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 assignment, the course content, course administration, or whatever. If appropriate, we will email an answer to you directly. Or, in many cases (where we think the question might be of general interest), we will instead post an anonymized version of the question+answer on this page instead (we'll chop your name off the posting).
Please email us your questions to steven.pollock (at) colorado.edu, or Oliver.DeWolfe(at)colorado.edu.
USEFUL SIMS for homework this week:
Phet has a nice converging lens sim
and here is another (which has diverging lenses too) This is quite helpful for the homework, maybe too much so! Just be careful not to become too dependent on it - you won't have the sim for the final exam! So, use it to *check* your homework, but not to do the problems for you!
Here is the sim to superpose two waves (go to "example 3") that I showed in class
And here is a nice sim (other than the Phet sim we used in lab) to show "two slit diffraction" (and a similar one for single slit diffraction)
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If you want to "see" your blind spot, check out the wikipedia page here.
Electrifying Images
Check out MIT's web page for their E+M course Their impressive collection of flash and java simulations are here. I showed this one and this one in lecture, Monday Nov 1.
Here is the "generator" sim that I also showed in lecture Mon Nov 1.
Here are some good "Snell's Law" sims, #1 (very simple "Snell's law" showing wave fronts instead of rays), #2 ("Snell's law" showing rays) and #3. (my favorite of these three, combining both! Be sure to click the "next step" button on the right to see increasingly interesting representations of what is happening!)
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Final exam review question:
Steve,
I am having a hard time understanding why PE=Wext. Is it because we need to put energy in (external work) in order to increase potential energy? So I moved same sign charges away from each other that would be negative work or PE, but if I moved them closer it would be positive work, so positive PE? Thanks!
It's *change* in PE which is given by Wext, and yes, just as you say - if you want to increase something's potential energy, you (the "external agent") have to put in some positive energy (work).
If you move same sign charges away from each other, they are naturally repelling, so yes, YOU would put in negative work (you'd be "holding them back"), and that means their PE is getting smaller. That's correct, PE = k Q1 Q2/r, and if r gets bigger, the PE of that pair of + charges goes down. (If you release them to run off to infinity, they can do LESS work if they start off farther apart from each other).
But if you moved them closer together, you would certainly be doing positive work on them, which means the *change* in their PE would be positive, the PE of the system would be getting bigger. (Again, PE = k Q1 Q2/r here, so smaller r means PE is going up).
This making sense?
Steve
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Solutions:
I remember hearing you say in class that there was a link to CAPA solutions somewhere. Where would I be able to find these? Thanks!
CAPA, written HW, and exam solutions are all on our CULearn page, just go to culearn.colorado.edu and log in to our 2020 course.
By the way, CAPA also provides answers (but not solutions) to your older CAPA sets - when you log in to CAPA with a recent set, you can always tell it to go back to an older set, it's on the "splash" page when you first log in.
Hope this helps!
Steve
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CAPA #14 question #3:
Steve, I am having trouble with capa question number 3. We are given the magnificantion of the image (hi) and the focal length, but without either distance from the image (di) or distance of the object (do) how can we use the lens or magnification equations? Thanks!
You have two useful equations here, the lens equation (1/d0 + 1/di = 1/f) and the magnification formula (m = -do/di)
You know the focal length f, and the magnification m (you'll need to think carefully about signs here, CAPA only gave you the magnitude of the magnification)
There are 2 things you don't know: d0 and di. But in general, 2 equations with 2 unknowns is always enough, you should be able to solve for everything. It's a little tricky algebra (because of those "inverses" in the lens equation), but maybe a little thinking will suggest a way to proceed... (Can you solve for ONE of the unknowns in one of the two equations, preferably the simpler of the 2 equations, then plug that back in the remaining equation, which leaves you hopefully with ONE equation in ONE unknown...)
Let me leave you with this idea as a starting point, and if you can't make progress, let me know early next week and we can talk some more, maybe in office hours?
Hope this helps you get started, though -
Steve
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CAPA question:
hi, i need some help with light (please). when light at a specific wavelength, let's just call it 3 meters, passes through a prism/glass, what happens? does light continue to travel at the speed of light or does it actually slow down? maybe it doesn't slow down but actually just loses energy in the form of heat. i have never seen a prism get hot but i have felt iron get hot when in the sun.
thanks for all your help,
The light does slow down. In principle, there is no need for the glass to get warm, there is no loss of energy. (When the light leaves again, it returns to the speed of light. ) It simply travels more slowly in the medium, at speed v = c/n.
