unit of capacitance- the farad). I see (below) that light current is not happy with this, so let me fill in some of the reasoning for keeping the intro as you modified it. First, the dielectric: There is no reason to mention a dielectric because it is not a required part of a capacitor. As in writing general claims, you throw out everything that is not absolutely necessary. Second, as for the zero net charge: There is no reason to mention this because it's a condition nature imposes. It would be like writing a claim wherein you state that the device operates within a gravitational field. Third, the terminals. Connection to external components is always part of using a capacitor, so failing to mention this key element would be a failing. Clearly I don't want to see a revert. This intro is far better than anything I seen so far. The way these things should decided is to go through each element of the intro and decide if it is necessary or if it is superfluous. So, are the terminals necessary? I say yes because that's the way you connect the device to the external components. Capacitors wouldn't be very useful if you had no way to connect to them. Are the plates necessary? Well, yes, mostly. One or both may not always fit that general description, but on the whole, two conducting surfaces form the basis of physical design of most capacitors. Is it necessary to discuss storage and release of energy, instead of just storage? Yes. Half the time a capacitor is storing energy and half the time it is releasing it. I don't want to assume that readers will divine that, if you're going to store the energy, you're going to release the energy. Is it necessary to discuss time-dependent effects? I think so, just to supply some motivation for why capacitors are such useful components. Of course in a minimalist world, this could be foregone. Is it necessary to state that the terminals are connected to the plates? Yes, because if you don't you're going to have some one think that the electrons enter the capacitor by one of the terminals during the storage phase, and leave the other terminal during the release stage. Must by one of Murphy's laws. I hope this is enough to convince light current of the validity of this approach. But should he decide to revert, I think it only fair that he write a long defense of his position so we can all understand what leads him to his conclusions. blackcloak 06:56, 29 July 2006 (UTC)

As in writing general claims, you throw out everything that is not absolutely necessary. Second, as for the zero net charge: There is no reason to mention this because it's a condition nature imposes. It would be like writing a claim wherein you state that the device operates within a gravitational field.

No it's not. Many (most?) people misunderstand and think that capacitors "store charge". They don't. They store electrical energy in the form of a charge imbalance. It's very important to clarify early in the article that capacitors move charge from one place to another, but the net charge stays the same. And no, there is no condition nature imposes that prevents components from having a net charge. So removing this would actually confuse people.

I'm sorry that I don't know how to pinpoint exactly what I dislike, but I think your writing approach (as demonstrated in this comment) is not going to be conducive to encyclopedia articles. Please remember that this is an encyclopedia; not a patent. For instance, a link to electronic component makes it unnecessary to clarify that capacitors have terminals. That is very obvious if you understand the concept of an electronic component. If you don't, you can read about them by following the link. We don't need to specify that the capacitors are made of physical matter, take up space, and have weight. These are obvious traits of physical objects.

Also, as mentioned above, I don't think there should be three articles about capacitors. I think capacitor (component) should be merged into this article, and capacitance can handle the stuff that doesn't necessarily refer to real-life components. — Omegatron 18:18, 29 July 2006 (UTC)

Partial agreement wth 'O'

Very strangely, I find myself agreeing with Omegatron's first two paragraphs. I have a problem with his third para however, in that it would make this article execeedingly long. I am willi ng , however, to listen to arguments on how this whole capacitor/capacitance/capacitor (component) thing can be resolved amicably! 8-) --Light current 20:37, 29 July 2006 (UTC)

Unclear text

What does this mean exactly?

By layering the insulator between two metal plates, von Kleist dramatically increased the charge density.

--Light current 00:05, 28 July 2006 (UTC)

Not a lot. See if you like my new version. --Heron 18:05, 28 July 2006 (UTC)

Sorry again - no! I thought it was still confused. I've rewritten it. See if you think my version is technically correct on Kleist. 8-|--Light current 18:50, 28 July 2006 (UTC)

Putting WP:REVERT policy into practice

Heron, Im sorry to say that I really dont care at all for the current version or the intro as I think it obfuscates things. 8-( I think my version was far clearer. Yes it may need things added to it, but we should start again from that version. I there fore intend to revert your changes to the last version by me. 8-|--Light current 18:30, 28 July 2006 (UTC)

Please be more specific. Saying that "it obfuscates things" without cogent justification for that position is like shaking a red flag in front of a bull. The only reason for taking your poistion that I've been able to figure is that you're a physicist (at heart, if not in fact) and that you see the practical aspects of real devices (like terminals) as implementation details not suited for a scholarly treatise. blackcloak 06:56, 29 July 2006 (UTC)

Well this:

A capacitor is a two-terminal electrical device that can store electrical energy and later release it, the speed at which it does so depending on the electric circuit to which it is connected. Under the influence of a voltage applied to the two terminals, energy is stored in an electric field that forms between a pair of closely spaced conductors called the capacitor plates, each plate being attached to one of the terminals. Capacitors respond to a varying applied voltage in a strongly time-dependent fashion, with high-value capacitors accumulating and releasing energy slowly, and low-value capacitors doing so quickly, all other factors being equal. This time-dependent characteristic is useful in making electrical filters and other circuits.

