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Unformatted text preview: homework 02 SMITH, TAYLOR 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 27000 N / C and directed upward (perpendicular to the Earths surface) exists on a day when a thun derstorm is brewing. A truck that can be approximated as a rectangle 6 . 2 m by 3 . 8 m is traveling along a road that is inclined 12 relative to the ground. Determine the electric flux through the bot tom of the truck. Correct answer: 622219 N m 2 / C (tolerance 1 %). Explanation: Let : E = 27000 N / C , = 6 . 2 m , w = 3 . 8 m , and = 12 . By Gauss law, = vector E vector A. The flux through the bottom of the car is = E A cos = E w cos = (27000 N / C) (6 . 2 m) (3 . 8 m) cos12 = 622219 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 = 2 E a 2 2. Q encl = 3 E a 2 3. Q encl = 2 E a 2 4. Q encl = 3 E a 2 5. Q encl = E a 2 6. Q encl = 0 7. insufficient information 8. Q encl = E a 2 correct 9. Q encl = 1 2 E a 2 10. Q encl = 6 E a 2 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 top = E a 2 (inward is negative) and the bottom bottom = 2 E a 2 , so the total electric flux is E = E a 2 + 2 E a 2 = E a 2 . homework 02 SMITH, TAYLOR Due: Jan 27 2008, 4:00 am 2 Using Gauss Law, the charge inside the box is Q encl = E = E a 2 . Question 3, chap 24, sect 2. part 1 of 1 10 points A charge of 9 . 1 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: 171294 N m 2 / C (tolerance 1 %). Explanation: Let : q = 9 . 1 C = 9 . 1 10 6 C and = 8 . 85419 10 12 C 2 / N m 2 . By Gauss law, = contintegraldisplay vector E d vector A = q . The total flux through the cube is given by tot = q = 9 . 1 10 6 C 8 . 85419 10 12 C 2 / N m 2 = 1 . 02776 10 6 N m 2 / C , so the flux through one side of the cube is = 1 6 tot = 171294 N m 2 / C . Question 4, chap 24, sect 2. part 1 of 1 10 points A point charge of 8 . 6 C is located at the center of a uniform ring having linear charge density 3 . 1 C / m and radius 4 . 01 m....
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This homework help was uploaded on 04/10/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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