The frequency is the frequency, that doesn't change in or out of the material. After all, "1 cycle / sec" is just *counting* wiggles, and if that's how you're "driving" the field, that's how it's wiggling, whether or not there is matter (electrons, protons) around you or not.
The wavelength is whatever it has to be to satisfy f * lambda = speed (in the medium). It adjusts, it's whatever it needs to be.
This helping?
Steve
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Just Curious:
Hello,
I was just wondering whether or not it's possible to accelerate, either positively or negatively, an EM wave. I know that c is supposed to be the universal speed limit, but is it possible to alter the speed of a EM wave? Also, I read that the universe approximately 13.5 billion years old, but I read in your notes that the farthest star we can see if 15 billion light years away. Is there a physical explanation for this or should I just not believe what I read on the internet about the age of the universe? I was just curious about these two questions, so if you have answers to them I would appreciate it, but if not I understand.
My "15 billion" number was just a sloppy rounding error, you are correct that the Universe is not quite as old as I implied. I should fix that in my notes!
As for your other question - there is no way to accelerate (positively) any EM wave, the speed of light is an absolute upper bound. This is one of those experimental facts that is as well grounded in experiment (and theory) as the "best of the best" in physics, like conservation of energy, conservation of charge, and conservation of momentum.
However, you can make the speed of EM waves *slower* than c, by running the wave through matter, of any kind. We'll be talking about this in class on Friday, in fact. It arises, microscopically, from the "lag" that occurs if an EM wave gets absorbed and then re-radiated, it effectively slows the propagation speed. There is no theoretical limit on how much *slower* than c you can get. There are some experiments over in the JILA tower (the "other" physics tower) which have succeeded in slowing light (of a very particular frequency) down by a factor of more than a million - the light is practically traveling at "running speed"! So, slowing EM waves below c is no problem, but there's no physical mechanism to make them go faster...
Happy to chat about this kind of thing, thanks for asking!
Steve
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Exam Review:
Steve
For question 5 on the practice exam I am confused of what equation includes mass and how to tell whether or not the particles are negative or positive. Any suggestions? Thanks!
Look again at the last question on Capa #7 (for the signs) or CAPA set #9 (several problems, including #1, #4, and #5, all of which involve the relevant formula/idea to think about the mass). Or, look back at Giancoli Section 20-4, which talks about this physics - charged particle in a uniform magnetic field.
Let me know if this doesn't help enough for you to figure it out for yourself!
Steve
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And more exam review - old CAPA questions:
Steve,
I was able to figure it out, thank you! I now have questions about CAPA. On CAPA set 8 I was looking at your solutions to #4 and I really do not understand how you found the third part of the force equation that is 2*E(kin)/r. I understand qvB=mv^2/r just not how that equals 2*E(kin)/r.
Yeah, that solution is a little cryptic, and looking at it I think it's even wrong!
Here's how I would go about it:
The "circular motion" formula (which I think you said you figured out) comes from Newton's law, F=ma:
F= qvB = mv^2/R because F is the magnetic force (=qvB), and acceleration when you are going in a circle is always v^2/R (that's an old 2010 formula, which we haven't rederived - but it's back in Ch. 5)
The other part of the story comes from conservation of energy. The ion starts at rest, and is accelerated by a voltage difference Delta V (given by CAPA.)
But we know change in PE (potential energy) is just q Delta V, that's straight from our very original, basic definition of voltage! (V = PE/q)
If we drop in voltage, than the change in PE (= q Delta V) is going to have to equal the change in KE (= 1/2 mv^2) You should think about this, convince yourself - it's a good review of older stuff, especially the basic definition of what voltage means!
So that means, q Delta V = 1/2 mv^2. (Since we started from rest, "change in KE" is the same as "final KE") Let me call "Delta V" just "V" for simplicity, and summarize:
Cyclotron motion says (1) qB = mv/R
Energy says (2) qV = 1/2 mv^2.
(Don't mix up big V = voltage with little v = speed).
We know ALL the symbols in these equations except for two: m and v.
2 equations in 2 unknowns! So we're pretty much done except for some algebra. There are several tricks, but one trick is to DIVIDE the two equations.
Left side of eq (2) / left side of eq(1) = right side of eq (2) / right side of eq (1)
(Does that make sense to you? )
So doing that gives me
V/B = 1/2 v * R (lots of stuff cancelled, nice!)
I can easily solve that for v (speed), because CAPA gave us voltage, B, and R.
