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Unformatted text preview: billing (cab4763) – hw 3 – opyrchal – (121104) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points An electric field of magnitude 2210 N / C is applied along the x axis. Calculate the electric flux through a rect angular plane 0 . 241 m wide and 0 . 84 m long if the plane is parallel to the yz plane. Correct answer: 447 . 392 N · m 2 / C. Explanation: Let : E = 2210 N / C , θ = 0 ◦ , L = 0 . 84 m , and W = 0 . 241 m . The area of the rectangle is A = L W = (0 . 84 m) (0 . 241 m) = 0 . 20244 m 2 . Since the field is uniform,the flux is φ = E · A = E A cos θ = (2210 N / C) (0 . 20244 m 2 ) (cos0 ◦ ) = 447 . 392 N · m 2 / C . 002 (part 2 of 3) 10.0 points 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 = (2210 N / C) (0 . 20244 m 2 ) (0) = 0 N · m 2 / C . 003 (part 3 of 3) 10.0 points Calculate the electric flux through the same rectangle, but now the rectangle contains the y axis and its normal makes an angle of 63 ◦ with the x axis. Correct answer: 203 . 112 N · m 2 / C. Explanation: Let θ = 63 ◦ φ = E · A = E A cos θ = (2210 N / C) (0 . 20244 m 2 ) (cos63 ◦ ) = 203 . 112 N · m 2 / C . 004 10.0 points A closed surface with dimensions a = b = . 578 m and c = 0 . 5202 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 = 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: 7 . 73788 × 10 − 12 C. Explanation: Let : a = b = 0 . 578 m , c = 0 . 5202 m , α = 3 N / C , and β = 3 N / (C m 2 ) . The electric field throughout the region is directed along the xaxis and the direction of billing (cab4763) – hw 3 – opyrchal – (121104) 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...
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This note was uploaded on 03/25/2011 for the course PHYSICS 121 taught by Professor Fayngold during the Spring '10 term at NJIT.
 Spring '10
 Fayngold
 Physics

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