Talk:Capacitance

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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 V_{max} \, (equal to E_{str}d \, where E_{str} \, is the dielectric strength), then the maximum energy it can store is:

W_{max} = \frac{1}{2} \epsilon_0 \epsilon_r \frac{A} {d} E_{str}^2

Contents

[edit] 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)

[edit] 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 μ 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 ε material so we get lots of electric field for a little electric flux. Is there any material for which ε < ε0? 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)

[edit] 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)

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)

[edit] 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:

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

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

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 ZC.

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

[edit] 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)

See my comment near the top of the page about the scope of this article. Nonetheless, here's what I believe will happen. Let's say we all agree that this article is about the the phenomenon of capacitance and that discussion of capacitors in circuits, applications etc. belongs in the capacitor article. Sooner rather than later, this article will once again start filling up with capacitor stuff. Someone will need to keep a vigilant eye on this article to keep it focused. Seems to me that you're up to the job. Alfred Centauri 17:57, 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)

[edit] 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:

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

where φ 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:

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)

[edit] '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)

What I was intending to say here before my brain got addled was this:

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 + dΦ/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)

[edit] 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)

[edit] 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)

[edit] 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 [2] and [3] 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:

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)

[edit] 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)
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