Massive hints for your imminent HW set:
Showing posts with label MASSIVE_HINTS. Show all posts
Showing posts with label MASSIVE_HINTS. Show all posts
Wednesday, September 14, 2011
Friday, September 2, 2011
HW2 number 8
For question 8, you will need to sum an infinite series. It is in fact a very famous and neat little result, which you can find here.
HW2 / today
This afternoon, from 2pm onward, I'll be in or around my Bevill office (room 2050) if you have homework questions. You can also email/text me if you like.
Some of the problems on HW2 are well known (i.e., you could potentially google solutions or hints), some I have asked before in either PH106 or PH126. Just throwing that out there.
Lastly, in general if a HW problem suggests a particular method of solution, but you think you have a better way to get the same result, you can use your own method if you like. For instance, on #2 of HW2, you might decide it is just easier to superimpose the fields of two lines an a semicircle, since the field from those objects are well-known results, rather than setting up the problem as suggested. In general, any consistent method is fine, solutions by any means necessary. This means writing code or numerical solutions are fair game in general [but you should turn in the code with the rest of your HW].
Some of the problems on HW2 are well known (i.e., you could potentially google solutions or hints), some I have asked before in either PH106 or PH126. Just throwing that out there.
Lastly, in general if a HW problem suggests a particular method of solution, but you think you have a better way to get the same result, you can use your own method if you like. For instance, on #2 of HW2, you might decide it is just easier to superimpose the fields of two lines an a semicircle, since the field from those objects are well-known results, rather than setting up the problem as suggested. In general, any consistent method is fine, solutions by any means necessary. This means writing code or numerical solutions are fair game in general [but you should turn in the code with the rest of your HW].
Friday, August 26, 2011
HW1
By the way, I should mention that professors in general have a habit of recycling problems that they like. Maybe from the same course, maybe from a similar one ... but if it were me, I would try googling key phrases plus the professor's name. That, and just checking last year's problem sets.
I won't reuse problems a lot, but given that I (a) assigned homework on the first day, and (b) made it due on the second day, I thought I should give you a fighting chance.
I won't reuse problems a lot, but given that I (a) assigned homework on the first day, and (b) made it due on the second day, I thought I should give you a fighting chance.
Monday, December 7, 2009
Actual hints (I)
I really think exams should be a learning experience. With normal exams, this should be so: you get to see the solutions and discuss. With finals, not so much. This is one more reason I like the take-home final: I can try to teach you a few last things, and I don't feel bad coaching you a little along the way, since most of the problems are brand-new for you. I'll be around campus until Thursday afternoon if you want to drop by. Anyway, some hints:
#1. Use the integral form of Faraday's law to get the first correction to the E field. Take a square contour which (looking from the sides) runs down the center, parallel to the plates, up the right side, and back to the center.
The original field E will have no contribution to the integral of E.dl around the line contour. The new contribution will. If the new contribution is due to time variation in B, you know its symmetry ... so all but one side of the square will give zero to the integral. Put another way, the flux of B only contributes to the new correction to the E field, so you can find the correction directly. After the exam, I'll tell you where I found this; brilliant discussion.
More massive hints to follow on this one later in the week; it is subtle.
#2. Build it out of rings. You know the field from a ring.
#3. If the network is infinite, one more element makes no difference at all. Terminate it at some arbitrary place, and the rest of the network continuing on can be represented by some Req. That Req has to be the same wherever you terminate, so pick some easy places: after just one instance of R1 and R2, and after none. The two have to give the same Req.
Next, imagine you're in the middle of the network somewhere. Now you can have a single R1 and R2 terminated on *both* sides by Req if it is an infinite network. Now you have a simple 4 resistor circuit; find the voltages. If the ratio holds for two arbitrary nodes like this, it holds for all.
#4. Download the final again so you get the correct equations without typos. Apply the curl equations for E & B in free space ... that's about it. Apply the divergence equations as a trivial sanity check. w/k should be the velocity of propagation, right? Energy density can be had from the field amplitudes.
#5. Just work it in one dimension until part c, it makes no difference really. Two dimensions if you like, one component of E is important, the other just gives a torque. For the last part, generalizing to three dimensions should not be too hard if you're careful.
