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Friday Quiz: Capacitors

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5/1/2015 00:04 AM EDT

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LarryBrasfield
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displacement field is not electric field
LarryBrasfield   6/15/2015 4:28:02 PM
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Quiz question 1 has only correct answer 'a'. Displacement is to electric field as charge is to voltage, one being the intensive quantity and the other being the extensive quantity. For capacitors made with dielectrics having no voltage dependence, (and otherwise constant C), displacement would be proportional to electric field intensity, (witth the constant of proportionality being the material permittivity), but that relation will not hold true in general for devices sold as capacitors.

For the same reason, quiz question 2 would be more accurately answered:

U = C * V^2 / 2 for constant capacitors

  and

U = integral from 0 to V of C(v) * dv for non-linear capacitors.

 

This refinement is all in the spirit of considering what is involved in using capacitors properly beyond what some might think.

MeasurementBlues
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Re: Next Friday Quiz Suggestion
MeasurementBlues   5/5/2015 4:49:57 PM
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@pseudoid

I have just the source for a quiz on inductors: Trilogy of Magnetics book, also by Wurth Electronik. Give new a couple of weeks or so. This Friday's quiz is already written.

Of course, you're always welcom to write a quiz. See About Us above right for contact info.

pseudoid
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Next Friday Quiz Suggestion
pseudoid   5/5/2015 12:44:27 PM
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@MeasurementBlues

Now that this Pop Quiz' replies are almost in the kTB; it would be only fair to have a Pop Quiz on C's brother L.

Bert22306
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Re: Capacitance of free space conductor
Bert22306   5/4/2015 4:06:18 PM
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Yes, we're in our space ship, it has some potential, we don't have to know what it is.

That voltage measurement is the relative voltage referenced to our ship. No other reference is required.

And no other is needed. That already makes two electrodes, for the device to behave as a capacitor. If you're referencing against spaceship ground, it could be that instead of charge wanting to flow from the sphere to the spaceship ground, the spaceship's ground would instead send more charge into the sphere.

Either way, makes no difference. The main thing is that to behave as a capacitor, i.e. to either absorb or release energy, that device requires the second electrode.

But the mass of the other object does not matter, as long as we know what our mass is.

Not true. To stay on topic, about energy stored, this is exactly the same as the charged-sphere-in-space problem. Whatever energy you store in the device, in this case the ball suspended in air while the earth existed, can only be retrieved when you supply some sort of new reference. And depending on that new reference, the energy may not be able to be retrieved at all.

To be pedantic, gravitational force F = GMm/r^2, for two masses, M and m, a distance r apart.

At the surface of the earth, the force acting on the ball

F = ma = m(GM/r^2) = mg, where g = GM/r^2 (~9.8 m/sec^2), r is the radius of the earth, assume the radius of the ball is negligible compared with r, and M is the mass of the earth.

Tossing the ball in the air, stored energy E = Fd = mgh, and it ONLY holds true close to the earth's surface, where d=h << r.

So, if the ball were tossed up an equal distance above Mars, or the moon, the energy stored would be less. Toss the ball in air, subtract the earth, and you have no way of retrieving that energy. Add some other object as reference, say the moon at an equal distance r+h, and now you can retrieve some energy from that ball, although not nearly as much as before. 

The real point is that the mass of the unknown was what it was before and after its value was known.

Can't parse this. The real point is that the mass of the ball, or the charge in the sphere, are not enough by themselves to determine how much stored energy there might be. You need to have a reference before you can determine whether, or how much, energy you can retrieve.

Les_Slater
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Re: Capacitance of free space conductor
Les_Slater   5/3/2015 11:28:23 PM
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@bert,

"How do you sample that voltage, Les? Has to be with respect to something."

Yes, we're in our space ship, it has some potential, we don't have to know what it is. We know what current drain of our voltage measurement instrument requires to measure a voltage, which times the time it takes to measure equals a charge transfer, which we will have to account for in our final calculations. That voltage measurement is the relative voltage referenced to our ship. No other reference is required.