is just is not up to the standard for any encyclopedia as its written in a pseudo scientific language and is just not a good introduction. We need the plain facts presented in as a simple a way as possible ( but no simpler!). THe present version actually will tend to confuse any reader who is ignorant on capacitors (it nearly confuses me!). Unless it can be improved substantially, I will revert to my version, then we can carry on from there. 8-|--Light current 12:25, 29 July 2006 (UTC)

I also have problems with the new intro. The statement "Under the influence of a voltage applied to the two terminals, energy is stored..." is just plain wrong. The voltage across a capacitor is a measure of the energy stored, not the influence for energy storage. A capacitor is charged by an electric current, not a voltage! Also, the statement "Capacitors respond to varying applied voltage in a strongly time-dependent fashion..." is an unusual use of the phrase time-dependent that is confusing at best. This sentence and the following sentence seem to imply that capacitors themselves are time dependent. I vote for the revert. Alfred Centauri 12:53, 29 July 2006 (UTC)

Ive changed it back to the old version. Maybe we can now move forward to discuss any improvements required to the lede.? 8-|--Light current 13:12, 29 July 2006 (UTC)

Well, for a start, it now mentions 'conductors' in the first sentence and 'each plate' in the second sentence. How is the reader supposed to know that these are the same things?

Secondly, 'when charged' contains a dangling participle. When what is charged? And what does 'charged' mean?

Thirdly, it talks about 'differing impedance'. This is too technical for people who don't know what a capacitor is.

--Heron 13:38, 29 July 2006 (UTC)

P.S. I acknowledge that the previous version wasn't that great, either. It was an attempt to simplify what was already there. --Heron 13:45, 29 July 2006 (UTC)

Heron, Ive amended the lede to address some of your comments--Light current 14:57, 29 July 2006 (UTC)

That's better. --Heron 16:33, 29 July 2006 (UTC)

Im glad. Im sure you know I have nothing against you personally! 8-)--Light current 20:30, 29 July 2006 (UTC)

Don't worry. I'm not feeling persecuted by you. We're all just guinea pigs in Jimbo Wales's cruel experiment. --Heron 21:27, 29 July 2006 (UTC)

Thats good to know!--Light current 00:10, 30 July 2006 (UTC)

Why especially in cars?

The problem is that there just isn't enough available current capacity from those puny little car batteries and 175 Amp alternators to produce the sort of bass that will levitate a car off the ground. Adding 4,000 Farads of super capacitors produces woofers with the equivalent power of your typical railgun, enabling people to maintain their self esteem in bass thumping contests.

Atlant 00:10, 11 August 2006 (UTC)

I know (sigh) 8-(--Light current 00:12, 11 August 2006 (UTC)

Do you reckon there actually might be a health hazard (apart from deafness or mushing the grey matter) from these boom boom boomers?--Light current 00:15, 11 August 2006 (UTC)

If you're not collapsing a lung, you're just not trying! [1] [2]

Atlant 00:27, 11 August 2006 (UTC)

Hey man, thats worrying. I dont boom in my car but I do play the (electric) bass!. Some SPL levels are needed here I think. Also maybe some legislation about max SPLs in car systems?--Light current 00:51, 11 August 2006 (UTC)

Its worth quoting this extract from [3] here:

A series of publications has focused on the effects of exposure to low frequency (<500 Hz) high intensity (>90 dB) noise, a range of structural lung changes having been reported mainly in rats12 but also in humans.13 These data suggest that low frequency high intensity noise may lead to structural and functional changes in the airways, pleural mesothelium, and lung parenchyma. With frequencies from commercial loud speakers typically in the range of 30 Hz to 20 kHz, it is the lower frequency band of 30–150 Hz which is usually boosted in big music venues for enhanced effect. These low frequencies could indeed be particularly damaging to the lung parenchyma if they coincide with its natural frequency of around 128 Hz.14 Furthermore, Mahagnah and Gavriely15 showed that, in normal humans, the lung acts as a low pass filter with flat transmitted energy up to 100 Hz or 300 Hz. In view of these experimental results, the most compelling observation in our four patients is that the lower frequency band, which tends to be boosted in the kind of music they listen to, corresponds with those frequencies, thereby ensuring maximal acoustic energy transfer to the lungs.

