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Unformatted text preview: Version 036 – midterm 01 – chiu – (56565) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider a long, uniformly charged, cylindri cal insulator of radius R with charge density 1 . 5 μ C / m 3 . (The volume of a cylinder with radius r and length is V = π r 2 .) The value of the permittivity of free space is 8 . 85419 × 10 12 C 2 / N · m 2 R 2 . 9 cm What is the magnitude of the electric field inside the insulator at a distance 2 . 9 cm from the axis (2 . 9 cm < R )? 1. 1185.88 2. 1468.23 3. 1027.76 4. 2456.46 5. 2399.99 6. 1106.82 7. 1411.76 8. 2591.99 9. 3049.4 10. 1016.47 Correct answer: 2456 . 46 N / C. Explanation: Let : r = 2 . 9 cm = 0 . 029 m , ρ = 1 . 5 μ C / m 3 = 1 . 5 × 10 6 C / m 3 , and = 8 . 85419 × 10 12 C 2 / N · m 2 . Consider a cylindrical Gaussian surface of radius r and length much less than the length of the insulator so that the compo nent of the electric field parallel to the axis is negligible. r R The flux leaving the ends of the Gaussian cylinder is negligible, and the only contribu tion to the flux is from the side of the cylinder. Since the field is perpendicular to this surface, the flux is Φ s = 2 π r E , and the charge enclosed by the surface is Q enc = π r 2 ρ . Using Gauss’ law, Φ s = Q enc 2 π r E = π r 2 ρ . Thus E = ρ 2 r = ( 1 . 5 × 10 6 C / m 3 ) (0 . 029 m) 2 (8 . 85419 × 10 12 C 2 / N · m 2 ) = 2456 . 46 N / C . 002 10.0 points Two dipoles are oriented as shown in the diagram below x A r r Version 036 – midterm 01 – chiu – (56565) 2 Each dipole consists of two charges + q and q , held apart by a rod of length s , and the center of each dipole is a distance r from location A . If q = 5 nC, s = 1 mm, and r = 7 cm, what is the electric field ( E y ) at location A ? Hint: Draw a diagram and show the direc tion of each dipole’s contribution to the elec tric field on the diagram (you do not have to turn in the diagram). The net electric field can be written in the form: E = E x , E y , E z where it is understood that E z = 0 by the planar nature of the problem. What is E y ? 1. 158.203 2. 324.0 3. 222.222 4. 81.0 5. 60.8565 6. 472.303 7. 210.937 8. 111.111 9. 316.406 10. 393.586 Correct answer: 393 . 586 N / C. Explanation: Now, we have to be more careful about determining the electric field. First, we remember that the electric field due to a point charge is F = k q r 2 ˆ r = k q r 3 x, y, z So, let’s consider the net field due to the left dipole first. First, note that the electric field due to both point charges has a positive y component since the electric field points away from positive charges and toward negative charges....
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This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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