Last minute exam questions.
hey professor! i figured I'd send an email because I'm not feeling too well and walking up
folsom hill when it's cold might not do me too great!
I had several questions about the practice test online and just some general
questions on some things I ahven;t understood in class.
1. #7 on the practice test..I'm having a problem with the theory--I know how
to spit what I know back out with the right answer, but I dont understand why.
It makes sense to me that the electric field of point "downhill" and so that v
woudl be greater higher up, furthest from the arrows. It also makes sense that
a uniform magnetic field will exert the same force everywhere...but how do you
reconcile the two? How can electric field be uniform and going "downhill"?
These seem like a contradiction. I know the SLOPE of the electric field is
uniform....is that what that means? If the slope is uniform, then the force
will be....or an i completely overthinking?
So you're right that E field lines point "downhill", from hi voltage to lower voltage. And also that a uniform field (Electric, in this case) is, well, *uniform*, i.e. it's the same everywhere. So how do you visualize it? I think of *voltage* as like "height", so I'm imagining a perfectly flat ramp (think of a flat, wide road) which is tilted downhill. Height measures/corresponds to voltage, and SLOPE measures/corresponds to E field. In this case, the slope of that road (which measures Electric field) is the same wherever you are, left, right, up the hill, down the hill, it's totally uniform. If you were a tiny little ball-bearing on that road, released from rest, you would start to roll downhill, exactly the same acceleration (force) no matter WHERE you were released. If I put you down "higher up the hill" you would not accelerate any faster, because the slope of the hill would still be same....
Is this helping? Feel free to ask if anything is still confusing!
2. for number 11, I remember working with you on choosing a circuit and
writing out what the different equations for each one. I read your email to
xxx and I understand that once you figure out that resistor 1 is
just affected by V1, that you see nothing changes, but I don't get how you
come to that conclusion. You could just have well have written the equaton for
the right hand side and gotten a different answer, it seems. Is it because you
can get R1 by itself with V1 that shows you they are directly related and
nothign else affects it??? I'm confused!
Yes, it's a tricky little circuit, isn't it? Well, I think we need to get specific. First, remember that I have to pick some direction to label my "current arrows", so let me draw a down arrow next to I1, a left arrow next to I2, and an up arrow (going through the battery) for I3. Those seem like good guesses, don't they? (Though it's totally arbitrarily, it really won't matter in the end if I guess wrong!)
Here are the two possible "loop" equations that Kirchoff's second law would have us write down:
1) (From the "small, left hand loop") +V1 - I1R1 = 0 (Do you see that? Start at the bottom, walk around that left hand loop clockwise, adding up the CHANGE in voltage as you go around) This is my favorite equation because it tells me the answer to the exam question right away and I'm done!
But let's just investigate the other loop to make sure it's all telling a consistent story!
2) (From the "small, right hand loop") Let me start at the top and walk around the loop clockwise.
+V2 - I2R2 + I1 R1 = 0. (Do you follow the signs on all three terms?)
So you *might* think, at first, that when you suddenly change R2 this equation changes. And it surely does! But what is free to adjust, what is the "solution"? I'm going to assume V1 and V2 are fixed, those are the batteries, all the problem is asking us is what happens if we muck around with resistor R2 (changing it to infinity, here) So if you CHANGE R2, what can change in that equation? We're stuck with V2, that's just the battery voltage. You're saying "maybe I1" changes?... and maybe it does, or maybe it doesn't, all we learn from this equation is that as R2 changes, *either* I1 changes, or I2 changes, or maybe both do... This one equation has TWO variables in it (I1 and I2) so by itself it simply does not tell us the answer! All it tells us is that
V2 = I2R2-I1R1, if you have one equation with two variables, there's just not enough info.
So if you had been "unlucky" and started with this equation, you might say "I can't answer the question" (but you could NOT conclude that changing R2 changes I1, it's simply not known yet!) We already know the answer, but let's look what happens if we COMBINE the equations. From 1), we get I1R1 = V1. (That's still all we need for this problem, but let's keep looking!)
From 2), we get V2 +I1R1 = I2R2. But check it out: from 1), we can replace I1R1 with V1. So this new equation also reads
V2 + V1 = I2 R2. Hmmm, now I see. If we change R2, then I2 will adjust itself, but I1 will not, it cancelled out of the equation! If R2 goes up, I2 goes down, and vice versa. (Physically, this is telling me that because I've hooked a battery, V1, DIRECTLY across R1, then nothing else I do can change the current through R1.. because I have a resistor with a definite voltage drop across it!)
If R2 goes to infinity, I2 will have to vanish... that's a slightly pathological case, because 0*infinity can be ANYTHING, which is why this particular equation is true but just not at all helpful in answering/thinking about the question I asked on the exam. (It's still correct, it just doesn't tell me what happens in this funny limit).
By the way, if R2 goes to 0, we would have big trouble. The equation says I2 will have to go to infinity, which will melt stuff! Physically, I can see WHY that's a bad situatio - it's sort of like you are trying to connect two DIFFERENT batteries across the SAME resistor. Putting two batteries in "parallel" with one another is bad news, if they don't have exactly the same voltage (and direction), because you start driving huge currents around! In real life, there are always internal resistances that let you do this, but stuff can still fry!