#6. The chain rule thing is key: d(fg) = f*dg + df*g. Also note that at certain points you'll want to write v in terms of gamma (to simplify the final result) and gamma in terms of v (to do an integral).
#7. Force is the gradient of the potential energy (with a minus sign). Write the energy of the capacitor for fixed charge ...
#8. See previous post. You may neglect atmospheric refraction, as it is essentially the same at the top and bottom of the cherry picker.
#1. Use the integral form of Faraday's law to get the first correction to the E field. Take a square contour which (looking from the sides) runs down the center, parallel to the plates, up the right side, and back to the center.
The original field E will have no contribution to the integral of E.dl around the line contour. The new contribution will. If the new contribution is due to time variation in B, you know its symmetry ... so all but one side of the square will give zero to the integral. Put another way, the flux of B only contributes to the new correction to the E field, so you can find the correction directly. After the exam, I'll tell you where I found this; brilliant discussion.
More massive hints to follow on this one later in the week; it is subtle.
#2. Build it out of rings. You know the field from a ring.
#3. If the network is infinite, one more element makes no difference at all. Terminate it at some arbitrary place, and the rest of the network continuing on can be represented by some Req. That Req has to be the same wherever you terminate, so pick some easy places: after just one instance of R1 and R2, and after none. The two have to give the same Req.
Next, imagine you're in the middle of the network somewhere. Now you can have a single R1 and R2 terminated on *both* sides by Req if it is an infinite network. Now you have a simple 4 resistor circuit; find the voltages. If the ratio holds for two arbitrary nodes like this, it holds for all.
#4. Download the final again so you get the correct equations without typos. Apply the curl equations for E & B in free space ... that's about it. Apply the divergence equations as a trivial sanity check. w/k should be the velocity of propagation, right? Energy density can be had from the field amplitudes.
#5. Just work it in one dimension until part c, it makes no difference really. Two dimensions if you like, one component of E is important, the other just gives a torque. For the last part, generalizing to three dimensions should not be too hard if you're careful.
#6. The chain rule thing is key: d(fg) = f*dg + df*g. Also note that at certain points you'll want to write v in terms of gamma (to simplify the final result) and gamma in terms of v (to do an integral).
#7. Force is the gradient of the potential energy (with a minus sign). Write the energy of the capacitor for fixed charge ...
#8. See previous post. You may neglect atmospheric refraction, as it is essentially the same at the top and bottom of the cherry picker.
Clarifications
There were a couple of typos on number four: the argument of the sin functions should have a + sign, not a - sign, and the B_z component should be divided by c. The posted version of the exam has been corrected.
Here are some clarifications I sent one of you by email, for all to see:
It is good to be certain before you start. The "e" is indeed Euler's number, the base of the natural logarithms. The "i" is the imaginary unit, so the formula for E is just writing a sine wave in complex exponential notation. Check the chapter in Griffiths on EM waves for similar notation & problems.
First, the distance d is the separation of the two charges, I should have noted that.
You can assume that the dipole is along the z axis, so the two charges are sitting on the z axis with the origin at the center of the two charges. You can also let the E field be along the z axis if you like. It isn't perfectly general then, but close enough; just work the first parts of the problem as if it is one dimensional, and generalizing for the last part is not hard.
There need not necessarily be a net force for the first part ... For the second part, you are to pretend that the E field is slightly different at one charge than the next. For instance, say E is a little bit bigger at the positive charge sitting at z=(d/2) than at the negative charge at z=(-d/2). You could say then that the field at the negative charge is just Eo, and the field at the positive charge is roughly that plus the gradient of the field times the separation distance: E = Eo + (dE/dz)*(d). If the separation distance d is small enough, it is pretty good to approximate the variation in the field as a constant plus distance times a gradient.
You are correct, they wait until the sun is level with the horizon and then rise up until they are again level with the sun & the horizon. You can ignore the atmospheric refraction, since we already did a problem on that. You can also assume that the cherry picker is sitting in a little hole, so they start out exactly at the ground level.
Ok, go ahead and look here if you like, but your solution should be a bit more thorough and elegant.
