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Unformatted text preview: oliver (jmo673) HW02 Tsoi (59090) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The diagram shows an isolated, positive charge Q , where point B is twice as far away from Q as point A . + Q A B 10 cm 20 cm The ratio of the electric field strength at point A to the electric field strength at point B is 1. E A E B = 4 1 . correct 2. E A E B = 1 1 . 3. E A E B = 8 1 . 4. E A E B = 1 2 . 5. E A E B = 2 1 . Explanation: Let : r B = 2 r A . The electric field strength E 1 r 2 , so E A E B = 1 r 2 A 1 r 2 B = r 2 B r 2 A = (2 r ) 2 r 2 = 4 . 002 (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. None of these 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 ) x 2 dx 4. integraldisplay x 2 x 1 k e Q ( x 2 x 1 ) x 2 dx 5. integraldisplay x 2 x 1 k e Q ( x 2 x 1 )( x x ) 2 dx correct 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 . 003 (part 2 of 2) 10.0 points If x 1 = 0 m, x 2 = 4 . 48 m and the charge Q = 3 . 07 C, what is the magnitude E of the electric field at x = 9 . 9 m? Correct answer: 514 . 216 N / C. Explanation: Let : k e = 8 . 98755 10 9 N m 2 / C 2 , x 1 = 0 m , x 2 = 4 . 48 m , x = 9 . 9 m , and Q = 3 . 07 C . oliver (jmo673) HW02 Tsoi (59090) 2 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 ) (4 . 48 m 0 m) (3 . 07 10 6 C) bracketleftbigg 1 9 . 9 m 4 . 48 m 1 9 . 9 m 0 m bracketrightbigg = 514 . 216 N / C . 004 (part 1 of 3) 10.0 points Consider the setup shown in the figure be low, where the arc is a semicircle with radius r . The total charge Q is negative, and dis tributed uniformly on the semicircle. The charge on a small segment with angle is labeled q . x y r x y I II III IV B A O q is given by 1. q = Q 2. q = Q 2 3. q = Q 4. q = Q 2 5. q = 2 Q 6. None of these 7. q = 2 Q 8. q = Q correct 9. q = 2 Q 10. q = Q Explanation: The angle of a semicircle is , thus the charge on a small segment with angle is...
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 Fall '09
 TSOI
 Physics, Charge

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