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Digix
2008-Sep-10, 12:29 AM
Lets say we have vacuum capacitor.
which is made of 2 plates and vacuum between them.
lets say we somehow charge it and leave

now according to e=mc^2 capacitor mass consists of mass of plates + mass of stored energy.

but where that energy mass is held? in the void between plates(vaccum somehow gains mass) or plates become more heavy?

NickW
2008-Sep-10, 01:46 AM
the dieletric of a capacitor is there as a way to control the speed of the discharge between the plates. The dielectric does not store any charge, but the discharge between the plates will pass through it.

I guess to answer your question, One plate would become heavier until discharge.

Paul Leeks
2008-Sep-10, 01:57 AM
A flying saucer acts a capacitator it stores energy (area51)

Digix
2008-Sep-10, 02:06 AM
the dieletric of a capacitor is there as a way to control the speed of the discharge between the plates. The dielectric does not store any charge, but the discharge between the plates will pass through it.
if there is dielectric we can expect it to store energy in the polarization effect


I guess to answer your question, One plate would become heavier until discharge.
I am not talking about mass of electrons, but about mass of energy itself

pzkpfw
2008-Sep-10, 05:11 AM
I am not talking about mass of electrons, but about mass of energy itself

Um, does a spring that has been compressed get heavier?

frankuitaalst
2008-Sep-10, 07:36 AM
Um, does a spring that has been compressed get heavier?
Uhum , I think you have the same kind of question about potential energy as I have in the OP : E=mc where I ask about the effect of potential energy .
My answer to your remark , till now , is , yes the spring should have more energy (and therefor mor mass ) after compression.

Digix
2008-Sep-10, 01:31 PM
Um, does a spring that has been compressed get heavier?

Without doubt answer is yes.

tdvance
2008-Sep-10, 03:08 PM
e=mc^2 does refer to total energy, including potential energy (which is relative to a ground state and hard to make absolute--but then mass isn't absolute either--but I'd guess you'd need to measure with respect to an ultimate ground state--say, the entire universe in a singularity). So, the plates (both, I'm sure--positive charges and negative charges both being forms of potential energy) would get unmeasurably heavier.

Not counting, of course, the small added weight to one plate from getting some extra electrons and the equivalent small subtracted weight from the other plate, which would probably be a bigger effect than the relativistic one.

AndreH
2008-Sep-10, 03:30 PM
Hmm, maybe I understood everythin wrong in my physics lesson. E=mc^2 does not mean Energy has mass. It means Enrgy and mass are equivalent. It is wheter energy or mass and can be converted into each other under some circumstances.

For example annihilation of matter/antimatter, before you have mass, after you have energy.
Same for the so called mass defect in nuclei. The particles (protons & Neutrons) have bigger mass when seperated. When togehter bound some of the mass is lost and goes into binding energy.

So to my opinion the capacitor does not gain any mass. But I may be completely wrong.

alainprice
2008-Sep-10, 03:53 PM
Maxwell would have said the energy is in the 'displacement current'. The way he decided to deal with electromagnetism is similar to this situation.

I feel the energy would be stored on the plates themselves as static charges. We could easily just say the energy is stored in the electic field between the plates though.

Digix
2008-Sep-10, 03:54 PM
Hmm, maybe I understood everythin wrong in my physics lesson. E=mc^2 does not mean Energy has mass. It means Enrgy and mass are equivalent. It is wheter energy or mass and can be converted into each other under some circumstances.

For example annihilation of matter/antimatter, before you have mass, after you have energy.
Same for the so called mass defect in nuclei. The particles (protons & Neutrons) have bigger mass when seperated. When togehter bound some of the mass is lost and goes into binding energy.

So to my opinion the capacitor does not gain any mass. But I may be completely wrong.

imagine black box where you have positron and electron they definitely have rest mass, and now they annihilate and increase temperature inside the box or we can use that energy to charge capacitor.

so does it mean that box weights less mow?
probably not, so that means that capacitor charge also has same mass like every other material energy form. which is also equally attracted by gravity

AndreH
2008-Sep-10, 04:18 PM
imagine black box where you have positron and electron they definitely have rest mass, and now they annihilate and increase temperature inside the box or we can use that energy to charge capacitor.

so does it mean that box weights less mow?
probably not, so that means that capacitor charge also has same mass like every other material energy form. which is also equally attracted by gravity

I would say the box weigh less now. The mass has been converted into energy. And after stored into the capacitor it is still energy, not mass.