If you move from left to right, it is probably easiest to say that a node starts at the left of an R1, and the next node starts after the R2 to its right. It doesn't matter too much, so long as you are consistent and move from left to right (along the top) as the figure is drawn. More hints on this one later. As an aside, this sort of circuit is useful for quick & dirty digital-analog conversion. Think about that: power of 2 ratio of voltages at every node ... just meant for binary.
Here are some clarifications I sent one of you by email, for all to see:
In the formulas for problem 1, what does the i term indicate? Also, I
think the little e is Euler's number, and just wanted to check and
make sure.
It is good to be certain before you start. The "e" is indeed Euler's number, the base of the natural logarithms. The "i" is the imaginary unit, so the formula for E is just writing a sine wave in complex exponential notation. Check the chapter in Griffiths on EM waves for similar notation & problems.
On problem 4, I'm not sure what the k and x stand for in the waveHere k is the wavevector, which relates to the spatial periodicity of the wave (k = 2pi/lambda). Omega is the angular frequency. You can basically treat them as constants. Since these are wave equations, you can guess that omega/k should give the velocity of the wave ... The energy density should be in joules per cubic meter, or energy per unit volume.
equations. Also, I think you left out the units you wanted the energy
density in.
For problem 5, are we supposed to assume the dipole is in a specific
orientation for the first two parts? It seems as though you meant for
it to be perpendicular to the z axis. I'm assuming total force for
part (a) is not the same as net force, because that would depend on
the direction of the field. Also, I'm not sure to what distance the
distance d in part (b) refers.
First, the distance d is the separation of the two charges, I should have noted that.
You can assume that the dipole is along the z axis, so the two charges are sitting on the z axis with the origin at the center of the two charges. You can also let the E field be along the z axis if you like. It isn't perfectly general then, but close enough; just work the first parts of the problem as if it is one dimensional, and generalizing for the last part is not hard.
There need not necessarily be a net force for the first part ... For the second part, you are to pretend that the E field is slightly different at one charge than the next. For instance, say E is a little bit bigger at the positive charge sitting at z=(d/2) than at the negative charge at z=(-d/2). You could say then that the field at the negative charge is just Eo, and the field at the positive charge is roughly that plus the gradient of the field times the separation distance: E = Eo + (dE/dz)*(d). If the separation distance d is small enough, it is pretty good to approximate the variation in the field as a constant plus distance times a gradient.
For problem 8, I have a certain vision for how this date would ideally
play out, and want to check that it is appropriate. I imagine that
the couple starts on the level of the earth, then, at the moment the
top of the sun disappears over the horizon, the cherry picker lifts
them until the bottom of the sun is level with the horizon. Also,
should we take into account diffraction due to the atmosphere?
You are correct, they wait until the sun is level with the horizon and then rise up until they are again level with the sun & the horizon. You can ignore the atmospheric refraction, since we already did a problem on that. You can also assume that the cherry picker is sitting in a little hole, so they start out exactly at the ground level.
Ok, go ahead and look here if you like, but your solution should be a bit more thorough and elegant.
I didn't recognize this until I got into the problem. On problem 3,
to which nodes exactly does "successive nodes" refer? Is it nodes
across the top, from top to bottom, across the bottom, or something a
bit more specific?
If you move from left to right, it is probably easiest to say that a node starts at the left of an R1, and the next node starts after the R2 to its right. It doesn't matter too much, so long as you are consistent and move from left to right (along the top) as the figure is drawn. More hints on this one later. As an aside, this sort of circuit is useful for quick & dirty digital-analog conversion. Think about that: power of 2 ratio of voltages at every node ... just meant for binary.
Friday, November 13, 2009
Friday's exam
You're ready. Get some sleep.
If you don't believe me, and want to cram anyway, I'd spend some time on Ch. 7, sections 2&3 in Griffiths, and then probably review the sections on Maxwell's equations (sans vector potential).
You will be rewarded if you can quickly recognize what to do with Maxwell's equations when (for instance) given an E field. You will also be rewarded if you have subjugated div, grad, and curl in spherical coordinates (formulas given).
Finally, you will be rewarded with bonus points if you remember what I said about tensors on Wednesday. Specifically, conductivity tensors.
PS - If you are unsure what a question means, or how to go about it tomorrow, don't hesitate to ask. More than likely, I will be willing to clarify the problem a bit or give you a hint to get you started. Also, show and turn in all your work, even if you think it illegible or unimportant. Partial credit is key.
