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For each homework question and lab report we will make a post, this will probably contain a few tips on what the problems are about and how to solve them. If you are stuck on something then instead of emailing us directly you should post a comment in reply to the relevant post. We will try to guide you through tough points and help you understand the problems and the concepts behind them.
Sunday, September 5, 2010
Chapter 16, Questions 3 through 6.
These questions are from SV8 sections 16.2, 16.3, and 16.4. Please try to look at them before the September 7th class.
14 comments:
Vito
said...
Hello,
I am having trouble understanding the mechanics of question 3 part (b) which states Find the total electric potential at the point having coordinates (1.50 cm, 0). I found part (a), but now I'm stuck. Any help is appreciated! Thanks!
Draw a sketch, noting that the two charges are on the y axis, and in part b, it is desired to find the potential at a point on the x axis.. I don't know the numbers you are seeing (and it would be helpful if you wrote them into your query), so I'll use the ones from the book. At observation point (1.50,0) (all dimensions in cm) you are sqrt(1.50^2+1.25^2) from the charge that is on the y axis at y=1.25 cm, and sqrt(1.50^2+1.80^2) from the charge that is on the y axis at y = -1.80 cm.
Query about penalties. Penalty for incorrect answers has indeed been removed. I have looked at the gradebook, and I didn't notice any 0.9, 0.8, etc. Half credit sounds like you only answered one part of a two-part question. Send me (pstephens@stonybrook.edu) an email with details so I can look at the grade book.
It is our intention to retroactively remove the 10% penalty for each incorrect answer from the Ch. 15 homework, but I'm having a technical issue with WebAssign over this. Please hold on while I work on it.
In the mean time, let's use this stream to discuss questions 3 through 6 of chapter 16, OK?
Hi, I am still struggling with question 3 part b with the (1.50cm, 0) coordinate! (A 3.5 µC charge is located at y = 1.40 cm, and a -2.26 µC charge is located at y = -1.60 cm. ) I did sqrt(1.6squared+1.5squared) to get 2.19cm for the bottom charge. For the top charge, i got 2.05cm. Then I applied these dimensions in meters to V=ke q/r and added the two electic potentials to get 4.46e5 as my answer! SOMEONE HELP PLEASE. ive only got two attempts left on this and about to pull my hair out. I must be making a really dumb mistake.
Your numbers for r, k and q seem to all be good. Are you putting the correct sign for the charge q? positive charges have positive potentials, and likewise, negative charges have negative ones.
Also, remember the charges are in micro Coulombs. So the charge of the positive one, in standard units is 3.5e-6 C, and the other is -2.26e-6 C.
Also, are you converting the distances from cm to m? The distance for the positive charge is .0205 m and for the bottom one is .0219 m. Check to see if you converted correctly.
If you do all this, you should get an answer different from 4.46e5...check your math on paper, your units on paper, and make sure you're putting numbers in the calculator correctly. Hope this helps.
Question 3 (SV8 Ch. 15, Problem 12) You may have to keep more than 3 significant figures in the two distances in this problem to get the right answer. For certain values of the randomization, the contributions from the two charges are nearly equal in magnitude (and opposite in sign) so the answer is the small difference between two large numbers. To get the difference accurate to 1%, each potential must be accurate to about 5 significant digits, at least with the numbers that one student had in the help room. You can use 9.0E9 for k_e, because that is within 1% of the correct value. But the distances must be kept much more accurately.
Q6. As promised for those that were in the help room earlier today, here is a brief explanation of this problem. At the point where the alpha particle comes to a halt, the electrical PE (aka "U")between said particle and the Gold nucleus is equivalent the loss in KE of the alpha particle, which can be derived from the givens in the problem (e.g. mass and velocity). Once you have this figured out use the the formula for elec. PE [U=k*Q1*Q2/r] and solve for r which is the same as distance d in the problem. Make sure you multiply fundamental charge e by the correct number of electrons for Q1 and Q2 and you should have no problems. I have entered it into web assign and it definitely works.