Now that I know v, I can plug it in to either (1) or (2) above to find v.
That was a slight pain in the algebra :-( I probably wouldn't ask you to do this level of algebra on a test, but I *would* like you to understand and be able to do for yourself some of the pieces (like, equation (1) , or conservation of energy! )
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Second I am confused on #7 on how you found the Force of gravity as
pi*d^2*h*density*g/4. I received help on this question in the help room but just have that equation written down - where does it come from?
The force between wires is given by Giancoli 20-6,
F= mu_0 I1 I2 l /(2 pi L) where "l" is the length of the wires, and "L" is the distance between them. We don't know the length l of the wires, but it really doesn't matter. (You can pick it to be one meter, or just leave it as a symbol. It will cancel out in the end).
But, Newton's II law says if the lower wire is "levitating", then it's not moving, the net force must be zero!
So F(up, from the magnetic attraction) = F(down, by gravity, also known as weight)
In other words,
mu_0 I1 I2 l / (2 pi L) = weight
The weight you have to hold up is m*g of the lower wire, and m = density * volume (right?)
And Volume = area*length, so putting it together:
weight = (density*area*length)* g (phew!)
So we have (mu_0) I1 I2 l / (2 pi L) = density * area * l * g.
(Notice there's an "l" on both sides, that's the length of wire we are considering. It cancels, it doesn't matter, as promised)
We're asked for I2, solving the above for it gives
I2 = density * area * g * 2 * pi * L / (mu_0 I1)
Area is pi*radius^2 = pi*diameter^2/4, everything has been given. (Capa gave us density, diameter, L, and I1) Just plug in the numbers, being sure everything is MKS.
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CAPA 10, Q 2 and 4:
Hi Professor Pollock!
I am fairly stumped on a couple of CAPA problems, and I thought I would see if you had any words of advice! I am looking at questions 2 and 4, which I believe are solved identically and both request a value of EMF, but use a different value for theta. I always get a bit confused when velocity comes into the problem, but I think an equation I used on last week's written homework could be useful. I can't remember the exact equation I used. I was curious if this was a good approach to assume when
looking at these problems. Thank you so much for your time!
In questions 2 and 4, yes, you're on the right track - check out your textbook, the discussion of section 21.3 (and Figure 21-12) are VERY close to this problem. (In my own handwritten lecture notes, it's on pages 5 and 6) The addition of a simple factor of "cos(theta)" is the only little twist which is not in the Giancoli example or in my notes (but if you look at the string of equations, like in Giancoli eq. 21-3, and you think about the true/basic/original definition of flux, eq 21-1, you should be able to see how/why the cos(theta) enters (it's nothing complicated, since it doesn't vary with time, it's pretty much just what you're thinking!)
Hope this helps - good luck. Oh and by the way - this "magic formula" you're using here that involves velocity - please be aware that although it feels like it might work in many *other* problems, it is quite specific to the details of this particular problem. The master formula is always Faraday's law, that's what you have to go back to!
Steve
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Written HW
Professor Pollock,
I'm just a little confused about the wording for written homework #5. So it says that the magnetic field of the Earth in Boulder is heading into the page, but that its at a slight angle (so its not headed straight down). In the diagram you have a compass where north is pointing to the right that is held parallel to the ground. I'm confused if it would be considered as going into the page or to the right. I understand that its not going directly into the page, but it still seems to be heading into the page and because its at an angle the part of the magnetic field that would be affecting current is smaller than if it was straight down. If its the case that I would draw X's on the diagram to show its still going into the page, how would I incorporate the angle? Thanks.
I think you understand the issue!
I showed you the direction of the Earth's B-field in class using a "dip" compass - here in Boulder, it is a vector with a Northward component and a downward component. So it points both north and down. If you look "from the top" (like our figure), you cannot see or show any "up/down" components, the B-field looks like it points northward. This is how it is on maps, too, they are 2-d, and the earth's field apparently points north. But in reality, the arrow does have a down component too.
And so, as you said, because it has these two components, the *part* of the earth's B field that affects our story IS indeed smaller than if it was purely down.
In a figure, I don't know a clever trick how to draw a "top view" of an arrow that has mixed components like this. Not sure if you really need to do that here - you might find a better representation. But if you want to do that, I agree it's not pure x, but perhaps you could draw an "X" and a "North" arrow, right next to each other, and label them clearly as the components?
