HW #2-solutions - nanni(arn437 – HW#2 – Erskine...

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Unformatted text preview: nanni (arn437) – HW #2 – Erskine – (56905) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A (8 . 82 m by 8 . 82 m) square base pyramid with height of 5 . 91 m is placed in a vertical electric field of 51 N / C. b 8 . 82 m 5 . 91 m 51 N / C Find the total electric flux which goes out through the pyramid’s four slanted surfaces. Correct answer: 3967 . 41 N · m 2 / C. Explanation: Let : s = 8 . 82 m , h = 5 . 91 m , and E = 51 N / C . By Gauss’ law, Φ = vector E · vector A . Since there is no charge contained in the pyramid, the net flux through the pyramid must be 0 N/C. Since the field is vertical, the flux through the base of the pyramid is equal and opposite to the flux through the four sides, so Φ = E A = E s 2 = (51 N / C) (8 . 82 m) 2 = 3967 . 41 N · m 2 / C . 002 10.0 points A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radius r with r < R is used to calculate the magnitude of the electric field E at a distance r from the center of the sphere. Which of the following equations results from a correct application of Gauss’s law for this situation? 1. E (4 π r 2 ) = 0 2. E (4 π r 2 ) = Q ǫ 3 r 3 4 π R 3. E (4 π r 2 ) = Q ǫ 4. E (4 π R 2 ) = Q ǫ 5. E (4 π r 2 ) = Q ǫ r 3 R 3 correct Explanation: Applying Gauss’s law, contintegraldisplay S vector E · vectorndA = Q inside ǫ ; i . e ., contintegraldisplay S E dA = Q inside ǫ . Because charge Q is uniformly distributed, so Q inside is just Q V olume ( r ) V olume ( total ) = Q r 3 R 3 , Where V olume ( r ) = 4 π 3 r 3 , and V olume ( total ) = 4 π 3 R 3 . So Gauss’s law gives E (4 π r 2 ) = Q ǫ r 3 R 3 . 003 10.0 points A closed surface with dimensions a = b = . 51 m and c = 0 . 663 m is located as in the figure. The electric field throughout the region is nonuniform and given by vector E = ( α + β x 2 )ˆ ı where x is in meters, α = 3 N / C, and β = 2 N / (C m 2 ). nanni (arn437) – HW #2 – Erskine – (56905) 2 E y x z a c b a What is the magnitude of the net charge enclosed by the surface? Correct answer: 5 . 13945 × 10- 12 C. Explanation: Let : a = b = 0 . 51 m , c = 0 . 663 m , α = 3 N / C , and β = 2 N / (C m 2 ) . The electric field throughout the region is directed along the x-axis and the direction of d vector A is perpendicular to its surface. Therefore, vector E is parallel to d vector A over the four faces of the surface which are perpendicular to the yz plane, and vector E is perpendicular to d vector A over the two faces which are parallel to the yz plane. That is, only the left and right sides of the right rectangular parallel piped which encloses the charge will contribute to the flux....
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This note was uploaded on 09/19/2011 for the course PHY 317 taught by Professor Gositz during the Fall '08 term at University of Texas.

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HW #2-solutions - nanni(arn437 – HW#2 – Erskine...

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