--Light current 04:59, 11 August 2006 (UTC)

Article looks good

So far as I can tell, the Capacitor article looks fine (big deal - who cares what I think anyway). I came across some discussion about this somewhere else that had some weird pseudo-science stuff that I disagreed with but it wasn't here. I'll cite it when I find it again. --hydnjo talk 02:03, 22 August 2006 (UTC)

Ah yes. THat could have been me and others. Did you check on talk:Capacitor ? Possibly on talk:displacement current also--Light current 15:15, 22 August 2006 (UTC)

latest revert by Light Current

Why the revert from the edit by 141.157.16.232? What was botched about it? It looked to me like a valid edit that made the sentence structure parallel. -Crunchy Numbers 14:51, 31 August 2006 (UTC)

The equations did not parse properly. Maybe it was a temporary error. Just looke like someone couldnt get the equation syntax right and decided to leave it!--Light current 18:09, 31 August 2006 (UTC)

Yeah just checked those revisions and they're OK. Wierd! Must have been a transitory Wiki problem.--Light current 18:12, 31 August 2006 (UTC)

How do wound film capacitors work?

Hi-

Is anyone out there knowledgable about the physics and electrical properties of wound film capacitors?

Carl Carlson Chitowncarl 13:45, 7 September 2006 (UTC)

If you want actual data its probably best to refer to the manf literature.--Light current 15:04, 7 September 2006 (UTC)

Um, I work for a capacitor company. The industry tends to publish specs. but doesn't offer much in the way fundamental capacitor technology.

CarlChitowncarl 14:21, 12 September 2006 (UTC)

Have you checked our page Capacitor (component)? --Light current 20:33, 12 September 2006 (UTC)

EverLife Flashlight

This thing uses a capacitor to power it. Power is obtained by causing a magnet to slide through a coil of wire, then the power is fed into a bridge rectifier into a capacitor, then to a switch, then to a white Light Emitting Diode. I have one of these lights. Martial Law 22:24, 8 September 2006 (UTC)

So?--Light current 20:34, 12 September 2006 (UTC)

Current flowing trough the capacitor?

If I have understod the meaning of the sentence under the DC title on the articel page this sentence is incorrect:

" In fact, the current through the capacitor results in the separation of electric charge, rather than the accumulation of electric charge."

Its the words "currents through the capacitor" that is incorrect. There are (almost) never current actually passing through the capacitor.

Lets say a battery is connected to a capasitor, at time = 0 nothing have happened (duh!) when time goes a current flowing from the minus pole goes towards the plate that is connected on that side,when it reaches the plate it stops, a voltage will build up on that plate as more current comes to that plate. The positive side will be unaffected by this(the total amount of electrons will be constant, although electron on the positive side will flow from the positive plate since the proximity from the electron on the minus pole will force the elctron on the positive plate to go elsewere, in this case to the wire from the battery), no electron will go or come on the positive side ( when saying the positive side i mean the positive plate and the positive wire from the battery to the plate).

The reason for the build up of voltage is just the electron going for the more positive plate, but the elctrons will never come to the other side of the capasitor. I apoligise for my poor english.-LeftSaidFred

What you have written above is not a correct description of the physics of the capacitor. A 'charged' capacitor does not have a net electric charge despite what you have stated above. The plates of a charged capacitor have equal but opposite electric charge so that, as a whole, the capacitor is neutral. Despite your claim above, when a capacitor is charging or discharging, electrons are added to one plate while being removed from the other. Thus, there is a current into one terminal of the capacitor and an equal current out of the other terminal such that we say there is an 'electric current' through the capacitor. Alfred Centauri 14:45, 13 December 2006 (UTC)

That is what I was saying, but I didnt know the term "electric current through the capacitor" dont actually mean that the current is going through the capacitor, I though that was litteerary, I though it was stated in the article that electric current was going through the capacitor! I think many will interpet that wrongfully 8 at lest I did). That is what I objected against.-LeftSaidFred

What I was describing is correct, I didnt say that the capacitor isnt neutral seen from outside. —The preceding unsigned comment was added by 83.108.136.92 (talk) 18:35, 14 December 2006 (UTC).

In capacitors with plastic dielectric, charges definitely flow within the plastic. As a voltage across the capacitor plates increases or decreases, this causes motion of electrons and ions in the dielectric. This motion is an electric current. This current is very important, since it marks the difference between a vacuum capacitor and a capacitor with material of higher dielectric constant. Of course this current in the dielectric cannot be continuous as with a conductive material. The current in the dielectric is transient, but it's nevertheless a genuine current.

Now with capacitors with vacuum dielectric, in that case there is a current in the vacuum, but no charges flow across the vacuum. This might not make sense until we realize that the word "current" includes charge-flow as well as including Maxwell's displacement current. Displacement current is composed of EM fields. But because Displacement current produces circular magnetic fields surrounding its path, we must include Displacement current under the definition of Electric Current. In other words, Electric Current is more than just charge-flow. If one electron moves forward, and this forces another electron to also move forward, we say that "electric current" exists in the space between the two electrons. --Wjbeaty 22:12, 14 December 2006 (UTC)

But it isnt the same elctrical current from the battery (in the original case), the capacitor doesnt take advantage of the current in the dielectric, maybe it will couse interrference, but we cant use that energy.

Has any one actually verified the flow of displacement in a vacuum capacitor? How is is measured?