Like I said, sounds to me like you understand the physics issue here. When people quote the value of the earth's B-field, they are telling you the magnitude of a vector, but for a particular problem (like this one with the loop "flat on the ground"), you only need a portion of that vector, the appropriate/relevant component. It's part of the HW to figure out how to get that - it clearly involves sin's or cos's of given angles :-)
Hope this helps -
Steve
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CAPA 9, Problem 8 question
Steve, I thought that the equation mentioned in the lecture notes (for the magnetic force between two wires) would be appropriate for this question but my answer is incorrect. Is there a better way to approach this problem? Thank you!
Well, the formula is correct for the force between any pair of wires. (Do you see where it comes from? It's basically just F=ILB, and then for B you use the familiar formula for the B field from another (long) wire. You will have to think about the derivation to decide what exactly the various symbols mean exactly...)
But in problem 8, you do have to be a little careful, because there are really FOUR wires in the loop, each feeling a force. Now, the two "vertical" legs won't matter (do you see why? There is a magnetic force on them, so why don't they matter here?...) But the two horizontal legs ("top" and "bottom" legs of the loop) certainly each feel a force (which you apparently know how to compute!) and you also need to consider how the forces on these two remaining legs combine (Which way does each one point? do they add? cancel? partially cancel?)
Hope this helps you!
Steve
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CAPA 9, Q7:
For CAPA 9 question 7, I felt I would use F=qVB=mv^2/r but we are given the
distance it travels, so there are no variables to find. I just don't understand which
equation we could use to find the distance it would be deflected. Thanks again
Yeah, that formula is only useful when a particle is going in a circular orbit. But in this CAPA problem... well, think about it, bullets in real life do not start spinning around in circles due to the earth's magnetic field! The numbers in the problem are randomized, but reasonable - those are pretty typical values for q, v, B for a bullet. The bullet goes MOSTLY straight, with a small sideways deviation due to magnetism.
So the idea is that the bullet is going pretty much in a perfectly straight line (neglect gravity for this problem - we only asked about the deviation due to magnetism!) There is a TINY sideways magnetic force, which you can compute (you know about magnetic forces now) From here, it's pretty much a beginning Phys 2010 problem: you have an object (with known, essentially constant speed in the "x" direction) which feels a tiny sideways force perpendicular to its motion. Just like "projectiles" in 2010, you can safely consider "x" motion and "y" (sideways) motion independently. In both directions, the motion is simple (but different) In x, nothing very interesting is going on at all, it's just cruising along with fixed speed, for the given distance? (Hmmm, how long does that take?) In "y", it feels a force (magnetic!) which is pretty small (and calculable!) and constant... So, can you remember/rederive your old 2010 "constant acceleration" formulas and figure out what is going on in the sideways direction during that period of time? That's what we're after.
Steve
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Studying - practice test
Steve,
This question regarding dielectric I cannot decide if the charge would increase or decrease. I know that because the capacitor is connected to a battery so Q is free to change but how do I approach whether it increases or decreases. Thank you for all your help.
"Connected to a battery" tells you "voltage is fixed", right?
Since V = E*d, I guess that means E must ALSO be fixed.
But Q=CV, so when you put in a dielectric, what happens to C?
(Since V is *fixed*, that then immediately tells you what happens to Q)
(And since stored energy = (1/2) Q*V, then you can also immediately figure out what happens to stored energy.)
This help?
Steve
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Studying - Capacitors
I am looking back at CAPA #5 question number 2 and 3. I thought from question number 2 as distance between plates increased, that U would increases proportionally as well. But in 3 I am asked, with the capacitor connected to the battery, increasing d increases U. I thought this would be true but its false. Why does the connection to the battery change this? Thanks Steve!
It's all about what is held constant.
Q=CV for capacitors, and U = (1/2) QV, which leads to 2 other equivalent expressions:
U = (1/2) CV^2 and U=(1/2) Q^2/C.
All three of those are always true. But which one is *useful* to you will depend on what you know is held constant, or fixed.
For an isolated capacitor, it is Q, the charge on the plate, that stays constant. (No matter what you do, if the plate is isolated, conservation of charge says... where can that Q go?) So, since U = (1/2) Q^2/C, the isolated capacitors energy is INVERSELY related to the capacitance.
But if you hook it to a battery, Q can flow on or off the plates as you make changes. It's not fixed! What is fixed now is Voltage. And the equivalent formula U = (1/2) C V^2 says the battery-connected capacitor's energy is now directly proportional to the capacitance. Very different result.
