# Talk:Capacitance

## Untitled

Max energy in a capacitor moved here from article page Not really relevant to subject of capacitance

If the maximum voltage a capacitor can withstand is ${\displaystyle V_{max}\,}$ (equal to ${\displaystyle E_{str}d\,}$ where ${\displaystyle E_{str}\,}$ is the dielectric strength), then the maximum energy it can store is:

${\displaystyle W_{max}={\frac {1}{2}}\epsilon _{0}\epsilon _{r}{\frac {A}{d}}E_{str}^{2}}$

— Preceding unsigned comment added by Light current (talkcontribs) 01:43, 6 September 2005

## Inconsistent use of Work / Energy

I think that the electrical equations should use E instead of W as well as the text should refer to Energy instead of Work. The concept electrical work is barely represented on Wikipedia, whereas electrical energy is much more detailed. But, being a Swedish engineer, I'm not confident enough about this linguistic issue to make the changes myself. Mumiemonstret (talk) 13:27, 18 December 2007 (UTC)

But looking at the page it seems that it would be best to use energy as the stored quantity (as work is actually the transfer of energy)--TreeSmiler (talk) 03:10, 19 December 2007 (UTC)

## Scope of Article

This article has been taken from article Capacitor to replace the former 'redirect to article Capacitor'. It has enabled the appropriate categorisation of the physical quantity capacitance, without introducing inappropriate categorisation for the physical objects, capacitors. I would hope that, in due course, it may be possible to remove the 'capacitance' aspects of the capacitor article to avoid duplication. Ian Cairns 15:03, 14 Nov 2004 (UTC)

The scope of this article is capacitance. There is a substantial amount of information on capacitors in this article that is, IMHO, outside the scope of capacitance. I propose that this information be deleted or merged with the Capacitor article. Alfred Centauri 22:49, 26 May 2005 (UTC)

I'm coming from a biological background. In neurones there is an important concept called the membrane potential. In studying the action potential a very critical determining factor is the capacitance of the membrane. Now I’m not particularly qualified to write much on the topic, but it would be good to include the biological perspective on membrane capacitance. -- Amelvin 14:53, 13 May 2006 (UTC)

## err

"Just as two or more inductors can be magnetically coupled to make a transformer, two or more charged conductors can be electrostatically coupled to make a capacitor. The mutual capacitance of two conductors is defined as the current that flows in one when the voltage across the other changes by unit voltage in unit time."

That's not a correct analogy, I don't think. A correct one would be something like "just as two inductors can be magnetically coupled to..., two capacitors can be electrostatically coupled to make a ___" - Omegatron 22:47, May 31, 2005 (UTC)
In other words, a capacitor is the dual of an inductor. This paragraph is trying to say that a capacitor is the dual of a transformer, which is wrong. - Omegatron 14:30, August 15, 2005 (UTC)
I was just thinking about this a couple of days ago. Essentially, I wrote down the matrix form of the equations for coupled inductors and then took the dual. The dual of the self inductances would need to correspond to the self capacitances of the conductors where the self capacitance is defined as the potential on the conductor w.r.t. the potential at infinity for unit charge place on the conductor. It seemed reasonable then that the dual of the mutual inductances would be the mutual capacitance which I would think would be the capacitance of the capacitor formed by the two conductors. However, I quickly became lost in the algebra and gave it up for the time being. Care to give it a try? Alfred Centauri 20:47, 15 August 2005 (UTC)
Ahh. Is this the difference between the "physicist's capacitor" and the "engineer's capacitor"? - Omegatron 21:06, August 15, 2005 (UTC)
Maybe so! Clearly, if the mutual capacitance mentioned above turns out to be the capacitance of the capacitor formed by the two conductors, then in general, the mutual capacitance is orders of magnitude larger than the self capacitances (the physicist's capacitance???). Whereas, the mutual inductance is, if I'm not mistaken given by the coupling coefficient k, where 0 <= k <= 1 multiplied by the geometric mean of the self-inductances. That is, the mutual inductance cannot be greater than the largest self inductance. I'm not quite sure what this leads to. Any ideas??? 00:34, 16 August 2005 (UTC)
I agree, the analogy is bad. Let's see - for magnetic circuits, we use high ${\displaystyle \mu }$ material so we can contain and get lots of magnetic flux for a little magnetic field intensity. But, for good electrostatic coupling, we would need a low ${\displaystyle \epsilon }$ material so we get lots of electric field for a little electric flux. Is there any material for which ${\displaystyle \epsilon <\epsilon _{0}}$? Alfred Centauri 04:52, 2 Jun 2005 (UTC)
"Nothing is going to have a relative permittivity less than that of a vacuum! All materials will therefore have a dielectric constant greater than 1." [1] - Omegatron 13:47, Jun 2, 2005 (UTC)
Removed comment to here for now. - Omegatron 14:27, August 15, 2005 (UTC)

## DC current through capacitor

It is actually quite incorrect to say that a capacitor blocks direct current. Consider the simple example of an ideal constant current source connected to an ideal capacitor. Clearly, the capacitor does not block this current. Instead, the voltage across the capacitor changes linearly with time at a rate given by I / C. That is, the voltage across the capacitor is unbounded. Of course, a real capacitor would eventually breakdown at some voltage. Or, a real current source would fall out of its compliance range. The phenomenon described in this article as justification for the claim that capacitors block direct current is in fact the observation that the DC steady state solution for capacitor current is identically zero.

Further, consider that a circuit that includes a capacitor or inductor is not, strictly speaking, a DC circuit! However, such a circuit may or may not possess a 'DC solution' otherwise known as the steady state solution.

All of this may seem pedantic but it is nevertheless true that capacitors do not block direct current. Alfred Centauri 15:27, 11 August 2005 (UTC)

If you suddenly apply an ideal current source across a capacitor, are you supplying dc, or ac? If instead a voltage source in series with a resistor is applied across the capacitor, then after an infinite time, current will be zero. At this time, (when you can consider this as a purely dc problem) the capacitor blocks dc.--TreeSmiler (talk) 00:22, 3 January 2008 (UTC)
"then after an infinite time... At this time...". Questions: What time is it after an infinite time has elapsed? After an infinite time, for how long a time does the capacitor block DC? Perhaps you should consider sharpening you idea of infinity. Alfred Centauri (talk) 02:33, 4 January 2008 (UTC)
Also I think your thought experiment is faulty. An Ideal current source does not exist in reality. And a practical current source can be converted into the equivalent (practical)voltage source model. When the (series) switch is closed (or shunt switch opened-- whichever), you have a transient situation. Transients are AC. The SS solution is ZERO. So the capacitor will accept charge until its terminal voltage equals the equivalent source voltage. Depending on the quality of your current source, this will take less than an infinite amount of time. Then it will stop accepting charge (passing current). What passes is ac. Capacitors do not pass DC!--TreeSmiler (talk) 04:03, 5 January 2008 (UTC)
I quote from the page on direct current:
Within electrical engineering, the term DC is a synonym for "constant". For example, the voltage across a DC voltage source is constant as is the current through a DC current source. The DC solution of an electric circuit is the solution where all voltages and currents are constant. It can be shown that any voltage or current waveform can be decomposed into a sum of a DC component and a time-varying component. The DC component is defined to be the average value of the voltage or current over all time. The average value of the time-varying component is zero.

