Welcome to blog help for PHY 122! Here's how it works.
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.
Wednesday, September 1, 2010
Chaper 15, questions 5-9
These five questions are from SV8 section 15.4. The whole assignment is due on Tuesday, September 7th at 3:30 PM.
10 comments:
Anonymous
said...
I'm a bit confused about question number 9 and want to make sure I'm doing it correctly: I found the height of the traingle and then used E=(ke*q)/r^2 to find the field on the point midway between due to the 3 nC charge. (Also- I can't quite see the figure, those are now nano columbs??). that is the only y component of the field. THen I used .25 m as the radius and found the field from both the leftmost (a positive number) and rightmost (a smaller neg number) charge, and added them together for the x component. My resultant field seems to make sense, but I don't think that my angle is "below" the x axis as the problem asks. Am I doing something wrong?
Remember that you are calculating the electric field observed at a given point, which arises from each of three charges. Electric field points away from a positive charge, towards a negative charge. The positive charge at the top of the triangle produces a field with negative y component at the observation point. The two charges along the x axis both produce fields with positive x components, and zero y component. So the vector sum has a relatively large positive x component, and a much smaller negative y component. So the resultant (vector sum) points a little bit clockwise from the x-axis. That's the sense in which it is below the axis.
Professor: I did the correction and got it correct with your help....but the charges are actually in nanocoulombs, not micro--and we're still loosing 10% per entry, so I lost a bit of credit thinking that they were in micro....
Your equation 8.5e-6/(6-Y)^2 = -5/(4+Y)^2 has a couple of problems. First, on the left side, you have coulombs, but on the right, microcoulombs. Second, you have probably figured that since both given charges are negative, you have to find a point between them for the field to cancel. So you want to equate the magnitude of each field, which should be a positive number on each side. Does that fix it?
I am having the same problem with number 8, no matter what I put in the answer comes up incorrect. I am also confused why the original poster starts with -6uC for the charge at -4m and has -5 in their equation. Was this just a typo? If not, where does the -5 come from?
Re Question #8: We are preparing and posting solutions to selected homework problems, worked out on video. You can find them in Blackboard, Course Documents. The only one posted now is #8 (SV8 Ch. 15 Prob. 26). So far, this is an experiment; we want your feedback.
10 comments:
I'm a bit confused about question number 9 and want to make sure I'm doing it correctly: I found the height of the traingle and then used E=(ke*q)/r^2 to find the field on the point midway between due to the 3 nC charge. (Also- I can't quite see the figure, those are now nano columbs??). that is the only y component of the field. THen I used .25 m as the radius and found the field from both the leftmost (a positive number) and rightmost (a smaller neg number) charge, and added them together for the x component. My resultant field seems to make sense, but I don't think that my angle is "below" the x axis as the problem asks. Am I doing something wrong?
Remember that you are calculating the electric field observed at a given point, which arises from each of three charges. Electric field points away from a positive charge, towards a negative charge. The positive charge at the top of the triangle produces a field with negative y component at the observation point. The two charges along the x axis both produce fields with positive x components, and zero y component. So the vector sum has a relatively large positive x component, and a much smaller negative y component. So the resultant (vector sum) points a little bit clockwise from the x-axis. That's the sense in which it is below the axis.
Forgot to add that the charges are given in microcoulombs.
Professor: I did the correction and got it correct with your help....but the charges are actually in nanocoulombs, not micro--and we're still loosing 10% per entry, so I lost a bit of credit thinking that they were in micro....
Oops - sorry.
I have a question about number 8.
I am using the equation " E1 + E2 = 0" to solve, but when I do that I never get the right answers.
My numbers are -8.5uC at 6m along the Y axis and -6uC at -4m along the Y axis.
I end up with 8.5e-6/(6-Y)^2 = -5/(4+Y)^2, cross multiply and solve for Y, but that does not work.
Am I missing something here? Thanks!
Your equation 8.5e-6/(6-Y)^2 = -5/(4+Y)^2 has a couple of problems. First, on the left side, you have coulombs, but on the right, microcoulombs. Second, you have probably figured that since both given charges are negative, you have to find a point between them for the field to cancel. So you want to equate the magnitude of each field, which should be a positive number on each side.
Does that fix it?
I am having the same problem with number 8, no matter what I put in the answer comes up incorrect. I am also confused why the original poster starts with -6uC for the charge at -4m and has -5 in their equation. Was this just a typo? If not, where does the -5 come from?
Yes, I had noticed that there was a problem with the units. I think will keep trying, perhaps my math is just skewd. Thanks professor!
-DM
Re Question #8:
We are preparing and posting solutions to selected homework problems, worked out on video. You can find them in Blackboard, Course Documents. The only one posted now is #8 (SV8 Ch. 15 Prob. 26). So far, this is an experiment; we want your feedback.
Post a Comment