I've a few more for ya...

chump

Answer correctly and I will reward you...

1.
Ted and Mary are 40m apart on ice (frictionless), Ted weighs 20kg and Mary 10kg (infants). In between them (20m away from either) lies a toy. The both pull on either end of a (light)rope that connects them and try to get to it.
What happens, who gets their first etc.?

2.
Make an estimate of the Electric Field strength at the surface of an Li nucleus.

3.
A bar magnet, axis in z-dir. is place infront of thin circular metallic disc that is in perpendicular plane to axis of magnet. If the disc is rotated about the axis what potential different/currents does it induce?
What happens when disc at rest and magnet rotates around it?

Ron DMC

chump wrote:

Answer correctly and I will reward you...

1.
Ted and Mary are 40m apart on ice (frictionless), Ted weighs 20kg and Mary 10kg (infants). In between them (20m away from either) lies a toy. The both pull on either end of a (light)rope that connects them and try to get to it.
What happens, who gets their first etc.?

Mary cos she'll be pulled towards Ted

chump

Ronny I want a detailed response of where they are end up with respect to start positions, and the physical explanation why

breadmonkey

2.) E = (1/4*PI*e0)*(q/r)

I think that's the formula. I'm not arsed googling for the radius of an li atom

chump

Electric field is force divided by charge. Fair enough.

That's the coulomb force that you mention there... Is that the force that is active in that region? I know the nuclear force acts more in the femtometer region. Is there also not some potential curve that varies with r, that starts and the origin (centre) nucleus and doesn't vary linearly with distance. At what point is the distance of half the nucleur radius there?



Do we have to use the Yakuwa model as it is the nuclear force at that distance? And ignore this Coulomb force...
I think I've some notes on the range of the Yukawa model. I'll get back to you

No1 is any use with the others?
USELESS

breadmonkey

Which nuclear force are you talking about, the strong nf? Anyway, where do you find this sh!t?

I'm in 1st engineering (not exactly a master phyics course), but that's the formula I (and the lecturer) was using to do the type of problem you mentioned.

chump

I nab them from tcd physics paper 5 Modership paper, ie. 4th year... I've to answer questions of this kind in a little over a weeks time...

I'm pretty sure you're meant to use the Yukawa model for that question.
I've no idea of the magnet one and I could make a stab at the 1st one, but am still unsure...

dudara

Use Gauss' Law here to determine electric field. I think that it is that simple

chump

Sorry dudara but I don't understand ... How would one do this?

dudara

Gauss' Law states that the closed integral of the electric field over a Gaussian surface is related to the charge contained within that surface.

Take the Li nucleus, radius r, containing a charge equal to +Ze, where Z is the number of protons and e is the charge on the electron. Draw a Gaussian surface around it, just snuggly fitting over it, this has surface area 4*PI*r^2.

From Gauss' Law, the Integral over the closed surface of (E.dA) = Q/epsilon

Thus
E (4*PI*r^2) = Ze/epsilon

Thus E = Ze / (4*PI*epsilon*r^2)

There's no need to make it any more complicated than that.

As you move away from the surface of the nucleus, the Gaussian surface increases in size, but the charge contained within it stays constant. Thus the electric field drops off away from the nucleus.

It's a different story inside the nucleus, as the amount of charge contained within the Gaussian surface will depend on your location in the nucleus. I'll try to explain this better if you'd like.

As to the infants on the ice, I think that they'll get there at the same time, as long as they both pull with equal force. Because there is no friction, the normal reactions and hence the weights, don't matter. All that matters for horizontal motion is horizontal force, and that will be the pulling force that they each exert. The only other possible horizontal force is friction and that is ruled out.

chump

Sorry dudara, I thought you were talking about question 3...

I'd never even considered than approach for question 2. It sounds feasable alright. When I get around to it, I'll use both approaches. I've a suspicion they expected people to go for the Yakuwa model answer as there's a latter part of the question that asks what would this field be in the 2s orbital which would probably revert to coulomb potential as it's far enough away from the strong force (not too sure)...

And as for the friction one, I was thinking the exact same myself just there about an hour ago (this stuff bounces around my head for a while ) ... I suspect though that if we were to give the ice a small coefficient of friction (say .1) this answer would change a lot, but yea if it's frictionless the forces should balance...

I think, need to consider a bit more...

And an after thought on your nucleus answer, I don't think it could work, as remember there are no electrons in there... they don't start until much further out...

Ishmael

dudara wrote:

Gauss' Law states that the closed integral of the electric field over a Gaussian surface is related to the charge contained within that surface.

Take the Li nucleus, radius r, containing a charge equal to +Ze, where Z is the number of protons and e is the charge on the electron. Draw a Gaussian surface around it, just snuggly fitting over it, this has surface area 4*PI*r^2.