Anyway, if you buy the above, that changing R2 will only change I2 (but not I1), then is *this* physically reasonable? We've already argued it is - let's also look at our third equation, Kirchhoff's current law, which says I1 + I2 = I3. So this says, if you adjust R2, then we know I2 changes, (but I1 is staying fixed...) so I3 changes too. Is that right? Let's look at the THIRD possible loop for Kirchhoff's voltage law, namely the "big outer loop". Start at the bottom, and go around clockwise, you get
3) +V1 + V2 - I2R2 = 0.
Oh, that's the same as what we just had, V1+V2 = I2R2. Note that I3 never enters any of the voltage equations, it's not running through a resistor anywhere, just through the battery V1. So I3 "adjusts" itself as needed, and has no other impact here! If you change R2, you change the current I2 through it (and correspondingly the current I3 that "leaves" from it), while I1 stays unaffected.
Like I said, a little tricky, although if you trust Kirchoff's laws, the solution is just staring right at you from the very beginning!
3. NUmber 18 did quite a number on me :). I knwo that there's an equation
ABNd/dt....but where does htis come from? And can I take the derivative for A
or B or N (though i dont see how N could ever have a derivative) and plug it
in, or will it always be B that is changing? I just a little confused on this.
This is all Faraday's law, EMF = - d(phi)/dt. That's a biggy, one of the "big four" equations of the term! It says that if you change phi (that's magnetic flux, which is B*A, maybe times a cos(theta) if they're not lined up, it's really B (dot) Area) then you generate an EMF. Actually, you need an "N" because flux is B*A for EACH loop, so if you start piling up loops, you just increase the flux. (Twice as many loops => twice as much flux, does that make sense?)
So Faraday says EMF = -N d(BA)/dt. In this particular problem, A is constant, and B is changing with time.
In one or two CAPA problems, it was the cos(theta) that's changing (that's how our electricity is really generated - you take a coil of area A, in a field B, and rotate it. So you take the derivative of cos(theta)! )
This EMF is a voltage, so if that is connected to a resistor, this is just like "V", then you could use V=IR to figure out the current (or power, or whatever)
So the big idea is EMF = -N d(phi)/dt (there, I've put the N in explicitly to remember it), where phi = B*area. The "area" would "the area over which the B field is acting". If the B field was not uniform, we'd have to add up little B*d(area)'s first! If the B field was constant but only over half the area, you'd only use that half area in the formula!
Making sense?
4. 22 and 23...will u ask us to find the current in an inductor at jsuta
random time?? How would that be done? You would have to give us L and R,
right? Also...What is the relationship with Voltage to time when a switch
first closes to infinity and when a switch first opens to infinity? I think
it's linear...but I dont really know since current is not a linear
relationship.
I think we won't ask for random times, just "right after" or "after a long time" (But, you could figure it out by writing Kirchoff's law, walking around a loop and adding up voltage drops, setting the sum to zero. If the loop goes past an inductor, it would involve the voltage across an inductor, EMF = -L dI/dt, and that would give you a differential equation for I! Indeed, for a simple "inductive discharge", it's a lot like capacitors, current would die away like an exponential, e^(-R/L* time) It's not that bad, but I think we won't ask you to do this on our exam :-) So no, with inductors, it won't be linear with time, it'll either be an exponential decay, or a kind of (1-exponential) buildup. But like I said, probably that's beyond what we need for this exam...
5. 24 also really confused me. Am I to understand that the Power of a signal
does not change when going through the air no matter how long it travels??
No, not really. Well... TOTAL power doesn't change (that's just TOTAL energy/sec, and energy is conserved). But the power passing any given area (like a receiver) certainly does change - right, signals get weaker as you get farther away. We know how they do this: intensity = power/area, and if the power is spreading out uniformly, then intensity = original power/(4 pi r^2). So as you get farther away, intensity drops like the square of the distance
I saw in 20 thats= I1/I2 was equal to r1^2/r2^2.....so the intensity of a signal
with power Po has a smaller intensity going into a bigger disk??? THis doesn't
seem right...coudl you explain why? Maybe it's because there's more space for
the power to "spread" in a bigger disk, so the intensity at one point would be
less? But it seems overall they;d be the same. I guess I dont understand the
concept or how to use that equation
OK, so let's just get it squared away :-)
Intensity of a signal (that's power/area) decreases like 1/r^2. This makes reasonable sense to me - the farther from a light bulb you get, the dimmer it looks. The sun looks brighter than other stars. It's just a "weaker" intensity where we stand, it's not that the bulb (or the other stars) are intrinsically weaker...
If you have the SAME SIZED receiver, then if you stand twice as far, the signal will be 1/2^2 = one quarter as strong. If you stand three times as far, the signal will be 1/3^2 = 1/9 as strong? So far so good?
The formula I = P/area is subtle. On the one hand, you can talk about "total power", or "power of the source", P0. In that case, your area has to be the TOTAL area over which the total power is spreading (a giant sphere, 4 pi r^2 is the area). On the other hand you can look at a receiver, and talk about I = P/area, and now you're talking about P = "power received by that detector" and "area of that detector". It's the same old I, I is a property of the signal... this is the trickiest part! Let's get specific -
If you stand in ONE place and change your receiver size, then what? Intensity is intensity, it doesn't matter if you are big or small, intensity if "power per area", it's just how "bright" the radiation is. It depends on where you stand (I=P0/4 pi r^2, where r is the distance from the source), not how BIG the receiver is. (You need to think about that one to make sure it makes sense!) So what does change if you change the receiver size? If you double the area of your receiver, you will receive twice the energy each second, right? This comes from "power received" = intensity* "area of receiver", or P = I*area. So if you stand at ONE spot, where the radiation has some GIVEN intensity, then the power you receive is directly proportional to your area. Double the receiver size, you double the power you get. (Double the area of the solar panel on your roof, you double the power you get, right?)