It's a little mysterious at first, but you can think about the work being done on the system and it makes some sense. (E.g, if you pull the plates apart, YOU are doing work on the system, so if it is isolated, the energy must be going up. But, if it's connected to a battery, charges can leave the system, which means the system is doing work on the battery (charging it up), and thus LOSING some of its energy.
This helping?
-S
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Circuit cartoons sent by students:
(we'll get to transformers in a couple of weeks)
and
http://xkcd.net/730/
I have to confess, I needed explanations of the jokes for the xkcd cartoon:
http://www.explainxkcd.com/2010/04/21/circuit-diagram/#comments
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Phys 2020 in the UMC:
Just thought I would share a quick story with you. Some kid in the UMC had his computer charger plugged into the wall. When he unplugged it, he said "Why is the light still on? I unplugged it already." I was able to use my knowledge of physics thus far and explain to him why! I thought it was great that I was able to use physics in a real life situation. Proof that physics is used every day!
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Exam question:
Professor Pollock,
I was wondering how much of AC currents were going to be on the exam. I am a little confused Vrms and Irms stuff. Thanks!
RMS won't be on the exam. See the "exams and grades" link on the upper left - many more details there!
Cheers,
Steve P
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Exam review:
I'm studying for the exam and wanted to work on practice problems. I've reviewed all the iclicker questions, CAPA problems, written homework, and labs. I was wondering if there are certain problems from the text book I should solve or practice problems I could work on from another source. Thank you for your time,
Lab this coming week is going to be all exam review, including a written sample exam with practice problems. Prof. DeWolfe should be posting that on the usual "lab download" page, I trust this weekend (i.e. tomorrow). You are of course welcome to work on it outside of lab if you like, whatever is most helpful to you.
Other sources - at the end of each chapter, after Giancoli's "summary" and *before* the many pages of "problems", is a page (or two) of "Questions". Those are good, I think they would be excellent for exam review. Another possibility is from the link at the bottom left of our course page, to "Giancoli's Physics". If you go there, pick a chapter, and then select "MCAT practice problems", you will find more good review questions. (And you can submit them to check yourself, which is nice) Just be aware that there are plenty of things in Giancoli and those practice problems that we do not cover explicitly in class, so just cheerfully ignore questions about things that seem totally unfamiliar!
Cheers,
Steve
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Tripping out on Physics class:
hey Dr. Pollock,
so i'm getting tripped out by electricity and magnetism. i realize that E-field and B-fields are two different thing. However, they must be linked. For starters, think of an electromagnet or the last clicker question on friday. Electric current can generate a magnetic field and thus B-field lines. So if you have a flow of charge (or current) then you must also have E-field lines, right? if this be true, then E-field lines can create B-field lines......? right?
happy friday. thanks for your time as always,
Yes, E&B are different, but they are intimately related to each other too. Just as you said, since *moving* charges make B fields, and B fields affect moving charges, it's immediately clear E&B have deep connections to one another. And yes, in a sense (we'll clarify how in class) E fields can create B fields. We'll get to this - this was one of Maxwell's great insights, and after him, in a deeper way, good old Albert E. What "appears" to be an E field in one reference frame will be interpreted as a B field in a different frame, and vice versa. So E&B are different, but like two sides of a coin. Hence the phrase "electromagnetism".
So, you're in good company in your musings :-)
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Solar link:
One of the students in our class has a website about solar panels, check it out:
Rubysolar.com . (Ruby is my cat and she loves to sleep in the sun).
Just finished my 9th solar project and somewhat surprisingly the first where the upfront rebates were more than the actual cost. Most folks however I just tell them the standard 5-7 years to break even but there are really too many variables in terms of size of system, components, ease of install, and how much actual work one does themselves to make this anything other than ballpark.
Obviously, I don't do this for a living, but my goal is only to demystify solar.
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CAPA #6, Q10:
I am confused whether the bulbs are rated at 6.3 watts given the usual 120 V supply that we have in the US or whether we are supposed to use the 110 V given as supply to try to back out the resistance from the givens. I tried it both ways, but still haven't gotten the answer and I'm on my last try. I had no trouble with the others (esp. #11) which makes me wonder what i'm doing wrong.
Thanks,
I agree it's a little ambiguous. But what we intended is that there are a bunch of identical bulbs in series, and you are given the total voltage (110 V in your case, this will vary for other students), and each of them is glowing, with an actual (measured) power which is also given. (In your case, each one is glowing with 6.3 W of power, again, the number is given). So the "6.3" Watts is not some hypothetical, it has nothing at all to do with 120 V: this number is what the power is (of EACH bulb), when you hook them up as described. That help?
