HT1_Sample_Test_Solution

HT1_Sample_Test_Solution - Department of Physics, Lehigh...

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Unformatted text preview: Department of Physics, Lehigh University Physics 21 — Introductory Physics II _ Spring 2010 Hour Exam | Closed Notes Student’s Name i February 16, 20/10/ 9:20 AM—10:10 Al‘s/l Recitation Section Number “Li g “5 Recitation Leaders Name Dig} (@SdéhSfi-{t' (a? The test is a multiple-choice examination. First, check to be sure that you have a six—page examination including the equation sheet. Please answer each question by writing the letter corresponding to your choice in the answer area. Problem 1. Consider three particles with a charge q=0.50 uC on the any plane as shown in the figure. The particles are located on the .9 4 3 vertices of an equilateral triangle of side a=l .0cm. 1. What is the magnitude of the net force on particle 3‘? (a) .0044N (b) 0.0039N (c) 39N Answer: C (d) 45 N 2. Which unit vector bests describes the direction of the net force on partiCle 3? (a) “'3 (b)i (on (d) -i ' Amen; 3. What is the magnitude of the electric field at the origin? (a) are 1.05“; (b) 5.2 "x 105g _ (c) 4.5 x 107'? (d) 0 Answer: A 4. Which unit vector bests describes the direction of the electric field at the origin? (a) i +3 (13) i (on (d) —3 Answer: 0 5. What would be the total flux through a Gaussian surface that contained these three point charges? a) if I b) i: c) E d) 0 Answer:__C/__ #1 [iii Symmmli’ .41 .3 bf FB'ICOSCM) +5) 505(30) ... kzgf} E A kg)? —-_. fi‘Z «5; PK 1 (1%“? a; 2 n __ a _ i :@J_ __ -5'C?»??x:o?)(0.5xm“)n‘ A Ne; 39Njf '53:} (Logo-91 3 <3 9; KBNJA @ TIfo rifi‘ OHKCHOFI [5 A“: 7%: (mm (QB wam Emu“ 5033119 05 #1:??? ’ 139 \C‘ A e .4 I, 11%: Emir '— E! 7t Ez+ 53 @‘fl’kH‘Q? \f—E. 1 ' 7}? EN =‘4PL-9 Caech OéJfiILO zit—Fae“ _ _ J Symwfy’ff @ Thy, CgffiEmLiffih 01F Nd; Hat’s? a?! paw f} (52352} [5 $3 'GQUEIQH L/IJV—‘\$UT€W£ @ % a \e’ Problem 2. The plane P (shown in the figure edge-on) at x = O has a uniform layer of charge, 6 = - 2nC/m2. Parallel to the plane and a distance 5 = 10 cm from it there is a long line, L (shown in the figure end-on), with a uniform charge I = +10nC/m. 1. Consider the electric field due only to the plane P. By symmetry, the direction of the electric field lines is (a) radially inward toward the origin of the coordinate system (b) out from the plane P and parallel to the X axis (c) into the plane P and parallel to the X axis (d) from any point on the plane toward line L. Answer: C 2. In order to calculate the electric field due only to plane P, a Gaussian cylinder of P length 2D and cross sectional area A is used, as shown. Which of the following equations expreSSes Ganss' Law applied to the cylinder? (a) 42:13:}? = 9-5— (b) 27rDE = 3:41 '90 50 D D (c) AE = 931 ' (d) 2AE = 9—31: ‘ Answer: 1) an 50 V 3. I In order to determine the electric field due to the line L only, what Gaussian surface should be chosen? (a) A sphere centered on the line L. (b) A cube with four sides parallel to the line Land centered on the line L. (c) A cylinder with ends parallel to the line L and centered on the line L. (d) A cylinder with axis of the cylinder along line L. ‘ I ' Answer: 4. Consider a point on the X axis, midway between plane P and line L. If a positive charge were placed at this point, what direction would the