As said before, I maybe understood relativity wrong. Even though I have a degree in physics, I have never been good in theoretical physics.

ETA: MAybe Publius shows up to sort this out.

Digix
2008-Sep-10, 04:38 PM
I would say the box weigh less now. The mass has been converted into energy. And after stored into the capacitor it is still energy, not mass.

As said before, I maybe understood relativity wrong. Even though I have a degree in physics, I have never been good in theoretical physics.

ETA: MAybe Publius shows up to sort this out.

that is obviously wrong, because we could use this to create perpetual motion engine:

imagine that you have matter + antimatter they have weight and pull the lever down, now you annihilate them and store energy in the capacitor, weight becomes less so lever goes up. then you convert capacitor energy back into matter+ antimatter, cycle repeats and it can even produce usable work.

korjik
2008-Sep-10, 04:59 PM
imagine black box where you have positron and electron they definitely have rest mass, and now they annihilate and increase temperature inside the box or we can use that energy to charge capacitor.

so does it mean that box weights less mow?
probably not, so that means that capacitor charge also has same mass like every other material energy form. which is also equally attracted by gravity

If you have a box with a positron and an electron in it, and they annihilate, then you have a box with two .511 MeV photons in it. Technically, it dosent have a temperature, it isnt in equilibrium.

The mass of the box (assuming the photons are still bound in the box) is unchanged.

At no time could you use the contents of the box to charge a capacitor, since there was never any net charge present to charge the capacitor.

To answer the OP: If you charged a cap, yes the mass would go up a very very very tiny amount, that amount would be dependent on the energy needed to charge the capacitor. To first order, the energy is in the electric field across the plates of the capacitor. Exactly speaking, the energy is bound into the potential between every two like charges forced into the cap.

this is similar to the slight increase in mass due to the binding energy of a molecule

AndreH
2008-Sep-10, 05:13 PM
that is obviously wrong, because we could use this to create perpetual motion engine:

imagine that you have matter + antimatter they have weight and pull the lever down, now you annihilate them and store energy in the capacitor, weight becomes less so lever goes up. then you convert capacitor energy back into matter+ antimatter, cycle repeats and it can even produce usable work.

Not sure if I understand your example completely, but for sure there would be unavoidable losses during the conversion of the enrgy into electricity to store it in the capacitor.

But anyway, as I said before I have really some doubts about my understanding of relativity. So I better should bow out and listen to what others have to say.

AndreH
2008-Sep-10, 05:18 PM
If you have a box with a positron and an electron in it, and they annihilate, then you have a box with two .511 MeV photons in it. Technically, it dosent have a temperature, it isnt in equilibrium.

The mass of the box (assuming the photons are still bound in the box) is unchanged.

At no time could you use the contents of the box to charge a capacitor, since there was never any net charge present to charge the capacitor.
To answer the OP: If you charged a cap, yes the mass would go up a very very very tiny amount, that amount would be dependent on the energy needed to charge the capacitor. To first order, the energy is in the electric field across the plates of the capacitor. Exactly speaking, the energy is bound into the potential between every two like charges forced into the cap.

this is similar to the slight increase in mass due to the binding energy of a molecule


bold mine: I think Digix meant to use the energy of the annihilation an transfer it by some means of "dynamo" into electric energy.
But as said in my previous post: I better stay out of this discussion.

Digix
2008-Sep-10, 05:22 PM
Not sure if I understand your example completely, but for sure there would be unavoidable losses during the conversion of the enrgy into electricity to store it in the capacitor.
theoretically it is possible to perform lossless process so that is not important



But anyway, as I said before I have really some doubts about my understanding of relativity. So I better should bow out and listen to what others have to say.
ok, but i don't think that it is necessary
also I have very bad experience trying to find at least someone who would think in relativistic mass change way instead of space time dilution.

there was one discussion what if some piece falls into black hole and hits singularity with extreme kinetic energy.

pzkpfw
2008-Sep-10, 08:45 PM
Just to complete the thought, where I was going was this:

I thought a capacitor stored energy as the different densities of electrons in the two plates.



| -|
|- -|
+ve ~~~~~~~~| -|~~~~~~~~ -ve
|- -|
| -|


...so I was wondering about the mass related effects on the electrons themselves.

