This how I think of voltage / resistance / current

[Ratch]

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Energy is relative, not absolute, so the correct question to ask is "How much energy does it take to fill the tank with fluid?"

If, for example, the tank were at the top of a tower (a water tower), then the water in the tank would have potential energy according to the formula E = Mgh, where M is the mass of liquid, g is the gravitational acceleration and h is the elevation above ground. It would take that much energy to pump the water up into the tower (plus friction losses), and you could get that much energy back again (minus friction losses) by letting the water down.

Agreed, but I specified an isolated tank.  How much energy to pressurize an isolated tank of fluid to a particular pressure?

Ratch

[IanB]

Agreed, but I specified an isolated tank.  How much energy to pressurize an isolated tank of fluid to a particular pressure?

If the tank is isolated, how do you propose to change its pressure?

[Ratch]

As I mentioned in the above post. What energy does a tank of fluid have at particular volume and pressure when it is isolated, and not part of a moving fluid stream?

Let's suppose you are thinking of a closed metal tank completely full of water at atmospheric pressure, and now suppose you want to add a small additional volume of water into the tank. The work required to do this is equal to the volume of water added times the pressure difference, which initially is zero. But after you have pumped in a bit of water the pressure in the tank has gone up (let's say the walls are elastic and have stretched a bit). So the next bit of water you want to pump in will require some work as the pressure difference is no longer zero. The next bit of water after that will require more work, and the next more work still. The work to keep pumping in more water will keep increasing as the pressure in the tank goes up and up.

Hopefully you will begin to see the similarity here with pumping charge into a capacitor.

The energy you can get back out of the tank after you have pumped water into it is essentially equal to the work you did pumping water into it. The water itself doesn't have energy, but the tank plus water system has stored some energy, just like the capacitor plus charge system has stored some energy when you charge it up.

This, and the recognition that both pressure and voltage are scalar fields of potential, goes further towards explaining why voltage is "electrical pressure".

Allowing the walls of the tank to be elastic is a backdoor way of saying the liquid is compressible. One way or the other, the volume of the liquid has to change for energy to exchange.

You might call it electrical pressure for an analog, but the it does not make sense for a real electrical circuit.

By the way, you don't pump up a capacitor with charge. You pump it up with energy.  The net charge of a capacitor is the same at 0 volts, 100 volts, or 1000 volts.  The amount of charge on one plate is exactly the same as the charge removed on the opposite plate for a net change of zero.  So it is wrong to say a capacitor is "charged".  It is instead energized.  Same for batteries.

Ratch
Hopelessly Pedantic

[Ratch]

Agreed, but I specified an isolated tank.  How much energy to pressurize an isolated tank of fluid to a particular pressure?

If the tank is isolated, how do you propose to change its pressure?

With completely rigid tank walls and a incompressible liquid, you cannot.  With gas, one can release the gas or pump in more.

Ratch

[helius]

You are calculating the divergence of an enclosed surface using vector calculus.  The result will be 50 times the area of the balloon surface in square inches.  How does that turn energy/volume into force/area?

You normalize the surface S to 1 unit of area, which is where the m^-2 comes from. I was showing you that the pressure is a scalar, even though it is derived from a vector. The force really needs to be calculated as a tensor field in this case, and I'll admit my tensor calculus is not very strong. But I hope I don't need to argue that this can't change the units of the problem.

In practice you would skip the tensors (since the air is isothermic and isobaric) and calculate the pressure using the ideal gas law, which is completely scalar. In \$ PV = nRT \$, R has units of joules per kelvin. Which means that the product of pressure and volume is energy 

[Ratch]

You are calculating the divergence of an enclosed surface using vector calculus.  The result will be 50 times the area of the balloon surface in square inches.  How does that turn energy/volume into force/area?

You normalize the surface S to 1 unit of area, which is where the m^-2 comes from. I was showing you that the pressure is a scalar, even though it is derived from a vector. The force really needs to be calculated as a tensor field in this case, and I'll admit my tensor calculus is not very strong. But I hope I don't need to argue that this can't change the units of the problem.

In practice you would skip the tensors (since the air is isothermic and isobaric) and calculate the pressure using the ideal gas law, which is completely scalar. in \$ PV = nRT \$, R has units of joules per kelvin. Which means that the product of pressure and volume is energy 

I agree with you.  My physics book says that pressure force is defined to be only in the normal direction from the surface.  Therefore no directional variation is possible, which makes pressure a scalar value.

Ratch
Hopelessly Pedantic

[IanB]

Allowing the walls of the tank to be elastic is a backdoor way of saying the liquid is compressible. One way or the other, the volume of the liquid has to change for energy to exchange.

