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Unformatted text preview: mcconnell (kam2342) homework04 Turner (60230) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A charge Q is distributed uniformly along the x axis from x 1 to x 2 . The Coulomb constant is 8 . 98755 10 9 N m 2 / C 2 Which of the following integrals is correct for the magnitude of the electric field at x on the x axis? Assume that x > x 2 > x 1 and k e = 1 4 . 1. integraldisplay x 2 x 1 k e Q ( x 2 x 1 )( x x ) 2 dx correct 2. integraldisplay x 2 x 1 k e Q ( x 2 x )( x x ) 2 dx 3. integraldisplay x 2 x 1 k e Q ( x 2 x 1 ) x 2 dx 4. None of these 5. integraldisplay x 2 x 1 k e Q ( x 2 x ) x 2 dx Explanation: For a continuous charge distribution, dE = k e dq r 2 . Recall that for a uniform linear charge dis tribution, dq = dx = Q L dx = Q x 2 x 1 dx . Furthermore, the point x is a distance ( x x ) from a charge element dq . Hence integrating from x = x 1 to x = x 2 yields E = integraldisplay x 2 x 1 k e Q ( x 2 x 1 )( x x ) 2 dx . 002 (part 2 of 2) 10.0 points If x 1 = 0 m, x 2 = 3 . 36 m and the charge Q = 2 . 44 C, what is the magnitude E of the electric field at x = 10 . 7 m? Correct answer: 279 . 223 N / C. Explanation: Let : k e = 8 . 98755 10 9 N m 2 / C 2 , x 1 = 0 m , x 2 = 3 . 36 m , x = 10 . 7 m , and Q = 2 . 44 C . E = integraldisplay x 2 x 1 k e Q ( x 2 x 1 )( x x ) 2 dx = k e Q ( x 2 x 1 )( x x ) vextendsingle vextendsingle vextendsingle vextendsingle x 2 x 1 = ( 8 . 98755 10 9 N m 2 / C 2 ) (3 . 36 m 0 m) (2 . 44 10 6 C) bracketleftbigg 1 10 . 7 m 3 . 36 m 1 10 . 7 m 0 m bracketrightbigg = 279 . 223 N / C . 003 (part 1 of 4) 10.0 points Consider a disk of radius 3 cm, having a uni formly distributed charge of +5 C....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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