midterm 01-solutions

midterm 01-solutions - Version 036 – midterm 01 – chiu...

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Unformatted text preview: Version 036 – midterm 01 – chiu – (56565) 1 This print-out should have 17 questions. Multiple-choice 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.

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midterm 01-solutions - Version 036 – midterm 01 – chiu...

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