Virtual Office Hours: Questions and Answers
Please send us your questions with "Phys2020" in the subject line of your email! They may be about anything - this week's homework, content, administration, whatever. If appropriate, we will email an answer to you directly. If we think the question might be of general interest, we'll post an anonymized version here (we'll chop your name off the posting).
Please email us your questions to steven.pollock (at) colorado.edu, or kyle.mcelroy (at) colorado.edu Also look at our discussion forum on D2L!
Professor Pollock, Hi there. I was in in your physics 2020 class last semester. You may remember my name because I sent you many emails with physics related questions over the semester. I wanted to start off by saying thank you for all of your assistance over the semester -- I very much appreciated your detailed responses. I also wanted to say just how much I enjoyed your course. I am a master's student in the school of education and found it very wonderful as how you incorporated so many different instructional methods in your lectures. In fact, sometimes we would discuss a certain pedagogical style that I would then observe you using during lecture. And, just like the good ole times, I do have a physics related question for you. Today I was reading an article about scientists constructing a 'tractor beam' with light that attracts microscopic objects. To do this, the scientists used bessel laser beams. Here is my question. The article described how a normal laser will spread out across distances (i.e. you aim it at paper a few feet away and it's a small point, you aim it at a wall across the room it it becomes a blurry point. You aim it at something a mile away and the point produced is very large.) However, a bessel beam, according to the reading, doesn't do this -- the central point stays focused regardless of distance. I was wondering why this happens for a bessel beam and not a normal laser pointer? Is it because a normal laser pointer has a small amount of diffraction from a 'one-slit interference' where as the bessel beam somehow does not?
I was also reading through some of the old virtual office hours and noticed you have a video lecture series on particle physics. However, the price tag to buy it is a little steep for me right now, so I was wondering if there were other ways of accessing it that are a little cheaper? Definitely no worries if not, as I'm sure I'll get it at some point in the future when I can.
I'm sorry for the lengthy email, but, more than anything, I just wanted to thank you for a great semester. I hope you have had a great start to the year!
Good to hear from you. I'm afraid I don't know about "bessel laser beams", so can't help you too much - I just looked at the Wikipedia page and it seems to confirm what you were describing - it's a beam which maintains its spatial profile over longer distances. Kind of interesting, I've never heard of that before.
The "tractor beam" idea in general is more commonly referred to as "optical tweezers", turns out there's even a PhET simulation page about that here It functions, in a sense, by something we had a clicker question on once. If you have a *nonuniform* electric field and put a dipole in it, the dipole will orient and then get "pulled in" towards the region of stronger field (because the non-uniformity means the E field, and thus the force, is not the same on the two opposite charges) We saw grass-seeds getting oriented and then migrating towards the stronger field regions in a lecture demo - a "macro" version of this phenomenon.
As for those video lectures - I bet you will enjoy them, but yeah, they're kind of pricey. I believe the Boulder Public Library might have a copy, you could check that out. For awhile there was a bootleg version on YouTube but I think the Teaching Company spotted it and took it down, but if you keep your eyes open it wouldn't surprise me if that will crop up again. Seems to be the norm on the web, and frankly I'm not that concerned about it. I think I might also still have a spare copy of DVD's in my office (I'm not there right now), so if you can't find it at the library and want to check it out, swing by some time and I can lend them to you.
Thanks for the nice feedback, hope your new term is starting off well! Cheers, Steve P.
From a student: here is a link for a fun XKCD comic on Rayleigh scattering. https://xkcd.com/1145/
enjoy
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Good morning Dr. Pollock,
I decided to not wait 10 years before asking you a question.
I'm curious about EMPs. Specifically, how do they damage devices that are off? Is being unplugged and off better? Are simple circuits more 'rugged' and better able to survive? Why? And of course, please feel free to elaborate on anything else related.
I don't know too much about EMP's. I've heard of them - the idea is that in some situations (e.g. a nuclear explosion) where there is a sudden large and rapid acceleration of charged ions, you generate a very short-lived electromagnetic wave. That means oscillating E fields traveling away from the source (at the speed of light) which are very large in magnitude. So any material with conductors will (for a very *brief* amount of time) have large oscillating currents set up in it, and large E field also suggest large voltage differences (so sparks, or overheating).
Being plugged into the grid is probably worst, because the grid lines serve as big "antennas" for the pulse, and once current flows, it can flow along the lines. So you would get brief but strong surges of current that flow into anything connected to the grid. Can you protect from it? I would think so - building circuits so they quickly dissipate short-lived current pulses is something that people know about (e.g, I used to plug my computer into an electrical "conditioning" box which would protect it in the event of an external lightning strike, which is another mechanism to dump a lot of current for a brief time into the lines) But, such systems are also expensive. I'm not quite sure what would happen to small-sized electronics (like cell phones or laptops) I think the worry is that some components "fry" when too much current runs through them, but I don't know if the small size of modern electronics would protect them (small size means small peak voltage for a given E field, right?)