If you don't believe me, and want to cram anyway, I'd spend some time on Ch. 7, sections 2&3 in Griffiths, and then probably review the sections on Maxwell's equations (sans vector potential).
You will be rewarded if you can quickly recognize what to do with Maxwell's equations when (for instance) given an E field. You will also be rewarded if you have subjugated div, grad, and curl in spherical coordinates (formulas given).
Finally, you will be rewarded with bonus points if you remember what I said about tensors on Wednesday. Specifically, conductivity tensors.
PS - If you are unsure what a question means, or how to go about it tomorrow, don't hesitate to ask. More than likely, I will be willing to clarify the problem a bit or give you a hint to get you started. Also, show and turn in all your work, even if you think it illegible or unimportant. Partial credit is key.
Wednesday, October 21, 2009
HW 7.1 hint
Check this. The first problem is an awkward one, but valuable for precisely that reason. The first time you see it, you scratch your head wondering why the ring would rotate, but once you understand why, you've learned something important.
Friday, October 16, 2009
Homework
Back to serious stuff. On number 9, remembering your calculus is a big help. For instance, note
and use the Lorentz transformations along with velocity addition. I'll spell this out more tomorrow in recitation.
For number 6, the x component of the velocity is just what you think it is. However: the y component of the velocity in the second reference frame would be
We'll go over the rest tomorrow, but you will find many of the other questions in my PH102 notes or previous PH102 homework sets. I'll give some hints on where to look ... but start with the problems at the end of Ch. 1.
\frac{dv_x^{\prime}}{dt^{\prime}} = \frac{dv_x^{\prime}/dt}{dt^{\prime}/dt}
and use the Lorentz transformations along with velocity addition. I'll spell this out more tomorrow in recitation.
For number 6, the x component of the velocity is just what you think it is. However: the y component of the velocity in the second reference frame would be
u_y^{\prime}=\frac{dy^{\prime}}{dt^{\prime}} Since y=y^{\prime}the numerator is trivial. However, there is still time dilation, so you'll need to use the Lorentz transformation to relate dt and dt'. Again, remember the calculus trick above, and the main point is this: the velocity along the direction of relative motion follows the addition formula we derived, but along the orthogonal direction, there is still a transformation because while distance is uncontracted, time is still dilated.We'll go over the rest tomorrow, but you will find many of the other questions in my PH102 notes or previous PH102 homework sets. I'll give some hints on where to look ... but start with the problems at the end of Ch. 1.
Monday, October 5, 2009
Wednesday, September 30, 2009
Thursday, September 24, 2009
And one more ...
Another is here. Be sure you know how to do the problems if you use these hints ... such problems will likely reappear at inconvenient times.
Tuesday, September 22, 2009
HW 4 ... more information
Problem 1: the power in a circuit element is in general current times voltage. Using Ohm's law, for a single resistor we can write this as
Problem 2: now the resistors are in series, the internal resistance r and the external "load" resistance R. You can easily figure out the current through both resistors in this case, and again find the power in each resistor using the formula above. Maximize with respect to R.
Problem 3: see previous post; we'll do some more examples of Thevenin equivalents tomorrow.
Problem 4: break the cone up into tiny disks. If the current is running along the axis of the disk, the resistance of a tiny disk of radius r and thickness dx is
P=I^2R The total power in the circuit is the sum of the power in the two resistors, P=I_1^2R_1+I_2^2R_2 Using the conservation of charge equation, you can put power in terms of the current in a single resistor. Minimize.Problem 2: now the resistors are in series, the internal resistance r and the external "load" resistance R. You can easily figure out the current through both resistors in this case, and again find the power in each resistor using the formula above. Maximize with respect to R.
Problem 3: see previous post; we'll do some more examples of Thevenin equivalents tomorrow.
Problem 4: break the cone up into tiny disks. If the current is running along the axis of the disk, the resistance of a tiny disk of radius r and thickness dx is
dR = \rho\,dx/\pi r^2 The radius is position-dependent for a cone, you might write it as r=a+(b-a)x/l in this case. Integrate over all such disks with resistance dR to find the total R. We'll set up a similar problem in class tomorrow.
