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Spring 11 Phys 121 - HW 03

Spring 11 Phys 121 - HW 03 - dzifanu(sd26397 Hw 3...

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dzifanu (sd26397) – Hw 3 – opyrchal – (121102) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of3)10.0points An electric field of magnitude 2170 N / C is applied along the x axis. Calculate the electric flux through a rect- angular plane 0 . 451 m wide and 0 . 7 m long if the plane is parallel to the yz plane. Correct answer: 685 . 069 N · m 2 / C. Explanation: Let : E = 2170 N / C , θ = 0 , L = 0 . 7 m , and W = 0 . 451 m . The area of the rectangle is A = L W = (0 . 7 m) (0 . 451 m) = 0 . 3157 m 2 . Since the field is uniform,the flux is φ = E · A = E A cos θ = (2170 N / C) (0 . 3157 m 2 ) (cos 0 ) = 685 . 069 N · m 2 / C . 002(part2of3)10.0points Calculate the electric flux through the same rectangle, if it is parallel to the xy plane. Correct answer: 0 N · m 2 / C. Explanation: The angle is θ = 90 ,cos θ = 0 , and φ = E · A = E A cos θ = (2170 N / C) (0 . 3157 m 2 ) (0) = 0 N · m 2 / C . 003(part3of3)10.0points Calculate the electric flux through the same rectangle, but now the rectangle contains the y axis and its normal makes an angle of 43 with the x axis. Correct answer: 501 . 028 N · m 2 / C. Explanation: Let θ = 43 φ = E · A = E A cos θ = (2170 N / C) (0 . 3157 m 2 ) (cos 43 ) = 501 . 028 N · m 2 / C . 004 10.0points A closed surface with dimensions a = b = 0 . 522 m and c = 0 . 783 m is located as in the figure. The electric field throughout the region is nonuniform and given by vector E = ( α + β x 2 ı where x is in meters, α = 2 N / C, and β = 3 N / (C m 2 ). E y x z a c b a What is the magnitude of the net charge enclosed by the surface? Correct answer: 1 . 03541 × 10 11 C. Explanation: Let : a = b = 0 . 522 m , c = 0 . 783 m , α = 2 N / C , and β = 3 N / (C m 2 ) . The electric field throughout the region is directed along the x -axis and the direction of
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dzifanu (sd26397) – Hw 3 – opyrchal – (121102) 2 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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