Do you disagree with this definition?--TreeSmiler (talk) 04:12, 5 January 2008 (UTC)
As I wrote the quoted section in the DC article, it is unlikely that I disagree with it. But, it's not my thought experiment that's faulty, it's your 'logic'. What you have claimed above is that a capacitor blocks DC because ideal current sources do not exist. Such 'logic' is just sloppy thinking. Consider that if I replace the fully charged capacitor in your circuit with a battery of identical voltage, the current through the circuit would remain zero. Do batteries block DC too?
Further, if you insist that only 'real world' sources and capacitors be considered in your 'argument' then be aware that DC doesn't 'really' exist. Look at the definition above closely: "The DC component is defined to be the average value of the voltage or current over all time". That is, a DC source has a constant output forever (all time). The moment you use the term DC (as defined above), you enter the world of ideal components and, in that world, capacitors do not block DC. Alfred Centauri (talk) 14:50, 5 January 2008 (UTC)
Actually, looking at your (self admitted) definition of DC, it does seem to be rather self contradictory. In the first sentence you say that DC is a constant (ie unchanging). But you weasel out of it by saying that dc is not necessarily constant, but is only an average value. Which is it? Maybe your thinking needs clarifying?--TreeSmiler (talk) 15:06, 5 January 2008 (UTC)
"you weasel out of it by saying that dc is not necessarily constant, but is only an average value". Really? It should be self evident that a time average over all time has no time dependence (do the math [2]) and thus is constant . There's no contradiction there. But, if you think that there is, please bring this to the attention of the authors of the textbooks and references that I have because they disagree with you. Alfred Centauri (talk) 15:53, 5 January 2008 (UTC)
I have no argument with the mathematical statement made in the link that you so kindly provided for my edification. However, I feel you may have misinterpreted what it is actually saying. It is saying that for the integral to be the DC value of a waveform, the waveform has to be periodic: your current signal into a capacitor is not periodic; it is just unidirectional. If you are saying that a dc signal is just one that is unidirectional (as your current signal is) then you have contradicted yourself by relying on the above mathematical definition of dc.--TreeSmiler (talk) 20:59, 5 January 2008 (UTC)
The misinterpretation is yours, TS. First, the link gives two equations for <x(t)>; one for aperiodic signals and one for periodic signals. Then the link clearly states "<x(t)> is the DC value of x(t)". Again, there is no contradiction. Alfred Centauri (talk) 22:50, 5 January 2008 (UTC)
The link is not very clear on that point (ie to which equation <x(t)> is the dc value applies). Anyway its obvious to me that if you have an ac signal riding on an offset, the dc value of the resultant is indeed the long term average value of the whole thing. Can you provide a book ref to back up your claim? Strangely, I cant find a defn of 'DC' in any of my books so far.--TreeSmiler (talk) 23:51, 5 January 2008 (UTC)
(after edcon) Alfred : on ruminating on this over a few libations, I dont think we really disagree. How about you compromise your statement to say 'a capacitor can pass unidirectional current only for a finite time (until it is fully charged)'?--TreeSmiler (talk) 22:56, 5 January 2008 (UTC)
Where in the capacitor equation is there a mathematical description of the notion of 'fully charged'? The capacitor equation clearly allows a constant current without regard to any concept of 'full charge'. Question: Does the capacitor itself or the circuit connected to the capacitor determine what 'full charge' is? In a first order RC circuit, does the capacitor become 'fully charged' in finite time? Alfred Centauri (talk) 23:55, 5 January 2008 (UTC)
The C is fully charged when its terminal voltage is equal to the terminal voltage of the source, obviously. Even though I feel you are avoiding my point, I will respond. In reality, the capacitor is fully charged in a finite time. In theory, it takes an infinite time. But this argument depends upon how you look at charge; is it quantized or not?--TreeSmiler (talk) 00:29, 6 January 2008 (UTC)
"The C is fully charged when its terminal voltage is equal to the terminal voltage of the source, obviously". Really? Consider a parallel combination of an ideal current source, a resistor, and a capacitor. Isn't the capacitor voltage at all times equal to the voltage across the current source even when the current through the capacitor is non-zero? I think you need to re-think your definition of 'fully charged'.
"is it quantized or not?" Introducing charge quantization doesn't change the fundamental result that capacitors do not block DC. Alfred Centauri (talk) 01:24, 6 January 2008 (UTC)
I do not accept ideal current sources as they do not exist. PS Were still waiting for your defn of DC--TreeSmiler (talk) 01:33, 6 January 2008 (UTC)
If you do not accept ideal components, then you do not accept the mathematical basis of electrical engineering. Go find something else to write about that you do accept, and leave this topic alone.71.240.58.110 (talk) 20:54, 17 January 2009 (UTC)
"I do not accept ideal current sources as they do not exist". Irrelevant to this issue. Further, the parallel resistor models, to first order, a non-ideal current source. The point remains that your definition of 'fully charged' is, shall we say, lacking.
"PS Were still waiting for your defn of DC". 'My' definition of DC is simply the accepted definition of DC that I have in my text and reference books and that definition you already have. Alfred Centauri (talk) 02:01, 6 January 2008 (UTC)

Another problem: it is stated in the AC Circuits part of the article about capacitive reactance: "Since DC has a frequency of zero, the formula confirms that capacitors completely block direct current." This is a case of extrapolating a concept beyond the point that it is valid. The concept of reactance in AC circuits is intimately tied to the notion of AC steady state where the peak amplitudes of the sinusoidal voltages and currents are constant. Thus, setting the frequency to zero is not valid for the reactance formula. Consider letting the frequency be arbitrarily close to zero. It is evident that a small peak amplitude sinusoidal current through the capacitor produces an arbitrarily large peak sinusoidal voltage across the capacitor. In the limit as f goes to zero, the peak voltage does in fact go to infinity. What must be understood is that this peak voltage is, by definition, the voltage in AC steady state which, due to the fact that the frequency is zero, is never reached in finite time. The correct interpretation of this formula is that for an AC current of zero frequency, the voltage increases with time and is unbounded as t goes to infinity - precisely the result I described above for a constant (DC) current source. Alfred Centauri 01:44, 12 August 2005 (UTC)

So your argument is that it doesn't block dc current--it just requires infinite voltage to pass dc. Similarly an open switch or a blown fuse doesn't block current, it just requires infinite (in theory, for an ideal switch) or very large (in practice, for physical element) voltage. I don't think we need to re-write all of Wikipedia taking into account thought experiments involving non-physical infinite contingencies. Ccrrccrr (talk) 14:26, 5 January 2008 (UTC)

"So your argument is that it doesn't block dc current--it just requires infinite voltage to pass dc". No, my argument is not that. Try again.
"I don't think we need to re-write all of Wikipedia taking into account thought experiments involving non-physical infinite contingencies".
Since no one has suggested such a thing, your comment seems particularly out of place. Alfred Centauri (talk) 16:06, 5 January 2008 (UTC)
Alfred I suggest you come up with your exact definition of DC. Then well all know what we are discussing.--TreeSmiler (talk) 21:03, 5 January 2008 (UTC)
You described a current source feeding the cap, and the capacitor producing the infinite/unbounded voltage. But the behavior of the capacitor would be the same if you provided a ramping voltage source--then there would be a constant current in the cap. In either case the source, whether it's ramping voltage or constant current, would need to be capable of infinite voltage in order to maintain constant current in the cap.
My comment about re-writing Wikipedia was intended to point out that I think it is absurd to remove the statement that a cap blocks dc based on contingencies that require infinite voltage. Sorry if that message got lost somehow, or more likely, my sarcasm may have been offensive and probably isn't appropriate here.
I don't think that any of us disagree about what happens in the thought experiments you've proposed. What we should be discussing is whether the statement that capacitors block dc is a useful description that would help readers. It's easy to find references in support of the idea that capacitors block dc. (e.g. 57 results in google books search.) The talk page isn't supposed to be a forum in which to conduct original research to then put into the article. So I think we should ask AC to find a reference that supports his view. 01:48, 6 January 2008 (UTC)
"would need to be capable of infinite voltage in order to maintain constant current in the cap". At what finite time would the source or cap produce an infinite voltage?
As I said, we don't disagree about what happens in your though experiment. It wouldn't need an infinite voltage at a finite time. But it wouldn't be dc in a finite time...that's the theoretical answer. The practical answer is that electrical engineers regularly use caps in circuits for the purpose of blocking dc. Ccrrccrr (talk) 03:28, 6 January 2008 (UTC)
"So I think we should ask AC to find a reference that supports his view". My 'view' was expressed in 2005. I put a short sentence in at the time to the effect that although it is often said that capacitors block DC, this is not technically correct. That sentence was removed a short time later and that was the end of that. What exactly is it you would like for me to do now in 2008? Alfred Centauri (talk) 02:08, 6 January 2008 (UTC)
I just happened across this debate. Most books say that caps block dc. Here's one WP editor who thinks otherwise. I don't think the article should match that minority view, unless I'm mistaken that it's a minority view. I'd be open to having something that says "although it is often said that capacitors block DC, this is not technically correct" if there is something we can cite for the "this is not technically correct" part. Otherwise, I think we should simply say that it blocks dc. But maybe I'm missing something in the history--perhaps there is more support for your point of view than I see here.Ccrrccrr (talk) 03:28, 6 January 2008 (UTC)
OK, so I just read the history. Turns out the part you (AC) wrote that you are quoting was in fact a huge improvement over the nonsense that preceded it. And it didn't get deleted from this article because someone didn't like it--rather it got moved to the capacitor article where it has continued to get improved and that's all fine and good. The other place there was stuff about that was in the discussion of AC impedance. I think that should be restored, even though it seems that the article has done without it OK for several years. But it's not that big a deal--it would be a problem if the article made an unverified claim that caps pass dc, but simply omitting the generally accepted fact that they block dc is not a major flaw.Ccrrccrr (talk) 03:59, 6 January 2008 (UTC)