Monday, September 21, 2009
Thevenin equivalents
The wikipedia has a nice page on this, including an example suspiciously like your homework. That is not by accident I guess.
More hints on Wednesday's problems will follow tonight or tomorrow afternoon.
More hints on Wednesday's problems will follow tonight or tomorrow afternoon.
Thursday, September 17, 2009
More last minute thoughts on the exam
UPDATE: I made corrections & additions to the HW3 solutions today, might be worth a last-minute check.
So I've actually made it now, and I think you will be fine. I am going to do the problems again myself just to be sure the timing is somewhat reasonable. A few of random thoughts:
So I've actually made it now, and I think you will be fine. I am going to do the problems again myself just to be sure the timing is somewhat reasonable. A few of random thoughts:
- As promised, the exam has 4 problems, you can solve any 2.
- There are no terrible integrals involved. Just polynomials and so forth, no weird arctans or anything.
- For most of the problems, there are at least two straightforward methods of attack. This is on purpose, with the hope that you'll see one of them quickly.
- Two problems will favor those of you that remember the basics of mechanics, two problems will favor those of you that like the math.
- The binomial approximation is just about the coolest thing ever. Know when you can get away with it.
- There are no numbers on the exam (except things like pi and 2, possibly a 3). A calculator is not useful unless it happens to do symbolic math.
- It is fine with me if you bring an complex calculator that does symbolic math. The basic rules are no pc's, no cellphone calculators, no PDAs that have wireless communication capabilities.
- Use your time wisely, and watch the clock.
- Don't get stuck on anything - if you find yourself stuck, see if you can make a simplifying assumption to move on (this might entail some lost points, but many less than not finishing the problem at all), or pick another problem
- In spite of the time pressure, spend a few minutes reading and thinking about all the problems before starting. Make sure you really know what is being asked, and have a physical picture in your mind before moving on. If you can't at least see the 'flavor' of the answer, math is probably not going to help.
- Please, please make sure you read the problems you choose to answer a second time, there are multiple parts to some of the problems.
Friday, September 11, 2009
Stray HW3 hints
(1) We set this one up in class. Put one charge at the origin, and write the distance from the second charge to the field point in terms of the distance to the first charge and the separation of the charges. E.g., if b is a vector pointing from one charge to the other,
Do not forget that your volume element in spherical coordinates has an r-squared, or you will get a nasty integral.
(2) The field is nonzero only between the two spheres ... the result is the energy of a spherical capacitor, if you want to check your answer.
(3) Set this up in class. Solve for the potential at an arbitrary point, and either (a) show that at least 4 points from the same ellipse give the same potential, or (b) plug in the equation for an ellipse and show that it results in V=constant.
(4) See previous hint ...
(5) You can use the boundary conditions we derived for this one. From one side of a given plate to the other, the difference in the normal components of E has to give you the surface charge density.
Above the top plate, or below the bottom plate, the field is zero. Thus, the field in between any two plates can be related to the surface charge on the upper or lower plate. How does that relate to the charge on a given side of the middle plate? Two parallel plates will have the same charge on the sides that are facing. Just like we talked about on Wednesday, if one plate has charge Q, it will induce -Q on the adjacent plate.
The only thing left is to realize that since the electric field must be constant inside (infinite plate, no spatial dependence!), the electric potential must just be
This gets you the surface charge. Once you have that, you know the field. Once you know the field, you can use the method of problem 1 to find the total energy. Which can then be optimized. I bet you can guess the optimum spacing already though.
(6) Capacitance is just total charge divided by potential for a given conductor. If you know the potential for a prolate spheroid with a charge Q (problem 3!), C=Q/V. You can fancy that up by noting that the eccentricity (epsilon) is just d/a. After that, note
7. Here's the basic scheme for finding the coefficients. The self capacitance terms C_ii is just the capacitance of the system when the two elements are joined together, so they have the same charge and the potential of both is zero. The C_ij off-diagonal terms are found by setting the potential of one of the conductors to zero, and having a charge Q on the other. You'll need the formula for the capacitance of a spherical capacitor, which you can find in your text or online.
