course phy 202
7/19 4PMThis assignment was very unorganized on the page. I think i sifted through it okay but you may want to look at it and correct its layout.
Question: `qBased on what you learned from Introductory Problem Set 2, how does the current in a wire of a given material, for a given voltage, depend on its length and cross-sectional area?
Answer: Increasing the length would increase the current as well as increasing the cross-sectional area.
Self-critique Rating:
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Question: `qHow can the current in a wire be determined from the drift velocity of its charge carriers and the number of charge carriers per unit length?
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Self-critique (if necessary):
You would use the equation 1.6 * 10^-19 * nElectrons * `dL / L amps
Self-critique Rating:OK
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Question: `qWill a wire of given length and material have greater or lesser electrical resistance if its cross-sectional area is greater, and why?
There will be less resistance since R=r(L)/A
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Self-critique (if necessary):
OK
Self-critique Rating:
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Question: `qWill a wire of given material and cross-sectional area have greater or lesser electrical resistance if its length is greater, and why?
R=r(L)/A so the reistance will be greater.
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Self-critique (if necessary):
OK
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Question: `qQuery Principles and General Physics 16.24: Force on proton is 3.75 * 10^-14 N toward south. What is magnitude and direction of the field?
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Self-critique (if necessary):
The positive proton has a force to the south, and the magnitude will be f/q which will be 3.75*10^-14/1.6*10^-19
Self-critique Rating:
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Question: `q`qThe direction of the electric field is the same as that as the force on a positive charge. The proton has a positive charge, so the direction of the field is to the south.
The magnitude of the field is equal to the magnitude of the force divided by the charge. The charge on a proton is 1.6 * 10^-19 Coulombs. So the magnitude of the field is
E = F / q = 3.75 * 10^-14 N / (1.6 * 10^-19 C) = 2.36* 10^5 N / C.
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Self-critique (if necessary):
OK
Self-critique Rating: OK
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Question: `qQuery gen phy problem 16.32. field 745 N/C midway between two equal and opposite point charges separated by 16 cm.
What is the magnitude of each charge?
E=kq/r^2
We can solve for E and then use what we already know to solve for q. E=(1/2)745 which gives 372.5. We then use the radius of 8cm and k to get q which would be 2.6E-10 for both charge.
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Self-critique (if necessary):
OK
Self-critique Rating: OK
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Question: `qIf the charges are represented by Q and -Q, what is the electric field at the midpoint?
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Self-critique (if necessary):
You would use the equation 2kQ/r^2 to find E at the midpoint
R is equal to .08meters and we use 2 because each would have magnitude of Q since they are equidistant.
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Self-critique (if necessary):
OK
Self-critique Rating: OK
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Question: `qQuery Principles and General Physics 16.26: Electric field 20.0 cm above 33.0 * 10^-6 C charge.
E = 9 * 10^9 N m^2 / C^2 * 33.0 * 10^-6 C / (.200 m)^2
Which would be 7.43 * 10^6 N / C.
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Self-critique (if necessary):
OK
Self-critique Rating: OK
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I just discovered the formatting anomalies today, and believe I've fixed them.
If you understand everything, you're OK. If you need to go back and look at the Open Queries, I believe the formatting is now correct. Sorry for the confusion.