net force on the charge have? (a) To the left (b) To the right. (b) No force exists at that point (d) None of the above. Answer: A 5. Along the X axis, to the left of plane P, there is a point where the net electric field goes to zero. How far to the left of the plane P does this happen? ' (a) 20in (b) 1.5 m (c) 15 cm (d) 10 cm Answer:B% ‘ «HAVE .4- o-- I ,- WLJ rt) D fizzy/.9911. Q Er’eczirr'c. Hag 059% , r L 1‘0 9: fiégm in’wff r; E v a a d ¥ _ m ~ m%_ :1: __€__ +.x In '3 O" a 52ch 5‘0 “'9 ;_::~———--— I l- {Ufa/3;: b» 5‘9:ng n 97‘ Mplgcm INTO we fr—u ‘ 1. Has. WSW- em 3—?“ J I §= I06??? Chafing, (a): 1M1? +1“ pim 1:} mp 1:0 ( Pmihfa +0 +& fig}; ' We chasfi a 8%? I ‘ Cyh'hfiécr who? was {5 Mimi {Alf—Iii; {ML 5 {‘1‘ (TIM, Faces Gare; 5 +5 ‘55»; [51% 5). Farm. (:5 422%}? await 2} 7% #K (a; lbxfflgé‘ mrfia‘afi - "- 9*“; 04:5qu Problem 3. Two charged particles are located on the y-axis. The first particle has charge +Q and is a distance :1 above the X- axis. The second particle has charge -2Q and is a distance d below the X-axis. 1. What is the potential at the origin? Q - "Q “Q2 (a) 0 (bums-0d (C) thread (d) 41rsod2 2. What is the potential at any point on the positive y-axis? Q 1 —2 (am 03) 4x50[y—d+y+d] —Q —Q .B (c) 4figo(y+d) (d) 47:80)} Answer. 3. At What point on the y—axis is the potential equal to zero? (a) y = 0 (b) y = 3d (0) y = d2 (d) nowhere Answer: 3 4. What is the electric field at any point on the positive y—axis? (a) 0 (b) "29 47550322 Q 1 _ 2 1‘ Q 1 r 2' '5 _ d " (C) 47:50 [0—002 (3"5‘df]z () Arm‘lhy—dy (y+d)2:!J Answer: Q I 5. A is located at (x,y,z) = (—2d, 0, 0) and point B is located at (32,352) = (2d, 0, 0). What is the potential difference VB —VA? 5' —Q (a) 0 (b) “work/E A 3Q “Q ~. . (c) threadfi (d) 4xgodfi } Answer.____ @ Emma WM“ m? m; . Q ariak‘v majfl3%0¢1 we knew 43% We mug?“ fittflrfi or! 4R $99.9?va rail} H = ~——-—- _> g by __E_ m —— M = ~ 2 50:14 V >66 \[ Vi flag - y NILE (b) T‘ d _. J ; L233; EE— 3m] 08 " -‘ .3 '“ EP_ 8(a))"— E k1}? 0U?“ Caesa- Iq :C"25’€ 02 O) ' J 17? A 4o, T ' 8%(20f10fi?) y+ 0 WM B gem ,4—98 Macaw. *pmmlur ilk/7" 6 £1015 Cgmfiggg X) M; yd; $5M. £3; 5: {Bi/mm? in a fiaflSkmmgfffi Omméflfi‘lgaf: . A:_.O ' 6M? 9210‘} Problem 4. Consider a charged isolated air—filled, parallel-plate capacitor with charge Q, plate area A, gap size at, and the dielectric permittivity of air is 80. Useful formula: 2 A? , electric field E = $9 , energy stored U =g _ _Q_ The capacitance C a — . A50 2 2C 1. If the plate area of both plates is doubled while the charges on each of the plates remain the Same, the electric field between the plates will be: (a) doubled (b) halved ' (c) unchanged (d) none of the above ‘ Answer-:3 2. If the gap distance between the two plates is halved while the area of the plate and the charges on the plate are unchanged, the electric field between the plates will be: (a) unchanged (b) doubled (c) halved (d) none of the above. A Answer: " 3. If the plate area is A and the charge is Q, the work one needs to do to bring the two plates fi‘orn a gap = d to gap “—' d/2 is: Qd (b)_Qd (C) Qd (CD—Qd Answer:© («3) 2.4.90 214.90 4.450 42150 4. With the charges on the plates, the gap and the areas of the plates fixed, inserting insulator of dielectric constant K in the gap originally filled with air between the two. plates (a) will increase the electric