Exam 1 Solution Sheet - EXAMI GENERAL PHYSICS 11 SPRING...

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Unformatted text preview: EXAMI GENERAL PHYSICS 11 SPRING QUARTER 2008 April 25, 2003 Name F ) Vi Student ID # Lecture hstructor Section Number Full credit will only be awarded to answers with all Work shown. Mango) = 9 x 109 N am:2 so = 8.85 x 10''2 (3%: m2 e=1fixlfimc llIJ PROBLEM SCORE 1. r 25' 2. x 25 3. z 25 TOTAL SCORE ! 100 4. z 25 W PROBLEMS ZZZ?” A small, nonccnducting ball of mass n1 = 1.9 mg and charge q = 2.0 it 10‘s C (distributed mafcrnfly through its volume) hangs from an insulating thread that makes an angle 0 = 30° with a vertical uniformly charged nonconducting chant (shown in cross Section). The ball my be treated as a point charge. Considering the gravitational force acting on the hall and that the sheet is an infinite plane, (a) calculate the magnitude and give the direction of the electric field that causes" 1 ' swing out by an angle 3. 1— ._.. 2a 25., E = LlIf? $5 0 " . r 2. Consider a mass attached to a spring moving on a horizontal, fi‘iction-free table. At a certain moment, t= 0, the mass is observed a distance L = 4.5 cm to the left of the position of equilibrium, and it is found mat the mass is moving towards the position of equilibrium with velocity Va = 2.0 mfsec. The spring constant is k = 250 me and the mass is 0.10 {a} Wl‘lat is the angular frequency {a ofthie oscillator? F Jesse/H M "7 Er; 6?.an I @ w =-* 5w?) E (b) How many times, N, does the mass pass though the position of equilibrium per second? F:;V;r:—Ef——/-—irlfljé, 2:11 at: 5 L“ Plffié’fi- Titangl. 1L2 €§Uilr Fa‘fifll'fla-l To; 3:31; £ng 1‘ 4w? “5"” ' T (c) What is the amplitude of the oscill ifth Janet- (911 913?? 1 a”: we a. Me A '5' fl HEHYJPM :‘(flfléfl by a?” U W" Ch . ms? (d) What is the highest speed achieved by the mass (1 g X: Arwsf’wf-tcfi _ g9 .42” := "AkflfiiNanfibD m t w ~=(®¢~>Ie/fil~? 2: 300 and; =- J :5 Hffi _ Hid-"FF.- (e) At a certain time, the kinetic energy of the mass is lightly one-third of the potential energy of the system. How far is the mass displaced from equilibrimn? m M, ‘l'i‘flua an”: mam} ' 1 u (a 1‘: LE 5&0”; 4/912: mitt") all}: i “L “V” - E: WA :: #5219095- «59 It : x= "—1 :— 55}: CM___.J 6 3:, ‘5 a i n if; P 3. A very leng sulid insulated cylinder nfradius R1 with a uniform linear charge density per unit length 2”! cf +3}. is surreunded by a concentric metal conducting cylindrical tube of inner radius R; and outer radius R3, as shown in the diagram. The metal tube has a charge per unit length uf Jl placed on it. Determine the vector electric field E(r) as a fimctien of distance r from the center of the cylinder fer: Direction; + r ._ jib..— N‘ “HY “at? E —- mag in elm? -—t‘ since he; cm!“ _ v (e) What are the linear charge densities (in terms of it.) an the inner an uter surfaces at the conducting tube? filhct 5‘6? £71,934: WeJav [7 €Y2Name Student ID Score (Zr/i last first I, [20 points total] A pointcharge +Q is placed a distance s from point P. as shown at Em A ' right. This situation will be referred to as case A. A. In this part you will be asked to determine how the electric field at point .P +0 P changes as various changes are made to the setup. i. [div-pro} in case B. a second charge +9 is placed a distance s above point P. as shown at right. Is the magnitude ofthe electric field at point P in case B greater than. less than. or equal to that in case A? Explain. r \ At point .P. the electric field due to the original charge points to the right. while that one to the second charge points down. Hy superposition. the net 5 electric field is the vector sum of the fields due to the individual charges. the magnitude of which is given in this case by the Pyhagorean theorem for by ‘. geometrical construction ) and is mm the magnitude of the netfield in case A. ii. [firm] In case C. the original point charge is replaced by a uniformly charged rod with net charge +Q. as shown at righL Is the magnitude of the electric field at point P in case C greater than. less than. or equal to that in case B? Explain. Tofind the electric field due to the rod, imagine breaking it up into very " short sections. The field due to the whole rod is the vector sum of the fields due to the small sections. The charges on the rod are farther from point P ' than was the original point charge. so the magnitude of the field of each piece is less than that ofa piece with the same charge at the center of the rod. Furthermore. the vertical components of the fields due to charges above the center cancel those from charges below the center. leaving only,' the horizontal components to contribute to the field due to the rod. Both of these efi'ects tend to reduce the horizontal congranent of the net electric field at P without afiecting its vertical component. Therefore. the magnitude of the net field at P is W that in case 3. iii. [ll-par] In case D. another point charge +Q is placed a distances cm D +9 to the left of the rod. as shown at right. Is the magnitude ofthe electric field at point P in case D greater than. less than. or equal to that in case C? Explain. +9 The net electric field in this case is the vector sum of the net | + electric field in case B and the field due to the new point charge. 9 | At point P. thcficld clue to the new paint charge is in the same l ' 3 3 P direction as the field due to the rod. The horizontal component i of the net field is therq’ore greater than in case C . while the vertical component is the same. So the magnitude of the net electric field is Wuhan that in case C. B. [said In case B. the charges added in part A are removed. and a thin neutral Case 5 metal rod is placed between the original point charge and point P. as shown at ri ght. [s the magnitude of the electric field at point P in this case greater than. F—“ '-—"i I:- less than. or equal to that in case A? Explain. Since the charges in the rod are free to move. the point charge will induce a charge separation in the rod. with positive charge at the right end. and the same amount of negative charge at the lq‘t. (The rod is neutral. J Since the positive charges are closer to point P than the negative charges. they will produce a strangerfield there. so the net field due to the rod at point P will be to the right. The field due to the point charge has not changed. The net field is then the vector sum of the field due the rod and the net field in case A; since they are in the sense direction. the net field at P is mention that in case it. 122C. Winter 2008 Exam 2 EM-UWARZCWIT-EZICJ-lfifiFFLsoldoc ...
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This note was uploaded on 10/12/2009 for the course PHYS physics 2 taught by Professor Kogan during the Spring '08 term at University of Cincinnati.

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Exam 1 Solution Sheet - EXAMI GENERAL PHYSICS 11 SPRING...

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