From Gauss' Law, the Integral over the closed surface of (E.dA) = Q/epsilon

Thus
E (4*PI*r^2) = Ze/epsilon

Thus E = Ze / (4*PI*epsilon*r^2)

There's no need to make it any more complicated than that.

As you move away from the surface of the nucleus, the Gaussian surface increases in size, but the charge contained within it stays constant. Thus the electric field drops off away from the nucleus.

It's a different story inside the nucleus, as the amount of charge contained within the Gaussian surface will depend on your location in the nucleus. I'll try to explain this better if you'd like..

I'm posting this question out of interest more than anything.
Does this account for the charges on the other electrons in the atom or do you have to take account of these at all?

dudara

No, it doesn't. It's a simple first approximation, which I think can be considered relatively valid. Why? Because at the nucleus surface, the electrons are sufficiently far away that their electric field is weaker than that of the nearby densely-packed protons.

chump

But dudara what charge are you saying is coming from the nucleons when you say e in your expressions is the charge of the electrons. And so Ze is the charge of the total no. electrons in the atom (number electrons = number protons fair enough), so the number of nucleons is 2Z, and what is their charge? That of the electron?


Ah wait I think I ge what you're saying...
The number of e-'s = no protons, only protons are charged in nucleus, their charge is = that of e-'s (opposite)... so thru it all into the equation and out pops a value for the E-Field...

dudara

There are Z protons in the nucleus, each contributing an amount of charge of +e each. (The proton is equally but oppositely charged to the electron). The neutrons contribute nothing.

the nucleus is itself positively charged. When surrounded by a cloud of electrons, with an opposite and equal charge, then the atom becomes charge neutral.

The fact that the nucleus contains positive charge means that the direction of the electric field surrounding it is radially outwards, as opposed to inwards.

Does that help?

chump

danke... now get working on the other ones

dudara

nah, I don't like magnetics!!

chump

who does eh... I dont like electricity or magnetics, or superconductivity or semi conductivity, which are I suppose in the same vein, cheers anyway for the above

ApeXaviour

1.
Ted and Mary are 40m apart on ice (frictionless), Ted weighs 20kg and Mary 10kg (infants). In between them (20m away from either) lies a toy. The both pull on either end of a (light)rope that connects them and try to get to it.
What happens, who gets their first etc.?

2.
Make an estimate of the Electric Field strength at the surface of an Li nucleus.

I concur with Dudara's answer for the electric field question.

I don't think ted and mary would reach the toy at the same time however. Wouldn't that break conservation of momentum rule?

Total initial momentum (w.r.t. the ice) = 0, therefore total final momentum would also be zero. If they were to reach the toy at the same time it would mean they would have the same average velocity. Vt=-Vm ('t'ed and 'm'ary)
But if that was the case then Mt*Vt would not equal -Mm*Vm. To make it equal, mary's velocity (w.r.t. the ice) would have to be twice that of ted's. Meaning she'll get there ~ twice as quickly. Maybe I'm wrong (criticism welcome) but that's how I see it at this very late hour.

Good luck in your finals chump.. I won't be sitting them for another 2 years yet cos I'm utterly failing 3rd year I adopted a complete dosser persona this year. Should know better being 22 and all.

ApeXaviour

Also I'm a little rusty on the magnetism, but from my understanding, an induced emf will only occur when there is a change in magnetic field. And from what I can tell from the question there would be no/marginal change in magnetic field (for either circumstance) if the pole of the magnet is facing the disc.

If I'm right then those questions (apart q 2) only require a leaving cert standard of physics. I dunno chump, from what I remember Weaire going over them in prob-solving I remember them being way more difficult.

Of course I could be just misinterpretating to fook cos I'm tired..

chump

Apex hope your exams go better than you expect! Although last year half my class failed as I'm sure you know, sad story.

Anyway about the cons. of mom. ... they aren't really the one system are they? and how are they transferring energy?

Maybe I've a warped sense of the law, but I thought there had to be some interaction? (I don't know if pulling on the string counts for this)

And yes, these questions are the easy enough ones! Although they all look fairly straight forward when you see the solutions, it's tough though when it's open season and they can throw any which law or concept into a question...
I'm fairly screwed overall tbh, not to worry though...



http://www.maths.tcd.ie/~pwalsht/paper5.html

Fair play to him... and good luck to you

chump

That paper 5 has hung me! Nooooooooooooooooooooooooooo

ApeXaviour

Sht, you sure chump? It's not worth a whole lot is it? After your JS results, project and the other four papers?

I'll definitely be repeating by the way. Better off, just didn't put in the work..

chump

It's a tough game. My mate whose who just repeated 3rd year, who actually used to get 1st's isnt even sure if he'll get the 50% to pass. Must have been rough papers.

Goin in2 last paper now, that paper 5 has knocked all the goodness out of my first 3 papers, and this one is gona be a mini disaster aswell...

Not to worry, I'll let ya know how I got on in a month. Good luck with yours