field strength by a factor K. (b) will not change the electric potential between the plates. (c) will increase the electric energy U stored in the capacitor by a factor of K. (d) will decrease the electric energy U stored in the capacity by a factor of K. 2 Answer: 5. For a parallel—plate capacitor with plate area A and with the top half of the gap filled with air and bottom half of the gap filled with an insulator with dielectric constant it, the capacitance is (a) ma“ (1+ K) (b) £41510 + K) d d A80 27c As If __ d 0 M__ (c) d 1+1: ( ) d 1+ K C: “LLW 9. Gig E: Asa ~V‘Q U 2Q? 3C 4- on; ** ET: (3235a: * ’ C3? L E l: E = E (vurecfmmigrfi) @2 Q1052 g‘gngvaitfan ogé‘v‘ U+W=O U: fig-:23: i [,L=-W T j Ma. (95 km rugst‘fgw; Cgpm'ag, mm saflsé bétimsfa yam MM #3 “MM em?! "#15 mm bwaujt W321i: 1% fr}?! H a: i - if"? '16 U§=3§ UE' #Tl/wifiuehslgmf i j [S ficfiqg‘fifig bulk-3:179; (Giana Q1 @1 ?< > 1 _ p T1 U’C - ‘ch : 5:5 _ ‘- ciffiiwfim L5 g’aSenLa-i @ FL 3:? 2}: (Bags in 5mg; 2. Air 7< 5777/3/75”? E094, .L z J. Jr J. _ Gag Cl CL 'I _ __I 0.01 U/ 6&5 GWCI. is; me, CI “(1" . = M i $45.: 2:59: Ca +6“? !‘ ‘,,§Tk Ml ffifiJO-éw) R I E: Problem 5. Consider the circuit draWn to the right. + I Use this figure to answer the following 5 questions. E ' The points A through H are used to denote specific _ B C points of interest in fire circuit and are NOT circuit elements- Switches s1, 32 and 33 are initially OPEN. Sl 4R E= 4-0 Volts S3 32 Pay close attention to whether or not the switches are L—n—F/ open or closed for each question. 1. With 51, $2 and s3 closed, which of the following ' H I points best represent junctious in the circuit? R a: 4.0 Voits R (a)AandD (b)AandB (c) D and C , (d) All of the above Answer: A 2. Wi‘di S 1, 82 and 5'3 open, What is the equivalent resistance (Reg) measured from points G to A to F? (a) Req = R (b) Rm; 2 5R/4 (c) Req = 5R ((1) Req = 4R Answer: C 3. With 81 and SP. open and SB closed, which of the following best describes the direction of the current flow? (a) A —> B (b) A 9 D (c) 81 and 82 open, NO current (d) C % B Answer: 3 4. By closing 81 and SB but leaving S2 open, what is the magnitude of the current measured at point B? Given that R 3 2.00 9.. (a) 1': 0.625 Amp (13) i= 0.000 Amp 3 (c) i= 0.250 Amp ((1) i: 0.875 Arnp Answer: 5. By closing Si, 82 and S3 , what is the magnitude of the current measured at point C? Given that R = 2.00 £2 and the current at point P is i= 1.08 Amps moving in a direction fiom A to F to D. (a) i=1.237Arnp (b) i: 0.00 Amp (c) 1‘: 0.154 Amp (d) i= 1.08 Amp Answer: f I! R 25: r. t “tic *6! as f , 3' ‘ ED 835-3 923 am C may? a! 2; é; Ti} I A _€</§: If: 313149415145 Om: Pagflfs .' A. fl ‘96P ' I 33 Jr; a; F H a _ m R ‘Iwfl‘ R (funcfiéhg) ,— LIR F 3‘ 315’253 095:»! *wrv—a 7 Am; , LAW—=- % %_R:: ‘1} 265 R 3 6 Q 81) 5:? Open 533 C/OSIQJ‘ r 3 - + E— §<Cmmi~ WEN flaw A m a: + 5 . ‘ 3 _.- 3” Che. SALE} Lam/<- gi “PM. Cufrmer E f *9 Bécamc 52 115 {533% 71,45“, [:5 'IO' 7‘0 Upflw jamming e" “ choke J: m I 5—; 81:52:83 doggy] i2: 11082431? [5- Wag: 3R5; v» ~ lé— 533' _ - {W L? 5‘ Cunard flows 2!: 2: M i E ‘ €3.33? GEE-W 3 :2 L! N 2: him? A *7" 3(2) if may! L035}? IDafiaEéA-a Fart} _, , (43 ...
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HT1_Sample_Test_Solution - Department of Physics, Lehigh...

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