Cheers,
Steve
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CAPA #5, Q9:
Dear Professor, I cannot figure out where to start on Number 9 on this week's CAPA. Can you help me please?
This one is a little tricky to try to describe/help by email, but the core physics idea is that resistance of real wires is proportional to their length.
(Remember, R = rho * L/A, so if you double length, you double the resistance).
In this CAPA problem, you have some wires, and some lengths, and some resistances. It's pretty much just a "proportionality" problem. For instance, if I had a 10 m long wire with resistance of 20 Ohm, then a 20 m long wire (of the same type) would have to have resistance 40 Ohm, and a 1 m long wire would have resistance 2 Ohm, etc.
This help you get started?
By the way - the problem is "real", in that utility workers do use this method to estimate where shorts occur in buried underground cables. Save a lot of digging if you know where the problem is ahead of time!
Cheers,
Steve
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Exam 1 blues
I did terribly on the first midterm and would really like to talk to you about the exam and what i can do to improve my scores in the future.
In order - see our "advice" page, look over the solutions on CULearn, visit us in office hours...
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Exam Crib Sheet?
Will we be getting a 'cheat' sheet for every exam, or only the first one? Just in terms of study habits for all the exams and what I should get used to doing.
Thanks!
Yup, one page (one side) cheat sheet is ok on each exam. You will be allowed to ADD one more side each exam. (So exam 2 you can have 2 sides. By the time we get to the final, that'll be 4 sides!) For most people, on exam 2 that means bringing their old one again, plus a new one for the new material. But whatever works for you.
Steve
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Practice Exam?
Hello, I have been trying to find some old practice tests to review...I know last semester our Physics professor had one old practice exam so we could at least see what he was expecting and could have a way to practice. Is there any way I could get old practice exams? Thanks
OK, check out my "exams.html" link again - near the bottom of the "Review suggestions" portion, I have posted (a chunk of) an old midterm of mine. Caveat emptor - old exams don't ever exactly reflect what my current one will look like (and this one is from 10 years ago!) but I think it'll give you some sense of how I write tests.
Cheers, Steve
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Question while studying:
Professor: Can you please explain the difference in these two items? Can I say Potential = U/q ?
Potential energy is the good old fashioned idea from Phys 2010, it is a form of energy, it is measured in Joules, the symbol for it is U.
"Potential" is the new term we have introduced this term, and it is an exact synonym for " the voltage".
It is measured in Volts, with 1 Volt = 1 Joule /Coulomb.
So *yes,* you can indeed say Potential = U/q,
which we write in class sometimes as V = U/q, (or more often V = PE/q). These are all equivalent expressions. (Voltage = potential, same units, same quantity, just synonyms!)
By the way, we usually write a "Delta" in front of these expressions, that's a little safer, since the "zero point" of potential energy is not defined, only differences in potential energy are meaningful, and ditto for only differences in voltage, i.e. "potential". When you have just a handful of "point particles" creating the voltage, then we typically like to define V=0 at infinity, and PE = 0 at infinity, and then you can quite safely write what you did, Potential = U/q,
or if you prefer writing it out: Potential = (potential energy)/charge
Steve
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Concept test question from Friday's class:
Dr. Pollock, As we sit here reviewing physics notes on a Friday night, we still find ourselves perplexed at one of the clicker questions from today. The questions which asked for the voltage given Q and r (and q) for one scenario and Q and 2r (and 2q) for the other.
Are you referring to the FIRST clicker question from yesterday? (See
http://www.colorado.edu/physics/phys2020/phys2020_fa10/notes/Reports/SessionSummary_11.html
That question was asking about the difference between the Voltage at point P in situation 1 (distance r from charge Q) and the voltage at point X in situation 2 (distance 2r from same charge Q) Although I perhaps "distracted" you by putting different test charges at those points, the intent was for you to realize that the voltage at a point is determined by the *source* charges (i.e. the charges everywhere ELSE in space, not at that point) and is definitely not determined by the test charge you choose to put there. Really, I suppose I should ask about voltage at *empty points* in space, but you are certainly allowed to put test charges wherever you want, they are called "test" charges because they're supposed to be so small they don't affect anything else in the world. (Not the other charges, not the fields, nothing. They're tiny, they're just "testers")
I think the answer is that because voltage = KQ/r,
q does not matter and therefore the first voltage of the first one will be larger because the second one will be divided by 2q.