--Light current 21:40, 12 January 2007 (UTC)

I think you're confused. Current doesn't "come from" a battery. And current is not energy. The charges that flow in a circuit are provided by the wires, while the battery is like a charge pump. And charge is an entirely different thing from energy (for example, charges flow slowly, while electrical energy moves at nearly the speed of light.) Think of a wire: whenever we force some charges into one end of a wire, DIFFERENT charges flow instantly out the other end. And with capacitors, when we force some charges into one terminal, DIFFERENT charges flow instantly out the other terminal. Same process: the first charge pushes on the next charge, which pushes on the next one. Capacitors do act different than wires: in order to create a constant current through a capacitor, you have to constantly *change* the voltage. (Or in other words, capacitors only behave as conductors for AC.) With wires, we can create a constant current by applying a constant unchanging voltage. --Wjbeaty 02:37, 16 December 2006 (UTC)

I really need help here. What do you mean by an imbalance of charge? I'm still confused. And while I'm here, does anyone know about capacitors and capacitance under different temperatures? And just to let everyone know, I'm not a user (yet).—The preceding unsigned comment was added by 207.63.251.238 (talk) 19:27, 12 January 2007 (UTC).

Yes, but anyone can sign their posts to "talk" pages by including four tildes (~~~~) after their posting. When you press "Save page", they will be replaced by your username or IP address in a handy Wikilinked format. A timestamp wil also be included.

Atlant 19:34, 12 January 2007 (UTC)

Sorry, I'll fix that. Thx. And can anyone answer my FIRST question? Thx in advance. 207.63.251.243 18:54, 12 February 2007 (UTC)

Charge imbalance? That's when you have more positives than negatives, or vice versa. For example, neutral matter is made of charge (made of positives and negatives,) so it contains an enormous amount of charge. Yet we often say that uncharged matter contains no charge. The word "charge" can lead to semantic problems: how can matter be full of charge, while it also has zero charge? Simple: matter can have ZERO CHARGE IMBALANCE, while at the same time it contains huge amounts of positive and negative charge. Where capacitors are concerned, we must notice that the charge within a capacitor never changes, and a "charged" capacitor contains just as many charges as an "uncharged" capacitor. Capacitors are not intended to store charge. However, if one capacitor plate has excess negatives, and the other plate has excess positives, then that capacitor contains an imbalance of charge. Or said differently: if we start with neutral capacitor plates, then we move some electrons out of one plate and into the other, then the number of electrons inside the capacitor has not changed. Only the amount of imbalance has changed. --Wjbeaty 20:40, 12 February 2007 (UTC)

I thank you, god of technology. 207.63.251.231 18:51, 16 February 2007 (UTC)

Capacitors in series

A wire is supposed to have capacitance zero. If I make a circuit with a battery connected to a capacitor of capacitance C2, that would essentially be a capacitor in series with 2 wires. Using formula C = 1/(Σ(1/C)), I get C = 1/(1/C1 + 1/C2 + 1/C3) = 1/(inf + 1/C2 + inf) = 0. Inf stands for infinity. Could someone explain why it is possible to get a capacitance in a circuit? Please don't laugh. I'm a beginner. 69.12.155.64 01:47, 22 April 2007 (UTC)

Whenever a contradiction seems to appear, check your premises. A capacitor is formed from two conductors separated by a dielectric. A single wire does not form a capacitor. The two wires in your circuit form a capacitor in parallel with the capacitor C2.

On the other hand, one might make the case, using a limit type of argument, that an ideal wire has infinite capacitance with an initial 'charge' exactly zero. With infinite capacitance and finite current, the time rate of change of voltage is zero so the voltage 'across' the wire remains zero always and further, the charge 'stored' is always zero. In this limiting case, the two 'infinite' capacitors (wires) in series with C2 combine to give an equivalent capacitance equal to just C2. Alfred Centauri 03:02, 22 April 2007 (UTC)

One of the external links is troublesome

http://www.uoguelph.ca/~antoon/gadgets/caps/caps.html

While this page does what it's said to do (discuss how to read capacitor markings), the text in that topic is confusing, the micro symbol is not to be seen (and might actually be encoded as 'f', of all things), and I'd drop the site from the links just for

You will find plenty of folks, especially here in the U.S. who don't like the Euro way of marking. But these same folks don't like change. As a matter of fact, a lot of these old clowns will still use a "33 em-em-eff condenser to set the frequency of their 4.5 megacycle" oscillator. Just ignore them!

Anonymous129.7.202.112 21:09, 30 April 2007 (UTC)

As far as I'm aware, there is no Euro way of marking, simply an (North?) American way of marking and an international way of marking that's use throughout most of the rest of the world in Europe, Oceania, Asia and I expect Africa and probably South America too. BTW, the site does use the µ symbol, just not in the section comparing American markings and a few other sections. I've written to them to inform them about the problem Nil Einne 13:38, 25 June 2007 (UTC)

"As a matter of fact, a lot of these old clowns will still use a "33 em-em-eff condenser to set the frequency of their 4.5 megacycle" oscillator. Just ignore them!"

