Homework 1 is due next Thursday (Sept 30). It includes 3 tutorials and a number of problems from Chapter 20. As you have probably done in previous physics 6 series classes, you enter your results online at the mastering physics site. (See the post: "Getting started..." for the link and other related info that will help you "get started".)
This assignment deals with the electric fields and forces associated with charge particles. A lot of the problems involve detailed numerical calculation involving things like k (in N-m^2/C^2) and q in coulombs. From personal experience I know that that can be pretty annoying and not fun. The answers rarely turn out right, and when you enter them it just says you are wrong with no explanation. At least you get 6 chances (the most they allow) and only a 1% penalty for incorrect entries. Some are more conceptual. My favorite so far is 48 (the answer is an integer).
The basic idea of the assignment is to get some familiarity and understanding of electric fields and forces, and their relationship to electric charge. I would suggest working the problems on paper, like you would on a test, before you try to enter your results online. One tip: remember to convert centimeters to meters, etc.. Unfortunately there are a lot of not small exponents, especially since we often start with charges of 1.6 * 10^-19 Coulombs (and then sometimes square that). Feel free to post comments and questions here, and remember that the primary goal of HW is to help you learn and to help you prepare for tests. Critiques of particular problems, their appropriateness, efficacy, etc., as well as questions and comments are most welcome here.
-Zack
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ReplyDeleteI used a friend's mastering physics key for my previous physics course to complete the online homework. I was able to change the email and user ID number but not the name. My 6A professor Sam Gweon was able to set up a system for this and was wondering if I could continue doing this for your course.
ReplyDeleteThank you,
Ariel Anders
Ariel,
ReplyDeleteI am not sure, but this sort of issue is probably better dealt with by e-mail. Please e-mail me at zacksc@gmail.com
There seems to be an error in example problem 20.3 of the text. Units should be N, not N/C.
ReplyDeleteI am a little confused about what a unit vector represents. The book says it helps to determine the direction of the force, but the force can go in the opposite direction. Is it just the "line" along which the force acts on between two charges; only relatively helping with direction by narrowing it down to two directions?
ReplyDeleteRegarding Jesse's comment: what do people think? Is there an error in the example problem in the book?
ReplyDeleteAlso, feel free to comment here on the unit vector issue.
In equation 20.2a ( E=F/q ) the charge does not have to be specified, but in Eq. 20.2b ( F=E*q ) the charge does need to be specified (+ or -). Why is this? Is it because electric fields (like charge distributions) combine the effects of two or more charges and so there is an overall neutral charge? And what is the difference between charge distributions and electric fields?
ReplyDeleteAbout the typo: There is a mistake in units for example problem 20.3. It should just be newtons.
Is there supposed to be an annoying amount of algebra involved in 20.47? The problem itself is so straightforward and easy to set up...
ReplyDeleteI'll check on that later. Has anyone else worked on that problem?
ReplyDeleteRegarding problem 20.47: first, thanks for bringing that up. It is a good question. I think
ReplyDeleteperhaps the key to avoiding annoying algebra is to take the square root of “both sides” at the right moment. That way you don't get a quadratic equation, just a linear equation in 1/r, or r, if you turn it upside down. (Where, in my notation, r is the distance from the smaller charge to the point in space that we are looking for (where the field is zero).)
The way that I set it up, was to have the smaller charge on the left, the bigger charge on the right, the distance between them I called d. I realized, using intuition, that the point where the field could be zero must be between them and that it should be closer to the smaller charge. So I drew that on a piece of paper and called the distance from the left-hand charge to that point, r, which is the unknown in this problem that we must solve for. So one distance is r, and the other is d-r.
I also drew two arrows: one pointing to the left representing the field from the charge on the right which is a distance d-r away; the other pointing to the right representing the field from the charge on the left.
Anyway, I grew to like this problem and to me it seems like a good sort of problem to have on a midterm. What do you think?
20.49 asks for the expression on the y axis, however, it does not denote where on the y axis.
ReplyDelete-Ariel
I think they want you to express it as a function of y. in that way you have a result that is valid everywhere on the y-axis. By the way, y-axis means x=0.
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ReplyDelete"Anyway, I grew to like this problem and to me it seems like a good sort of problem to have on a midterm. What do you think?"
ReplyDeleteIt was a good problem, but it's very easy to go in circles and never arrive at the answer if you miss the part where you can square root both sides as I did. In my opinion (if this problem were on the midterm), if someone got to the part where there is one equation with one unknown they should get >half credit for it...
wasting time trying to enter the answer for 20.37... It wants it as one sig fig. 10^xx or 1*10^xx should work. I've checked my work against similar problems and am confident I'm doing the problem correctly. any suggestions?