I *think* I'm following, and agree. voltage = kQ/r, yes, and q (the value of the test charge) does not matter when you are computing voltage. Voltage is a number at each point in space - it is a property of the space located a distance r from the SOURCE charge Q. You can put ANY test charge "q" or "2q" or "10q" you want at that point, the voltage is just the voltage, it's just this ONE number, kQ/r. In this problem, situation one was at "r" and so had kQ/r voltage, situation two we were at distance "2r" and thus had half the voltage there.
If you focus your attention JUST on situation 1 from here on out, and imagine putting different test charges there (q, or 2q, or 100 q) what we're saying is, never mind the test charge, the voltage at that point is kQ/r, it's determined by the source. It doesn't matter what test charge you put there, voltage is a property of the point in space.
Now what *would* be different with those different test charges (all at the same spot) would be their potential energy. (not potential, but potential energy!) Their potential energy is given by qV, or in other words kQq/r. In this case (asking about PE now) the value of the test charge certainly matters: big test charges have lots of potential energy at a given point in space, compared tosmaller test charges. (But again, they do not sit at a "location of higher voltage". If we're talking about just this one point in space, it has one value of voltage there, and any test charge of any magnitude placed there will "feel" the same voltage. Again, more test charge will mean more PE right there, but not more voltage. The voltage is determined by all the OTHER charges in the universe. )
I hope this is helping, I know it's confusing!
What we do not understand is how q plays no part in this equation. How does the charge of q not matter? I understand a test charge is usually insignificant but if the problem were 300q instead of 2q, wouldn't that have to make some difference? We are just having some problems wrapping our heads around this. Thanks so much,
If the "test charge" gets REALLY big, it's perhaps not a "test" charge anymore. (If it is big enough to, say, polarize the source or something!) But if you aren't worried about that, the value of the test charge certainly influence the total energy (in this case, the potential energy) just as you'd expect: 10 times the test charge would have 10 times as much total PE. But when you divide PE by the value of the test charge (to get back to voltage, V = PE/q), you always get back to the same number, the voltage! That's what a voltmeter tells you. It cannot know what test charge you will choose to put at that point at a later date, and thus it cannot know how much work you will do bringing in your test charge (that what PE tells you!) It tells you V. If you want to know PE, you use PE = qV, and you get a different number for PE depending on the test charge q. But you get the SAME number for V, no matter what test charge you used!
Making sense? I think I'm starting to repeat myself, so if it didn't help the first time, it probably isn't helping more now:-( Let me know if you're still confused!
Steve
Clicker question answers: Can you please put those concept tests and answers up when you get a chance? Sometime today or tomorrow PLEASE!
The histogram and answer to all concept tests that we do in class get posted (typically within about an hour of class finishing!) The link to them is at the very top of the "concept tests" page (which is the 3rd link under "Lecture" on the left) Did you notice the bold comment at the top of that page?
I don't have explanations there, but the correct answer is indicated in the clicker summary table (at the top right of each question, it tells what the correct answer is. If it's a question mark, it's because it's not really a question with a "correct" answer.)
I haven't been generating additional "solution slides" for all the clicker questions, I figured the q&answer plus class discussion would be enough. Let me know if you have specific questions, though. Also, there will be occasional extra clicker questions in the files that I post for each class period that we didn't get to in class - those I don't have any written solutions for, so you'll need to use office hours if you want to check on those!
-Steve
CAPA #4, Question #8
[I think] the Work should be negative, because you're moving a negative charge away from a negative and towards a positive..
In the figure, the negative test charge is moved from "i" to "b". (It may be randomized for different people). Look carefully at the figure and ask yourself what voltage you START at and what voltage you END at, and think about the difference, are you going up or down in voltage? (Going from + to - is going DOWN, going from - to + is going UP, right?) Then, given that, what does that mean for the sign of the work done on an negative test charge? These minus signs are tricky, but you'll get the hang of it, you can always go back to your intuitions about "which way do + charges want to go", and "which way do - charges want to go"?
-Steve
CAPA #4, Question #4
Professor Pollock: A lightning flash transfers 3.94 C of charge and 4.43 MJ of energy to the Earth. Between what potential difference did it travel?
I thought we needed to know the distance in order to compute this. What is another way to approach this besides using PE= -q*E*d
What is our basic, starting definition of Delta V? (Go back to where we *first* defined voltage, in class, or text, or notes) I think you'll find that once you think about that basic definition, it's a very quick question!
