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03homework03 - Husain Zeena Homework 3 Due Feb 9 2004 4:00...

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Husain, Zeena – Homework 3 – Due: Feb 9 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 26 questions. Multiple-choice 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 0 . 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

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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 ) Δ V Δ 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 0 E What is the magnitude of the potential dif- ference between A and B ? 1. E l 2. 2 E l 3 3. E l 2 4. 2 E l 5. none of the above. 6. 3 E l 2 correct 7. 2 E l 8. E l 9. 0 10. E l 2 Explanation:
Husain, Zeena – Homework 3 – Due: Feb 9 2004, 4:00 am – Inst: Sonia Paban 3 The potential difference between two points A and B is Δ V = Z B A ~ E · d~s. In this case, we can choose a path which goes vertically up from A , then horizontally to B . Along the vertical path, E is perpendicular to the path, but along the horizontal part of the path, they are parallel, so that Δ V = E s = E l cos 30 = 3 E l 2 .

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