Exam 1-solutions

Exam 1-solutions - Version 004 – Exam 1 – Chiu –(58270 1 This print-out should have 16 questions Multiple-choice questions may continue on

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Unformatted text preview: Version 004 – Exam 1 – Chiu – (58270) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points + Q − Q A test charge is placed an equal distance away from the two charges of a dipole, as shown in the figure. In which of the following directions will a negative test charge experi- ence a force? a e c g h b d f i zero magnitude 1. i 2. b 3. d 4. e 5. a 6. f 7. h 8. g correct 9. c Explanation: By inspection the field direction is along c or the force on the electron is along g . 002 10.0 points Consider the experiment using two tapes The setup is shown in figure 1. The Tapes A and B are oppositely charged. The electric attrac- tive force lifts it up to the location shown with a separation d. Following the approximation of the text, the attractive force is estimated as the attraction between two point charges Q A and Q B separated by distance d. The upward pull on the lower tape due to the electric at- traction is balanced by the weight of the lower tape. Now consider a new case where the lower tape is replaced by a new tape that weighs 2 times as much as the original tape, and the charges are the same as before. Assuming the same approximation is still valid , which expression gives the new separation distance d ′ so that the lower tape is again lifted? 1. d ′ = 2 d 2. d ′ = d 2 Version 004 – Exam 1 – Chiu – (58270) 2 3. d ′ = √ 2 d 4. d ′ = d 3 . 5. d ′ = d √ 3 6. d ′ = √ 3 d 7. d ′ = d √ 2 correct 8. d ′ = 3 d Explanation: Original set up gives k Q A Q B d 2 = mg . (1) New setup gives k Q A Q B d 2 new = 2 mg . (2) Dividing eq(1) by eq(2) leads to, d 2 new d 2 = 1 2 d new = d √ 2 . This is reasonable, since a stronger electric force requires a smaller separation distance. Furthermore, the 1 r 2 dependence of the force tells us that a decrease in the distance of 1 √ 2 leads to an increase of the force by a factor of 2. 003 10.0 points Two dipoles are oriented as shown in the fig- ure. The horizontal distance is a = 25 cm and the vertical distance is b = 16 . 3 cm. Each dipole consists of charges held by a short rod (not shown to scale). What is the magnitude of the electric field (in N/C) at location A? 1. 2 . 76 N / C 2. 4 . 83 N / C 3. 8 . 05 N / C 4. 43 . 5 N / C 5. 9 . 66 N / C correct 6. 33 . 8 N / C 7. 19 . 3 N / C 8. 1 . 93 N / C Explanation: let : k = 9 × 10 9 N · m 2 / C 2 , q 1 = 7 nC = 7 × 10 − 9 C , q 2 = 18 nC = 1 . 8 × 10 − 8 C , s 1 = 0 . 3 m m = 0 . 0003 m , s 2 = 0 . 4 m m = 0 . 0004 m , a = 25 cm = 0 . 25 m , and b = 16 . 3 cm = 0 . 163 m . At A, the electric field due to the dipole 1 located the lower right is E 1 = + 2 k p 1 b 3 ....
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This note was uploaded on 02/24/2010 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas at Austin.

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Exam 1-solutions - Version 004 – Exam 1 – Chiu –(58270 1 This print-out should have 16 questions Multiple-choice questions may continue on

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