{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

5. Gauss's Law 1-solutions

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

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

gilvin (jg47854) – 5. Gauss’s 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(part1of3)10.0points A uniform electric field vector E = a ˆ ı + b ˆ , 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. 0 Explanation: BasicConcepts: Φ = contintegraldisplay Surface vector E · d vector A For a uniform electric field and a flat sur- face, this simplifies to Φ = vector E · vector A . Solution: The electric flux through a surface is given by Φ = 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 Φ = vector E · vector A = ( a ˆ ı + b ˆ ) · A ˆ ı = a A . E E E fig 1 fig 2 fig 3 x y z y y x x z z 002(part2of3)10.0points 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 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 Φ = ( a ˆ ı + b ˆ ) · A ˆ = b A . 003(part3of3)10.0points 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. 0 correct 6. ( b a ) A 7. ( a b ) A

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

View Full Document
gilvin (jg47854) – 5. Gauss’s 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 Φ = ( a ˆ ı + b ˆ ) · A ˆ k = 0 . 004 10.0points 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 10.0points A cubic box of side a , oriented as shown, con- tains an unknown charge. The vertically di- rected electric field has a uniform magnitude E at the top surface and 2 E at the bottom surface.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

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

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

View Full Document
Ask a homework question - tutors are online