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hw2 solutions

# hw2 solutions - homework 02 SMITH TAYLOR Due 4:00 am...

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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 Earth’s 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 θ = 12 . 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 = (27000 N / C) (6 . 2 m) (3 . 8 m) cos 12 = 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 ǫ 0 a 2 2. Q encl = 3 E ǫ 0 a 2 3. Q encl = 2 ǫ 0 E a 2 4. Q encl = 3 ǫ 0 E a 2 5. Q encl = E ǫ 0 a 2 6. Q encl = 0 7. insufficient information 8. Q encl = ǫ 0 E a 2 correct 9. Q encl = 1 2 ǫ 0 E a 2 10. Q encl = 6 ǫ 0 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 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 .

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homework 02 – SMITH, TAYLOR – Due: Jan 27 2008, 4:00 am 2 Using Gauss’ Law, the charge inside the box is Q encl = ǫ 0 Φ E = ǫ 0 E a 2 . Question 3, chap 24, sect 2. part 1 of 1 10 points A charge of 9 . 1 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: 171294 N · m 2 / C (tolerance ± 1 %). Explanation: Let : q = 9 . 1 μ C = 9 . 1 × 10 6 C and ǫ 0 = 8 . 85419 × 10 12 C 2 / N · m 2 . By Gauss’ law, Φ = contintegraldisplay vector E · d vector A = q ǫ 0 . The total flux through the cube is given by Φ tot = q ǫ 0 = 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 Φ = 1 6 Φ tot = 171294 N · m 2 / C . Question 4, chap 24, sect 2. part 1 of 1 10 points A point charge of 8 . 6 μ C is located at the center of a uniform ring having linear charge density 3 . 1 μ C / m and radius 4 . 01 m. q R a λ Find the total electric flux through a sphere centered at the point charge and having radius R < a as shown. The premittivity of free space is 8 . 85419 × 10 12 C 2 / N m 2 . Correct answer: 971292 N m 2 / C (tolerance ± 1 %). Explanation: Let : q = 8 . 6 μ C = 8 . 6 × 10 6 C , λ = 3 . 1 μ C / m = 3 . 1 × 10 6 C / m , ǫ 0 = 8 . 85419 × 10 12 C 2 / N m 2 , and a = 4 . 01 m .
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hw2 solutions - homework 02 SMITH TAYLOR Due 4:00 am...

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