So in this exam problem, we went three times farther away (making the intensity drop by 9), but then we used a receiver which had 3 times the radius (which means 9 times the area), so we compensated! The power received is the same. (So if you go out to Mars, you will need bigger solar panels to get the same power as you get here on earth! If you go twice as far, you need twice the area to compensate...
Phew.
6. number 28, I tried a few ways of doing this, but nothing seemed to work. I
tried using P=IV, by finding the peak voltage and peak power, and using that
to find the max instantaneous current. I got 8.333 kW.
THen I multiplied that by SQRT(2) to see what it got me and it was 11.985...is that the 12 what was
got in the problem or did I just happen to get close by accident? How do you
do this?
Average power = (Vpeak)*(Ipeak)/2 = V(rms)*I(rms)
Note the factor of 2 in one, but not the other.
Given power = 100 MW, and V(rms) = 12,000 volts, I get
I(rms) = (ave power)/(V(rms)) = 100E6 W / 12E3 V = 8,333 Amps.
So I(max) = I(rms)*Sqr[2] = 11,785 A, which is (roughly) 12 kA.
The other way to get it is to first argue Vpeak = V(rms)*Sqrt[2], and then solve
I(peak) = 2*(average power)/Vpeak. Should be the same thing!
7. I'd venture to say that there will be optics on the real test, though there
isn't any on the practice test, correct?
I'd say there should be at least one very basic lens/ray diagram, no doubt!
8. also, there wasn't much about elctromagnetic waves and polarization and
their orientation and what not--would it be expected for there to be a lot
more of that on the test tonight or is the practice test representative of how
much of the new stuff will be asked?
No, it's just an old test from a different semester, (and not one that Murray taught, it was from my class!) So, I'd probably expect *something* about polarization for sure too, yeah...
sorry for all the questions, so late...it's been hectic trying to study for
everything and I'm just getting these important questions in...thank you very
much and see u tonight
Good luck!
More last minute questions!
When you have a electric or magnetic field, is the other automatically in every
case, induced? How do you tell what direction the electric field is going from
the magnetic field and vice versa?
If you have a changing magnetic flux, than you induce an E field. That's Faraday's law.
If you have a changing electric flux, then you induce an B field. That's the "Maxwell-modified Ampere's law" (which we haven't worked with so much in this course)
The direction of the induced E field is the one we ask about a lot, that's "Lenz' law". The induced E field (or resulting current, if you happen to have a conducting wire there!) is always directed to "fight the change".
also, im not sure with polarizers how to tell what the angle is to put into the
I=Io(cos(theta))^2....how do i find the right one from whats given in the
problem (like number 31, i dont get where I"m seeing the view from and how
it's oriented
Theta is defined to be the angle between the polarization axis of the filter and the direction of polarization of the light hitting THAT filter.
Once light goes through any filter, it is thereafter polarized in the direction of that last filter (until it hits another one!)
also, with electric potential and potential energy--what should I focus on for that subject?
Both! Electric potential is "voltage", and potential energy (which we've usually called U) are directly related,
U = q V, or Voltage = potential energy per charge. We definitely want you to be able to distinguish these two different things!
One of the TAs made a nice "course review" for his students - I'm posting it for everyone.
Here is is, from Noah Fitch. (Feedback welcome, if you find typos let me know.) (I think the pdf problems we had before have been fixed, if not let me know)
Question about a big recent chunk of the course.
Hey Steve,
I feel like we really rushed through the whole section on inductors, transformers and rms voltage, current...etc. I have all the formulas from that capa and I understand how to plug the capa problems into the formulas, but I don't know what I was supposed to take away from the whole section. I never saw a conclusion. If at some point this week we could have a "you should understand..." or "...should be able to..." "...should have taken away..." for that whole section that would be a big help to me and I'm sure to many other students.
Thanks,
Yeah, I agree that part of the class was just a little rushed. I just typed up some extra notes on this, kind of a summary piece. Let me know if this still doesn't help...
Final exam crib sheet:
Hi Professor Pollock: Is there a cheat sheet allowed on the final? If so, can we use two pages front
and back? Thanks!
Yup, 4 sides total (on 2 sheets, basically you can add one more side to what you already had) Handwritten please, no photocopies!
Grades/Extra Credit:
This email is concerning a few questions I have about our CAPA homework. First, is our lowest CAPA score going to be dropped? So far on CAPA I have done every one and gotten an overall 100%, so I was wondering if I can just opt to not do this weeks CAPA and still recieve a 100% CAPA grade? Also I heard that there is going to be an optional CAPA set to do after this weeks CAPA that would be for extra credit, is this true and how much extra credit will it be worth? If you could clear up these questions for me it would be much appreciated. Thanks a lot for your time.
I will drop the lowest CAPA score for everyone. So in principle (if you're SURE you haven't missed one yet!) you could blow off the last one... but I really wouldn't recommend it, though, since that material will definitely be on the final!
As for the optional CAPA - we had toyed with the idea of doing that, but instead we made CAPA #14 be due later (2nd to last day of the term), so I do NOT think we'll have any extra credit set, sorry!
There is an extra credit survey up, which will be worth a couple of clicker questions (which is not much! But we do appreciate it when people do that one)
And of course, please don't forget the regular weekly pretest (usual place, due the usual time - Tuesday AM) This week's is another kind of survey, very helpful for us as we consider what's working well for you and what could still be improved in this course.