We'll go over that in class on Friday, but here's what you should come up with:
8. Textbook problem. Try checking in some textbooks :-) Feynman (Vol. II) does a great job.
\vec{r}^{\prime}= \vec{r}-\vec{b}\\
\hat{r} \equiv \frac{\vec{r}}{|\vec{r}|}\\
\vec{E}_1 = \frac{kq_1q_2}{r^2}\,\hat{r} = \frac{kq_1q_2\vec{r}}{r^3}\\
\vec{E}_2 = \frac{kq_1q_2}{r^{\prime\,}^2}\,\hat{r}^{\prime} = \frac{kq_1q_2 \left(\vec{r}-\vec{b}\right)}{|\vec{r}-\vec{b}|^3}
Do not forget that your volume element in spherical coordinates has an r-squared, or you will get a nasty integral.
(2) The field is nonzero only between the two spheres ... the result is the energy of a spherical capacitor, if you want to check your answer.
(3) Set this up in class. Solve for the potential at an arbitrary point, and either (a) show that at least 4 points from the same ellipse give the same potential, or (b) plug in the equation for an ellipse and show that it results in V=constant.
(4) See previous hint ...
(5) You can use the boundary conditions we derived for this one. From one side of a given plate to the other, the difference in the normal components of E has to give you the surface charge density.
E_{\mathrm{above}} - E_{\mathrm{below}} = \sigma/\epsilon_0
Above the top plate, or below the bottom plate, the field is zero. Thus, the field in between any two plates can be related to the surface charge on the upper or lower plate. How does that relate to the charge on a given side of the middle plate? Two parallel plates will have the same charge on the sides that are facing. Just like we talked about on Wednesday, if one plate has charge Q, it will induce -Q on the adjacent plate.
The only thing left is to realize that since the electric field must be constant inside (infinite plate, no spatial dependence!), the electric potential must just be
V = Ed = \sigma/\epsilon_0
This gets you the surface charge. Once you have that, you know the field. Once you know the field, you can use the method of problem 1 to find the total energy. Which can then be optimized. I bet you can guess the optimum spacing already though.
(6) Capacitance is just total charge divided by potential for a given conductor. If you know the potential for a prolate spheroid with a charge Q (problem 3!), C=Q/V. You can fancy that up by noting that the eccentricity (epsilon) is just d/a. After that, note
\ln{\left(\frac{1+x}{1-x}\right) = \ln{\left(1+x)} - \ln{\left(1-x)} \approx 2x
7. Here's the basic scheme for finding the coefficients. The self capacitance terms C_ii is just the capacitance of the system when the two elements are joined together, so they have the same charge and the potential of both is zero. The C_ij off-diagonal terms are found by setting the potential of one of the conductors to zero, and having a charge Q on the other. You'll need the formula for the capacitance of a spherical capacitor, which you can find in your text or online.
We'll go over that in class on Friday, but here's what you should come up with:
C_{11}=\frac{ab}{k\left(b-a\right)} = -C_{21}=-C_{12} \\
C_{22}=\frac{b^2}{k\left(b-a\right)}
8. Textbook problem. Try checking in some textbooks :-) Feynman (Vol. II) does a great job.
Wednesday, September 9, 2009
Cube of charge
Two things:
1) The MIT course analogous to ours is 8.022. Open courseware is awesome.
2) Problem 4 on your homework is the same as Purcell problem 2.30.
Actually, most of your problems so far have been from the Purcell book, as it turns out, which is also favored for 8.022 at MIT during certain semesters. Looking through the 8.022 content on the open courseware site is highly recommended. Typically very thorough and lucid solutions.
Not all their content is on the open courseware site yet, but it can be found. try googling "purcell 2.30 MIT 8.022" and look at the first couple of links.
1) The MIT course analogous to ours is 8.022. Open courseware is awesome.
2) Problem 4 on your homework is the same as Purcell problem 2.30.
Actually, most of your problems so far have been from the Purcell book, as it turns out, which is also favored for 8.022 at MIT during certain semesters. Looking through the 8.022 content on the open courseware site is highly recommended. Typically very thorough and lucid solutions.
Not all their content is on the open courseware site yet, but it can be found. try googling "purcell 2.30 MIT 8.022" and look at the first couple of links.
Tuesday, September 8, 2009
HW3 #3
Here's a journal article which covers question 3 on homework 3. It is a bit terse in the beginning, but it should be big help if you can get through it ... (link will only work on campus).
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