Cheers,
Steve P.
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My last clicker question:
Flight vs Invisibility
(With John Hodgman on "This American Life")
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Book recommendations:
I actually really enjoyed physics quite a bit more than I thought I would and I was was wondering if there are any books about the history of physics or just physics in general you recommend? I'm not looking for necessarily text book dense style reading, but I'm more interested in this field than I thought I would and I would like to know more about it...I'm not sure if there are more "light readings" on physics, but if you know of any I'd be delighted to get the titles.
"Empires of Light" (the Tesla vs Edison story), was strongly recommended to me (but I haven't read it yet, so caveat emptor)
I enjoyed David Bodanis' "E=mc^2", a personal tale of the equation, focused on stories, characters, and history. (Not technical stuff)
I enjoyed James Gleick's "Genius: The life and science of Richard Feynman" a biography of an amazing physicist. There's another bio of Feynman that's less scholarly but more personal, Ralph Leighton's "Tuva or Bust" that I enjoyed too.
Brian Greene wrote "The Elegant Universe" about string theory. It's meant to be a "pop" book, but I still get a little dizzy myself about halfway through. In a similar vein, longer ago, Stephen Hawking wrote "A brief history of time", but it's also sort of a coffee-table book (that people buy but rarely read through) - it "feels easy" but is hard to really make sense of.
A better one in this genre is Lawrence Krauss' "Fear of Physics". Especially the opening chapters are cool. (It gets harder about halfway through, but I think the beginning is worth checking out)
George Gamow (of "Gamow tower" fame!) wrote a couple of fun books (a LONG time ago!) , one called "1-2-3-infinity" (which I haven't read!) and the other "Mr. Tompkins in Paperback" (which I have). The latter is frankly more fun as an idea than in actual implementation- it's whimsical stories of a fellow who goes to popular physics talks, falls asleep, and has dreams in which the modern physics is exaggerated (like, the speed of light becomes 30 miles/hr, or the quantum constant becomes huge, so quantum effects become "ordinary world" things. It's very clever… just not sure how helpful it would be if you didn't already know some of the physics)
And last but not least, and even more local :-) you might enjoy watching my "Particle Physics for Non-Physicists" lecture series from the Teaching Company (they are 24 thirty-minute videos). It's sold like hotcakes, so apparently somebody likes them!
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From some of you:
Playing with our diffraction gratings, we mean studying.
(Shalaya and Courtney)
In that image you can see something neat about the way computer screens are backlit. Do you see how the light seems to come in discrete bands rather than a continuous rainbow? This is because light like this use a ultra violet or blue color to excite fluorescent molecules. These in turn emit longer wavelengths of light like cyan, yellow and red. Lights like this (which include flourescents and many white LEDs) waste much less energy than standard Edisonian filament ones. -KM
And another image:
(Lauren)
Here we see clearly the m=0, and m=+/- 1 patterns very well separated. m=0 looks "normal", because all colors do the same thing (same angle, theta=0, for all wavelengths). But for m=+/-1, the red light "spreads out" more than the blue, so red is at the bottom in the bottom image, and the top of the top image. The bulb itself remains white in those images, I presume because you have multiple colors from the physically spread-out object (the bulb) overlapping. (The more pointlike the source, the more distinct the rainbow pattern) -SJP
And another:
(Lauren)
Notice the dim spectra to the sides of the reflection of the candle up at the top of the jar? Nice! -SJP
(Jordan)
Looks to me like the gas used in the "OPEN" letters is different than what's in the two swooshes, because the "OPEN" seems to be mostly yellow/orange, the swoosh more greens and blue!
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Holiday Present ideas:
If you're looking for a geeky cool science-y present for someone (or yourself :-), check out my favorite Physics "toy" store: Educational Innovations.
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Exams and stress:
Ariel Paul send this link to a TED talk about "power posing" as a simple (2 minute!) strategy for dealing with stress. I'm not completely sure what to make of it, but found it interesting and inspiring!
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CAPA #14 and Phet Lens sims:
I am having trouble with the phet to understand diverging lenses. The simI am using now is the phet demonstrated in class but do not know how to covert it to the diverging lens. I do not know where the + sign is. Is it on a different phet? Please let me know.
Thanks,
http://phet.colorado.edu/sims/geometric-optics/geometric-optics_en.html
The Phet sim does not have a button to go to diverging lenses, unfortunately!
The one I showed in class (with the +/- sign you can toggle) is a different sim - it's linked at the bottom of the course web page, but also here:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=48
It should pop up a little window with the sim after about 10 seconds, and the "+/-" switch is top center. Let me know if you cannot get it working!
You can pop up a second window if you want them side by side, scroll down that web page to item 8 and click start again.
Cheers,
Steve P.
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Studying for Midterm #3 question:
Hi Professor Pollock,
I was reviewing the histograms and answers on the webpage and for the same clicker question, you list two different answers. The question is the final clicker question for lecture 26 and the first clicker for lecture 27. I believe the answer is B, but I'm not positive. If you could please let me know the correct answer, I'd greatly appreciate it!