14 comments:
Hello,
I am having trouble understanding the mechanics of question 3 part (b)
which states Find the total electric potential at the point having coordinates (1.50 cm, 0). I found part (a), but now I'm stuck. Any help is appreciated! Thanks!
PS.
I did ke[q/r] for part (b) using 2.0e-6 as q, and radius .0265m (1.15cm + 1.50cm) I got 677735.84 and that isn't correct.
Draw a sketch, noting that the two charges are on the y axis, and in part b, it is desired to find the potential at a point on the x axis..
I don't know the numbers you are seeing (and it would be helpful if you wrote them into your query), so I'll use the ones from the book. At observation point (1.50,0) (all dimensions in cm) you are sqrt(1.50^2+1.25^2) from the charge that is on the y axis at y=1.25 cm, and sqrt(1.50^2+1.80^2) from the charge that is on the y axis at y = -1.80 cm.
I thought the penalty was gone for this quiz! its still there! for one submission i got .5 points taken away! Am i the only one?
Query about penalties. Penalty for incorrect answers has indeed been removed. I have looked at the gradebook, and I didn't notice any 0.9, 0.8, etc. Half credit sounds like you only answered one part of a two-part question. Send me (pstephens@stonybrook.edu) an email with details so I can look at the grade book.
Prof. Stephens, if I got all the questions correct on the 1st WebAssign, but don't have 100% credit, are you going back and making them 100% ?
It is our intention to retroactively remove the 10% penalty for each incorrect answer from the Ch. 15 homework, but I'm having a technical issue with WebAssign over this. Please hold on while I work on it.
In the mean time, let's use this stream to discuss questions 3 through 6 of chapter 16, OK?
Hi,
I am still struggling with question 3 part b with the (1.50cm, 0) coordinate! (A 3.5 µC charge is located at y = 1.40 cm, and a -2.26 µC charge is located at y = -1.60 cm. ) I did sqrt(1.6squared+1.5squared) to get 2.19cm for the bottom charge. For the top charge, i got 2.05cm. Then I applied these dimensions in meters to V=ke q/r and added the two electic potentials to get 4.46e5 as my answer! SOMEONE HELP PLEASE. ive only got two attempts left on this and about to pull my hair out. I must be making a really dumb mistake.
Dear Anonymous (who posted at 5:50 PM),
Your numbers for r, k and q seem to all be good. Are you putting the correct sign for the charge q? positive charges have positive potentials, and likewise, negative charges have negative ones.
Also, remember the charges are in micro Coulombs. So the charge of the positive one, in standard units is 3.5e-6 C, and the other is -2.26e-6 C.
Also, are you converting the distances from cm to m? The distance for the positive charge is .0205 m and for the bottom one is .0219 m. Check to see if you converted correctly.
If you do all this, you should get an answer different from 4.46e5...check your math on paper, your units on paper, and make sure you're putting numbers in the calculator correctly. Hope this helps.
Thanks muntazimm! I've found the error in my calculation! :)
Good!
Question 3 (SV8 Ch. 15, Problem 12) You may have to keep more than 3 significant figures in the two distances in this problem to get the right answer. For certain values of the randomization, the contributions from the two charges are nearly equal in magnitude (and opposite in sign) so the answer is the small difference between two large numbers. To get the difference accurate to 1%, each potential must be accurate to about 5 significant digits, at least with the numbers that one student had in the help room. You can use 9.0E9 for k_e, because that is within 1% of the correct value. But the distances must be kept much more accurately.
Q6. As promised for those that were in the help room earlier today, here is a brief explanation of this problem. At the point where the alpha particle comes to a halt, the electrical PE (aka "U")between said particle and the Gold nucleus is equivalent the loss in KE of the alpha particle, which can be derived from the givens in the problem (e.g. mass and velocity). Once you have this figured out use the the formula for elec. PE [U=k*Q1*Q2/r] and solve for r which is the same as distance d in the problem. Make sure you multiply fundamental charge e by the correct number of electrons for Q1 and Q2 and you should have no problems. I have entered it into web assign and it definitely works.
Good explanation - thanks.
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