This is perfectly valid terminology (if you correct your odd spelling first), even if old fashioned. A competent engineer needs to understand all the standards used in their field of work. If they don't, thats their lack of knowledge. Not much use blaming everyone else who uses a standard that isnt the only one they've learnt. Tabby (talk) 07:33, 19 January 2008 (UTC)

Polar and Non-Polar

Can non-Polar capacitor be used where Polar is shown in circuit diagrams. Can we interchange. zaheershamsi1894@hotmail.com

Yes. Polarised capacitors have some advantages - lower cost, smaller size - but unpolarised ones work just as well, and usually better. You cannot generally use a polarised capacitor when an unpolarised one is specified, though. --Heron 13:03, 30 June 2007 (UTC)

Net charge on a capacitor

Yes Alfred, wee all know the net charge on any capacitor is zero. Its the charge separation that counts.--SpectrumAnalyser 21:36, 11 August 2007 (UTC)

We also all know that the voltage across a capacitor is proportional the magnitude of the charge Q 'on one of the plates'. It is not proportional to the difference in charge which is Q - (-Q) = 2Q. Why did you put this misleading statement into the caption, SA? Alfred Centauri 01:02, 12 August 2007 (UTC)

Reason :Say you have a discharged capacitor with one plate connected to ground. Then you put a charge of Q coulombs on the other plate. The charge difference is Q =CV.--SpectrumAnalyser 10:59, 12 August 2007 (UTC)

Nonsense. Placing Q coulombs of charge on the other plate will induce a surface charge density on the grounded plate equivalent to a charge of -Q on that plate. You'll have to do better than that, SA. Alfred Centauri 14:07, 12 August 2007 (UTC)

Consider a capacitor with one plate connected to an infinite charge source or sink (like the earth maybe) and place a charge of Q on the unearthed plate. Charge on the earthed plate cannot change by definition (its an infinite source/sink). Charge cannot be induced into an infinite source/sink. So total charge difference is Q = CV :-)--SpectrumAnalyser 18:25, 12 August 2007 (UTC)

SA said "Charge on the earthed plate cannot change by definition (its an infinite source/sink)". Poppycock! By definition, it is the potential of an infinite source/sink cannot change when any amount of charge is added or removed. Bottom line, -Q Coulombs of charge flows to the 'earthed' plate from the infinite source/sink. Alfred Centauri 19:15, 12 August 2007 (UTC)

How you gonna tell the diff between infinity and (infinity +/-Q). Is there any diff? I agree charge can flow to/from earth, but it cant substantially alter the amount of charge (or the voltage) on an infinite source of/sink for charge.--SpectrumAnalyser 19:26, 12 August 2007 (UTC)

You can't and we agree on what you have just said. But, I disagree with you when you say "Charge on the earthed plate cannot change by definition (its an infinite source/sink)". As is well known, we can induce a charge on a conductor as follows: (1) electrically connect a conductor to earth, (2) place a charged object close to the earthed conductor, (3) disconnect the earthed conductor from earth. Now, the conductor is charged. It received that charge from the earth when the charged object was placed close by. By disconnecting the conductor from earth before the charged object is moved away, the charge on the conductor does not have any way to flow back to earth [[4]]. Note that the earth is not left with a net charge because, just as you said, infinity minus some finite number is still infinity. Also note that this implies that the earth is still ground - 0V - no change in potential between the earth and other charged objects.

Now, if you carefully consider this example, you will see that it is exactly the same as the capacitor. The earthed plate received charge from the earth due to the nearby presence of the charged other plate. Disconnecting the earthed plate lead would reveal an electrically neutral capacitor with equal (Q) and opposite charges on it's plates. Alfred Centauri 20:03, 12 August 2007 (UTC)

So, if one plate of an uncharged capacitor is earthed, and charge +Q then pumped into the other plate, (from the earth ultimately of course), the amount of charge moved is Q = CV and the earth becomes negatively charged by the amount Q (even though you cant tell because its voltage remains the same)? I didnt consider the amount of charge (re)moved from the earth because it has no effect on the earths voltage and it therefore appeared to be negligible.. Hence the solution to the apparent paradox. I see my error now--SpectrumAnalyser 20:20, 12 August 2007 (UTC)

I'm not sure you can say the infinite source/sink becomes negatively charged. The infinite source/sink has infinite positive charge and infinite negative charge. Addding or removing charge of either type doesn't change this. The net charge apears to remain zero. If instead, the amounts of positive and negative charge were merely very large, the net charge does change. But, assuming the earth is still very large, the charge density would still be very small. Do you see? The infinite source/sink is infinite spatially as well as chargically. Alfred Centauri 20:35, 12 August 2007 (UTC)

Well those were my thoughts initially, which is why I said

it (charge flow) cant substantially alter the amount of charge (or the voltage) on an infinite source of/sink for charge.

ps Well it cant alter the charge at all- can it?

