12 April 2006

WebTA signing off

Well, I've had an interesting semester experimenting with this WebTA project. I have to say I missed having contact with you the students in tutorials, but it's been interesting to get your questions and challenging to try to answer them effectively over the web. Watching the traffic on the site it seems many of you are using the material I've been providing. I hope this means you have found all this helpful and have learned something from what I've posted here. I'll take this last opportunity to remind you that you can still provide feedback by filling out my survey and returning it to my mailbox (see post WebTA survey), or by emailing me or commenting here.

I wish you all the best with your courses and exams. Try to stay clear-headed for exams. Get sleep. Breath deep. Give yourself the chance to think on the exam... very often it is more effective that those last couple of hours of cramming.

Good bye and Good luck!

Sarah

Question: Max current for an electromagnet (Ch 20 #80)

I am confused about Ch 20 #80. I know Power is the rate of work and I can use that to calculate current . I would also used the magnetic field equation B = N (uI)/2r to calculate the number of turns required to run power at max power (i.e use the calculated Imax). But I don't know B and I am not sure how to calculate that from V.



Ahh... ok, it took me a minute to figure this one out, but the trick is that it is using things you know from previous chapters (tricky folks these text book writers are). Do not be distracted by the equation they give you immediately! Think back to when you looked at electrical power:

P=IV

So, you know V, and Pmax... you can directly determine Imax from this. Then you can use the equation they give to determine the maximum magnetic field that can be generated.

Oftentimes half the trouble in figuring out a problem is determing what information you need to use and when. Most of the problems in textbook are divided up into the appropriate sections where you learn the material so it is easy to determine how to approach a given "type" of problem. The trouble is in the real world there are no "types" of problems. You have to take each situation and consider what factors are present and relavent. Every now and again textbooks (and exam writers!) throw in questions that test your ability to discern what is needed to help train you to solve problems more generally. Learning how to do this takes some experience (which unfortuantely just requires time and practice), and requires that you think about the concepts and how they apply, not just classify the problem as a particular type.

Hope that helps you out.

Question: Two resistors, Two capacitors (Ch 19 #50)

Ch 19 #50 part b

I am not sure how to distinguish whether a capacitor is charging or discharging from this problem. Sometimes, I can't tell through which resistor will a capacitor will discharge...is it through the one in parallel or in series?



Let's just look at simplifying what we have when the switch is open, then closed.

When the switch is open there are two resistors in series which are in parallel with two capacitors in series. The voltage across the set of resistors will be the same as the voltage across the set of capacitors (24 V in this case). At point "a" there will be some voltage/potential between 0 and 24 V due to the voltage dropped (IR) across one of the two resistors. Likewise at point "b" there will be some voltage/potential between 0 and 24 V as there is some voltage dropped (Q/C) across one of the two capacitors. (Note that to determine this you will have to use the equivalent resistances and capacitances of each branch of the circuit to determine the net current and charge, respectively.)

Now, when the switch is closed, charge will flow from a to b or vice versa because the potential at a and b were not equal initially. Once the switch is closed this potential difference cannot be maintained. Without working through to see which is greater and thus which direction the charge will flow I can't say whether the capacitors are charging or discharging. However, I can tell you if the potential at "a" is greater than at "b" the current (charge) will flow toward the capacitors and thus the capacitors will charge (partially). Conversely if the potential at "b" is greater than at "a" the current (charge) will flow away from the capacitors and the capacitors will discharge (partially).

Good luck. This is a tricky problem.

10 April 2006

Correction: Example Ch 16 #61

Sorry, there is an error in the example posted on the main page: Electrostatics / Ch. 16 #61

In the first part, I forgot to square the velocity, so instead of 5.4e-9 m, I should have gotten a distance of 0.115 m. The error is also carried forward into part b, instead of 1e-12 s, I should have gotten 2e-8 s for the time.

I will try to fix the original page soon.

WebTA survey

To get some feedback on the usefulness and usage of the webTA project I have developed a short survey. Paper copies should have been (will be?) available in class at some point, however, if you missed your chance to have your say and are interested in providing feedback I will give you another chance here.

You may download a PDF version of the 2006 webTA project survey HERE [35 kB PDF].

Print it out, fill it out, and return it to my mailbox (Sarah Burke, in the Grad students section) in the mailroom of the Rutherford Physics building.

(Location of the mailroom: Enter Rutherford through main doors, turn right and go through a big brown door. At the end of the hall is a staff lounge. The mailroom is on the right just before this. It's a room with a photocopier and a bunch of labelled mailboxes. Mine's on the left hand side near the beginning as they are more or less alphabetical.)

Of course any creative ways you think of to send it back electronically are good too... sorry I don't have a way to make a form-fillable pdf. :)

End of term looming...

Just a quick warning in case you were not already aware. The webTA site will not be maintained after the end of classes. I will do my best to answer as many of the questions you ask as possible until then.

Question: Stable and unstable equilibria

How do we tell the stable and unstable equilibrium mathematically? cuz using the definition in the text to determine what equilibrium i'm dealing with is ambiguous and obscure...



This is a very good question which is actually very general. Let's start with some definitions...

Equilibrium is a state of a system in which the variables which describe the system are not changing (note that a system can be in a dynamic equilibrium where things might be moving or changing, but some variable(s) which describe the system as a whole is(are) constant). One example you are all familiar with is a mechanical system in equilibium where positions of objects are not changing (ie. no net forces acting).

In a Stable equilibrium if a small perturbation away from equilibrium is applied, the system will return itself to the equilibrium state. A good example of this is a pendulum hanging straight down. If you nudge the pendulum slightly, it will experience a force back towards the equilibrium position. It may oscillate around the equilibrium position for a bit, but it will return to its equilibrium position.

In an Unstable equilibrium if a small perturbation away from equilibrium is applied, the system will move farther away from its equilibrium state. A good example of this is a pencil balanced on it's end. If you nudge the pencil slightly, it will experience a force moving it away from equilibrium. It will simply fall to lying flat on a surface.



Ok, now that we know more about equilibria it will be easier to determine what "kind" we have. Strictly speaking, mathematically we determine whether a mechanical equilibrium is stable or unstable by looking at the second derivative of the energy with respect to the coordinate of interest. If the equilibrium is at a minimum (second derivative is positive) the system is in a stable equilibrium. If the equilibrium is at a maximum (second derivative is negative) the system is in an unstable equilibrium.

However, there is a simpler way to quickly test whether an equilibrium is stable or unstable. If we think about the definitions, each involves the response to a small perturbation. So let's "apply" a perturbation. If we say change the position slightly, is there a net force? in what direction? does the perturbation result in a force which drives the system back toward equilibrium (stable) or away from equilibrium (unstable)?

If you can't easily picture the situation in your head (as with the pencil and the pendulum), what this means in practical terms is that you re-calculate the forces at some position near, but not at equilibrium and determine whether they are driving the system back towards equilibrium (again, this means stable) or away from equilibrium (this means unstable).

Hope that clarifies things!