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exam1solution - Name Instructor Louisiana State University...

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Name: _______________________ Instructor:__________________ Louisiana State University Physics 2102, Exam 1, February 24, 2011. Please be sure to write your name and class instructor above. The test consists of 3 questions (multiple choice), and 3 problems (numerical). All numerical quantities must have appropriate units. Points will be deducted if units are absent. For the problems: Show your reasoning and your work – no credit will be given for an answer without explanation or work . Note that in many of the problems, you can do parts (b) or (c) even if you get stuck on (a) or (b). You may use scientific or graphing calculators. Cell phones cannot be used as calculators. Feel free to detach, use, and keep the formula sheet pages. No other reference material is allowed during the exam. Good Luck!
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Question 1 [6 points] Two point charges are fixed in place on an x -axis, separated by a distance L as shown in the figure. The charges have equal magnitudes and opposite signs, q 1 = +Q , q 2 = -Q . For a net force of zero to be exerted on a third charge it must be placed: (a) To the left of +Q (b) Exactly half way between +Q and –Q ( x = L/2, y = 0 ) (c) Half way between +Q and –Q, but not on the x-axis ( x = L/2, y 0 ) (d) To the right of –Q (e) There is no place where the net force is zero Since the charges in place have equal magnitude but opposite sign (they are a dipole), the electric field will be non zero everywhere (except at infinitely far away). To the left of +Q, it will point towards the left; in between the charges and to the right of –Q it will point towards the right. On the y-axis, the electric field will point to the right.
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Question 2 [12 points] The figure shows three different orientation of an electric dipole (with dipole moment p ) in a uniform electric field E , which points in the positive x-direction. p (1) E p (2) E (3) E p (a) [6 pts] Rank the scenarios according to the potential energy of the dipole in the electric field, greatest first (circle the right answer): U 1 > U 2 > U 3 U 2 > U 1 > U 3 U 2 > U 1 = U 3 U 1 > U 2 = U 3 U 1 = U 2 = U 3 The potential energy of a dipole in an electric field is U = p E = pE cos θ , so U 1 = 0, U 2 > 0 (largest), and U 3 = pE < 0 (smallest).
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