What does happen though is that charge is transferred from the very large bucket of charges(earth) to the other plate of the capacitor without substantially altering the earths net charge (although it must do somewhat as the earth is not actually infinite). The amount of charge transferred Q = CV. Anyway doesnt it all depend on whether you define the earth as having zero net charge? If you do define it as such, then Q IS the charge difference, isnt it? Curious.--SpectrumAnalyser 21:57, 12 August 2007 (UTC)

So, taking the argument one step further: if we have an electrical system where we define our 0v rail or chassis to have net zero charge and to be an infinite repository of both negative and positive charges (we usually do otherwise we would be putting a limit on the current we could draw from the supply), then it is true that the charge Q is now only on one plate and Q=CV, where Q is the charge difference. Do you agree?--SpectrumAnalyser 22:20, 12 August 2007 (UTC)

I can't agree for the reason I gave above (I also linked to another explanation). There is a charge density on the earthed plate. That is, there is more of one type of charge on the earthed plate than the other type. That doesn't mean that the system formed by the plate, the lead, and the earth are charged. But, the earthed plate is charged. Think about the method of images in electrostatics. A charged plate above an infinite conductor induces a charge density in the conductor below it. The infinite conductor as a whole is not charged but, below the charged plate, there is a non-zero surface charge density. If you integrate this charge density over the area below the plate, you get the equal but opposite charge on the charged plate above. Alfred Centauri 22:32, 12 August 2007 (UTC)

---

The caption says that the voltage is proportional to the charge transferred. In your model, with +Q on one side and - Q on the other, the amount of charge transferred from one side to the other must have been Q. So the statement is now correct.--SpectrumAnalyser 11:07, 12 August 2007 (UTC)

Suits me. Alfred Centauri 14:07, 12 August 2007 (UTC)

Etymology

Why did we start using "capacitor" instead of "condenser", anyway? — Omegatron 03:43, 12 August 2007 (UTC)

So we wouldn't have to define condensetance??? Alfred Centauri 14:10, 12 August 2007 (UTC)

Would it not have to have been 'condensation'?--SpectrumAnalyser 18:27, 12 August 2007 (UTC)

Interesting. Other languages use a variant of "condenser" for the device, but use a term like "capacity" or "capacitance" for the property. es:Capacidad (condensador)— Omegatron 23:05, 6 September 2007 (UTC)

"Why did we start using "capacitor" instead of "condenser", anyway?"

Ego. Think about it. Tabby (talk) 07:36, 19 January 2008 (UTC)

Lead section

I believe that the following sentence:

"When current is applied to the capacitor, electric charges of equal magnitude, but opposite polarity, build up on each plate."

is more accurate than this version:

"When charge is made to flow into the capacitor, electric charges of equal magnitude, but opposite polarity, build up on each plate."

You said in your edit summary that "charge flows, current is a flow", implying that current doesn't flow (verb). You are correct, so it's a good thing the original version didn't claim that "current flows"!

Also, the phrase "charge is made to flow into the capacitor" isn't really accurate, as overall, charge neither leaves nor enters the capacitor.

Regards, Oli Filth 22:45, 7 September 2007 (UTC)

OK How about: when the capacitor is charged....

--212.139.83.93 22:48, 7 September 2007 (UTC)

The only problem with that is, it doesn't mention the method by which charging is performed. Oli Filth 22:51, 7 September 2007 (UTC)

Does it matter? Current is never 'applied'. Voltage may be applied- not current —Preceding unsigned comment added by 212.139.83.93 (talk) 22:52, 7 September 2007 (UTC)

I think it's important to address one of the fundamental characteristics of the capacitor in the lead section, i.e. how it's actually used.

I'm not sure what's wrong with the word "applied" here. If one has a current source, one can apply it to a component or network. (See this Google search for "current is applied".)

Incidentally, the word "applied" in this sentence is a hangover from when it used to read "When voltage is applied to the capacitor...". That's definitely wrong, so I transposed "current" for "voltage". Oli Filth 23:05, 7 September 2007 (UTC)

Yes now you see the difficulty...that why I said 'when charge is made to flow'. I suppose it could say when charge is separated on the capacitor plates....--212.139.83.93 23:24, 7 September 2007 (UTC)

I'm not sure which "difficulty" you're referring to? Oli Filth 23:35, 7 September 2007 (UTC)

The difficulty is deciding just what is applied to a capacitor for it to charge--88.110.43.183 23:47, 7 September 2007 (UTC)

Well, it's not a voltage. If one could apply an ideal voltage source to a capacitor, an infinitely large current would result. However, if one applies a current source, then the capacitor charges, and the potential difference grows linearly with time. I'm not sure what the problem is there! Oli Filth 23:50, 7 September 2007 (UTC)