## Edits by Machtzu

I have reverted the edits by Machtzu for the following reasons:

(1) Reactance is the imaginary part of the impedance and as such, is real:

${\displaystyle Z=R+jX\,}$
Capacitive reactance is a real number:
${\displaystyle X_{C}={\frac {-1}{\omega C}}}$
Please do not put the imaginary unit in the formula for ${\displaystyle X_{C}}$.

(2) Impedance is complex in general but capacitive impedance is pure imaginary:

${\displaystyle Z_{C}=jX_{C}={\frac {-j}{\omega C}}={\frac {1}{\omega C}}e^{-j{\frac {\pi }{2}}}}$
Please do not add a real part to the formula for ${\displaystyle Z_{C}}$.

Thank you. Alfred Centauri 18:04, 31 August 2005 (UTC)

## Electric cct stuff inappropriate

All the 'electric cct' stuff in this article is really to do with practical capacitors and not with the concept of capacitance. I therefore propose moving all this stuff to the capacitor page. Any comments/objections etc? Light current 17:33, 3 September 2005 (UTC)

Thanks for your vote of confidence!! ;-) Light current 18:28, 3 September 2005 (UTC) Done it! Light current 18:42, 3 September 2005 (UTC)

## Alternate definition of capacitance

'LC': I'm not so sure about your alternate definition of capacitance. For inductance, we find the magnetic flux by integrating (B dot dS) along a closed contour that defines a bounded surface. We find that this flux is determined by the total electric current through the surface. If we have a coil of wire, the current through the wire pierces the surface N times where N is the number of turns in the coil. Thus, for a given amount of flux, the required current through the coil divided by N. The inductance is defined as the ratio of the flux to the current through the coil:

${\displaystyle L={\frac {\lambda }{i}}={\frac {N\phi }{i}}\,}$

where ${\displaystyle \phi }$ is the flux produced by the current i alone.

If you are trying to find an analogous equation for capacitance, you would start with finding the emf by integrating (E dot dS) along a closed contour that defines a bounded surface. Find that this emf is determined by the negative of a total current through the surface. What current? Let's call it the magnetic flux current in analogy to the electric flux current (displacement current). Now multiply this result by the permittivity of the medium and then find that the ratio of the electric flux to the magnetic flux current has units of capacitance. This is your analogous equation:

${\displaystyle C={\frac {\phi _{E}}{\frac {d\phi _{M}}{dt}}}\,}$

Regardless, I do feel that such and alternate and non-standard 'definition' needs to be at the end of the article as an interesting footnote. Alfred Centauri 19:49, 3 September 2005 (UTC)

Thanks for your comments. I am trying (as you can probably guess) to link the ideas of 'capcitance' and 'inductance' together so that the understanding of one concept helps the understanding of the other. However I may have got my alt defn for capacitance wrong. I'll look at that. Light current 20:11, 3 September 2005 (UTC)
As Q = Øe, (where Øe is the electric flux) according to Gauss Law, I'm not sure where my second equation is wrong. (unless its the differential form you disagree with) Light current 20:17, 3 September 2005 (UTC)

Give me an example of how you would apply your formula to some physical system to calculate a capacitance. Alfred Centauri 20:31, 3 September 2005 (UTC)

Well, unless you could measure the electric flux (Øe) directly then it would be difficult. But the electric flux is equal to the enclosed charge by Gauss Law and electric charge is easy to measure with an electrometer So my equation is just another interpretation of the standard equation but one that mentions electric flux in order that 'parallels'/comparisons with the inductance case can be drawn. Also, I dont think the dimensions of your proposed analogy are correct (Q=CV)(although I may be wrong on that) :-) Light current 20:46, 3 September 2005 (UTC)

I'll answer your last comment first. The numerator contains the electric flux in Coulombs. The denominator contains the time rate of change of magnetic flux which has units of weber/second. But a weber is Volt-seconds. Thus so 1 weber/second = 1 Volt. Thus, the ratio is Coulomb/Volt = farad.

Now, consider this. Imagine a parallel plate capacitor with +Q on the top plate and -Q on the bottom and separated by a distance r. The electric flux that exists between the plates depends on the value of Q but not on r. On the other hand, the voltage between the plates depends on the value of Q and r. Thus, dphi/dv = dphi/dr dr/dv = 0. So, this formula for capacitance doesn't give the correct answer. If you use C = phi/V, you get the right answer for a parallel plate capacitor if phi is the flux out of a surface containing one of the plates and V is the voltage across the plates. If you have an isolated conductor, phi is the flux through a surface enclosing the conductor and V is the potential with respect to infinity. But, it is the application of Gauss Law here that gives you the form C = Q/V, not the other way around. BTW, compare the definition of inductance I gave above to the wording in the new opening sentence for the Inductance article. Alfred Centauri 21:46, 3 September 2005 (UTC)

Yes you are quite correct on your units (dimensions)Its just that it looked strange to me to be bringing in the magnetic field when I was trying to look at things from the electrostatic point of view . My mistake - my apology! I'm still considering your other statements.BTW Im thinking of an isolated sphere when I talk about capacitance (not a parallel plate cap. does that make any difference to the argument? Light current 22:33, 3 September 2005 (UTC)
Ive changed my new equation at the top of page. Now since Q is identically equal to Ø, you surely must agree that this equn is correct?? :-) Light current 23:33, 3 September 2005 (UTC)
Just a mischevious thought based on my wild general induction theory -- is Ø somehow induced by Q?? Light current 23:18, 5 September 2005 (UTC)

The alternate definition in terms of the flux is incorrect. Q is not identically equal to the flux, not in cgs units, and not in SI units. At best they differ by a factor of 4pi, and even then there would need to be an explanation of what gaussian surface was being referred to.--24.52.254.62 16:11, 3 November 2006 (UTC)

## 'Back Current' in Capacitors

Never heard of it?? No neither have I. Yet this is what we are to infer exists if we believe in back emf for inductors (since capacitors and inductors are duals of each other). If a capacitor has been charged and then the leads are short circuited, a large current will flow in the opposite direction to the original charging current. This is never called a back current yet is is the exact analogy of an open circuited inductor with a magnetic field surrounding it (due to previous current in the inductor). Note that the emf generated in an open cct inductor is caused by the collapsing magnetic field previously established (or in deed any external changing magnetic field). The fact that the mag. field may have been, in the first place, generated by the inductor is neither here nor there. The inductor is not aware of this.

Also, when a capacitor is being charged, there is no mention made of the electric field generated producing a back emf, yet this is exactly what it does do. The more voltage on the plates, the more back emf generated thus giving the expected capacitor charging response by opposing the applied voltage. Why have we not heard of this? Any comments?? Light current 04:59, 4 September 2005 (UTC)

'LC': First, let me say that I think it is great that you are asking such questions and making bold claims. I wish more of my students would even start to think like this and ask these kind of questions. However, I do recommend thinking through some of your arguments a little longer because some of the statements above are clearly based on flawed logic. Examples:

I'm not neccesarily advocating the proposed point of view on back current. I was engaging in sarcasm trying to debunk the ideas of 'back' anything and using a capacitor as a 'dual' example of an inductor. The point is, current does flow out of the capacitor but we dont call it a back current!.Light current 21:24, 4 September 2005 (UTC)

I see what you're saying. The voltage across an inductor when it discharges quickly is sometimes called a kickback voltage (that's why relay coils have snubber diodes, right?). OK, I get it now. Alfred Centauri 22:18, 4 September 2005 (UTC)

Yes, sorry. Over here people call this 'kickback' a 'back emf'. I think it's an induced voltage due to collapsing magnetic field-- this is obviously in reverse direction to the initial applied voltage, but this is OK because it is created by an 'external' field (ie the field that the coil created before it was switched off). This obeys Amperes law of induction? I think that the direction of energy flow may help us out in terms of sign of voltage/current here-- Any thoughts? Light current 22:46, 4 September 2005 (UTC)

(1) The discharge of a capacitor or an inductor is some kind of 'back' something or other - wrong! It is the rate of both charge and discharge that is affected by a back something!