The answers are not inconsistent!
In lecture 26, the answer was A) Theta_1.
In lecture 27, I changed the figure!! So, the answer was physically the same, but it became B: Theta_2.
So the answers look different, but in BOTH cases, the currect answer was "the angle with respect to the perpendicular to the surface", or "the angle with respect to the normal to the surface" or, in these figures, "the upper of the two angles".
Seem ok? Thanks for checking!
Steve
Midterm #3 question:
Hi Professor Pollock,
I am wondering if we need to know the EM spectrum in your lecture notes and which wavelengths and frequencies belong in each group.
Please let me know.
I don't expect you to memorize facts and figures, but the most basic *idea* (e.g. that red and blue light are different frequencies, as are radio waves, etc) is nice to have. But I don't need you to memorize the names of the various portions of the spectrum, or what wavelengths correspond to microwaves, or things like that. You should know how wavelength, frequency, period, and speed all inter-related, though, and be able to connect them to a graph of E field vs time, or vs space (like we did in a bunch of clicker questions)
Steve
What about how radios work? Should we be familiar with AM and FM frequency modulation?
We didn't cover that in lectures, so no, we'll leave it off the test. (But it's kind of cool, you might still want to read about it :-)
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Thanksgiving break curiosity:
Professor Pollock,
I had a question about how light reflects off of mirrors. I was wondering what
happens at the atomic level? I have an understanding of how photons are absorbed by electrons if the energy of the photon matches the next energy level of the electron, etc. etc. But was wondering what exactly happens with a mirror? Do the photons give up their energy to the material, but then a photon of identical energy is emitted? Please feel free to be as detailed as possible -- I would really love to have a deep understanding of this concept.
I hope you had a great Thanksgiving.
Yes, it's pretty much as you hypothesized - a photon of identical energy is emitted!A mirror is just a conducting sheet. (Think of a layer of silver). A conductor, recall, is a material with free electrons, lots of them. To the extent that the conductor is "ideal", those electrons are completely free to move when an E-field is present. That's why we said, much earlier in the term, that E=0 inside conductors (and why you are safe inside a metal car or plane even if it's struck by lightning). It's because the electrons in the conductor are free to move/rearrange, and will continue to do so in response to an imposed E field, until E=0 (at which point there is no reason for them to move, and you have reached equilibrium)
That was what we said for "nearly static" E fields. But what happens when a traveling EM wave (light) comes rushing towards a conductor (at the speed of light?) Well, in a sense, nearly the same thing. The wave is just an alternating E field, which is reaching the conductor's surface, but the electrons in the conductor can and will respond, wiggling at exactly that same frequency. If the frequency is not TOO high, then they respond ideally, and as usual, they cancel the E field out INSIDE the metal. Of course, in doing so, they are oscillating theselves, and this *generates* another EM wave which travels outward. That new wave is cancelling the wave INSIDE the conductor, but not outside. Outside, in fact, you have generated a wave essentially identical to the incoming wave, but in a different direction.
The physics terminology here is that it is "Boundary conditions" at the surface of the conductor (that you must have E=0 there, even when an oscillating EM wave approaches) which result in the reflection. Mathematically, you solve Maxwell's equations with a traveling wave (in vacuum) on one side, and ideal conductor on the other, and what you find is precisely the law of reflection pops out (it's an exercise in junior-level E&M) So too does Snell's law pop out mathematically, if you solve the equations in the presence of a sheet of "dielectric" rather than a conductor. The math involved here is basically Calc III, so if you've had that, I can point you to a text that explains it in nice detail in just a few pages. (But if you haven't seen div, grad, and curl, it won't make much sense)
By the way, if the conductor is real (not ideal), then the electrons cannot respond perfectly without dissipating their energy through Ohm's law, so there is some absorbtion of energy (lost as heat), so real mirrors are never 100% reflective. Also, in a real conductor, there is a certain maximum frequency at which the electrons can "respond" freely. Above that, they can't keep up, and you DO get penetration of the EM wave into the conductor (hence, X-rays or Gamma rays can pass through metals - at least, partially)
At lower frequencies, you don't have to be a very good conductor to keep out EM waves. So, e.g., submarines submerged in salty sea water (a conductor) cannot pick up FM or even AM radio waves, those don't penetrate through the conducting salt water, they reflect off the surface like it was a mirror! (So, submarines require other ways to communicate, including VERY low frequency, and thus long wavelength, radio waves)
And thanks, I had a great T-giving. Hope you did too!