Cheers - Steve P.
CAPA 13 question:
Hi, I am having touble finding a good equation for number 5. I have searched through my notes and the book. I was going to go into the help lab but there were too many people and I didn't get help before I needed to leave for class. If you could help that would be great.
Thank you,
Well, the "equation" would be "Faraday's law" - and, I hope you're searching for an explanation, not just a formula! :-) Although it may not look like it at first glance, the physics of #5 is quite a bit like #4! (After all, both have a solenoid "in the middle", with a changing current, which generates an EMF around loops, real or imaginary, *outside* of it. ) Question 5 is worked through in the text in pretty good detail around page 1097-8, the section on "induced electric fields" (which explains WHY there's a force on that electron, that's what E fields do!) The only big difference is that in the text, they are working out the induced e-field INSIDE the solenoid (at arbitrary points), so you have to think about where in that derivation something would change if you are OUTSIDE the solenoid...
Hope this helps - if not, let me know, maybe we can meet in the help room Monday, we can figure out a time..
Cheers,
Steve P.
I hope you appreciate the incredible coolness of these two problems, 4 and 5. Question 4 is what I've been calling the "electric toothbrush" question in help room - how can a SEALED, smooth plastic case of an electric toothbrush slip into a SEALED, smooth plastic base, and still charge up? Where's the electrical connection if there are no metal wires sticking out? The answer is basically question #4 - put one coil in the base (plugged into the wall, so there's an alternating current in it, which means an alternating B-field) and then the coil in the other part (the toothbrush) will have an EMF (which means voltage, which means current flows, charging up a little rechargeable battery hidden in there). It's "wire free" electricity transmission - wireless! And it's not just a *signal*, it's really power.
This is also what's going on INSIDE those little black boxes you plug into the wall, and then plug the other end to charge your computer or cell phone. Inside that box, is a small transformer (basically, something like the picture in #4). The voltage from the wall is 120 V, but the voltage in the second coil can be ANYTHING, you choose the numbers (number of turns) and you can get the answer to be, say, 1.5 V on the second coild, which might be just right to charge a small battery which would be FRIED by 120 V directly. (And, you get an added bonus: if your transformer is LOWERING the voltage, then you INCREASE the current going into your battery, letting it charge more quickly)
Question #5 is the most amazing of all. Just think about this. There's an electron sitting out in empty space. There's no magnetic field OUTSIDE a solenoid, so you might think the electron would be oblivious to what goes on, far away, inside that solenoid. But Faraday's law says that changing flux DOES produce electric fields at a distance. Changing the current in the solenoid (which is here acting like an antenna) results in a real electric force in EMPTY SPACE far away - you can wiggle electrons on the MOON by changing currents in a solenoid here on earth (that's how we talked to the astronauts when they were up there - that's what all radio communication is!)
When you look at a star in the night sky, think about what's happening. Millions, possibly billions of years ago, an electron in a hot star bumped into a neighboring atom, changing its path. That's a changing current, which makes a changing magnetic field. But Faraday tells us changing magnetic fields make electric fields (question #5 again!) and Maxwell tell us changing electric fields make magnetic fields. This "dance" travels across empty space, never ending, an electromagnetic wave generating and propagating itself, for millions of light years... Until it happens to strike your eyeball, making an electric field at your retina, which accelerates an electron in a nerve cell. If the frequency is just right (about 10^18 wiggles/sec), it sends an electrical signal to your brain which says "nice stars tonight"...
You just gotta love this stuff :-)
Last CAPA:
Hello-I was wondering if CAPA 14 is going to be released this weekend at all or is its release being pushed back according to the new deadline for 13? Thanks for all of the quick responses, I'm sure fielding e-mails can be time consuming.
No worries! I'm still working on CAPA 14, it will get distributed next week (probably Monday or Tuesday) and you will again have more than the usual week to do it - it will probably not be due till the very end of the term (like, Thursday evening, Dec 14). So it'll be the last set!
Grades:
Several people have been asking me about grades in the course. It's linked in the syllabus, but you can go straight to our detailed grades page, to see how final course scores are calculated.
CAPA 12 question:
Professor Pollock, A question on the CAPA for this week:
Question 6 is stumping me;.... I used the flux I found in Question 5 as the final flux, and then I found the initial flux assuming the current was the current at t=0. I then calculated the difference and divided by the time of 0.5 seconds. I even tried my solution as a negative and positive but it isn't correct. I'm sure there's something simple I'm missing, can you tell what it is? Thank you for your help!!
I think the problem may be that if you just use "final -initial flux"/time, you are finding some sort of *average value* of EMF over an entire time interval, but since the current is not linear with time in this problem, the average value is not the same as the value at any randomly chosen time! What you need to do is treat it symbolically - find a *formula* for the flux (as a function of time), and then take the actual derivative (d PHI/ dt), and THAT will depend on time (because you're taking the derivative of a quadratic). So then you figure out its value at whatever particular time CAPA is asking about.
Does that make sense? If not, let me know!
A question about conductors:
I have a few questions...This first few are about the properties of charged conductors. I understand that all the charge exists on the surface of the conductor,
In static equilibrium, this is all right...
that the E field inside is =0,
ditto, yup
that the voltage inside is constant,
If by "inside", you mean physically INSIDE the metal, then yes indeed. The voltage will be linearly changing in the empty space between one plate and the other one, which I suppose you *might* call "inside the capacitor" (but it's not inside the actual metal... Make sense?)
and that there is no Net Force inside the conductor.