Yes I = Cdv/dt. I is rate of charge affected by back emf in a capacitor.Hence slowing its charge rate up. Yes?? Light current 21:24, 4 September 2005 (UTC)

No! What if a capacitor is connected directly to a voltage source? The charge rate is unlimited. The voltage (not the rate of change of voltage) that builds on a capacitor (is this what you are calling the back emf?) can only affect the current if there is resistance in the circuit since the current is then given by Ohm's Law (Vs - Vc)/R. Alfred Centauri 22:18, 4 September 2005 (UTC)

Sorry, I thought I mentioned a cap charging thro' a resistor-- but I may have forgotten to mention it in the excitement! (Bear in mind I'm trying to debunk the notion of 'back emf', 'back current' by saying that it can all be explained by the simple familiar equations. So I'm not putting up bait for you, I'm simply playing Devils advocate. Light current 22:52, 4 September 2005 (UTC)

Afterthought: There is current between the plates of a capacitor - the displacement current. Note that the displacement current produces the magnetic field as that of the electric current. So in fact, the charge rate IS limited - by the self inductance of the capacitor!!! Alfred Centauri 22:32, 4 September 2005 (UTC)

Yes, you are of course strictly correct, but I dont think we should be sidelined by considering the capacitor inductance-- its complex enough as it is! Light current 23:01, 4 September 2005 (UTC)

(2) What is this about a 'back emf' when a capacitance is charging??? If you are taking the dual, voltage goes to current and current goes to voltage and L goes to C. Thus, the dual would be:
"The induced current through a capacitor due to a changing electric flux between the capacitor plates produces an electric field that opposes the change in electric flux that induced the current".
If a capacitor is being charged from a voltage source via a resistor, the changing electric flux generates a 'back current' in the capacitor in opposition to the charging current so there is actually no current passing between the plates of the capacitor. Right? My brain is starting to hurt now. Having short rest! Light current 21:45, 4 September 2005 (UTC)

If a capacitor is being charged from a voltage source via a resistor, the changing electric flux between the plates generates a magnetic field in opposite? direction but exactly equal in magnitude to to the magnetic field that would be caused by a real current passing between the plates of the capacitor. This real current passing between the plates we call 'displacement' current. Now the question is, is this displacement current real or is it just imagined to exist because the magnetic field between the capacitor plates indicates it does? If this current actually exists, one would assume that it was in the same direction as the conduction current and gives a complete circuit for the battery to drive charge around. If it does not really exist, then charge will stack up on the top plate with nowhere to go. Now in reality, we know that charge does indeed stack up on the top plate (assuming bottom plate is earthed/ common/ 0v etc). So what does this say about displacement current -- is it a lie? No I dont think so, but the only solution to this anomaly (that I can see) is that there must be a current generated inside the capacitor in the reverse direction to that flowing in the wires outside the capacitor. So there is current apparently flowing between the plates but its cancelled by an equal but opposite current generated by the aforementioned magnetic field. This seems to me to satisfy three apparently contradictary arguments:

a) there is currrent flowing between the plates,

b) there is no current passing between the plates and

c) charge is forced to stack up on the top plate becuase it cant pass thro the dielectric. Now my brain really hurts! Light current 00:18, 6 September 2005 (UTC)

Forget the voltage source and resistor - it is simpler to just use a current source. Further, I recommend that you generalize you concept of current. Think of electric current as (dQ/dt + d${\displaystyle \Phi }$/dt). That is, electric current is the flow of electric charge plus the flow of electric flux. When you do this, your brain will stop hurting for then you can say:
There is an electric current through the capacitor but there is no charge flow through the capacitor.
As long as the current source provides a constant current, there is no 'back current' because a constant displacement current does not generate emf. On the other hand, if the current source changes with time, there is a 'back current' due to an induced emf but this is an inductive phenomenon that is miniscule unless the change in current is very large. Alfred Centauri 01:54, 6 September 2005 (UTC)
Changing electric flux doesn't generate an electric current! Didn't you read my comments below??? A changing electric flux could generate a magnetic current (a current of magnetic charges). Take another look at Maxwell's equations. dD/dt generates a magnetic field that could drive magnetic charges (if they existed) just as dB/dt creates an electric field that drives electric charges. Alfred Centauri 22:18, 4 September 2005 (UTC)
Thus, you should be referring to a 'back current'. However, the real problem is this: an electric current doesn't produce an electric field!!! To get the dual that you are looking for, you need a magnetic charge current density term to make Maxwell's equations symmetrical. Then you could say:

"The induced magnetic current through a capacitor due to a changing electric flux between the capacitor plates produces an electric field that opposes the change in electric flux that induced the magnetic current".

Alfred Centauri 21:02, 4 September 2005 (UTC)

## capacitor not physical dual of inductor

LC: Sorry, but I had to revert your statement that a physical capacitor is dual to an inductor because it is not true. This is the insight I came upon this afternoon. A physical capacitor is governed (mostly) by the divergence equations in Maxwell's equations while a physical inductor is governed (mostly) by the curl equations. Thus, they are not physical duals. They operate on different physical principles. The duality in circuit theory is a mathematical duality. The amazing (to me anyhow) duality I discovered this afternoon is that there are actually two measures of inductance - one has units of henries and the other has units of farads! Similarly, there are two measures of capacitance - one in farads and the other in, you guessed it, henries. I'm still working on the presentation of these ideas in my sandbox. Alfred Centauri 02:10, 6 September 2005 (UTC)

OK Alfred. L&C Mathematical duals but not physical. Yes I am very eager to learn more of this. This is now getting very interesting. Light current 02:18, 6 September 2005 (UTC)

Here's the short version. The origin of the physical non-duality is in the nature of the fields but not in the sense you might be imagining. I'm sure you heard it said that "a capacitor stores energy in the electric field" and "and inductor stores energy in the magnetic field". So yes, the fields are different in this regard. But the difference that I'm referring to is whether the field is conservative or not. In the case of the capacitor, the electric field is conservative - that is, the field is generated by a charge density. Thus, the capacitance C = Q/V refers to charge per potential difference. Further, if you exchanged electric charge for magnetic charge, the capacitor would store energy in a magnetic field and a conservative magnetic field at that. In this case, the 'capacity' to store a given amount of magnetic charge would given by weber/amp otherwise known as the henry! Alas, magnetic charge has not been found so it might not exist.

Conversely, the inductor stores energy in a non-conservative field. A non-conservative field is not associated with charge but is instead 'induced' by a current. What we usually think of as an inductor stores energy in a non-conservative magnetic field produced by an electric current (which includes charge and displacement currents). Since the closed contour integral of this non-conservative magnetic field is not necessarily zero, we refer to this integral as the magneto-motive force (mmf) measured in amps. The inductance turns out to be the ratio of the magnetic flux integral to the magnetic field integral which gives units of weber/amp - the henry.

Now, replace electric current with magnetic current and you get energy stored in an electric field. Using the same logic as above, you get that the inductance is the ratio of the electric flux integral to the electric field integral which gives units of Coulomb per volt - the farad! But - and this is crucial - the Coulomb in the numerator is not a quantity of charge - it is the total electric flux along a closed path. Likewise, the volt in the numerator is not a potential difference - it is the total electric field along a closed path otherwise known as the emf!