-S
Transformer question:
Professor Pollock,
I had a quick question over the material we covered in class today, specifically
transformers. My question is this: In a transformer, say a step-up transformer, why does the current decrease when the voltage increases? I have a practical
understanding of this (conservation of energy) which makes complete sense, but I am lacking a more intuitive understanding. From a purely intuitive sense, you
would think that current would increase with voltage (for example, Ohm's Law
states that for a given resistance, current increases with voltage). I understand that ohm's law doesn't apply here, and that the conservation of energy says otherwise, but it does seem rather interesting current would decrease with larger voltages (especially considering the water pipe example where voltage is the pressure pushing a current). As always, thank you so much for your help
It's a great question, and I'm not sure I have a totally satisfying answer. The first comment is that Ohm's law is a really limited law, it's approximately true for conductors, but there are LOTS of physical situations where Ohm's law just doesn't work at all - like here, where unfortunately our "Ohm's law" intution is just not applicable for transformers. Transformers are wildly NON-Ohmic devices, they are just really different. The reason is the underlying physics - Ohm's law arises because of an effective kind of "friction" internal to conductors, (and because it's a kind of friction, it dissipates energy in the form of heat) but transformers work by Faraday's law which is quite different in origin - it is fundamental (not approximate), and it is definitely not dissipative. Faraday's law is almost the opposite of dissipative, it's conservative, it is closely tied in to energy conservation. So here you just have to let go of Ohm's law in this case, and grab hold of conservation of energy!
We have seen other failures of Ohm's law - for instance, you can have ANY current flow out of a battery, yet V is always the same V of the battery. So a battery is a device where the relation between current and voltage is radically non-Ohmic. (Constant V, and I you demand up to a max) For a battery, the operating "rule of thumb" is "Constant V", and I will figure itself out based on the load (resistance) you hook up, following Ohm's law for THAT. For a transformer, the operating "rule of thumb" is "Power in = Power out", followed by Faraday's conclusion "Vout/Vin = Nout/Nin". Those two come from laws of physics, and then the conclusion (that current out goes down if voltage out goes up) follows…
But I agree with you that it is interesting (and curious) that we get LESS current with higher voltages. I have no idea what the "water-pipe" analogy of a transformer is, but it would be some crazy mechanism to get water flowing "over there" based on what is flowing "over here" WITHOUT a direct water connection! (Right? The current out of a transformer is physically NOT connected to the current coming in, in any direct way. Electrons over here wiggle, creating fields, that make electrons in a totally separate "tube" start to wiggle). I don't know if there is any mechanical analogy, because Faraday's law, the new and critical element of the story, is not something that I believe has an "water manifestation".
This help at all?
Steve
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Written HW 5 question:
I am writing regarding a tricky subtle point in the written homework this week. In part C I need to calculate the magnitude of the Force of the B field of the earth on the sliding rail. However, Because the field is coming in at an angel of 66 degrees I can't assume that the B field and the Current are perpendicular. So. my question is what angle is required to perform this calculation? My gut says that it is the same 24 degrees that I used in part b because by taking the sin of this angle, I achieve a component of the B field that is normal to the current carrying slider. Am I on the right track?
I agree that this is a tricky point! It requires very careful visualization in 3D (and is thus quite difficult to do by email) I've been finding that many guts today have not been correct :-)
If you find me at the start of class Friday, I'll be able to show you more clearly. What I have been doing in help room is suggesting that people take pencils or pens and use them to represent arrows. E.g, in this case, lay one pencil down on the table to represent the current in the sliding rail, (I think that should be pointing East, and lying flat on the table. Do you agree?) and hold another pointing "down and north" to represent the B field. You need to "cross these" for the force, because F = L I x B. So the question is, what is the angle between those two arrows? Sometimes staring carefully at the physical objects provides the *aha* moment you are looking for. (One suggestion that seemed to help people today was to consider DIFFERENT angles other than 66 degrees, and as you sweep the magnetic field arrow (the pen, for example) from "totally flat, 0 degrees to the ground, but pointing purely north" all the way to the other extreme, "totally straight down into the earth, 90 degrees to the ground, no more North component left behind", and as you sweep through all those angles, ask yourself "what is the angle between I and B here?" You may be surprised at the answer! Look at the physical objects, that's really the key for me… When you go to the extremes I described, it is often visually more obvious what the angle is, and that's usually where the aha moment comes.
Hope this helps - if not, ask again!
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Exam 2 question:
There weren't many questions on Kirchhoff's laws in the practice test. Where can I find some?
Good point! Of course, the CAPA problems we did involving 2 batteries are good ones - if you found yourself "guessing" (or plug-n-chugging), go back and make sure you understand the concepts behind those.Or, take a look at Giancoli's Kirchoff homework questions at the end of Ch. 19. Or go to his website (have you spotted the link, it's the 2nd to last one on all our course pages on the bottom left!) and go to Ch 19, choose either "Practice Problems" or "MCAT problems", he's got some good ones. And, that system will tell you the answers too...
Exam 2 question:
Quick question, in your lecture notes you touch on RC currents.
Would you recommend us knowing about those for the exam? They
look awfully complicated..
RC circuits, nope, we are skipping that!