If you had a (hypothetical) perfectly incompressible fluid contained within a (hypothetical) inelastic vessel, then the pressure inside the vessel could be any value you want it to be. In such a case no energy change is required for the pressure to change.

You might call it electrical pressure for an analog, but the it does not make sense for a real electrical circuit.

Everything has matched exactly so far. For instance, if I take an insulated metal sphere and move it in an electric field from a position of lower potential (let's say 0 V) to a position of higher potential (let's say 300 kV), then I have increased the voltage on the sphere by 300 kV. How much additional energy is stored in the sphere after I have done this? (This is the analog of the incompressible fluid in the inelastic vessel.)

By the way, you don't pump up a capacitor with charge. You pump it up with energy.  The net charge of a capacitor is the same at 0 volts, 100 volts, or 1000 volts.  The amount of charge on one plate is exactly the same as the charge removed on the opposite plate for a net change of zero.  So it is wrong to say a capacitor is "charged".  It is instead energized.  Same for batteries.

When a conventional two plate capacitor is charged up, charge gets moved from one plate to the other, and it takes work to do this. A capacitor can then do work on an external circuit by allowing the accumulated charge to flow back to the lower energy state. So it is quite legitimate to say the capacitor has been "charged up".

If you had a single pole capacitor (a metal sphere in free space), then you would actually have to add charge to it (or remove charge from it) when wanting to change its electrical potential.

[Ratch]

Allowing the walls of the tank to be elastic is a backdoor way of saying the liquid is compressible. One way or the other, the volume of the liquid has to change for energy to exchange.

If you had a (hypothetical) perfectly incompressible fluid contained within a (hypothetical) inelastic vessel, then the pressure inside the vessel could be any value you want it to be. In such a case no energy change is required for the pressure to change.

The pressure could only be zero because no volume change or compression is allowed.

You might call it electrical pressure for an analog, but the it does not make sense for a real electrical circuit.

Everything has matched exactly so far. For instance, if I take an insulated metal sphere and move it in an electric field from a position of lower potential (let's say 0 V) to a position of higher potential (let's say 300 kV), then I have increased the voltage on the sphere by 300 kV. How much additional energy is stored in the sphere after I have done this? (This is the analog of the incompressible fluid in the inelastic vessel.)

As I said before, I am not into analogs too much.

By the way, you don't pump up a capacitor with charge. You pump it up with energy.  The net charge of a capacitor is the same at 0 volts, 100 volts, or 1000 volts.  The amount of charge on one plate is exactly the same as the charge removed on the opposite plate for a net change of zero.  So it is wrong to say a capacitor is "charged".  It is instead energized.  Same for batteries.

When a conventional two plate capacitor is charged up, charge gets moved from one plate to the other, and it takes work to do this. A capacitor can then do work on an external circuit by allowing the accumulated charge to flow back to the lower energy state. So it is quite legitimate to say the capacitor has been "charged up".

No, the charge is added to one plate and deleted from the other plate.  The effect may be the same as a charge shift, but the net change of the charge is still zero.  You could say one plate is charged up, but then it is also legitimate to say the other plate is charged down.  It is unambiguous to say the cap is charged with energy, so one might as well say it is energized.

If you had a single pole capacitor (a metal sphere in free space), then you would actually have to add charge to it (or remove charge from it) when wanting to change its electrical potential.

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

Ratch
Hopelessly Pedantic

[rs20]

The pressure could only be zero because no volume change or compression is allowed.

Wrong, pressure is defined as Force/Area. There is no requirement for changes in volume for a force to be applied, particularly in the hypothetical inelastic regime that IanB is referencing.

No, the charge is added to one plate and deleted from the other plate.  The effect may be the same as a charge shift, but the net change of the charge is still zero.  You could say one plate is charged up, but then it is also legitimate to say the other plate is charged down.  It is unambiguous to say the cap is charged with energy, so one might as well say it is energized.

You can say a capacitor is storing a certain amount of charge, and you can say that a capacitor is storing a certain amount of energy. It is pointless to try and claim that one view is better than the other, because each is useful in different circumstances. I think about the charge stored in a capacitor far more often than the energy; and in particular, if one end of the capacitor is connected to circuit ground (which is a very common use-case), the "charge on the capacitor" becomes a blindingly obviously unambiguous and useful concept.

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

Dielectric: space. Voltage reference: theoretical sphere of infinite radius. Link: https://en.wikipedia.org/wiki/Capacitance#Self-capacitance . For example, planet Earth could be considered a single-pole capacitor of value 710uF.

Btw, I encourage you use the "Preview" feature to make sure your quotes are matched up properly, makes it less confusing to read.

[IanB]

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html#c2

[Ratch]

The pressure could only be zero because no volume change or compression is allowed.

Wrong, pressure is defined as Force/Area. There is no requirement for changes in volume for a force to be applied, particularly in the hypothetical inelastic regime that IanB is referencing.

