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Unformatted text preview: Husain, Zeena Homework 3 Due: Feb 9 2004, 4:00 am Inst: Sonia Paban 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points A 12 V battery does 1635 J of work transfer ring charge. How much charge is transferred? Correct answer: 136 . 25 C. Explanation: Given : W = 1635 J and V = 12 V . The work done is W = q V q = W V = 1635 J 12 V = 136 . 25 C . 002 (part 1 of 1) 10 points A voltmeter indicates that the difference in potential between two plates is 36 V. The plates are 0 . 57 m apart. What electric field intensity exists between them? Correct answer: 63 . 1579 N / C. Explanation: Given : V = 36 V , and d = 0 . 57 m . The potential difference is V = E d E = V d = 36 V . 57 m = 63 . 1579 N / C . Dimensional analysis for E : V m = J C 1 m = N m C 1 m = N C 003 (part 1 of 2) 10 points Consider the figure + Q #1 + + + + + + + + + + + Q #2 A B C D x y Of the following elements, identify all that correspond to an equipotential line or surface. 1. line CD only 2. neither AB nor CD 3. line AB only correct 4. both AB and CD Explanation: Consider the electric field + Q #1 + + + + + + + + + + + Q #2 A B C D x y An equipotential line or surface ( AB ) is normal to the electric field lines. 004 (part 2 of 2) 10 points Consider the figure A q + B + q C D Husain, Zeena Homework 3 Due: Feb 9 2004, 4:00 am Inst: Sonia Paban 2 Of the following elements, identify all that correspond to an equipotential line or surface. 1. both AB and CD 2. line CD only correct 3. neither AB nor CD 4. line AB only Explanation: Consider the electric field: A + B C D An equipotential line or surface ( CD ) is normal to the electric field lines. 005 (part 1 of 1) 10 points Through what potential difference would an electron need to be accelerated for it to achieve a speed of 48 % of the speed of light ( c = 2 . 99792 10 8 m / s), starting from rest? Correct answer: 58867 . 1 V. Explanation: Given : s = 48% = 0 . 48 , c = 2 . 99792 10 8 m / s , m e = 9 . 10939 10 31 kg , and q e = 1 . 60218 10 19 C . The speed of the electron is v = 0 . 48 c = 0 . 48 ( 2 . 99792 10 8 m / s ) = 1 . 439 10 8 m / s , By conservation of energy 1 2 m e v 2 = ( q e ) V V = m e v 2 2 q e = ( 9 . 10939 10 31 kg ) ( 1 . 439 10 8 m / s ) 2 2(1 . 60218 10 19 C) = 58867 . 1 V . 006 (part 1 of 1) 10 points Consider two points A and B in a constant electric field ~ E as shown. B l A 30 E What is the magnitude of the potential dif ference between A and B ? 1. E l 2....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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