... I now have a question about what we need to know on chapter 20 for the exam. In your summary it only mentions the first right hand rule will be on the test, but I feel like we have used the second right hand rule in several lectures as well. Will the second right hand rule (and the third one we learned recently) actually not be on the test, or was it just something that was left out and we should still know it?
What I wrote in that web page is correct - I agree we've practiced RHR #2 many times in class, but since the Friday lab section did not have a chance to do the lab which involved magnets and wires, I just decided not to put that (magnetic forces) on this exam. It will be on Exam #3 for sure! But for this test, no, I won't ask for "right hand rule #2" (forces.) The only magnetism that you should expect to appear would be something like what you had to do for the *first* part of the written homework due last week, i.e. understanding the pattern of Magnetic field around a magnet, or around a long wire (RHR #1), and the idea of superposition. I think everyone in class has had ample opportunity to practice that…
Let me know if this is not clear -
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Interesting application of what we are studying this week - how animals can "see" magnetic fields! http://en.wikipedia.org/wiki/Magnetoception
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CAPA SET 8 #6.
So, for this CAPA I am either severely misled or the answer CAPA has is wrong and won't accept mine. So, for number 6 it asks about the power delivered from battery 1 in the circuit. I got the earlier question about the current in that loop correct, so it should just be I*V(around that loop)=P. I do that and get 133.333 W which CAPA says is incorrect.
Then I used the I^2*R1 equation since R1 is a given value and I had
solved for current around that loop in the previous question (and CAPA said I was right.) Again the answer came out to 133.333 W. I used less sig figs and answers came out to 133.467 and 133.4, both of which were still wrong.
So, in closing here is what the loop consists of, a 20 V battery, and a 3 Ohm resistor. This caused the current to be 6.666 A (which CAPA said is correct).
If I am doing something wrong please let me know,
but I have a feeling this may actually be CAPA's fault, at least I hope.
Alas, I'm quite sure CAPA isn't wrong on this one :-)
When you say "current in that loop", I'm a little confused, because there is one current going up the left side (through the resistor R1), but a totally DIFFERENT current I running through the battery E1! (It's different because there is a junction at the top, where current from the right hand loop can join or exit, depending on your numbers) Does that make sense?
I1^2 R1 is definitely not what you want, because that's power dissipated in the resistor, but some of that power might be coming from or shared with the other battery! So, you really do want I*V, but it needs to be I through the battery in question.
The remaining issue, then (which many students had today in the help room) is to look very carefully at the DIRECTION of the current through R1, and (more interesting, and depending on your battery numbers, tougher) is also figure out the DIRECTION of the current through R2. That latter really matters when you then use Kirchhoff's 1st law (current in = current out) to try to deduce what current is flowing into/through the battery E1. (At that point, armed with the current through E1, it's nice and simple, power = I*V....)
This help? Good luck!
Steve
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CAPA Set 8, Q9:
Dear Professor Pollock,
Hello. I am a student in your PHYS 2020 class this semester. I was working ahead on the CAPA and came up with a correct answer to a conceptual RHR#2 question but only by making an assumption that I was unsure about. In problem 9, I assumed that the B field was pointing out of the page due to the common notation of dots. However the problem does not explicitly state this and without knowing the direction of the b-field I can't say that I am able to make a conjecture about the TRUE (before flipping it if it is negative) direction of the force on any moving charge.
Thanks so much for your time!
Sincerely,
Good question! Indeed, the problem doesn't explicitly state the direction of the B-field (I think because there are 5 figures which it selects randomly from) Some figures will have dots, and some have x's. Your working assumption is exactly correct, that's the convention we (and the text use): dots means "out of the page", x means "into the page".
This help clarify for you? Let me know if anything is still confusing!
Cheers,
Steve P.
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Humorous (and topical) cartoon, sent from a student in class.
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Written HW #3 tip:
The hardest/oddest question on Written HW #3 is the 2nd one. (What happens when S3 is closed? )The reason it is strange is that we do NOT usually wire our homes this way, with bulbs in series! As you work this problem, you will figure out why not.
Here's the crux: when I give you a bulb labeled "100 W", you must NOT assume that it always dissipates 100 W of power. (When I hand it to you in the box, it is dark, certainly not 100W yet!) At the start of the homework problem, in big bold font, we point out that this bulb will ONLY have a power of 100W if you hook it up (properly!) to a 120 V power supply. Look at bulb R4 in your homework - when you close the switch, is there 120 V across it? (I would say NO, try to understand why not! It has to do with the fact that it is in series with R3)
When I worked this homework, the very first thing I did was figure out what the resistance of all 4 objects was. (How? Well, I know their power rating, and the power rating tells you power IF you hook it straight up to 120 V. That's enough to figure out R.) Once you know R for a given object, that's R, it doesn't change just because you happen to wire it up in a screwy way! Once you know R4 (say), you know it. Note that you do NOT know that that object will *always* dissipate 100 W (it will not unless it has 120 V across it!) and, for question #2, you should certainly not assume (because it is not true!) that the bulbs will glow with their rated powers! Something else happens - what is that?