Again, "inside" meaning in the metal itself, this is quite right. (In equilibrium)
I tried to prove that mathematically (through all the given equations), but I was unable to do so. If v= - (integral of (E dot dr)), and E=(kq/r^2),
The integral gives Delta V, not V (we'll come back to this point in a sec, it's important!)
E = kq/r^2 is from Coulomb's law for a point charge, which we definitely don't have here! We have an infinite *sheet*, in fact TWO of them, the negative sheet on one plate, and the positive sheet on another. Each of those sheets all by itself creates a uniform field (given by sigma/(2 epsilon0) - do you remember how we got that from Gauss' law?
So, in different regions, you'll have to "superpose" the E field arising from BOTH sheets. In the region BETWEEN the plates, they'll add up (so you get E = sigma/(epsilon0), adding the two "halves" gives you one.) But everywhere else (including INSIDE the metal, and beyond, outside the capacitor) the two E fields will be in opposite directions and cancel.
the E-field is =0 because q=0. But, the integral of zero is zero, not a constant,
So now although your reasoning wasn't right for why E=0, the result itself was indeed correct. Inside that metal conductor, E=0, and so as you say, the integral of zero is zero... which means DELTA V is 0. That means there is no change in V as you move around inside the metal. (And "no change" is another way of saying "constant".)
so due to that reasoning, v=0 inside a conductor (and I know that is not the case).
So, I would argue V= constant. It COULD be 0 (if you "grounded" that conductor), but it does not HAVE to be zero. If you have two capacitor plates, you are free to call one of them (typically the one charged up negative) V=0, but then it's not possible for the other one ALSO to be V=0, it will be V=some constant. (In fact, the constant is determined for a single, simple capacitor - it would be E*d, where d is the distance between the plates, and E = sigma/epsilon0. Do you see why this is?)
Also, I understand that if the E-field in the conductor is =0, there is no net force. BUT, in class you proved that the E-field is zero BECAUSE there was no net force (and I think of it the other way around, there is no net force because there is no E-field.) Can you explain that reasoning?
Well, the argument was that IF there was any force, the electrons in the conductor would MOVE (because that's what forces do!) And electrons are pretty much free to move, that's what a conductor *means*. So this would not be equilibrium, things are still accelerating around! The electrons will move, but as they do, they start to pile up, and start to cancel out the E field that was there before (draw a picture, see if you can see qualitatively that this seems reasonable). So as time goes by, they keep moving UNTIL E=0, and then they stop moving, and everything is stable- that's static equilibrium. It takes a tiny fraction of a second for most charged up metals to achieve this state!
So bottom line - if you TRY to set up an E in a metal somehow, the charges nearly instantly re-arrange, and KEEP doing so until finally E=0 everywhere inside, and then everything is stable.
Does this help? This may be tough via "writing" - if it doesn't help enough, let's just get together during office hours! If you let me know you're there down in the help room, I can always put 1120 folks to the front of the "line" down there! And if regular hours don't work, it's easy enough to schedule another time...
CAPA 7 notation question:
hey professor,
i was just doing capa 7 and i found #11 to be confusing. You give the current
as i=.360 Amps. i thought little "i" and big "I" were two different variables
and after much trouble i found that i should not use it in the I=ei respect,
but just as big "I". so im not sure if that was a typo on your part, or are
they interchangable?!?!?!?
Alas, everyone uses different symbols - in the text you're right that i and I are considered different (in different chapters!) but yes indeed, most people use them both completely interchangeably (and some use j for current, instead of current density, argh!) . But it should usually be clear from context... and in this case, the UNITS make it completely unambiguous! Amps is the metric unit for current, 1 Amp = 1 Coulomb/sec... so there's nothing else "i" can be here but plain old current. Sorry for any confusion, though, glad you sorted it out.
__
Tutorial Homework question on Capacitors:
I was wondering on the tutorial hw... If the battery stays the same, there would still be the same potential difference across the plates?? Also,
the charge has to stay the same if the battery stays the same , correct?? So if the area increases the charge just spreads out, yes??
Batteries basically do one thing - they keep a fixed potential difference across themselves. If a battery is labeled "12 Volts" (like a car battery) it does whatever it needs to do (moving charges around is no problem for it!) to make sure those two plates stay 12 Volts apart... If that means *adding* charges to the plates it will, if it has to take charges away, that's what it does. The battery is labeled "V" because those two sides are always V volts apart!
So when is charge "fixed"? Not often - but it would be fixed when it has nowhere to go, and nowhere to come from. So an isolated plate has a fixed charge. ("Isolated" means it's not connected to anything, so the charges have no place to go!) But when you're hooked to a battery, you're definitely not isolated, so no reason to believe charge would be fixed then.
Now, if area were to increase and charge were fixed, then you bet, the charges would spread out. But if the the area increases and the charge is NOT fixed, then what happens? Hmmm... what would NEED to happen in order to keep the voltage difference fixed between the plates?
(You might think about "Q = C delta V"... if delta V is held fixed, and C changes, what would have to happen to Q? ...
Hope this gets you rollin' on the Tutorial homework!
______________
CAPA 6 question
A student wrote me, carefully explaining his work on CAPA 6, #8 (the one about the Van de Graaf). In the end, he did everything right, computed the charge (which came out to be 3.11E something or other) , but he INPUT 3E something or other... and CAPA said no. Turns out his calculator (TI89) just rounded the # without his realizing it!