But then I realised any effect on the -ve plate would be "balanced" by the +ve plate.

So that went nowhere.

frankuitaalst
2008-Sep-11, 05:17 AM
Lets say we have vacuum capacitor.
which is made of 2 plates and vacuum between them.
lets say we somehow charge it and leave

now according to e=mc^2 capacitor mass consists of mass of plates + mass of stored energy.

but where that energy mass is held? in the void between plates(vaccum somehow gains mass) or plates become more heavy?

I would say the E=mc equation doesn't apply here for the total energy in a physical sense for the following reason :
Suppose you have an uncharged capacitor , enclose it by a box and call its mass m . So you have an internal energy of E=mc.
Now you charge it by applying a delta V over the poles . What happens is that some electrons move from one plate to another . The number of electrons missing on one plate will be found on the other plate , so delta m will be zero , meaning there is no delta energy contributed to mass.
What happens is that an electric field is created by the opposite charges.
The energy of this field is Efield=1/2*Q/C , where C is the capacitance :
C= eta*A/d , so Efield = 1/2Q*d/(eta*A). One can write this in terms of the voltage also .
The Energy simply comes from charges with are stored at different potential level , not from an increase of mass , und thus has an electrical ground .
Best way to represent the energy from my point of view is :
Etot = mc + Efield.
Of course one can write the equivalent mass to be : meq= Etot/c , but this lacks a bit of physical sense .
Writing this way means information is lost about where the energy really comes from : electrostatic energy .
I think in standard calculations where electromagnetic fields are involved , such as Eel, Bmagn the mass , charge are always kept apart and are not mixed in a relativistic way : Finertia = d(mv)/dt ; Fel=qE; Fmagn=q(vxB)+ Maxwell equations

sirius0
2008-Sep-11, 06:13 AM
What sort of a dielectric is a vacuum? To my mind the electrons will just drift from one plate to the other. Capacitors store electrical energy in the dielectric and express the potential at the plates. The plates could be removed (in air) from the dielectric and placed against another at which point the original dielectric retains a charge and the new one will have a similar charge and so on and on. Yes work has been done in the dielectric to displace the charges and the temperature, mass and potential energy will have increased. Now why am I right? If you are interested I can write a very strange instance in industry that stemmed from exactly this principle (spontaneous fires in a paint booth)

Grey
2008-Sep-11, 06:31 AM
Digix is essentially right here. A system with greater energy behaves like a system with more mass in every way that we can measure. Or, putting it another way, in general relativity, the thing that gets put into the equations for gravity and the like isn't actually the mass, it's the stress-energy tensor, which includes the total energy of the system, including contributions from both rest mass and any other form of energy.

So, yes, a spring that's compressed gets heavier, and a charged capacitor weighs more. But the size of this change is really, really, really, really small. :) And in the case of the capacitor, I'm not sure where that extra "mass" would be localized. I'd be inclined to suspect it would be the capacitor as a whole, even if the dielectric is a vacuum, since the energy is being stored in teh electric field here.

frankuitaalst
2008-Sep-11, 07:09 AM
Digix is essentially right here. A system with greater energy behaves like a system with more mass in every way that we can measure. Or, putting it another way, in general relativity, the thing that gets put into the equations for gravity and the like isn't actually the mass, it's the stress-energy tensor, which includes the total energy of the system, including contributions from both rest mass and any other form of energy.

So, yes, a spring that's compressed gets heavier, and a charged capacitor weighs more. But the size of this change is really, really, really, really small. :) And in the case of the capacitor, I'm not sure where that extra "mass" would be localized. I'd be inclined to suspect it would be the capacitor as a whole, even if the dielectric is a vacuum, since the energy is being stored in teh electric field here.
I can join your statement about the system considering having more mass when the capacitor is loaded , in a mechanical viewpoint , ie in mechanics .See my post nr19.
If however one is interested in the behaviour of the system in a E,B field then I believe the mass should be considered unaffected , as if one does so the energy may be counted double .
Where is the energy stored ? As both plates and the dielectrical space between are necessary to store the energy my believe is that the energy in stored in the system as a whole .

mugaliens
2008-Sep-11, 07:56 PM
As excess electrons on the plates.

hhEb09'1
2008-Sep-11, 09:24 PM
As excess electrons on the plates.There is no excess.

