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Unformatted text preview: Flores, Michael – 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 20 V battery does 1344 J of work transfer ring charge. How much charge is transferred? Correct answer: 67 . 2 C. Explanation: Given : W = 1344 J and V = 20 V . The work done is W = q V q = W V = 1344 J 20 V = 67 . 2 C . 002 (part 1 of 1) 10 points A voltmeter indicates that the difference in potential between two plates is 58 V. The plates are 0 . 32 m apart. What electric field intensity exists between them? Correct answer: 181 . 25 N / C. Explanation: Given : V = 58 V , and d = 0 . 32 m . The potential difference is V = E d E = V d = 58 V . 32 m = 181 . 25 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. line AB only correct 3. both AB and CD 4. neither AB nor 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 Flores, Michael – 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. line AB only 4. neither AB nor CD 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 46 % of the speed of light ( c = 2 . 99792 × 10 8 m / s), starting from rest? Correct answer: 54063 . 7 V. Explanation: Given : s = 46% = 0 . 46 , 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 . 46 c = 0 . 46 ( 2 . 99792 × 10 8 m / s ) = 1 . 37905 × 10 8 m / s , By conservation of energy 1 2 m e v 2 = ( q e )Δ V Δ V = m e v 2 2 q e = ( 9 . 10939 × 10 31 kg ) × ( 1 . 37905 × 10 8 m / s ) 2 2(1 . 60218 × 10 19 C) = 54063 . 7 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. √ 3 E l 2 correct 2....
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This note was uploaded on 02/18/2010 for the course PHYS 303l taught by Professor Panab during the Spring '04 term at North Texas.
 Spring '04
 Panab
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