So here's the deal: CAPA checks your number by seeing if it agrees with its calculation to within +/- 3% agreement. Note that 3 and 3.11 differ by 3.6%, greater than that threshold, so CAPA said "no". Normally, this is quite reasonable - you should be able to do calculator work to better than 3%, (that just means getting 2 or 3 digits accurate...)
But in this particular case, I realize that (although this wasn't what the student was doing!) the problem was actually little ambiguous about sig figs (It said the voltage was "about 80 kV" in my case.) So, it sure looks like CAPA is only giving you ONE sig fig in the voltage, which means that you could, technically, legitimately insist on rounding your final answer to one sig fig! And alas, in this case, rounding 3.11 to 3 moves the answer past CAPA's threshold of agreement. (I just fixed the CAPA code so that it will now give a "yes" even after rounding on this particular one...)
So the bottom line is this - it's always safest with CAPA to go ahead and assume that when CAPA gives you any numbers, it INTENDS for them to be at least 2 or 3 sig figs (so, you should simply read "80 kV" as meaning "80.0 kV") That way, you'll always be nice and safe with its numerical error checking! (When you're working with numbers in the lab, you should definitely think about sig figs and answer accordingly!)
By the way, this student realized he needed his calculator to be FORCED into giving more sig figs. Here's what he came up with:
I just figured it out if you'd like to tell other students that may be having this problem. Here are the steps on a TI-89 Titanium:
1. Press the [Mode] button
2. Go down to Display Digits (It's the 3rd option)
3. Press the right arrow button
4. Scroll down to Float 12
5. Press [Enter]
6. Press [Enter] again
I hope this can help other students who may be getting problems wrong and aren't asking about it
_________________________
another CAPA 6 question
Professor Pollock,
I had a question about the notation on one of the questions in Capa #6. It has to do with the question about finding the voltage difference across a capacitor. The question gives the value 175 uF, and I wanted to know if this was a typo which should have read 175 μF - the greek letter Mu (for micro), not the alphabetical "u".
Yup, it's definitely meant to be micro, sorry for any confusion. Is this typo happening on the printed out sheet you get from the bins, or on the web page? (or both?) Because CAPA doesn't really know about modern html and can't produce greek fonts on the web, but it should get it right on the printout!
Hope this helps, let me know if there's still any confusion
CAPA 5 question
A student asked about the CAPA question with the contour lines, and said Regarding the question about "the electric field at "j" points east" . I'm thinking the field from the point charge points west at "j" so it would be false, but with superposition i am not sure...
So the thing is, the equipotential lines have *already* taken superposition into account. The electric field lines will always be perpendicular to the equipotentials (it's like the "fall line" on a mountain). So if you are *given* thos equipotential lines, then you can visually SEE the true direction of E, it's always (locally) perpendicular to the contour! (that tells you the total E, due to all charges *everywhere*, not just the nearby ones!) Now, there is an ambiguity in direction (there are TWO ways to move perpendicular to a given contour line), and here the rule is "E always points from higher voltage towards lower voltage". (Note that when I say "lower", I might mean "less positive" or "more negative"!)
Hope this helps!
Late Clicker registration?
Professor Pollock
I was looking at my physics grade on WebCT last night and I noticed in dismay that despite the frequent reminders at the beginning of class, I somehow neglected to register my clicker, or at least I don’t think I’ve gotten every single clicker question wrong… I have since registered it but I was wondering if the computer will back date the score, if there’s anyway I can make up the points, or if I’m out of luck and will just need to study a little harder!
No worries, the computer will "back date", it keeps track of all your clicker's clicks all term, and merely associates them with YOU once we get the registration information input. This should be reflected when I next update WebCT, probably not till the next midterm.
Cheers,
Steve P.
Exam question - getting at some good physics!
Dear Prof. Pollock,
I took the yellow exam, and question #5 was about a dipole in a non-uniform electric field. When I first read the problem, I realized the point charge (or similar object creating the field) must be positive and that the nearest side of the dipole to that charge was also positive, so I thought it would have net force to the right. I remembered I had something on my crib sheet about dipoles, so I read it and it said something like, “the positive end of a dipole points with the field lines.” I realized the dipole was faced in the opposite direction that it should align itself to. Then, I concluded the dipole needed to flip, thus giving it torque but no net force for that first moment. Just to be sure I hadn’t written something silly, I took out my book after the solutions to the exam had been posted. On the bottom of page 840 (of Knight Vol. 4), I found this statement, “Suppose that a dipole is placed in a nonuniform electric field…[it gives a little more detail]…The first response of the dipole is to rotate until it is aligned with the field, with the dipole’s positive end pointing in the same direction as the field.” If the dipole’s first response is to rotate, then shouldn’t it have no net force? I would also find it hard to believe that the dipole would maintain a stable enough equilibrium to have a force and consequent acceleration to the right when aligned incorrectly as it was.
Please, tell me where my reasoning is breaking down. Is the book not necessarily correct?