ReplyDelete20.37 follow up: when I view the "my answers" record it says that my answers were "evaluated" as "1.00*10^xx" which is 3sig fig. How do you make it evaluate your submitted answer as 1 sig fig? I've tried 1.*10^xx as well without success.
ReplyDeletehmmm. let me look at that one.
ReplyDeleteJesse: Yes, I totally agree.
ReplyDeleteThat problem, 20.37, is problematic. Really the only meaningful part is the exponent (which is about 10^-11 i think). Thanks for pointing that out. Much appreciated.
It is interesting in the sense that 1 part in 10^11 removal of electrons will, roughly, lead to a charge of a few coulombs in an object that size,(so that is why I assigned it), but it rejects answers that are totally fine.
anyway, i'll make sure you get credit for that.
ReplyDeleteI was wondering if I could get some help on 20.48. I am pretty sure i am doing it correctly but I wasted about 3 tries trying to express it in an integer... Maybe i did the problem wrong, but i've done the problem several times to check my answer...
ReplyDeleteI'll take a look at that (20.48) pretty soon.
ReplyDeleteAlan, are you using a positive or negative integer?
ReplyDeletei caught my first attempt at a positive and tried a negative. got them both wrong. should the integer just be the 1 sigfig times the exponential value?
ReplyDeletecould you review 20.48 in class?
ReplyDelete20.37 doesn't really make much sense to me. I know how many electrons get removed but without a way to know how many electrons a ping pong ball will normally have I can't state the fraction remaining.
ReplyDeleteI am also having trouble with 20.48 I have tried it many times and I am certain of my answer of -1.44*10^-18 but the computer disagrees. I noticed at least one other person is having trouble on it. Could I get some help with this problem?
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ReplyDelete"i caught my first attempt at a positive and tried a negative. got them both wrong. should the integer just be the 1 sigfig times the exponential value? "
ReplyDeleteAlan, no not at all. It should just be an integer like: -1, -2, -3, ... where -1 means the same things is -1.6 x 10-19 C in this case. The answer is not in coulombs!
(-2 means the same things is -3.2 x 10-19 C ...
Eric: let's start with this:
ReplyDeleteHow many electrons get removed??
Then: how many protons are there in one gram?
Further: what is the typical ratio of protons to neutrons in a ping pong ball, (or most anything)?
20.48: they want the result in terms of e. Just an integer between 1 and 100 or so. Not Coulombs!!!
ReplyDeleteLike when people say that oxygen has a charge of -2 --what does that mean???
Were we actually required to do the first section (Introduction to Mastering Physics)? I just skipped it because I already used mastering physics before.
ReplyDeleteyea there was another assignment due yesterday (Introduction to Mastering physics) its the assignment that gets you familiar with the website. I skipped it because i've already taken a masteringphysics course and you said our first HW assignment was the chapter 20 HW. It said it's worth 8 points though. can you either make the assignment not worth anything or open it up to be submitted again(right now no late credit is possible for that assignment)
ReplyDeleteNo, don't worry about that one. That is just a Mastering Physics "built-in" assignement. You do not have to do it for this class.
ReplyDeleteJust do the real assignment. The one that has problems and tutorials from Chapter 20.
i was wondering if i could get 20.28B explained. I was following the example problems you did in class the other day but i still can't get to the right conclusion. i adjusted for positive and negative direction, but i still don't understand how i got it wrong. is there a huge difference amongst the equations you gave us in the class and the one we need to utilize for 20.28B?
ReplyDeleteThe equations you need to use for this problem are exactly the same as the ones we have used in class. One thing to note is that you need to convert the distances given in centimeters to meters.
ReplyDeleteDo that, and draw a nice picture, and carefully calculate the magnitude of the electric field from each charge, and then the component you are looking for––using trigonometry. I think that should work. please let me know with a comment here.
More on that: 20.28 b is just like what we did in class yesterday. One thing to note is that the distance from the center point, P., to a charge is only 2.5 cm. So I don't think the angle is 45°. Am I getting that right? What did you get for the sine or cosine or tangent of the relevant angle?
ReplyDeletefor 20.49 I got part A, it was just like what we did in class, but i still don't understand where to begin on part b. Would you want to take the derivative and find the maximum points or is it much simpler than that.
ReplyDeleteJust what you said sounds right. It is not simpler than that. It's a hard problem. It's difficult conceptually, and the calculation to find the maximum is difficult. Your method sounds correct to me.
ReplyDeleteThere is one other way. Use a calculator to find the maximum by guessing and zero in on the key reason where it is largest. That might be easier?
Thanks. I finally worked it out with the help from the ta. Just forgot my calculus.
ReplyDeleteok. good
ReplyDelete