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Unformatted text preview: segovia (cs39966) – Homework 2 – Spurlock – (14441) 1 This printout should have 58 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charge of 5 . 6 μ C is at the geometric center of a cube. What is the electric flux through one of the faces? Correct answer: 1 . 05412 × 10 5 N · m 2 / C. Explanation: Let : q = 5 . 6 μ C = 5 . 6 × 10 − 6 C . By Gauss’ law, Φ = contintegraldisplay vector E · d vector A = q ǫ The total flux through the cube is given by Φ tot = q ǫ = 5 . 6 × 10 − 6 C 8 . 85419 × 10 − 12 C 2 / N · m 2 = 6 . 32469 × 10 5 N · m 2 / C . The flux through one side of the cube is then Φ = 1 6 Φ tot = 1 . 05412 × 10 5 N · m 2 / C . 002 10.0 points A closed surface with dimensions a = b = . 269 m and c = 0 . 2421 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 = 4 N / C, and β = 6 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 . 26022 × 10 − 13 C. Explanation: Let : a = b = 0 . 269 m , c = 0 . 2421 m , α = 4 N / C , and β = 6 N / (C m 2 ) . The electric field throughout the region is directed along the xaxis 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. The net electric flux through the cube is ΔΦ = integraldisplay right side E x dA ⊥ integraldisplay left side E x dA ⊥ = a b bracketleftbig α + β ( a + c ) 2 α β a 2 bracketrightbig = a b β (2 a c + c 2 ) = a b cβ (2 a + c ) = (0 . 269 m) (0 . 269 m) (0 . 2421 m) × [6 N / (C m 2 )] [2 (0 . 269 m) + 0 . 2421 m] = 0 . 0819975 N m 2 / C , so the enclosed charge is q = ǫ ΔΦ = [8 . 85419 × 10 − 12 C 2 / (N m 2 )] segovia (cs39966) – Homework 2 – Spurlock – (14441) 2 × (0 . 0819975 N m 2 / C) = 7 . 26022 × 10 − 13 C . 003 10.0 points Consider a long, uniformly charged, cylindri cal insulator of radius R with charge density 1 . 6 μ C / m 3 . (The volume of a cylinder with radius r and length ℓ is V = π r 2 ℓ .) The value of the Permittivity of free space is 8 . 85419 × 10 − 12 C 2 / N · m 2 R 2 . 8 cm What is the magnitude of the electric field inside the insulator at a distance 2 . 8 cm from the axis (2 . 8 cm < R )?...
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This note was uploaded on 11/10/2011 for the course PHYSICS 303L taught by Professor Hoffman during the Fall '11 term at University of Texas.
 Fall '11
 HOFFMAN
 Charge, Work

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