This is why the capacitor and inductor are not physically dual. Despite the fact that the units work out, the meaning of the units are different, e.g., using volts to measure potential difference AND emf which are quite different ideas. Alfred Centauri 03:03, 6 September 2005 (UTC)

## User :Sidam- permanent block?

Isn't this guy asking for a permanent block for repeated vandalism and refusing to discuss on the talk pages? Could admins please take note?--Light current 20:20, 24 September 2005 (UTC)

## Mutual capacitance and self-capacitance

There's a contradiction in our article, and I wonder if anybody thinks it is significant.

1. Capacitance exists between any two conductors insulated from one another.
2. The capacitance of the earth, if considered as an isolated conducting sphere, is about 680 µF.

In statement (2), where is the second conductor implied by statement (1)? I think I know the answer to this question, and it will require the article to be reworded slightly.

I've been reading articles like these two [3] and [4] by Fred Erickson, which explain the difference between the self-capacitance of a conductor and the mutual capacitance of two conductors. Perhaps we need to make this distinction in our article. The value printed on a parallel-plate capacitor is the mutual capacitance between its plates, and this is good enough for most circuit designs. However, to be precise, each plate also has its own self-capacitance, which could be thought of as the capacitance between the plate and the rest of the universe. Normally, these self-capacitances are much smaller than the mutual capacitance, so they can be ignored (although they sometimes need to be considered as parasitics). However, as you increase the plate separation, the mutual capacitance drops and the self-capacitances eventually become dominant. An example of the widely separated case is the capacitance between the Earth and the Moon, as discussed in the above-mentioned articles. Here, the mutual capacitance is small (about 3 μF) and the self-capacitances (about 680 μF for the Earth, as our article says, or 710 μF by my reckoning; and 193 μF for the Moon) dominate. The total capacitance is that of the mutual capacitance in parallel with the two self-capacitances:

${\displaystyle C_{total}=C_{mutual}+{\frac {1}{{\frac {1}{C_{self1}}}+{\frac {1}{C_{self2}}}}}}$

(Imagine an equivalent circuit with C_mutual between the two bodies, and a C_self between each body and the universe, which acts as a ground plane.) Plugging the Earth/Moon values into this, I get a C_total of about 152 μF, which is close enough to Erickson's value of 159 μF.

I think our article blurs the difference between these two types of capacitance. Any thoughts? --Heron 19:32, 12 November 2005 (UTC)

## reciprocal of capacitance

c is acapacitance expressed in FARAD. 1/c is invrse of capacitance & its unit is 1/F = --------

Yes, this is mentioned under 'elastance'. --Heron 17:50, 29 September 2006 (UTC)

## Articles should not be merged

Diffusion capacitance is a change in charge with voltage, but that is all that it has to do with capacitance in the normal sense. In semiconductor device with a current flowing through it (an ongoing transport of charge by diffusion) there is necessarily some charge in the process of transit. If the applied voltage changes and the current in transit chges, a different amount of charge will be in transit. This change in the transiting charge is the diffusion capacitance. I think this is a different enough phenomena that it is only confusing to merge the articles.Brews ohare 03:36, 5 November 2007 (UTC)

Also, Capacitance has nothing to do with nominal capacity and the latter should not redirect to the former. -Nathan24601 (talk) 18:32, 25 February 2008 (UTC)

## Capacitance and Displacement Current

Brews, I've just had another look at this. Yes indeed, Maxwell's equation (138) would more or less exactly correspond to our modern equation Q = CV. But the point which we must not overlook is that Maxwell takes his displacement current concept from equation (105), whereas nowadays, we have shifted the emphasis to his equation (138). Would this help as regards your clarification tag? David Tombe (talk) 15:11, 18 November 2008 (UTC)

Brews, I've looked at your latest edits. Here's a few points to bear in mind.

(1) Equation (105) is the beginnings of Maxwell's displacement current. But it seems that since Heaviside in 1884, the concept has got tied up with equation (138) instead.

(2) Equation (138) leads us straight down the road into Gauss's law. In fact Maxwell led us down that road straight after equation (112).

(3) When Maxwell used the displacement current in his 1864 paper to derive the EM wave equation, he only derived the H equation. If he'd tried to do the E equation too, Gauss's law would have tripped him up.

(4) The point is that Maxwell's displacement current term may only be useful in wireless telegraphy where the E term can match up with the E term in Faraday's law (the -(partial)dA/dt term). In that case div E will always equal zero. Your section contains good information. I wouldn't want to delete any of it. But ask yourself 'what is the focus?'. When you have answered that, you may wish to play around a bit with the sequencing. At least you've already debunked the myth that Maxwell invented displacement current in connection with capacitors. David Tombe (talk) 18:59, 18 November 2008 (UTC)

Brews, why have you got that V in brackets in the displacement current term? CdV/dt should be sufficient, but you have written C(V)dV/dt. David Tombe (talk) 08:41, 19 November 2008 (UTC)
You've given me a lot of points to consider. As for C(V), the (V) is meant to show a functional dependence. Brews ohare (talk) 16:27, 19 November 2008 (UTC)

Brews, does capacitance depend on voltage? Anyway, keep in mind these points. Maxwell's equation (105) is all about wireless telegraphy. However his equation (138) is all ultimately about cable telegraphy, but it was Heaviside who developed that aspect later on. For the former, you want to have div E = 0, with E = -(partial)dA/dt, because that is the term in Faraday's law (or the Lorentz force) which is needed to derive the EM wave equation. For the latter you get into the realms of Gauss's law where dive E does not equal zero in general. David Tombe (talk) 17:36, 19 November 2008 (UTC)

Your point about telegraphy is interesting, but probably a digression for this article. I have modified the discussion of C(V); hope it is clearer. I also rewrote the "elasticity" reference. See what you think of it. Brews ohare (talk) 17:48, 19 November 2008 (UTC)

Brews, You've explained Maxwell's views on the elasticity very well. As regards C being a function of V, it's OK. I personally wouldn't have bothered introducing that extra complication for that particular section, but it's OK. And yes, cable telegraphy would indeed be a bit of a digression. But do bear in mind the close connection between capacitance and cable telegraphy which doesn't appear to be so with wireless telegraphy. Maxwell's displacement current dealt with the latter, but the modern equations around Q = CV deal with the former. You're making good progress with Maxwell's 1861 paper. What do you think of the centrifugal term in equation (5)? Equation (77) is actually the Lorentz force term as per equation (D) in the 1864 paper. There is no centrifugal force term in it. But I'll be interested to hear what you make of equation (5). Between the two equations, do you not see the inverse square law attractive force, the Coriolis force, the centrifugal force, and the Euler force staring out at you? Kepler's second law gets rid of the curl bits of Faraday's law and leaves us with the radial orbital equation including centrifugal force. Maxwell uses the centrifugal force to account for magnetic repulsion between like poles. David Tombe (talk) 18:25, 19 November 2008 (UTC)

I followed your advice on C(V) in that section. I rephrased the "elasticity" sentences. I'll mull over your other remarks. Brews ohare (talk) 19:53, 19 November 2008 (UTC)

## Capacitance and charge

A volt is one joule of energy per coulomb of charge. The energy that results from volts is volts times the number of charges (${\displaystyle E=VQ.}$) The work required to charge a capacitor equals one-half of VQ:

${\displaystyle W={1 \over 2}VQ.}$

VQ is twice a capacitor's charging energy. Within a capacitor, Q does not seem to represent electrons and V does not seem to be energy per Q. Otherwise, the energy due to a capacitor's electrons would be V joules per coulomb of electrons, regardless of any particular value of Q or V.

Volt was originally defined as 1/1.434 of the emf of a Clark cell. This definition assumed that one electron per volt producing event occurred in a Clark cell.