Charge doesn't flow into a capacitor. Instead, charge is 'pumped' from one plate to the other which is exactly what a current source does if it is 'applied' (attached) to a capacitor. But, from the perspective of the current source, there is a current through the capacitor. I know that most will object to this since (ideally) no charge flows between the plates but, at least in EE, we say there is a current through a capacitor. —Preceding unsigned comment added by Alfred Centauri (talk • contribs) 01:06, 8 September 2007 (UTC)

Current is not applied and neither is voltage. Actually it is charge that is fed into one of the plates (from the other plate- or whatever is connected to the other plate: like earth for instance) by whatever means. Current doesnt flow: current IS (a flow of charge). A current appears to pass between the plates, but we know that charge CANNOT pass between the plates; so we are left with the only true statement;

A capacitor is charged by robbing charge from one plate and feeding that exact same quantity of charge (Q) to the other plate via the external circuit.

this brigs up an interesting question though: Is it possible to apply charge to only one plate of a capacitor without robbing it from the other plate (or whatever is connected to the other plate)?--88.110.43.183 13:52, 8 September 2007 (UTC)

I answer my own question:

(Positive) charge from plate B is transported via the external circuit to plate A. This leaves plate A more (positively) charged than plate B. Or conversely, plate B is more negatively charged than plate A. Now if we define plate B as having zero charge in our system (like we define earth to have no net charge), then plate A accumulates charge and the whole system (inc earth) becomes charged. Clearly impossible. Conclusion: Is is the whole SYSTEM (including the capacitor and the earth)has zero net charge. Talking of the charge on the earth separately from the whole system is meaningless.--88.110.43.183 14:16, 8 September 2007 (UTC)

Now if we define plate B as having zero charge in our system. Sorry, but you can't do that. Charge density is physical and measurable. I suppose you're thinking of re-defining the potential of plate B, right? Alfred Centauri 14:56, 8 September 2007 (UTC)

Yes I just proved my hypothesis wrong by getting an absurd result from it: violation of the [[conservation of charge

]] —Preceding unsigned comment added by 88.109.100.75 (talk) 00:36, 9 September 2007 (UTC)

Creating a new thread

This is my first post, and I don't know how to create a new thread, so I thought I'd join in here since everyone seems passionate about capacitors. I don't like the sentence about constant currents through capacitors being "not possible." Surely everyone knows that a linear voltage ramp will produce a very nice constant current until either the capacitor blows or the voltage ramp hits its limit. I'm not that familiar with the hydralics analogy, but as long as you didn't rupture the membrane, a constant current is possible. the lower the current the longer you can maintain it. How does one go about modifying such a misconception?Tjmozdzen 06:36, 20 September 2007 (UTC)

Let's keep this capacitor discussion going. What if you had a capacitor suspended in space, initially uncharged. A simple set of two parallel plates floating in outer space with vacuum between them. They are kept in place by artistic insulating arms, or pretend someone aligned them exactly and let them go so there are no leakage paths between them.

Let's also say that one of the plates, A, occasionally gets hit by an alpha particle with net charge of +2. The plates are so massive that we can ignore any motion associated with this hit for the duration of our measurement. Surely plate A will have a net positive charge of +2q and the other plate B will have an induced charge of -2q on the side nearest the plate A and +2q on its outer side. Perhaps the +2q on plate A will migrate from the outer face to the inner face, leaving 0q on its outer face. The net delta here is +2q, but only because I picked a goofy example of an alpha particle with charge +2q. If we had a beta particle of charge -q, then we could repeat the whole thing and talk of differences of +/- 1q.

this should mean that when we place a resistor on it, we'll get half the current as opposed to the case where we had equal and opposite charges on both sides.

I'm convinced that when we hook up a capacitor to a battery, we get +Q on one side and -Q on the other. When we short this capacitor out with a resistor, the excess electrons (-Q) simply redistribute to the other end of the plate (flow) so that if we integrate I(t) from t=0 to infinity, we get Q. Think of it as only the electrons flowing. If the other side had only 0Q and not -Q, then only half the current would need to flow to even things out.Tjmozdzen 08:11, 20 September 2007 (UTC)

NO charge carriers in vacuum

Alfred I am intending to revert change to reinstate the the statement that there are no charge carriers in a vacuum unless you can explain your revert.--88.109.53.38 14:39, 2 October 2007 (UTC)

Vacuum#Quantum-mechanical_definition Alfred Centauri 16:22, 2 October 2007 (UTC)

These are not charge carriers. And even if they were, they exist for infinitesimal time. Conduction current cannot exist in a vacuum. I intend to change the page back unless you can come up with a more sensible answer PDQ!--88.111.81.231 01:08, 3 October 2007 (UTC)

'They' are charge carriers. See Vacuum polarization. Alfred Centauri 02:14, 3 October 2007 (UTC)

How long do these charge carriers last? You cant get conduction current thro a vacuum! If you can, refer to an experimental result that proves it.--88.109.121.209 01:39, 4 October 2007 (UTC)

Google "conduction current in vacuo" —Preceding unsigned comment added by 88.109.121.209 (talk) 01:42, 4 October 2007 (UTC)

"How long do these charge carriers last?" How long do they not last (exist)?

