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# FinalExamSolutions - PHYSICS 9C FINAL EXAM I certify by my...

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PHYSICS 9C FINAL EXAM June 13, 2007 I certify by my signature below that I will abide by the UC Davis Code of Academic Conduct. This includes not copying from anyone else’s exam not letting any other student copy from my exam not discussing this exam with any student who has not yet taken it, nor providing any information, written or oral, that might get to a student who has not yet taken it. Anyone suspected of cheating will be automatically reported to Student Judicial Affairs. LAST NAME: FIRST NAME: STUDENT ID: (P R I N T) (P R I N T) (LAST 4 DIGITS) Signature: Problem 1 (two points) . Three very large parallel uniformly charged non-conducting sheets with the surface density =+8.85x10 -12 C/m 2 are placed at the distance L=5cm one from each other. What is the value of electric field between the sheets, outside the sheets? A) 0 N/C between all sheets, 1.5 N/C outside. B) 1.5 N/C between all sheets, 0 N/C outside. C) 1.5 N/C before first sheet, 1 N/C between first and second sheet, 0.5 N/C between second and third sheet, 0 N/C beyond third sheet. D) 1.5 N/C before first sheet, 0.5 N/C between first and second sheet, 0.5 N/C between second and third sheet, 1.5 N/C beyond third sheet. E) 0 N/C before first sheet, 0.5 N/C between first and second sheet, 1 N/C between second and third sheet, 1.5 N/C beyond third sheet. Solution: Each sheet produces electric field directed to the left from its left, and directed to the right from its right. The value of the electric field from each sheet is σ /2 ε 0 =0.5 N/C. From the very left and very right, all three sheets contribute to the field, therefore E=3*0.5=1.5 N/C. In the region between sheet 1 and 2 as well as between 2 and 3 contributions from 2 sheets cancel, therefore E= 0.5 N/C. Ε? Ε? Ε? Ε? 1 2 3

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Problem 2 (two points). If 1000 J of work are required to carry a 20 C charge from one point to another the potential difference between these two points is: A) 20,000 V B) 50 V C) 0.02 V D) 1000 V E) dependent on the path Solution: Work is equal to the change in potential energy therefore W= U=qV, V=W/q=1000/20=50 V. Problem 3 (two points). When switch S is open, the ammeter in the circuit shown reads 2.0 A. When S is closed, the ammeter reading will approximately: A. increase four times B. double C. decrease four times D. remain the same E. halve Solution: When switch S is open, the total resistance is 35 Ohm. To have the current 2.0 A in this circuit, the battery’s EMF should be equal to 70 V. When switch S is closed, the new resistance in the circuit is 1/(1/20+1/3)+15=17.6 Ohm, and the new current is 70/17.6=4 A. Therefore the ammeter reading will double. Problem 4 (two points). In the sketch, all bulbs have equal resistances. The brightness of bulbs B and C added together, compared with the brightness of bulb A , is A) four times as large B) twice as large. C)
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## This note was uploaded on 04/07/2008 for the course PHY 9C taught by Professor Zieve during the Winter '08 term at UC Davis.

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FinalExamSolutions - PHYSICS 9C FINAL EXAM I certify by my...

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