Physics problem...

jumpguy

An instrument attached to a mast outside a research station must be maintained at a temperature of 10degrees Celsius using a small heater. The instrument and the heater are completely enclosed in an insulating cylindrical foam casing of length 70 cm, diameter 25 cm and thickness 1.2 cm. The thermal conductivity of the foam is 0.033 W m^−1K&−1. If the air temperature is −30degrees Celsius determine the power output required for the heater.

Some help with this question would be appreciated, I simply can't spot where I'm going wrong! :mad:

What I did was calculate the amount of heat lost per second using the thermal conductance formula (k*Area*temp difference)/Thickness and I come up with 65.9W (or 65.9 Joules per second). Shouldn't the heater have to only put out 65.9W in order to keep the space at 10degrees Celsius? That would keep the energy within in the container constant, or am I misguided? The answer is 71.3W!

Morbert

jumpguy wrote: »

Some help with this question would be appreciated, I simply can't spot where I'm going wrong! :mad:

What I did was calculate the amount of heat lost per second using the thermal conductance formula (k*Area*temp difference)/Thickness and I come up with 65.9W (or 65.9 Joules per second). Shouldn't the heater have to only put out 65.9W in order to keep the space at 10degrees Celsius? That would keep the energy within in the container constant, or am I misguided? The answer is 71.3W!

You are forgetting the area on the lid.

I.e.

Area = 2 Pi*r*h + 2*Pi*r^2

You were using:

Area = 2 Pi*r*h + Pi*r^2

jumpguy

Ugh, you might notice it's turning into a dismal day of study for me...

Two parallel plate capacitors are identical, except that the space between the plates of one is empty and the other is filled with dielectric (permittivity=4.50). The empty capacitor is connected to a 12 V battery. Determine the potential difference that is required across the plates of the capacitor filled with the dielectric so that it stores the same amount of electrical energy as the empty capacitor.

Right, so I used the formula E=CV^2/2 for this. Because the energies will be the same, I let CV^2/2 for the empty capacitor equal the CV^2/2 for the dielectric capacitor. The halves cancel and I have

Dielectric capacitor Energy = Empty capacitor energy
CV^2 = CV^2
C is broken into permittivity*Area/distance, but because the capacitors are same in every way except for the dielectric area/distance cancels, so I've only permittivity left. Let P=permittivity from now on...

PV^2 = PV^2
(4.5)V^2 = (8.85e-12)(12^2)

Solve for V^2...I get 1.682e-5. Which is off by a mile, the answer given is 5.7V.

*sigh*

Morbert wrote: »

You are forgetting the area on the lid.

I.e.

Area = 2 Pi*r*h + 2*Pi*r^2

You were using:

Area = 2 Pi*r*h + Pi*r^2

Ah...utter facepalm of a mistake there. Thanks so much!

Morbert

jumpguy wrote: »

Ugh, you might notice it's turning into a dismal day of study for me...

Right, so I used the formula E=CV^2/2 for this. Because the energies will be the same, I let CV^2/2 for the empty capacitor equal the CV^2/2 for the dielectric capacitor. The halves cancel and I have

Dielectric capacitor Energy = Empty capacitor energy
CV^2 = CV^2
C is broken into permittivity*Area/distance, but because the capacitors are same in every way except for the dielectric area/distance cancels, so I've only permittivity left. Let P=permittivity from now on...

PV^2 = PV^2
(4.5)V^2 = (8.85e-12)(12^2)

Solve for V^2...I get 1.682e-5. Which is off by a mile, the answer given is 5.7V.

Dielectric Capacitor Energy = Empty Capacitor Energy
4.5 A/d V^2 = A/d 12^2
4.5 V^2 = 144
V^2 = 144/4.5 = 32
V = Square root of 32 = 5.7

jumpguy

Morbert wrote: »

Dielectric Capacitor Energy = Empty Capacitor Energy
4.5 A/d V^2 = A/d 12^2
4.5 V^2 = 144
V^2 = 144/4.5 = 32
V = Square root of 32 = 5.7

Thanks very much! I have yet another question though, why is the permittivity of free space ignored in the empty capacitor?

Morbert

jumpguy wrote: »

Thanks very much! I have yet another question though, why is the permittivity of free space ignored in the empty capacitor?

It's not technically ignored. When they say the dielectric has a permitivity of 4.50, they mean it has a permitivity of 4.50 times the permitivity of free space. I.e. One has a capacitance of eA/d, and the other has a capacitance of 4.50 eA/d (Where e is the permitivity of free space).

So more explicity, the calculation is
4.5 e A/d V^2 = e A/d 12^2
4.5 V^2 = 144 (e A/d is cancelled out on both sides)
V^2 = 144/4.5 = 32
V = Square root of 32 = 5.7

krd

jumpguy wrote: »

Thanks very much! I have yet another question though, why is the permittivity of free space ignored in the empty capacitor?

Because it's considered to be 1.

Free space is considered to be the vacuum. In other words there's nothing for the electric field to permeate.

Any material between the plates will offer a resistance to the electric field - I really have no idea why it does that. I don't know - maybe the dielectric is holding that charge. I know if you make the current too high, eventually a current will flow through the dielectric and burn it. I don't really know what current is either.

I read something recently that said current is electrons exchanging photons - is this true?

And if that is so - what is it when you get a current to flow between two highly charged plates in a vacuum?

