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Unformatted text preview: Kapoor (mk9499) oldhomework 03 Turner (60230) 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 4) 10.0 points A uniformly charged circular arc AB of radius R is shown in the figure. It covers a quarter of a circle and it is located in the second quadrant. The total charge on the arc is Q > 0. x y + + + + + + + + + R x y I II III IV B A O s R The direction of the electric field E due to the charge distribution at the origin is 1. along the negative xaxis. 2. in quadrant IV. correct 3. along the negative yaxis. 4. along the positive yaxis. 5. in quadrant I. 6. in quadrant II. 7. in quadrant III. 8. along the positive xaxis. Explanation: The electric field for a positive charge is directed away from it. In this case, the electric field generated by each q is going to be in quadrant IV, so the total electric field will be in the same quadrant. 002 (part 2 of 4) 10.0 points Determine E x , the xcomponent of the elec tric field vector at the origin O due to the charge element q located at an angle sub tended by an angular interval . 1. E x = k Q R sin 2. E x = k Q R 2 2 sin 3. E x = k Q R 2 2 cos correct 4. E x = k Q R 2 2 cos 5. E x = k Q 2 R 2 cos 6. E x = k Q R 2 cos 7. E x = 0 8. E x = k Q R 2 sin 9. E x = k Q 2 R cos 10. E x = k Q 2 R 2 2 cos Explanation: E = k q R 2 . The charge q per unit arclength s is Q 2 R = 2 Q R , so q = s = R = 2 Q R R = 2 Q . On the other hand, the vector vector E makes an angle with the xaxis, so E x = k Q R 2 2 cos . 003 (part 3 of 4) 10.0 points Find E x , the electric field at the origin due Kapoor (mk9499) oldhomework 03 Turner (60230) 2 to the full arc length for the case where Q = 3 . 7 C and R = 1 . 74 m . Correct answer: 6992 . 38 N / C. Explanation: Let : Q = 3 . 7 C and R = 1 . 74 m . E x can be found by integrating the contribu tions of all the q s in the arc. We get E x = integraldisplay / 2 k Q R 2 2 cos d = 2 k Q R 2 = 2 k (3 . 7 C) (1 . 74 m) 2 = 6992 . 38 N / C . 004 (part 4 of 4) 10.0 points The charge on an electron is e . If at O we have E x , then the magnitude of the force bardbl vector F e bardbl on an electron at this point is 1. bardbl vector F e bardbl = E x 2 e . 2. bardbl vector F e bardbl = 4 eE x . 3. bardbl vector F e bardbl = eE x . 4. bardbl vector F e bardbl = 2 E x 3 e . 5. bardbl vector F e bardbl = 3 eE x 2 ....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Charge, Work

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