Gauss's Law 1-solutions

Gauss's Law 1-solutions - gilvin(jg47854 5 Gausss Law 1...

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? = E E A = (a?? + b?) E A? gilvin (jg47854) - 5. Gauss’s Law 1 - meyers - (21235) This print-out should have 21 questions. y y E Multiple-choice questions may continue on the next column or page - find all choices 1 y E E before answering. This homework is due Tuesday, February 8, at midnight Tucson time. 001 (part 1 of 3) 10.0 points A uniform electric field E = a ˆı + b ĵ, inter- sects a surface of area A. What is the flux through this area if the surface lies in the yz plane? 1. (a − b) A 2. (b − a) A 3. A a 2 + b 2 4. a A correct 5. b A 6. (a + b) A 7.0 Explanation: Basic Concepts: Φ = E · dA Surface For a uniform electric field and a flat sur- face, this simplifies to Φ = E · A . Solution: The electric flux through a surface is given by Φ = E · A , where E is the electric field and A is the vector which is directed perpendicular to the surface and has magnitude equal to the area of the surface. When the surface is in the yz plane, it is perpendicular to the x-direction. Therefore, the resulting flux is due to the x-component of the electric field only. This can also be written as ı = a A . x x x z z z fig 1 fig 2 fig 3 002 (part 2 of 3) 10.0 points What is the flux if the surface lies in the xz plane? 1.0 2. (a + b) A 3. (a − b) A 4. (b − a) A 5. b A correct 6. A a 2 + b 2 7. a A Explanation: When the surface is in the xz plane, it is perpendicular to the y-direction. Therefore Φ = (aˆı + bĵ) · Aĵ = b A . 003 (part 3 of 3) 10.0 points What is the flux if the surface lies in the xy plane? 1. A a 2 + b 2 2. (a + b) A 3. b A 4. a A 5.0 correct 6. (b − a) A 7. (a − b) A

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gilvin (jg47854) - 5. Gauss’s Law 1 - meyers - (21235) 2 Explanation: When the surface is in the xy plane, it E is perpendicular to the z-direction. There is no component of the electric field in the z- direction, so there is no component of the elec- tric field perpendicular to the surface. This a means that the electric flux through the sur- face is zero. Mathematically, this is written 2 E Φ = (aˆı + bĵ) · Ak How much charge Q is inside the box? = 0 . 1. Q encl = 3 ϵ 0 E a 2 004 10.0 points A 34 cm diameter circular loop is rotated in a uniform electric field until the po- sition of maximum electric flux is found. The flux in this position is measured to be 6.2 × 10 5 N · m 2 /C. What is the electric field strength? Correct answer: 6.82879 × 10 6 N/C. Explanation: Let : r = 17 cm = 0.17 m and Φ max = 6.2 × 10 5 N · m 2 /C . Flux is Φ = E A cos θ . θ = 0 for maximum flux, so Φ max = E A E = Φ max π r 2 = 6.2 × 10 5 N · m 2 /C π (0.17 m) 2 = 6.82879 × 10 6 N/C . 005
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Gauss's Law 1-solutions - gilvin(jg47854 5 Gausss Law 1...

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