${\displaystyle C={\frac {Q}{V}}.}$

In a point charge simulation of voltage, capacitors contain an equal number of plus and minus charges. Capacitor discharge converts pairs of opposite charges into other energy. Therefore, in the definition of capacitance, each Q represents a pair of charged particles. --Vze2wgsm1 (talk) 13:41, 24 May 2009 (UTC)

You are very confused, but this is not the place to explain why you are so mistaken, take it to the ref desk. SpinningSpark 16:29, 10 June 2009 (UTC)

## Capacitance inductance duality

The statement in the article is wrong. Capacitance is the reciprocal of inductance only in the case of parallel wires (with arbitrary cross section), see Jackson, "Classical Electrodynamics". —Preceding unsigned comment added by 88.64.74.16 (talk) 09:06, 24 May 2009 (UTC)

## Weird section

The section Capacitance#Capacitance and 'displacement current' that intervenes between the lead and the more intelligible content sections seems to be the work product of Brews ohare under the influence of David Tombe, per the discussion above. It is sourced only to Maxwell's primary writings, and is all about interpretations thereof. Can't we have a simple section, later instead of here, on a modern statement of the relevant EM theory and displacement current, with links to the main articles? Does anyone understand a reason why it would be appropriate to have the Maxwell quotes there, or why we might want a lengthy diversion about displacement current before talking about the topic of the article? Dicklyon (talk) 05:10, 10 September 2009 (UTC)

It is unnecessary to speculate over the origins of this section, which could be taken as an attempt to impugn the section on the basis of its authors, rather than upon its merits.
There is nothing wrong with the section; it quotes Maxwell's work accurately, puts it into context accurately, and is germane to the notion of capacitance and to its history.
At most it should be moved to a later position in the article. I see no need to change it, or to eliminate it. It is well sourced and relevant. Brews ohare (talk) 05:46, 10 September 2009 (UTC)
It is sourced to a single primary document by Maxwell; how is that well sourced? What is the basis for judging it to be germane and relevant? Isn't that what secondary sources would tell us? Dicklyon (talk) 06:21, 10 September 2009 (UTC)

The section is about Maxwell's work on capacitance and displacement. He's a pretty good source. How about rolling up your sleeves and looking further if you are interested? Brews ohare (talk) 13:05, 10 September 2009 (UTC)

But why do we have a section about Maxwell's work on capacitance and discplacement, rather than a section about capacitance and displacement? That approach would belong in a history section in this article, or an article about Maxwell's work. --Ccrrccrr (talk) 14:21, 10 September 2009 (UTC)
I like history sections, too; they need to be motivated by the analyses of historians, though, hence secondary sources. And this strange section goes through a lot of stuff before even mentioning capacitance, dielectric constant, permittivity, or such concepts, so it would an impossible distraction to a non-expert reader. Dicklyon (talk) 18:56, 10 September 2009 (UTC)

For now, I've taken it out. If the selection of quotes and the interpretive comments can be sourced and tied directly to capacitance, something like this could go back in. I'll put it below as a talk section for now, and tagged a few places where citations to secondary sources would be appropriate:

## Capacitance and 'displacement current'

The physicist James Clerk Maxwell invented the concept of displacement current in his 1861 paper in connection with the displacement of electrical particles:Template:Cn[1]

He then added displacement current to Ampère's law.Template:Cn[2] Maxwell's correction to Ampère's law remains valid today, and is expressed in the form:

${\displaystyle \nabla \times \mathbf {H} =\mathbf {J} _{f}+{\frac {\partial \mathbf {D} }{\partial t}}\ ,}$

with Jf the current density due to motion of free charges and the displacement current density given as D/∂t with the electric displacement field D related to the electrical polarization density of the medium P as:

${\displaystyle {\boldsymbol {D}}=\varepsilon _{0}{\boldsymbol {E}}+{\boldsymbol {P}}\ .}$

Here ε0 is the electric constant. The polarization is the contribution described by Maxwell in the quotations above, and is due to the separation and alignment of charge in the material that is not free to transport, but is free to align with an applied electric field, and to move over atomic dimensions, for example, by stretching of molecules. (This polarization in response to the field actually screens the dielectric from the electric field, resulting in a lower field the greater the polarization of the medium.Template:Cn See the figure.) In simple materials, the polarization is proportional to the electric field and an adequate approximation is:

${\displaystyle {\boldsymbol {D}}=\varepsilon _{r}\varepsilon _{0}{\boldsymbol {E}}\ ,}$

with εr the relative static permittivity of the material. When there exists no material medium, εr = 1, so there still exists a displacement field when there is no medium present.

A capacitor with charges represented by blue and red circles. Left: Capacitor with unpolarized dipole (the two charges are not separated); electric field unscreened. Right: Capacitor with polarized dipole (the two charges are spread apart); electric field screened. The dotted box is a Gaussian surface Σ.

#### Gauss's law

{{#invoke:main|main}} To connect the displacement to charge, Gauss's law is used, which in integral form relates the charge in a region to the surface integral over an enclosing surface Σ of the component of D normal to the surface:

${\displaystyle Q(t)=\oint _{\Sigma }{\boldsymbol {D}}({\boldsymbol {r}},\ t)\ {\boldsymbol {\cdot }}\ d{\boldsymbol {\Sigma }}\ ,}$

where a vector dot product is indicated by the "·".

To relate this expression to a capacitor, the surface Σ is made to enclose the dielectric medium and one of the two electrodes of the capacitor. The electrode contains the net charge upon the capacitor, and the dielectric medium is charge neutral. Referring to the figure, suppose initially the dipoles in the dielectric are unpolarized, as on the left side of the figure. The electric field due to the charge on the capacitor plates is the same as though the dielectric were not present. Next, suppose the dipoles are able to respond to the applied field and become polarized, as on the right side of the figure. Then the field from the extended dipole opposes that of the electrodes and the electric field inside the dielectric decreases. Suppose the left panel corresponds to an initial time just after the field is applied and the dipole has not had time to respond, while on the right is a later time when the dipoles are in the process of becoming extended. During this extension of the dipoles, a displacement current flows across the Gaussian surface. The more polarizable the medium, the more current for a given voltage, and the greater the capacitance. The net displacement current I through the region Σ is related to the displacement current density through the equation:

${\displaystyle I={\frac {dQ}{dt}}=\oint _{\Sigma }{\frac {\partial }{\partial t}}{\boldsymbol {D}}\cdot d{\boldsymbol {\Sigma }}\ .}$

(The partial time derivative is meant to emphasize that the spatial variables in D(r, t) are held fixed.) This equation includes current through the region Σ related to polarization of the medium, and is connected to capacitance and an applied voltage:

${\displaystyle I={\frac {dQ}{dt}}=C\ {\frac {dV}{dt}}\ ,}$

where C is capacitance, Q is charge, and V is the applied voltage responsible for the field causing the polarization of the medium inside the capacitor. For some materials represented by complicated behavior of D, the capacitance can be a function of voltage and may exhibit time dependence related to the ability of the medium to respond to the signal (see subsections below).

It should be mentioned that when there is no material medium in the capacitor, the displacement is not zero, but D = ε0E. Consequently, a capacitance still is present. For example, a system of metal electrodes in free space may possess a capacitance.

Maxwell never used the term electric charge,Template:Cn but he did refer to the "distribution of electricity in a body" and to the "quantity of electricity". Capacity C was stated in his equation (138) for two surfaces bearing equal and opposite quantities of electricity e and electric tensions or potentials ψ1 and ψ2 as the ratio C = e / (ψ1 - ψ2). Then the effect upon C of inserting a dielectric between the plates was determined.[3]

Today, capacitance is viewed primarily in terms of the capacity for storage of charge, whereas Maxwell's paper stressed the current that flowed through a capacitor.Template:Cn He calculated this current focusing upon the specific calculation of polarization for an "elastic sphere" distorting under an applied field and resisting deformation by virtue of its elastic properties, and the current that flowed when this state of polarization altered.Template:Cn The modern approach attempts to treat the polarization of materials by modeling the microscopic events contributing to the displacement field using quantum theory: for example, see below.Template:Cn

## Unwanted Stray Capacitance

I firmly think that stating stray capacitance is unwanted is misleading and a biased view. There are several instances when it is useful. Maybe it just needs to be switched from such a blanket statement to it is often undesired. Tempust (talk) 18:30, 17 November 2009 (UTC)

I went ahead and put "often" in front of "unwanted". But I think a more extensive revision could be helpful.
However, before we attempt to agree about what to say about whether it's unwanted or not, we'd need to settle on a clear definition. One definition is simply that stray capacitance is any unwanted capacitance. If we use that definition, there's certainly no problem! Right now the article seems to defined it as capacitance between two conductors, which is not very useful, because that would define all capacitance as stray. How would you define stray? Ccrrccrr (talk) 12:24, 18 November 2009 (UTC)
I saw a use from the farad article suggesting it was useful, but I did some reading on the internet and it seems that stray capacitance is an umbrella term for any unwanted capacitance so I feel that I must reverse my argument and suggest that the brackets are dropped. That said though I think that section is written poorly, but unfortunately I don't know enough about the topic myself to write it.Tempust (talk) 04:29, 19 November 2009 (UTC)