We are talking about an incompressible liquid and inelastic tank walls.  If those two entities cannot be expanded or compressed, then how can a pressure be applied?

No, the charge is added to one plate and deleted from the other plate.  The effect may be the same as a charge shift, but the net change of the charge is still zero.  You could say one plate is charged up, but then it is also legitimate to say the other plate is charged down.  It is unambiguous to say the cap is charged with energy, so one might as well say it is energized.

You can say a capacitor is storing a certain amount of charge, and you can say that a capacitor is storing a certain amount of energy. It is pointless to try and claim that one view is better than the other, because each is useful in different circumstances. I think about the charge stored in a capacitor far more often than the energy; and in particular, if one end of the capacitor is connected to circuit ground (which is a very common use-case), the "charge on the capacitor" becomes a blindingly obviously unambiguous and useful concept.

No, I made it very clear that a capacitor has the same net charge when it is energized at 0, 100 volts, or 1000 volts.  That is because the charge added to one plate is removed from the opposite plate for a net change of zero.  Only the energy stored in the capacitor changes when a voltage is applied.

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

Dielectric: space. Voltage reference: theoretical sphere of infinite radius. Link: https://en.wikipedia.org/wiki/Capacitance#Self-capacitance . For example, planet Earth could be considered a single-pole capacitor of value 710uF.

The article mentions a Van de Graaff generator and the planet Earth. No explanation of what constitutes the plates and dielectric. Neither is a single metal sphere.

Ratch
Hopeless Pedantic

[Ratch]

Google did not have much information on a single-pole capacitor.  What they did show was not a metal sphere in space.  Where is its dielectric?  What is its voltage reference?  Show me a link describing one.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html#c2

OK, I looked at the link.  Too bad the reference the link gives is missing.  A single-pole capacitor (SPC) does not look too practical, because it would take a a static generator to charge up its one plate.  Why does it have to be a metal sphere.  Why not a big piece of sheet metal?  I will concede, however, that a SPC does "charge up".  Unlike a two-pole capacitor.

Ratch
Hopelessly Pedantic

[Stephan_T]

A single-pole capacitor (SPC) does not look too practical, [...]

You find them practically everywhere. All you need is isolation (e.g. pair of rubber soles will do). To charge them, it just takes some friction (e.g. walk over a carpet of a certain material will do) and you yourself can get charged substantially. This has the potential (pun intended) to hurt you.
But the EMF of such a charge can also interfere with your multimeter and oscilloscope. I usually use a PVC-Installation tube as test device to experiment with that effect.
Depending on the impedance of the measurement circuit, it can easily cause voltage readings of hundreds of mV for hundreds of milliseconds just exposed to the tiny tip of an otherwise shielded oscilloscope probe. If you use unshielded wires it gets much worse.

[rs20]

We are talking about an incompressible liquid and inelastic tank walls.  If those two entities cannot be expanded or compressed, then how can a pressure be applied?

[/quote]

Depending on your point of view and your level of trust in infinities and infinitesimals:
- Pressure can be applied by adding an infinitesimal amount of incompressible liquid. The point here is that the pressure in a tank with incompressible liquid and inelastic tank walls is undefined, not zero -- try taking limits as compressibility approaches infinity, you'll find that the pressure does not converge to zero.
- Pressure can be applied having an (inelastic but moveable) sliding piston in the side of the box. Pressure inside the box equals normal force on the piston / area of piston.

No, I made it very clear that a capacitor has the same net charge when it is energized at 0, 100 volts, or 1000 volts.  That is because the charge added to one plate is removed from the opposite plate for a net change of zero.  Only the energy stored in the capacitor changes when a voltage is applied.

I agree that the net charge on a capacitor remains unchanged (neglecting self-capacitance of the whole assembly, of course). However, the turn of phrase "charge up a capacitor" remains obviously meaningful and useful, even if it is only referring to a charge imbalance within the capacitor.

[Ratch]

A single-pole capacitor (SPC) does not look too practical, [...]

You find them practically everywhere. All you need is isolation (e.g. pair of rubber soles will do). To charge them, it just takes some friction (e.g. walk over a carpet of a certain material will do) and you yourself can get charged substantially. This has the potential (pun intended) to hurt you.
But the EMF of such a charge can also interfere with your multimeter and oscilloscope. I usually use a PVC-Installation tube as test device to experiment with that effect.
Depending on the impedance of the measurement circuit, it can easily cause voltage readings of hundreds of mV for hundreds of milliseconds just exposed to the tiny tip of an otherwise shielded oscilloscope probe. If you use unshielded wires it gets much worse.