We'll do the demo in class on Friday. Try to work it out before then, and the demo will make a lot more sense to you!
And another Written HW question:
Hi Professor Pollock,
For question number 4 on the homework asks if the analysis would change is the short-circuit was on the left side rather than the right. By this do you mean that the short circuit is still in parallel, but just moved schematically? Or do you mean that ... (details removed...) Thanks for your time and consideration.
Yes, exactly, I did mean what if the short circuit is still in parallel, but simply moved over to the left side, rather than the far right side, in the diagram.
-S
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Fun stuff sent by a student:
and
David Blaine goes high voltage
(My comment: It's just a stunt! Now that you know about the physics of conductors, you realize there's no likely danger here, just whatever amusement factor comes from seeing lots of impressive sparks near a person. Well, that and standing without eating for 72 hours!?)
And, on that same note, a video (from another student):
Prof Pollock -
I thought you might appreciate this article from the NYT that I just found...though it bugs me that people think the electricity is actually going through his body. They should ask our physics class to check their facts!
Sincerely,
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Exam review question:
Hi Professor Pollock,
I have been looking over the review clickers and had a couple questions about the last two. For the second one, when given V it asks if you can solve for E.
I said no because E=kQ/r2 and V at a point= kQ/r. We need r to solve for Q.
As for the first one, I am not sure. I think you need to have delta V.
Am I on the right track? Please let me know.
E = kQ/r^2 is only true if you have a *single* point charge. So, although I agree with your answer (knowing E doesn't tell you V) I think it's deeper than the use of that particular formula.
In general, E field and voltage are related but not "proportional", the relationship is subtler. But as you are pointing out, knowing one at ONE POINT is not enough to know the other.
So….
- If you only know V at one point, I claim you can't really know ANYTHING I can think of about E. (If I tell you that voltage is 1 V at the period at the end of this sentence, I don't see how you could make any statement whatsoever about E there. It could be anything, any value, any direction! It depends on what else is around!).
- And similarly, if you only know E at one point, you don't really know ANYTHING I can think of about the value of V there. (If I tell you that E is 1 N/C to the East at that period, I believe V could be anything! Again, it depends on what ELSE is going on elsewhere)
Other related thought on this:
- If you know E at one point, and you also know V at that same point, I would argue that you *can* figure out V nearby (but not farther away), by using Delta V = -E dot Delta d. (which is only true for small Delta d)
- If you know E *everywhere*, then you can figure out V everywhere. (Well, except for the arbitrary choice of where to call V=0, but besides that!)
- Conversely (and more surprising, since V is a scalar), if you know V everywhere, you can figure out E everywhere! (Think of going from an equipotential diagram and sketching E field lines, like we did in the watery lab)
This helping?
-S
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THEREMIN:
Professor Pollock,
After the theremin demo in class today, I thought you might like to see the
following youtube clip (if you haven't already).
Very beautiful music indeed.
Nice clip, I really enjoyed how she explained it. (Makes me want to practice more!)
-S
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Check out this phet sim about neurons (although, it's focusing on somewhat different physics than our HW #2 question. I had not noticed it until just now. In the sim, they are not focused on energetics like we were, nor on the "pumps", but rather on the action potential, which is the brief shift in voltage as the neuron fires). The sim suggests to me that the final result of our homework problem is (of course) oversimplified, and that the energy required to "fire" is not nearly at the scale suggested by our homework numbers. (Real biology can get deliciously complicated! )
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Written HW #2 question:
Hi Dr. Pollock,
I hope your afternoon is going well.
I've been working on the written homework, and after
encountering #4 I have a few questions. First off for the
formula for Capacitance is the A (Area) just for one plate or
is it the sum of both areas? Also, since the tubes, in the
written homework, nest within each other do we consider them
both approximately the same size since the space between them
is tiny tiny?
Also: would you only consider the area of the sides of the cylinder
(2*pi*r*h)...or do we also consider the ends of the cylinder
as well as the sides (2pirh+2pir^2)??
-S
These are good questions, since I did not derive the formula!
A is the area of one plate.
And yes, it's a good point about the tubes, but you're exactly right - because the gap is so tiny, the difference in area (due to the SLIGHTLY different circumferences) will be totally negligible.
For the last question - we are pretending that the plate is VERY large (that's the approximation that goes into the formula in the book and in class today). So, in that limit, it's just the "sides" that matter, the ends are "far away" and are assumed to not matter much.
Hope this helps!
Steve
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CAPA #4, Q#4
Hi Professor Pollock,
I'm having a hard time with numbers 4 and 5 on the current HW set. Here is the Problem:
An alpha particle is a naturally occuring and somewhat common particle. ... How much kinetic energy would an alpha particle acquire in the same apparatus starting from rest at B and moving to point A?