## "Consider a capacitance C, holding a charge +q on one plate and −q on the other."

In the leading section, where the article reads "Consider a capacitance C, holding a charge +q on one plate and −q on the other.": surely this should read "Consider a parallel plate capacitor C, holding a charge +q on one plate and −q on the other." ? JMatopos (talk) 10:24, 23 March 2011 (UTC)

It doesn't need to be parallel plates for this to make sense. Maybe instead of "plate" you could say "terminal" or "electrode"? Dicklyon (talk) 21:50, 23 March 2011 (UTC)
For a capacitor to work the plates must have area. A terminal is an end-node and has no dimensions. Wouldn't be too keen on electrode either, which does not give the sense of area. SpinningSpark 00:34, 25 March 2011 (UTC)

## "Separated, spherical plate capacitor."

In my introductory physical class "Physics - The Science of Absolute Truth" we were asked solve this problem: Two separated solid spheres comprise the "plates" of a capacitor. They are made of the same material (pick a conductor of choice, aluminum, i.e.) and they have the same density.
Their electrical inputs are center-fed at the inside the spheres. The first sphere has radius s, the second has radius t. s is constrained to be >= 2t. The centers of the "plates" (spheres) are separated by a distance > s + t but < 2s. Call that separation u (between the sphere centers).
I have no clue. Can you help me calculate the capacitance? My professor, Dr. Jack, refuses to help. — Preceding unsigned comment added by 72.95.47.115 (talk) 00:54, 22 November 2011 (UTC)
• Article talk-pages are only for discussing the article itself and are not general forums for discussing the subject. Questions like this are better taken to the science reference desk. But...
• Centre-fed isn't entirely clear; the capacitor can only be charged by transferring charge from one sphere to the other - along some actual path external to the spheres.
• While u > s + t is a necessity, s >= 2t and u < 2s seem irrelevant - likewise the density.
• Have you seen the article bipolar coordinates? What you actually want is bispherical coordinates, but the first-mentioned article has a much clearer picture. The spherical surfaces are your equipotentials (and two of them correspond to your capacitor plates - note that these are not centred on the poles of the coordinate system). The flux paths are along the toroidal surface (which meet the spherical surfaces at right-angles). Find V(r), thence E and Vc, thence D and Q --catslash (talk) 02:22, 22 November 2011 (UTC)
"s >= 2t and u < 2s seem irrelevant - likewise the density." Relevant or not, they were part of the problem statement given to me.
The "external" charge, so called was imparted to the sphere at its center. For example, if it was coax, it would run inside the sphere and terminate in the center. Center-fed I guess is the best phrase I could come up with.
So you gave the formula for the solution, or not? — Preceding unsigned comment added by 72.95.47.115 (talk) 10:27, 22 November 2011 (UTC)
Not. On closer inspection, it seems that these coordinate systems do not give you an exact solution in 3D (it works in 2D - cylinders rather than spheres). Perhaps s >= 2t and u < 2s are relevant because an approximate solution is expected. Anyway take your question to the science reference desk. --catslash (talk) 15:23, 22 November 2011 (UTC)
Thanks! I appreciate the effort and your honesty. — Preceding unsigned comment added by 134.223.116.201 (talk) 17:28, 22 November 2011 (UTC)
While we were sitting around jawboning about it, somebody actually built it. Where's the fun in that ? http://www.sensorsmag.com/sensors/electric-magnetic/epic-a-new-epoch-electric-potential-sensing-8961 — Preceding unsigned comment added by 134.223.230.200 (talk) 18:52, 18 June 2013 (UTC)

## Definition of self-capacitance

The following discussion has been moved to this page from a user talk page

Hi Materialscientist, I took a look at the reference you cited and looked at how it defined self-capacitance, and I can only say that I disagree with that definition. The electric charge required to raise the electric potential by one volt is still an electric charge, and therefore it is not a capacitance.

My formulation may be using bad English, but the current formulation is incorrect. If you do think my formulation is bad, you may instead help me to improve it. —Kri (talk) 05:08, 4 December 2011 (UTC)

It can be rephrased, and I see your point, but have a look here: "capacitance, property of an electric conductor, or set of conductors, that is measured by the amount of separated electric charge that can be stored on it per unit change in electrical potential". Unit change here is a more accurate equivalent of 1 V. Basically, this definition can be placed into the article for self-capacitance. Somehow, I like it better that the "added charge raises the potential" than the "added potential allows to store more charge". Materialscientist (talk) 05:20, 4 December 2011 (UTC)
Yes, I believe "unit change in electrical potential" would be more correct to use than "one volt". Except from part that said "one volt", I actually liked the other (the current) formulation better than my own. I just couldn't find any other way to get around the problem of including a seemingly arbitrarily chosen change in potential in the formulation, but this seems to do the trick. So should we change "one volt" to "one unit change in electrical potential" and then leave the rest as it is? —Kri (talk) 05:44, 4 December 2011 (UTC)
I tried [5] - I see what should be said, but the nice formulation doesn't come to mind. "Unit change" is a possibility, but there is this nice correspondence between adding charge and raising the voltage. Materialscientist (talk) 05:53, 4 December 2011 (UTC)
To be correct, one unit in this context means 1, i.e. the unitless constant 1. The current formulation is still incorrect, since it says one unit can be the same as one volt, which is not true and would bring us right back to where we were, i.e., we would be defining a current and not a capacitance. I suggest that what is now written in parenthesis is removed. —Kri (talk) 15:22, 4 December 2011 (UTC)
I agree it needs polishing, but note that if we meant to say 1 in the math sense, we would say 'unity' instead of 'one unit'. Actually, I added (one volt) to specifically clarify this 'one unit', i.e. unit=volt. Materialscientist (talk) 21:38, 4 December 2011 (UTC)
Is it okay if I rename this discussion to "Definition of self-capacitance" and move it to Talk:Capacitance instead? I realized that it probably makes more sense to have it there than here. —Kri (talk) 00:47, 5 December 2011 (UTC)
Sure. Materialscientist (talk) 00:51, 5 December 2011 (UTC)

## Inconsistency in Capacitance of Parallel Wires / Wire and Wall

Stated capacitance for this problem uses an approximation ${\displaystyle \left({\frac {d}{2a}}\gg 1\right)}$ applied inconsistently between the argument of the ${\displaystyle \cosh ^{-1}}$ and the ${\displaystyle \ln }$. — Preceding unsigned comment added by 50.131.141.62 (talk) 19:43, 11 March 2012 (UTC)

It's not an approximation; the condition d > 2a just means that the two wires aren't touching (d is the distance between the wire centres and a is the radius of each wire). In the case of the wire above a ground plane, d is the distance from the wire centre to the plane, so d > a is likewise the condition for no contact between the conductors. It is somewhat confusing because the ds in the two cases do not really correspond. --catslash (talk) 21:51, 11 March 2012 (UTC)

In the cited reference (Jackson, 2d Ed.), the capacitance is given as ${\displaystyle \propto 1/\cosh ^{-1}({\frac {d^{2}}{2a^{2}}}-1)}$. This appears to disagree with the value given on the Wikipedia page, ${\displaystyle \propto 1/\cosh ^{-1}({\frac {d}{2a}})}$, unless some approximation is made to get rid of the -1. This approximation does not appear to have been made within the ${\displaystyle ln}$, however. — Preceding unsigned comment added by 50.131.141.62 (talk) 06:58, 12 March 2012 (UTC)

It only appears so. Unfortunately Jackson doesn't use the simplest expression. The expressions are equivalent without approximation because of ${\displaystyle \operatorname {arcosh} (2x^{2}-1)=2\operatorname {arcosh} (x).}$ radical_in_all_things (talk) 17:19, 12 March 2012 (UTC)

It'd be nice to add such hyperbolic trig identities to the wiki page on inverse hyperbolic functions. — Preceding unsigned comment added by 50.131.141.62 (talk) 17:19, 14 March 2012 (UTC) I use of Image Theorem for sol this problem.(A straight conducting wire of radius a is parallel to and at height h from the surface of the earth). I have replacement alone D=2h in formula.thereupon do not to change coefficient of 1 to 2 in numerator. — Preceding unsigned comment added by Ghorbanib (talkcontribs) 10:39, 15 November 2013 (UTC) The explanation for the factor 2 is the same as in Inductance#Method of images.radical_in_all_things (talk) 16:54, 15 November 2013 (UTC)

## First sentence

Should the first sentence say that capacitance is the ability to store charge? or should it say energy? I am under the impression that a capacitor stores energy and charges are not actually stored. Thanks, Vokesk (talk) 00:44, 10 April 2012 (UTC)

The new text is:

a 1 farad capacitor when charged with 1 coulomb of electrical charge will have a potential difference of 1 volt between its plates

The old text was something like:

1 farad is 1 coulomb per volt.