-Steve
CAPA #3, Question #7
Hi, I keep getting the wrong answer for question 7 on the capa.
An electron starts from rest 68.2 cm from a fixed point charge with Q = -0.213 uC. How fast will the electron be moving when it is very far away?
I first solved for PE using PE=KQ/r*q, I then set that equal to KE(final) Where am I going wrong? Is this not the way to approach the equation or am I continually making the same calculating error?
Looks like the right physics idea to me, conservation of total energy, before and after! Although I'm a little unsure about your notation (when you write k*Q/r*q, do you mean for that second little q to be in the denominator, or the numerator?) Did you check units, make sure everything is in standard MKS metric (being careful of e.g. converting cm to m, and microCoulombs to Coulombs)? Did you use the proper mass for the electron? If you still can't get it after checking all that, you'll need to give me more details, maybe we can sit down in office hours.
Going in to CAPA and looking at the old set, it should now tell you what it thinks the answer is, so you can still check yourself, even if it's not for credit anymore)
Cheers,
-S
CAPA #3, Question #8:
I realize that it is asking for the potential and not the potential difference which means I can't use the equation V=-Ed. But I don't know where to go from here, although I assume I use the potential at point one to help me but I can't figure out how.
You're given the potential at point 1 and I claim you can figure out the potential difference between 1 and 2, so at that point do you think you can figure out the potential at point 2? That's the essence here,...
OK, but how do you find potential difference? Your equation Delta V=-E.d is good (that's from class, it's the correct and appropriate result when E is uniform. Do you remember where the formula came from? It's worth your while going back over how we got that!) The key thing is that it's a vector equation, that is really
Delta V =
- E dotted with d.
So, you need to remember how to "dot" two vectors, we talked about it a lot in class. (There's a nonzero angle between E and d here, you ned to take that into account)
This help you get going? Let us know if you're still stuck!
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Written HW #1:
...I am confused on part 7.... I was wondering if I am missing
something, because I can't seem to figure out why diameter and density of the ball are given. If you could point my thinking in the right direction, I'd really appreciate it. Thanks for your time and help!
Written homework problems are meant to be a little "real world" in a variety of ways - including the topic, but also in the style of the problems. I would say they tend to be a bit more vague than CAPA, giving some more opportunity and need for YOU to decide what and when to approximate, what to ignore, what to use. In most problems (beyond intro physics), we are generally faced with an overload of factual information and data (like the diameter and density here) and we have to decide: do I care about this here? why or why not? ( You may very well decide that you do not need the information in one part, but are happy to have it in a later part :-)
Hope this helps -
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Link from a student in our class:
Hello professor,
I thought you might like this youtube clip. It is a band that was on America's got talent that uses Tesla Coils and lightning in their performance.
http://www.youtube.com/watch?v=pBx2wkg9nhk
Sweet!
Another student mentioned a video in class and sent me the link - Chris Angel,
http://www.youtube.com/watch?v=OV3etThWyXI
Here were my comments about this second one: the video is trying to make it sound like some kind of epic, dangerous stunt, but as you've seen in class, if your (wimpy?) physics prof is willing to do it, it shouldn't really be that scary or dangerous :-)
He's clearly well protected, and it's *nothing* remotely like being hit by lighting as they imply throughout the video . First because real lightning strikes do not involve you being protected in a metallic suit. Second and much more important, the amount of charge transferred by real lightning bolts in storms is many many orders of magnitude greater than the charge transferred by a Tesla coil. The bolts *look* impressive from that device (but if you could see a real lightning bolt instead that close up, the difference would be unbelievable). Third, there's yet another difference - lighning strikes are "DC" (direct current), while Tesla coils are AC (alternating current) at very high frequency. Turns out that the AC case also has physiological differences, the *outside* of your body tends to conduct that better than DC, so you are significantly less likely to risk injury to the heart using the Tesla coil.
All in all, they're pretty much a bunch of flamers in this video :-( I only hope that nobody watching the video ever walks away thinking that you could try this "out in the field" with real lightning. Still, gotta admit, the final seconds of the video with him getting sparked are cool looking! Thanks for sending this.
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Class discussion on Wed Sep 8:
We talked about lightning safety (and the physics value of being entirely enclosed in metal) Check out this website for some dramatic pictures:
http://www.crh.noaa.gov/pub/?n=/ltg/plane_japan.php
And here, for more about lightning in Colorado.
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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