First I thought the KE of the alpha particle would be four times that of the electron but that was wrong. Now I am trying KE=1/2mv^2 which gives me velocity of the electron but still no ideas on how to find KE of the alpha particle. Can you maybe point me in the right direction? I've been in the helproom twice and there hasn't been a TA either time. Thanks!
Let's focus on question #4. It asks for the KE of an alpha particle, compared to the electron. (I suspect once you get Q4, you can get Q5, which just requires thinking about/applying KE = 1/2 m v^2 )
You said you tried "four times", but I'm wondering why. (It's true we told you the mass of the alpha is 4 times the mass of a *proton*, but I don't see your chain of logic as to how that would give the alpha particle four times the KE of an *electron*) .
We're looking for the basic principle here - what is the relationship between "moving between two points under the influence of electricity" and energy gained? This has been the focus of Ch 17: points in space have "voltage" associated with them, and there is a direct connection between change in voltage, and change in energy. It's a formula (and idea) that was emphasized for the last couple of classes. It's (unfortunately) not the FIRST formula in your text in Ch 17 (which is a particular special case example), but it may be the 2nd and/or 3rd formula! In particular, voltage difference is not exactly equal to change in energy. There's one other element in the equation. What is that element, what else matters? (It's something given in the problem statement too.) Of course, those formulas are all about "work done" and/or "change in potential energy", which is still one step removed from change in *kinetic* energy. That goes back to a big Phys 2010 idea, conservation of energy. (If PE goes down, then KE goes up - by the same amount!)
Does any of this help you get started? Might be easier in person - do you have time after class tomorrow to chat?
-Steve
And, a followup to the above:
Thanks for directing me to the FAQs, virtual office hours and I think I am approaching this the right way, but the eV is throwing me off. So I used the equation
delta PE = q delta V
However, the KE is given so I was thinking conservation of energy KEi + PEi = KEf + PE f
so I set my kinetic energy (should I use joules or eV here?)in the equation and solved for V since its constant in a capacitor. Then using my V value and the charge on the alpha particle I plugged it in and solved for the change in energy. I left the answer in eV and I'm not sure if i should convert to joules, and I'm not sure of the sign. I hope this makes sense.
Thanks,
I agree the big idea here is delta PE = q delta V.
And,I also agree that conservation of energy is the other big idea here. It helps if you rewrite your conservation of energy formula like this:
PEf - PEi = - (KEf - KEi)
(That's the other way of thinking about conservation of energy : LOSS of potential energy corresponds to GAIN of kinetic energy)
So that's the physics of this CAPA question. A particle moves from A to B, losing PE, and gaining KE.
A particle of the opposite sign moves from B to A (same magnitude of change in voltage, right? Because voltage merely depends on the points where you start and end, not on what your particular charge is. Though, I suppose that means it has the opposite SIGN change in voltage, since it goes B to A)
I think in the end perhaps the problem is a little easier than you are making it out to be. You might not even need a calculator for it!
Let me pose a related but slightly different question that might help you. Suppose an electron released from rest at A travels to B and gains a certain amount of kinetic energy. (I could tell you the number, give it to you in Joules, or eV's, or just call it "KE") Then, suppose a PROTON (opposite charge) is released from B. Since opposite charges go "the other way", it will travel from B to A. But, how would its change in KE compare to that of the electron? And once you've answered that, what do your two formulas at the top of your email tell me if, instead of a proton of charge e, it's an alpha of charge 2e? (What changes, what stays the same?)
As usual, doing physics by email is tough! Probably easiest if we just chat - you free to stay a couple minutes after class today?
-Steve
P.S. Capa does not care if you enter an answer in Joules, milliJoules, eV, kiloEv's, or whatever unit you want - as long as the unit is *energy*, and it's the right number.
So if the answer to a question is 1 eV, CAPA will say yes if you enter
1 eV
1000 meV
1E-3 keV
1.6E-19 J
1.6E-16 mJ
1.6E-21 kJ
etc.
They're all equally good!
And, as for the sign… well, try to visualize it. CAPA is asking for the amount of KE the alpha "acquires". Is KE ever negative? (What's the formula for KE again?)
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Steve,
there was one item that caught my interest, in something that you said regarding voltage...if my recollection of Friday's class is correct, I believe you said, that it was just a number, that voltage is a property of space...it what way, was that meant? I think, I might be reading more into that statement, then in what was said. Sometimes, one can misunderstand a statement regarding it's connotation versus it's denotation. That is how words are used or statements made in everyday language, might be quite different in what is meant in the language of physics!
I agree that that statement is an odd one, and hard to interpret. It's an abstract idea, and hard to articulate "carefully".
When physicists think about fields (like E fields, or voltages) they are associating something (a vector, or a scalar) with every point in space. What is E, or V? It's not a *thing* in the traditional sense, it's not a particle at that point in space, but I do think of it as a real thing, just a rather abstract thing. And, every point in space has "values" associated with it. (A value for voltage, a vector value for E-field, and more, a vector gravitational field, etc…) This voltage is generated by charges elsewhere, but given those charges, empty space has this "scalar field", the voltage, filling it. So in this sense, voltage is a property of space (but created by source charges elsewhere).
