hw2 solutions - homework 02 SMITH, TAYLOR Due: Jan 27 2008,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 13

hw2 solutions - homework 02 SMITH, TAYLOR Due: Jan 27 2008,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online