"Let's say you left an object in space. How much does it weigh? You don't know, until you bring that object up close to some other mass. Same thing."

But the mass of the other object does not matter, as long as we know what our mass is. We'll assume that two objects, one with the unknown mass can be brought together in a mechanically rigid connection. We don't need to know velocity or direction of the rigidly connected objects. We just need a calibrated thrust and the ability to measure CHANGE in velocity along the thrust line with respect to time. Taking into account any fuel burn, the mass of the connected masses can be calculated. All that is then necessary is to subtract the known mass from the total mass to find the mass of the unknown. The real point is that the mass of the unknown was what it was before and after its value was known. Again, no other reference required.

Bert22306
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Re: Capacitance of free space conductor
Bert22306   5/3/2015 9:51:00 PM
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Yes, I know everybody, everybody except me, is saying you need a reference point. But a reference to what?

Well, for capacitors, that's simple. A capacitor is of no use if it doesn't provide a voltage. And a voltage, by definition, is a potential difference. So that's pretty clear, Les. "Difference." You can't have a "difference" without a reference.

Let's say we are on a space ship with a mission to deliver a negative one volt to an isolated conductor in deep space, no other conductors around except it, and us approaching. As we approach, we sample its voltage, with respect to us, and store it for reference.

Negative one volt is measured how?


As we approach, we sample its voltage, with respect to us, and store it for reference.

How do you sample that voltage, Les? Has to be with respect to something. Let's say you left an object in space. How much does it weigh? You don't know, until you bring that object up close to some other mass. Same thing.


Then we add electrons to it until its one volt less than we first measured. We count the electrons delivered, put it in the log book along with the calculated capacity (capacitance derived from Coulombs Law), then head off on our next mission with a slightly less negative, or more positive charge.

For a capacitor, charge isn't enough to determine how much energy is stored that you can retrieve. Capacitors are supposed to transfer energy. The voltage created by that charge will vary, depending on the reference.


Les_Slater
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Re: Capacitance of free space conductor
Les_Slater   5/3/2015 9:32:23 PM
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@bert,

Yes, I know everybody, everybody except me, is saying you need a reference point. But a reference to what? I deliberately took that wiki description of self-capacitance out of context because it stands on its own. Let's say we are on a space ship with a mission to deliver a negative one volt to an isolated conductor in deep space, no other conductors around except it, and us approaching. As we approach, we sample its voltage, with respect to us, and store it for reference. Then we add electrons to it until its one volt less than we first measured. We count the electrons delivered, put it in the log book along with the calculated capacity (capacitance derived from Coulombs Law), then head off on our next mission with a slightly less negative, or more positive charge.

Talking about the necessity of a conductive sphere at an infinite distance, especially requireing the said isolated conductor must be at its center... is complete nonsence!

Bert22306
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Re: Capacitance of free space conductor
Bert22306   5/3/2015 9:04:14 PM
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This is how the Wikipedia reference continues, which is what I described in my previous post. You always need a reference point, even though this text is confusing (my refrernces made it much clearer):

http://en.wikipedia.org/wiki/Capacitance#Self-capacitance

The reference point for this potential is a theoretical hollow conducting sphere, of infinite radius, centered on the conductor. Using this method, the self-capacitance of a conducting sphere of radius R is given by:
C=4\pi\varepsilon_0R \,


Les_Slater
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Re: Capacitance of isolated spherical conductor
Les_Slater   5/3/2015 8:56:02 PM
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traneus,

The more I think about it the less sure I am that Maxwell knew what he was talking about.

Bert22306
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Re: Capacitance of free space conductor
Bert22306   5/3/2015 8:55:41 PM
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Yes, but it can be ANY other conductor with different potental, not necessarily one that it MAY have been previously paired.

So the fact remains, in order for the device to behave like a capacitor, in your circuit, it needs to have two electrodes. Same as the two mechanical analogies I gave.

You can remove the earth and replace it with a much smaller mass, and the potential energy stored in that ball will be way less than what it was originally. Or vice versa, replacing the earth with a larger mass.

Just puts this supposed conundrum in perspective.

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