The energy is stored in the system. That does create a new question though. What is the gravitational field inside a capacitor? :)

publius
2008-Sep-12, 05:57 AM
There's quite a few good threads here of late in Q&A I should've jumped into, but I've had my little mind on other things (politics.....say no more :) ).

Where is the energy stored in a capacitor? Well, it's apparently stored in the electric field, that thing that exists in the vacuum. The EM energy density is 1/2(D dot E + B dot H) expressed in the most general form to account for non-linear media and so forth. In simple media, including the vacuum, that can be written 1/2(eE^2 + uH^2) where u and e are mu and epsilon.

As far as the gravitational field of a capacitor, or any EM field, the source for the EFE will be the EM contribution to the stress-energy tensor, which is the EM stress-energy tensor. T_00 is the energy density above (which gives us the mass part in the Newtonian limit), T_0i and T_i0 is the Poynting vector, and the T_ij, the 3x3 lower part is the Maxwell Stress Tensor (momentum content of the EM field, which the most general -- and difficult -- way to calculate EM forces).

Now EM stress-energy curves space-time like any other stress-energy. However, that curvatures "feeds back" into Maxwell. Maxwell against curved space-time (and even in non-inertial frames in flat space-time) gets complicated, and involves the metric.

The EFE and Maxwell thus become coupled in a complex way. These are "electrovacuums". Exact solutions in some highly symmetrical, relatively simple, cases are known. But the general case is quite complex.

And EM field gives us mass-energy, mass-energy currents, and momentum and its currents, giving us the whole shebang of GR gravitational effects, including frame dragging. And curved space-time can do weird things feeding back and shaping the field itself.

However, these effects are very small via E = mc^2. A large amount of energy is a small amount of mass (well, according to our scales), so it would take a humdinger of an EM field to make measurable gravitational effects. Imagine converting the entire mass of the earth to "pure energy" as Mr. Spock would say in the form of EM radiation/photons. That would be one heck of a ball of energy. However, in the brief time before it blew itself to the four winds, it would be making the same external gravitational field as the earth did.

The earth's field is woefully weak as far as space-time curvature goes, but imagine how much energy that would be. That should give you an idea of a how big an EM field you'd need to make significant gravity.

-Richard

frankuitaalst
2008-Sep-12, 11:39 AM
There is no excess.

The energy is stored in the system. That does create a new question though. What is the gravitational field inside a capacitor? :)
In the meantime Publius already gave an expanation .
There should be a ( very very small ) "gravitational filed" when a capacitor is loaded if this is what you mean. :-).

In the classical way ( rudimentary ) .
Suppose N elektrons move from one pole to another at loading .
delta Egrav = G. (N*Mel)^2/d ( lets suppose first order formula also applies to a plate ) .
Plugging in the elektron mass , number of electrons ... in the case of a capacitor of 1m , 1mm distance at 230Volt gives : 1.3e-6 kg moved electronmass , which gives a Egrav of 1.0e-19 Joule !
The Electrical energy of the capacitor ( having C=1000 Farad ) is then 2.6e+7 Joule .
So : Eelectrical /Egrav ~2.5e+26 .
This explains why we use capacitors in order to store electrical energy instead of gravitational energy I think :-) .

mugaliens
2008-Sep-12, 06:22 PM
There is no excess.

The energy is stored in the system. That does create a new question though. What is the gravitational field inside a capacitor? :)

Yaa... Ok, excess electrons on the plates attached to one lead, and a lack thereof on the plates attached to the other lead.

The mass of the capacitor is the same, charged or not. Mass is energy-equivalent, but it's the energy which changes, not the mass, as opposed to chemistry, where energy changes form to increases or decreases in mass.

Hornblower
2008-Sep-12, 08:12 PM
Yaa... Ok, excess electrons on the plates attached to one lead, and a lack thereof on the plates attached to the other lead.

The mass of the capacitor is the same, charged or not. Mass is energy-equivalent, but it's the energy which changes, not the mass, as opposed to chemistry, where energy changes form to increases or decreases in mass.

E = mc2. Unconditionally. At least that is my understanding of the meaning of Einstein's famous equation.

If we forcibly redistribute the electrons in a capacitor, the energy we put into it to do so should manifest itself as a slight increase in mass. Immeasurably small, yes indeed, but still predicted by theory.