As far as Phys 2020 is concerned, here's what I'm getting at. It doesn't make sense to talk about "the electric force at a point in space" if there is no charge there, (force on what?) but it DOES make sense to talk about the E field at that point. It's a unique and perfectly well-defined vector, even in empty space. Similarly, I cannot talk about the "electrical potential energy at some point in space" if there is no charged object there, (potential energy of what?) but it DOES make sense to talk about the voltage at that point. It's a number, a single well-defined, unique number at each point. If you know it, then you can figure out what the potential energy of a charged test object WOULD be if it were placed there.
Not sure if this is helping! Perhaps easier to chat in person, maybe after class some time?
Cheers,
Steve P.
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(Regarding the Week 4 web site picture: Here is a somewhat infamous comic site that has recently raised $1M to support a new museam for Tesla. Warning: it's more than a little crude, some viewers may be offended, but it's a heartfelt defense of a brilliant and eccentric scientist. )
And, from another student:
You mentioned today in PHYS 2020 lecture, Tesla and alternating current. It made me think of this youtube video I had seen just a few weeks ago on the history of Tesla and his relationship with Edison. The video is somewhat crude but I don't think I will ever forget the history because of it.
http://www.youtube.com/watch?v=3gOR91oentQ
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I'm in your PHYS 2020 lecture MWF. I came across this picture
today, and thought you might get a kick out of it. We're not
quite to the Tesla yet, but still...
https://sphotos-b.xx.fbcdn.net/hphotos-snc7/s720x720/392521_523006474395477_1360708425_n.jpg
Enjoy your weekend!
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CAPA SET #3, question 6
Hello there, I seem to be having a bit of trouble with problem 6 on the CAPA.
[Many details here outlining the method used, which sounds great....]
I've probably made a mistake somewhere, but let me know. Thanks!
Sounds like you have the physics right, but you need to look again at your calculator work. CAPA requires answers to be correct to 1% accuracy, which means you should include at least 2, and probably 3, "significant digits". The number 3.0E-9 is DIFFERENT from 2.9E-9 or 3.1E-9, by more than 1%, so you have to be careful to see if your calculator is rounding for you. (It's common when numbers come out this small for your calculator to say, e.g. .000000003, and you might THINK that's exact, but maybe it's rounding. Is there any simple way you can think of to check to see if there is another digit or two "hiding" after that 3 that might change your answer enough for CAPA to care?
Good luck, let me know if you still have troubles.
Steve P.
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Professor Pollock,
... I am in your physics 2020 class. I have a quick question over the material covered in class today. (Sep 5)
One of the clicker questions involved placing a positively charged rod next to a metal sphere that was grounded and then quickly ungrounded. We determined the rod would cause the sphere to polarize (negative charges would go towards the rod, and positive charges would go away from the rod) and then, when the sphere was grounded, the positive charges of the sphere would head towards the Earth resulting in the sphere having a negative net charge when the grounding wire was removed. However, immediately after this example, we discussed how electrons are what flow -- that positive charges (protons) do not as they are locked in the nucleus of the atoms. With this being true, then how did positive charges flow away from the sphere when it was grounded? Shouldn't the sphere have stayed neutral?
Thank you for your help.
Great question, you are quite correct in everything you wrote - including that positive charges are mostly locked in place and cannot flow. So what happens in reality during the grounding process is that negative electrons FROM the ground travel up through the grounding wire, attracted to that positive rod. They flow onto the sphere, leaving it with a net negative charge.
The practical end result (that the sphere becomes negative) is the same whether it was "really" positives running AWAY to ground (which is not what happens in practice), or negatives running in FROM ground (which is what happens). Either way, the positive rod has induced a negative charge on the sphere during this process.
This make sense? Feel free to ask further if it's still bugging you! Giancoli section 16-4 also covers this, although he shows the picture in Fig 16-8 for the opposite case (where a negative rod is brought close, and so negative charges LEAVE through the ground wire, leaving the obect charged by induction.)
Cheers,
Steve P.
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Here's an email from a student with a (literally) very cool video.
Hey Professor Pollock,
I don't know if you've heard of TED Talks, but some of the stuff we are going over
in class reminded me of an interesting talk that I saw a while back. I thought you
might find the video interesting if not just aesthetically amusing.
http://www.youtube.com/watch?v=PXHczjOg06w
Enjoy,
For fun: check out some sites below. (If you find a link that you think is relevant for Phys 2020, let us know, we can add it here)
Many fun (and potentially useful) electromagnetism links: http://www.colorado.edu/physics/phys1120/phys1120_fa07/links/links.html
The CU "physics simulators" page: http://phet.colorado.edu
A crazy demo: http://drmegavolt.com/index.html
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