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Unformatted text preview: homework 02 – LIN, KEVIN – Due: Jan 27 2008, 4:00 am 1 Question 1, chap 24, sect 1. part 1 of 1 10 points An electric field of magnitude 21000 N / C and directed upward (perpendicular to the Earth’s surface) exists on a day when a thun derstorm is brewing. A truck that can be approximated as a rectangle 6 . 8 m by 3 . 8 m is traveling along a road that is inclined 15 ◦ relative to the ground. Determine the electric flux through the bot tom of the truck. Correct answer: 524150 N · m 2 / C (tolerance ± 1 %). Explanation: Let : E = 21000 N / C , ℓ = 6 . 8 m , w = 3 . 8 m , and θ = 15 ◦ . By Gauss’ law, Φ = vector E · vector A. The flux through the bottom of the car is Φ = E A cos θ = E ℓ w cos θ = (21000 N / C) (6 . 8 m) (3 . 8 m) cos 15 ◦ = 524150 N · m 2 / C . Question 2, chap 24, sect 2. part 1 of 1 10 points A cubic box of side a , oriented as shown, contains an unknown charge. The vertically directed electric field has a uniform magni tude E at the top surface and 2 E at the bottom surface. a E 2 E How much charge Q is inside the box? 1. Q encl = 1 2 ǫ E a 2 2. Q encl = 0 3. Q encl = ǫ E a 2 correct 4. Q encl = 2 ǫ E a 2 5. Q encl = 3 E ǫ a 2 6. Q encl = 3 ǫ E a 2 7. Q encl = 2 E ǫ a 2 8. Q encl = 6 ǫ E a 2 9. Q encl = E ǫ a 2 10. insufficient information Explanation: By convention, electric flux through a sur face S is positive for electric field lines going out of the surface S and negative for lines going in. Here the surface is a cube and no flux passes through the vertical sides. The top receives Φ top = − E a 2 (inward is negative) and the bottom Φ bottom = 2 E a 2 , so the total electric flux is Φ E = − E a 2 + 2 E a 2 = E a 2 . homework 02 – LIN, KEVIN – Due: Jan 27 2008, 4:00 am 2 Using Gauss’ Law, the charge inside the box is Q encl = ǫ Φ E = ǫ E a 2 . Question 3, chap 24, sect 2. part 1 of 1 10 points A charge of 18 . 3 μ C is at the geometric center of a cube. What is the electric flux through one of the faces? The permittivity of free space is 8 . 85419 × 10 − 12 C 2 / N · m 2 . Correct answer: 344470 N · m 2 / C (tolerance ± 1 %). Explanation: Let : q = 18 . 3 μ C = 1 . 83 × 10 − 5 C and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 . By Gauss’ law, Φ = contintegraldisplay vector E · d vector A = q ǫ . The total flux through the cube is given by Φ tot = q ǫ = 1 . 83 × 10 − 5 C 8 . 85419 × 10 − 12 C 2 / N · m 2 = 2 . 06682 × 10 6 N · m 2 / C , so the flux through one side of the cube is Φ = 1 6 Φ tot = 344470 N · m 2 / C . Question 4, chap 24, sect 2. part 1 of 1 10 points A point charge of 9 . 4 μ C is located at the center of a uniform ring having linear charge density 14 . 6 μ C / m and radius 3 . 38 m....
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This note was uploaded on 03/03/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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