Morbert

krd wrote: »

Because it's considered to be 1.

Free space is considered to be the vacuum. In other words there's nothing for the electric field to permeate.

The electromagnetic field permeates a vacuum. The permittivity, in standard units, is 8.85x10^-12.

Any material between the plates will offer a resistance to the electric field - I really have no idea why it does that. I don't know - maybe the dielectric is holding that charge. I know if you make the current too high, eventually a current will flow through the dielectric and burn it. I don't really know what current is either.

The material responds to the electric field, taking energy from the field.

I read something recently that said current is electrons exchanging photons - is this true?

Electrons interact by exchanging virtual photons. But a current is simply a net flow of charge. (I.e. Moving electrons in this case).

And if that is so - what is it when you get a current to flow between two highly charged plates in a vacuum?

It is simply electrons passing through the vacuum from one plate to the next.

krd

Morbert wrote: »

The electromagnetic field permeates a vacuum. The permittivity, in standard units, is 8.85x10^-12.

If there's nothing there, what is it permeating?

Is it some kind of covariant related to the speed of light?

The material responds to the electric field, taking energy from the field.

The material itself becomes charged?..........

I read recently, that the reason electric charge resides on the surface of a charged body, is the electric field lines cancel each other out internally - if that's true. Why if I raise the charge too high, will the interior of the dielectric burn?

Electrons interact by exchanging virtual photons. But a current is simply a net flow of charge. (I.e. Moving electrons in this case).

Passing current through any conductor, will cause it to heat - so that heat, I imagine, is real photons.....Like very real and observable photons are emitted from a tungsten filament in a light bulb.

Are virtual photons transferring charge in a conductor when current is flowing through it?.......Are these virtual photons becoming real photons and releasing heat?

Do electrons ever actually flow in a conductor - do they ever go anywhere or is the energy actually passed from electron to electron?

My studies never went as far as virtual photons. My understanding of how photons are created - electrons are energised, they're unstable, and fall back releasing the energy as a photon. Is the virtual photon, the same idea, but instead of releasing the energy as a photon, the receiving electron is raised in energy?

It is simply electrons passing through the vacuum from one plate to the next.

Is the electron something that can exist discreetly by itself - or is that electron in the vacuum, actually a ripple in the electric field between the two plates?

Morbert

krd wrote: »

If there's nothing there, what is it permeating?

Is it some kind of covariant related to the speed of light?

The EM field doesn't need a medium to permeate through. This is why light from the sun can reach us, even though it must travel through empty space.

The material itself becomes charged?..........

Close. The material becomes polarised.

I read recently, that the reason electric charge resides on the surface of a charged body, is the electric field lines cancel each other out internally - if that's true. Why if I raise the charge too high, will the interior of the dielectric burn?

Static charge, as the name implies, involves no moving charge. When you build a voltage across a medium, on the other hand, charge travels through the medium. The electrons "scatter" off the material, creating heat. It is the motion of electrons that causes heat. It is analogous to your hands heating up if you rub them together.

Passing current through any conductor, will cause it to heat - so that heat, I imagine, is real photons.....Like very real and observable photons are emitted from a tungsten filament in a light bulb.

The emitted photons are a consequence of the heat, not the cause of it.

Are virtual photons transferring charge in a conductor when current is flowing through it?.......Are these virtual photons becoming real photons and releasing heat?

No. Virtual photons are carrying the interactions between electrons, but they themselves have no charge. It is the electrons themselves that are responsible for current.

Do electrons ever actually flow in a conductor - do they ever go anywhere or is the energy actually passed from electron to electron?

They move, though they are highly de-localised in a metal. A current is a moving charge, and an electrons are charge carriers, not photons.

My studies never went as far as virtual photons. My understanding of how photons are created - electrons are energised, they're unstable, and fall back releasing the energy as a photon. Is the virtual photon, the same idea, but instead of releasing the energy as a photon, the receiving electron is raised in energy?

A a real photon is an excitation of the electromagnetic field, that satisfies certain equations and conditions. A virtual photon is a transient excitation that carries the electromagnetic force. http://en.wikipedia.org/wiki/Virtual_particle

Is the electron something that can exist discreetly by itself - or is that electron in the vacuum, actually a ripple in the electric field between the two plates?

I think I see the confusion. An electron is not made of the electric field, just as a skydiver is not made of the gravitational field. Instead, electrons interact with the electric field, since they have charge. What you could say is an electron is a quantised excitation of the electron field, which is distinct from the electric field.

Wh1stler

Morbert wrote: »

It is simply electrons passing through the vacuum from one plate to the next.

But in a functional capacitor, current does not flow between the plates; the electrostatic charge on one plate causes electrons to be attracted to, or repelled from, the other plate.

If electrons cross the space between the plates then the capacitor has broken down hasn't it; both plates would be at the same potential; a short-circuit would have occurred?

Morbert

Wh1stler wrote: »

But in a functional capacitor, current does not flow between the plates; the electrostatic charge on one plate causes electrons to be attracted to, or repelled from, the other plate.

If electrons cross the space between the plates then the capacitor has broken down hasn't it; both plates would be at the same potential; a short-circuit would have occurred?

Essentially yes. Current flows across plates if, for example, voltage exceeds the breakdown voltage. The important point is, if there is a current, that current is characterised by moving electrons, not photons.