I have been thinking about that.  Statically charged particles are a nuisance, but are they really miniature  capacitors?  Most are not hollow conducting shells, so their geometry is not such that their charge is separated from their electric field.  It is something to think about

Ratch
Hopelessly Pedantic

[Ratch]

We are talking about an incompressible liquid and inelastic tank walls.  If those two entities cannot be expanded or compressed, then how can a pressure be applied?

Depending on your point of view and your level of trust in infinities and infinitesimals:
- Pressure can be applied by adding an infinitesimal amount of incompressible liquid. The point here is that the pressure in a tank with incompressible liquid and inelastic tank walls is undefined, not zero -- try taking limits as compressibility approaches infinity, you'll find that the pressure does not converge to zero.
- Pressure can be applied having an (inelastic but moveable) sliding piston in the side of the box. Pressure inside the box equals normal force on the piston / area of piston.

I have to apologize.  I said "pressure", but I meant energy.  I guess I got stuck on automatic.  The piston will not be able to transfer any energy to the incompressible liquid, because it will not be able to move further, and no more energy can be exchanged.  This is similiar to a heavy force being applied to a immovable object.  No work is done.

No, I made it very clear that a capacitor has the same net charge when it is energized at 0, 100 volts, or 1000 volts.  That is because the charge added to one plate is removed from the opposite plate for a net change of zero.  Only the energy stored in the capacitor changes when a voltage is applied.

I agree that the net charge on a capacitor remains unchanged (neglecting self-capacitance of the whole assembly, of course). However, the turn of phrase "charge up a capacitor" remains obviously meaningful and useful, even if it is only referring to a charge imbalance within the capacitor.

Charging a cap or battery is an example of technical slang prevalent the tech world.  Another two examples are "current flow", and NASA referring to their astronauts as "walking" in space.  If their tether breaks, would they walk away?  They should say "external excursion".  Even though folks understand what the slang means, it is not accurately descriptive of what is really happening.  Understanding what action is being performed is not the same as being precise about what is happening.   

Ratch
Hopelessly Pedantic

[rs20]

I have to apologize.  I said "pressure", but I meant energy.  I guess I got stuck on automatic.  The piston will not be able to transfer any energy to the incompressible liquid, because it will not be able to move further, and no more energy can be exchanged.  This is similiar to a heavy force being applied to a immovable object.  No work is done.

Sure, Energy = F * d; even if F is huge, d = 0 --> Energy = 0.

Charging a cap or battery is an example of technical slang prevalent the tech world.  Another two examples are "current flow", and NASA referring to their astronauts as "walking" in space.  If their tether breaks, would they walk away?  They should say "external excursion".  Even though folks understand what the slang means, it is not accurately descriptive of what is really happening.  Understanding what action is being performed is not the same as being precise about what is happening.   

That is just:

Hopelessly Pedantic

But more seriously; yes, there are lots of instances of English being not literally true. It is a fool's errand to try and fix this though; and the attempt can be truly harmful and distracting when it derails a pedagogical discussion.

P.S./ You're not nesting quotes correctly which makes things confusing; perhaps ignore my earlier request please. I do appreciate the effort thus far

[Ratch]

P.S./ You're not nesting quotes correctly which makes things confusing; perhaps ignore my earlier request please. I do appreciate the effort thus far

I don't understand the problem.  Every quote in post #65 was labeled with the author.

Ratch
Hopelessly Pedantic

[Stephan_T]

I have been thinking about that.  Statically charged particles are a nuisance, but are they really miniature  capacitors?  Most are not hollow conducting shells, so their geometry is not such that their charge is separated from their electric field.  It is something to think about

Why do you say "particles"? That term usually refers to small objects. Do you consider your self to be a particle?
And how do you define the term "capacitor"? I prefer a pragmatic definition and only call objects that intentionally have a capacitance a capacitor. The nuisance is usually called "parasitic capacitance" and there is a lot of that around us.

[Ratch]

I have been thinking about that.  Statically charged particles are a nuisance, but are they really miniature  capacitors?  Most are not hollow conducting shells, so their geometry is not such that their charge is separated from their electric field.  It is something to think about

Why do you say "particles"? That term usually refers to small objects. Do you consider your self to be a particle?
And how do you define the term "capacitor"? I prefer a pragmatic definition and only call objects that intentionally have a capacitance a capacitor. The nuisance is usually called "parasitic capacitance" and there is a lot of that around us.

Because dust, lint, powders, and other small particles are attracted by static electricity.

No, I don't consider myself a particle.  I might, however, possess a small amount of capacitance.

A capacitor is an object that can store electrical energy in a electrostatic field.  By that definition, I surmise that small, statically charged particles are miniature capacitors.

I was thinking of unwanted static electricity as a nuisance, not the nuisance of unwanted capacitance.

Ratch
Hopelessly Pedantic