Thanks for writing! First of all, I do stick by our answer - the key thing is that we wrote "At the instant shown, what direction is the net force on the dipole? " Since we said "at the instant shown", there is no ambiguity about the answer at all - at the instant shown, there is ZERO torque (it's lined up antiparallel to the field, there is no "moment arm" but there is definitely NON-zero force. (I'll come back to Knight's statement in a sec)
But I appreciate that you're looking back at the exam, wrestling with the physics, and thinking about it appropriately. In particular, what you said is getting at some interesting and (mostly) correct physics. Your intuition that you find it "hard to believe" this position would be maintained is good - this is an unstable position! In real life, here's what would happen. At the moment shown, there *really* is a non-zero net force to the right (like you said, the positive end as in the stronger field) and at this instant NO torque... So, this little dipole would start to accelerate directly AWAY (to the right). In some ideal "perfect" world, it would just continue to the right... but in reality, there would surely inevitably be some little "twitch", a random bump, even a quantum fluctuation, that would *tip* it just the tiniest bit. (Nothing from the GIVENS in the picture would make this happen, but you're quite right that something ELSE in nature would make this happen) Then there would be a torque, because the + and - would no longer be lined up perfectly. But it would be TINY at first - even so, that would surely make it start to rotate. As you commented on, it's in equilibrium (as shown), but it's an UNSTABLE equilibrium - like a pencil balanced on its tip! It would still be accelerating to the right at this point, but the torque would keep growing as it rotated more and more. When it reached 90 degrees orientation, the torque would now be maximized, and the net force at THAT instant would be zero. (Up until then, it was accelerating to the right the hole time) Of course, with zero force, it would still be *moving* to the right (!!) but now, as it continues to twist further "into alignment", the force would now reverse, and it would start to slow down (but, still moving to the right for a short while!) What would happen next would depend on even more details we didn't give. If there was friction, then the oscillations would dampen, the thing would line up the other way (with - charge to the left), and ultimately start moving/accelerating left. How much time this would take would depend on many things, could be seconds, could be microseconds. (If there was no friction, it would overshoot and "wobble", moving right and left, twisting CCW and then CW.) But all of this is well AFTER "the instant shown"...
The question tried to avoid all this interesting but complicated physics by taking a snapshot at one instant in time! At the instant shown, none of the above has come into play, it's just a simple force diagram, nothing else...
Back to your comments, I just can't resist replying to your statement "Is the book not necessarily correct?" Well, I'd have to say NO, the book is not "necessarily correct"! The book is doing its best to pull all of physics into a freshman-level synopsis. (Nature is correct, not the book!) When Knight says "the first response of the dipole is to rotate until it is aligned with the field", he's referring to a dipole which is already partly tipped. (Just work out the torques. You'll see that it goes to zero in both the aligned and anti-aligned cases) And, when he says "the first response", you can't take him too literally - he's simply not bothering to mention the OTHER "first response", the one which we asked about, which is the direction of initial net force. He doesn't worry about it because he knows that as soon as the seed flips around, the force will also flip around, and from then on the story will be dominated by the attraction... But you knew that, that's why you wrote :-)
Hope this helps. See you in class!
Cheers,
Steve P.
Now we're getting to the good stuff! Send my your fun questions, I'll try to answer!
Dear Steven Pollock,
While reading the Chapter 26 on E-fields in Knight, I am having a hard time understanding why the field strength does not depend on the distance from an infinite plane of charge, but the field strength does depend on the distance
from an infinite line of charge. Like the book says on page 832, "What does it mean to be 'close to' of 'far from' an infinite object?" It seems to me that the field strength would be dependent on the distance for both the infinite line and the infinite plane. Hopefully you can help me grasp the concept a little better.
It's a great question. Not sure I can do justice to it in an email, you might want to find me after lecture or office hours some time! There are a couple of answers.
The least satisfying (but most rigorous) might be the physicist's "work it out"! I.e., starting from Coulomb's law (force drops off as 1/distance^2 from a point of charge), use the techniques of Chapter 26 to *calculate* the force in these other cases, and you find that it drops off like 1/distance from an infinite rod, and doesn't drop off at all from an infinite wall. (When we get to Gauss' law next chapter, we'll be able to derive this without doing any nasty integrals at all...) So there's the answer, but like I said, not very satisfying.
So how do we make physical sense of this? I can *sort* of make sense of the infinite wall thing... (well, knowing the right answer, anyway :-) If I walk away from a little spot of charge, it seems reasonable to me that the E field gets weaker, fast - the spot is getting far away! But try to picture a HUGE wall of charge in front of you, so big that you can't even see the edges. Take a step back... there is STILL a huge wall of charge in front of you, as far as you can see. As you walk farther and farther back, it STILL looks the same - there's no distance scale, there's no real sense of "being farther" from such a huge thing.... (I just looked at Knight, I hadn't read that particular section yet, and it's funny how Randy says basically the same thing!)
Ok, here's another thought, a slightly different perspective. When I "look" towards the wall of charge, I see a bunch of charges. "Looking" is figurative, but imagine that you hold your hands out, forming some 'solid angle' that you're peering out towards. Now let me step back, twice as far. It's true, all those individual charges I'm looking at are now twice as far (and so contribute only 1/4 as much E field, EACH), but on the other hand, because I've stepped back, I "see" more wall. So although each given piece of the wall contributes less, there are effectively *more pieces* of wall we "see", we have to add up! Geometrically, if you look at a section of wall which is twice as far away, you might argue that (looking out at some given angle, like the angle formed by my outspread arms) there is now FOUR times as much wall area out there that I'm "catching". That increase by four just cancels exactly with the 1/2^2 decrease in force, meaning the far away parts count JUST AS MUCH as the near ones, when added together. So as a result, the field is independent of distance.
With the infinite rod, it really is "intermediate". As you step away from it, the force of individual parts drops like 1/r^2, but you "see more" rod. If you go twice as far, you see twice as much. That's not SQUARED because it's only a linear object, not a 2-D wall. So we cancel out one power of r, but not both, and the force from the rod drops like 1/r (not 1/r^2, but not constant either).