Both are correct. The old text was a bit terse, but it wasn't backwards, and it did have the considerable merit of including the words is and per (or as I reinstated it for each). The new text on its own is ambiguous - it could equally well apply to reciprocal-farads. The notion that 1F & 1C => 1V rather suggests 2F & 1C => 2V. If you don't like my compromise wording, could you take another stab at this definition yourself? --catslash (talk) 15:51, 4 May 2012 (UTC)

The definition is exactly as per the reference. If it were changed then the reference would no longer support it. Wikipedia likes references. If you feel that some form of expansion is necessary, then I can see no objection to adding something suitable after the definition. I'm finished for the day, but if you don't get there first, it will have to wait until next week. 109.152.145.86 (talk) 16:36, 4 May 2012 (UTC)

## Frequency-dependent capacitors: Inconsistent notation

In the section on frequency-dependent capacitors, there is an inconsistency in the use of the term C(ω). In Eq. (3) (the third equation in this section), where we see the term G + jωC(ω), this frequency-dependent capacitance value is real. It comes only from ε' (the real part of the complex ε). In Eq. (4), the term C(ω) is complex, and it comes from ε (which is complex in general). Therefore, the same notation is being used to describe two different things. Elee1l5 (talk) 17:06, 24 September 2013 (UTC)

## Missing units

'Capacitors' section with formula was missing units; I put them in. Between people using cgs instead of mks, microns instead of meters, Farads per centimeter instead of Farads per meter, it's easy to get mixed up without rigorous definition. 71.139.169.27 (talk) 04:26, 14 October 2013 (UTC)

It is not a good idea to say, for instance, that A is the area in square meters. A is an area and as such has a value and a unit, the unit may be square centimeters or anything else.radical_in_all_things (talk) 19:24, 14 October 2013 (UTC)

## Capacitance or energy density?

What is the difference between capacitance and energy density?--Wyn.junior (talk) 03:01, 26 February 2014 (UTC)

Capacitance = energy in As/V (Ampere second per Volt) or Joule per Volt, energy density = energy (As or J) per volume or weight --Elcap (talk) 14:42, 26 February 2014 (UTC)
Why are supercapacitors measured in capacitance only and not energy density?--Wyn.junior (talk) 14:47, 26 February 2014 (UTC)
One argument I would give regarding supercapacitors, while admitting that I am not an expert in this specific area, is that many dielectric materials considered in the world of supercapacitors are "composite" materials -- having charge defects, charge traps, or what we could essentially call quantum dots, embedded throughout the dielectric material. Each of these charge traps can be treated as a unique capacitor, and given their size, typically will trap only a few electrons at any time. Each of these capacitors do not have well-defined equipotential surfaces, but instead are governed by their local potential landscapes (defined by the electrostatic configuration of electrons contained in each). In the extreme case of a since electron trapped in each defect/trap, the total amount of stored energy in that region of the bulk dielectric can be up to twice the amount of stored energy if each of them were "connected" as they would be between two metal plates. Instead, you have many self-capacitances of these charge traps throughout the volume of the bulk dielectric. Each of these regions have varying degrees of the amount of stored energy. So, 1) not knowing the actual distribution of these charge traps within the volume, 2) not knowing how many charges are "on" each trap, and 3) not knowing the local potential landscape of each of these traps, makes the use of an averaged "energy density" throughout the bulk device a very inaccurate way to describe the capacitance. TJ LaFave (talk) 22:23, 26 February 2014 (UTC)
I would argue that energy density is generally an insufficient replacement for capacitance given the nature of discrete electron charges. If you look into my published work concerning "monophasic capacitance" you will see my perspective on this -- which I believe represents an exceptionally general approach to the concept of "capacitance". Conventionally, we have come to think of all capacitors as having metallic electrodes that define two unique equipotential surfaces within each capacitor (the parallel plate capacitor is certainly the most famous example). In fact, the standard graduate-level textbook by Jackson on this subject only hints at the general formulation, but never considers it. Instead Jackson segues the discussion into descriptions of stored energy in dielectrics with metallic electrodes! This has muddied the concept of capacitance quite a bit as we haven't really looked at the general case which is that of a dielectric object in the absence of metallic electrodes. In these conventional cases, the equipotential surfaces needed to define capacitance are governed by the electrostatic locations of excess electrons within the device (instead of the enormous number of electrons in the metallic electrodes). The definition of a "ground" potential is typically taken to be the case when there are no excess electrons stored in the device. In order to precisely define capacitance you would need to first determine the electrostatic configuration of all charges. If you simply assume that you know how many charges you have put in the device (charge density) you have not defined capacitance very well. Only in cases where the number of electrons is very large (such as large dielectric objects and/or dielectric objects in the presence of metallic electrodes) could you make an argument by which you might replace capacitance with energy density -- and even then, it is an approximation (a good approximation, but still an approximation). In this large-N limit, the total number of excess electrons washes out all the spatial symmetry properties that we find in few-electron systems. With all this in mind, you generally want to define the electrostatic configuration of all the charges. This defines the capacitance. You could attempt to determine the actual volume of the capacitor (which, notably is generally not the volume of the dielectric!) which is the volume encompassed by the extent of the resulting electrostatic potential field, but this is typically an exceptionally complex three-dimensional problem, and completely unnecessary since you have already defined the electrostatic charge configuration! Consequently, energy density would, in general, only give you an approximation of capacitance. That approximation becomes better as the number of charges grows exceedingly large (e.g. within metallic electrodes). Put yet another way, I might have five charges in a unit volume, but if I haven't defined the electrostatic locations of those five charges, then the charge density is just an approximation of the total energy stored in the system. The electrons could very close to each other within that unit volume (highly confined) and have a much higher energy than when they are very far apart (very low energy, less confinement). TJ LaFave (talk) 22:14, 26 February 2014 (UTC)
My favorite way to think of capacitance is to consider the expression
${\displaystyle C={Q^{2} \over 2U}}$
Here, capacitance is defined in terms of the total charge (or charges, if we recognize that in general we have many electrons so Q = Ne) stored in the capacitor and U is the total potential energy stored in the capacitor as a result of that stored charge. Nowhere in this framework does the volume of the capacitor come into play except when we start talking about conventional capacitors with metal plates. Then we could talk about how the volume enclosed by the metallic plates is constant and how the capacitance might be defined as being constant by the shape and size of the plates. This is why energy density could be confused with capacitance (and a great reason why the conventional approach to capacitance is a bad pedagogical approach!). But, in general, we do not have metallic plates and the volume is not well-defined. This is why "energy density" is not the same as "capacitance". (Note, this expression is not general, see capacitance section about capacitance of nanoscale capacitors for more. But the discussion here doesn't change at all.) TJ LaFave (talk) 22:37, 26 February 2014 (UTC)
Another view: Supercapacitors are electric/electronic components, Hardware! They bridge the gap between electrolytic capacitors and rechargable batteries. To compare this components in applications you have to specify capacitance value as well as power and energy density. So far as I know most SCs for power applications do have a specification for power and energy density in their datasheets. --Elcap (talk) 09:49, 27 February 2014 (UTC).
1. See On Physical Lines of Force: Part III–The Theory of Molecular Vortices applied to Statical Electricity
2. See Eq. (112), p. 19 in Maxwell's paper.