5. Gauss's Law 1-solutions

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

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Unformatted text preview: gilvin (jg47854) 5. Gausss Law 1 meyers (21235) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices 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 vector E = a + b , 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 radicalbig a 2 + b 2 4. a A correct 5. b A 6. ( a + b ) A 7. Explanation: Basic Concepts: = contintegraldisplay Surface vector E d vector A For a uniform electric field and a flat sur- face, this simplifies to = vector E vector A . Solution: The electric flux through a surface is given by = vector E vector A , where vector E is the electric field and vector 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 = vector E vector A = ( a + b ) A = a A . E E E fig 1 fig 2 fig 3 x y z y y x x z z 002 (part 2 of 3) 10.0 points What is the flux if the surface lies in the xz plane? 1. 2. ( a + b ) A 3. ( a b ) A 4. ( b a ) A 5. b A correct 6. A radicalbig a 2 + b 2 7. a A Explanation: When the surface is in the xz plane, it is perpendicular to the y-direction. Therefore = ( a + b ) A = b A . 003 (part 3 of 3) 10.0 points What is the flux if the surface lies in the xy plane? 1. A radicalbig a 2 + b 2 2. ( a + b ) A 3. b A 4. a A 5. correct 6. ( b a ) A 7. ( a b ) A gilvin (jg47854) 5. Gausss Law 1 meyers (21235) 2 Explanation: When the surface is in the xy plane, it 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 means that the electric flux through the sur- face is zero. Mathematically, this is written = ( a + b ) A k = . 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 max = 6 . 2 10 5 N m 2 / C . Flux is = E A cos ....
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This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

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

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