I've got one more argument for you. Think about drawing field lines. For a point charge, by symmetry, they have to spread out in all directions. And since spread-out field lines means "weaker field", you clearly have less field when you get farther away. But now try to imagine drawing the field lines from a wall. They can't spread out, because the wall is infinite in all directions. There's nowhere to spread out to! And, what point would they spread out FROM?? There's no center, no special spot on the wall to "diverge" from. If the field lines can't spread out, they just have to run straight away from the wall (constant density of field lines), which means uniform field, it never gets weaker!
OK, I'm petering out on ideas here. Any of these help? It's probably not the biggest deal in the world if you still find this puzzling, it's a fairly deep point, but kind of cool to think about! (And to be fair, it's not completely artficial or impractical. If you were an electron, a 1 cm^2 capacitor plate would sure as heck LOOK pretty infinite to you!)
Cheers,
Steve P
WEBCT resource: Our 1120 WebCT page has a "threaded messages" list just for you - you are welcome to ask each other questions about the course, even the homework there, and respond to one another. I will be monitoring this occasionally, so please use good judgement and common sense. If anyone uses this to copy answers, tell formulas, or otherwise sabotage the learning goals of the course, we'll just turn it off. But I think it's great for you to ask questions, discuss the problems, give hints, talk about where you went to get help or how you figured something out, and so on. Use it as a resource to collaborate (but not to cheat!) When in doubt about how far you can go in "giving help" to other students' questions: if you feel comfortable knowing that I'll be reading it, then it's probably fine!
Weekly Pretests:
I must have missed something at some point. I did not realize we had a pre-test due over the labor day weekend or that we had weekly pre-tests due every week for that matter. Is it possible to make up the first? From what I have heard from others I know who are in PHYS 1120, a lot of people missed this, and we're all the overachiever types, so it's not as if we were trying to slack off or anything.
Murray announced it at the start of class Wed and again Friday, and it was on the main web page... but I kind of figured there'd be some folks who'd miss this first week, due to the holiday. I plan to make this first one count considerably less than the rest, so don't sweat it, it really won't matter much! I'm also planning on automatically dropping one additional pretest for everybody in the end, so if you don't miss any more it *literally* won't matter at all! (We don't grade 'em, full credit just for any attempt)
Normally, they'll be up from Friday till Tuesday 8 AM, basically something for you to do over the weekend or Monday. Spread the word - just be sure to do them from here on out! Let me know if you have any questions.
Also, will there be a way to access old pre-tests for study purposes?
Yep, I'm planning on trying to post the old ones in some version for you to study from, just working on how to do that. (I may have to do screen shots or something?)
Late enrollment, missing CAPA:
I tried using the CAPA pin getter but i did not work and my hard copy was not available today.
If you signed up for 1120 AFTER the Saturday-before-classes-started, you won't be in the CAPA system. No worries, just go to the CAPA late enrollment page, fill it in, and a human being will get you into CAPA. Be patient, we only do that about twice a day, so you may have to wait a few hours (So don't wait till Friday afternoon, or you're likely out of luck for this week!) Once you're in the system, pin-getter will then work for you, and you can log on and do the set (or even print it out, off the screen). From then on, your sets will print out with everyone elses and appear in the bins. If this fails, email me (steven.pollock@colorado.edu), but it should all work like a champ! You will only have to do this once, and you should NOT have to do it if you registered normally for Phys 1120.
(Waitlisted people who were already waitlisted before the term started should already be in CAPA even if you're not yet officially in the class - CAPA is independent of the university PLUS system)
Clicker question:
Prof. Pollock -
I'm in PHYS 1120 at 10am. I'm having problems with my clicker, I registered it and recieved an email confirming the registration, but when I try to use it in class I never recieve a green light on the clicker and it never shows on the screen. Any ideas on what the problem could be? I'd appreciate any suggestions or solutions. Thanks
It might be just fine - it won't show a green light on your clicker, the only way to know if it's working is to find your number appearing on the "matrix". If you come up after class (or before), we can test it out so you can see where your number appears (once you find it, it's always in the same spot on the screen no matter what order people click) If it's really not appearing on the screen, the next thing to try is getting a new battery. (Borrow a friend's first, just to see if that's the problem?) If THAT fails, you should probably take it back to the bookstore and see if they'll swap it for a new one (and if so, you'll need to register the new one, don't forget!)
Hope this helps!
Steve
CAPA SET 1 question:
My question is regarding the T,F physics 1 review (question 6 on my form). On my form, one statement reads "Statement 4 variation c". Could you please clarify for me .
This was a CAPA bug, it is fixed now! Check your set online. I bet that one now reads "If a light car and a heavy truck collide, the force of the car on the truck might in some circumstances be greater (in magnitude) than the force of the truck on the car." (I fixed this bug and reprinted ALL sets at about 1 PM Monday, so you can also just go back down to the bins and find your new set - there's a fresh copy with this statement filled in properly)
For the question regarding vector operations on this week's CAPA, what are you defining as Ax (subscript x). Is this the A vector in a certain direction? The notation that I am familiar with is i, j, and k. Similar to an x,y,z plane. Could you please clarify for me.
Yes, I think you've got it - Ax means "the x component of A" (which in your notation would also be the coefficient of i) So if A = 3 i + 2 j - 4 k, then A_subx = 3, A_suby = 2, and A_subz = -4. Does that help? Let me know if you're still confused!
Physics 1120 home page.