"You cant get conduction current thro a vacuum!" Does that imply that charged virtual particles don't exist in the vacuum? See Unruh effect Alfred Centauri 01:57, 4 October 2007 (UTC)

No experimental evidence offered. Page will be changed back. You really must quote normal physics in WP, not your version of meta physics--88.109.198.32 22:24, 8 October 2007 (UTC).

The fact that charged carriers exist in the vacuum is not my version, LC. Please review the literature before offering such nonsense on the talk page. Alfred Centauri 23:41, 8 October 2007 (UTC)

Im not disputing the idea that possible carriers of charge may exist momentarily in vacuo. What Im saying is they cannot be used in a conduction mechanism. Is this your latest argument in favor of the existence of displacement current? BTW what nonsense are you referring to? 8-?--88.109.198.32 13:15, 9 October 2007 (UTC)

If I had the lab facilities, I would do an experiment with a couple of closely spaced conducting plates inside a bell jar pumped down to about 10^-6 torr and measure the current due to a dc voltage of 1kV. I would expect absolutely FA current! What would you expect?--88.109.198.32 00:07, 9 October 2007 (UTC)

Given that displacement current is proportional to the derivative of electric field strength, I guess Alfred would expect zero current as well! Oli Filth^(talk) 07:15, 9 October 2007 (UTC)

My 2 previous posts are not connected. The phantom particles do not exist (for long enough) to enable conduction current but I was wondering if Alfred thought they also might have something to do with displacement current.--88.109.198.32 13:15, 9 October 2007 (UTC)

Hello, I removed

Because there is nothing at that link in the Wikibooks

AC capacitor

I see capacitors for sale (e.g. at halted.com) listed as "AC capacitors". What are these? The article should say, I think. --Treekids (talk) 07:24, 5 December 2007 (UTC)

Certain capacitors (generally, "electrolytic capacitors") can only work on direct current circuits; they will fail catastrophically if you try to charge them "backwards" from their intended polarity. The rest of the capacitor types can work on alternating current. But a few capacitors are specifically designed to work on line (mains) voltage ac power circuits. These are commonly used in induction electric motors and the like. This class of capacitors is what is usually being referred to when you hear someone discussing "AC capacitors".

Atlant (talk) 20:43, 5 December 2007 (UTC)

These may also be capacitors intended for use across the ac line (ie mains rated). Here they are called class X or class Y capacitors. I cant recall the diff off hand. It appears that the capacitor article does not mention this type (yet).--TreeSmiler (talk) 02:19, 6 December 2007 (UTC)

Could also mean it can handle largish amounts of ripple current (as you may get when connecting across the mains)--TreeSmiler (talk) 02:26, 6 December 2007 (UTC)

Several design differences distinguish "AC capacitors". Yes, they are designed to handle large currents so they tend to be physically large. They are usually ruggedly packaged, often in metal cases that can be easily grounded/earthed. They also tend to have rather high DC breakdown voltages; I think you'll find that 1600 volts is a common breakdown voltage for an AC Capacitor used on a 120 VAC mains circuit. And finally, they tend to use designs where a brief fault in the capacitor will self-heal rather than lead to a catastrophic short-circuit fault. So they use materials like metalized plastic where, in the case of a small short-circuit fault, the metalization will burn away from the faulted area, leaving the capacitor ever-so-slightly reduced in value but otherwise operational. This allows them to survive brief transients (voltage spikes) on the power line.

Atlant (talk) 12:21, 7 December 2007 (UTC)

You be talking Class X/Y capacitors methinks. [5]--TreeSmiler (talk) 01:28, 8 December 2007 (UTC)

Removed from page

by --TreeSmiler (talk) 21:05, 31 December 2007 (UTC)

I think you'll find that 1600 volts is a common breakdown voltage for an AC Capacitor used on a 120 VAC mains circuit. And finally, they tend to use designs where a brief fault in the capacitor will self-heal rather than lead to a catastrophic short-circuit fault. So they use materials like metalized plastic where, in the case of a small short-circuit fault, the metalization will burn away from the faulted area, leaving the capacitor ever-so-slightly reduced in value but otherwise operational. This allows them to survive brief transients (voltage spikes) on the power line.

Why remove valid information? If everything imperfect is removed, wiki will cease growing. Tabby (talk) 07:41, 19 January 2008 (UTC)

Nonsense. If everything imperfect is removed, it will leave room for improvement. But nobody is going to remove everything imperfect; the process is incremental. Just help. Dicklyon (talk) 07:54, 19 January 2008 (UTC)

Paper dielectric

This type of capacitor is still being manufactured eg [6] and many other refs on google. Im therefore reinstating my edit.--TreeSmiler (talk) 18:24, 4 January 2008 (UTC)