sp09test2s

Sp09test2s - PRINT Your Name 50 L UT 0 N Instructor Louisiana State University Physics 2102 Exam 2 March 5th 2009 Please be sure to PRINT your name

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Unformatted text preview: PRINT Your Name: 50 L UT/ 0 N Instructor: Louisiana State University Physics 2102, Exam 2, March 5th, 2009. - Please be sure to PRINT your name and class instructor above. ° The test consists of 4 questions (multiple choice), and 4 problems (numerical). ' For the problems: Show your reasoning and your 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, but you must derive and explain your answer fully on paper so we can grade your work. - Feel free to detach, use, and keep the formula sheet pages. No other reference material is allowed during the exam. - May The Force Be With You! Question 1 [8 points] A parallel plate capacitor has a dielectric material of dielectric constant K > 1 between the plates. The capacitor is charged by first connecting it to a 12 V battery and then disconnecting it from the battery. After it is disconnected, the dielectric is pulled out. Compared to before the dielectric is removed, what happens after it is removed? (Circle the correct answer for each part.) 1/ (i) The electric potential across the capacitor: "——--_. decreases stays the same increases V ! N6; Q (ii) The charge on the capacitor: decreases stays the same increases G CO N 5T C (iii) The capacitance: stays the same increases C l7 E C U u (iv) The potential energy stored in the capacitor: DiSCeNNECTED @ Q ": Con/5T C’__ 2 Kaufi/d— .— decreases stays the same FORMULA SHE ET -' K>ZL Jo REMOVING DIELECTfl/C C‘. DEC/7.54563 C ( C Fear- AFTEQ BE “ 2. u: MQL > Q : Ul3F-‘r-Ofl, ‘ U ’NC‘ EFL LC LC RF?" A B Wrrcazga>§czve :\/‘ LNC AWE B Problem 1 [18 points] Consider the circuit in the figure below. The capacitors have capacitance values C, = 0.75 pF, CZ: 0.50 pF, and C3 : 1.00 pF. The battery voltage is 12 V. Parallel q ' “ 6L Emmettsmv % V C123 C d (a) (7 PtS) What is the equivalent capacitance of this circuit? Cam:- Ci't'C; :- 0.?§44F+0.50{)F :: [~2ng :54 _‘ f _ C113 - C11 C3 : Ct.2b/37F)((,03dpp‘) : 0.551”: C¢2+C3 Cl-Zj‘qu)-{~ CLO'UVF) a fig}! '" CE . (b) (6 pts) Long after the switch is closed, how many electrons have traveled through pointa? C32 :_ vce:%3= Lag-5 x IO-HF)C I’LV) 123 “Z = 6- 66 x 10' C .41. “‘71 #E’iec‘t‘ranj = CQ : 969‘”: C" =[H‘J7XlOfilec‘lY0f3 (.6 ex to “‘C I? t - 4N0! Chapae par BIQLMVI (c) (5 pts) After the capacitors are charged, what is the voltage drop across capacitor C]? Ser‘aBJ: CQ‘z—s Z 62.1 2 Q3 CHMCGE SAME ' “’2 —H. =3) via: Qw/C'L ___ G-Ge x10 Q /l-Z.S‘x to F PAMLLEL: \/I :_. Va: VIZ: 1313'3V AFN-3‘. Question 2 [9 points] The figure shows a current-canying wire consisting of three sections, which have different radii, as marked on the figure: A B C (a) (3 pts) Rank the sections according to the magnitude of the current, greatest first (circle one): i>i>i i>i>i i>i>i A B C C B A B C A . . . " RUE)? m>lc>13 CURRENT £5 CotqéE (b) (3 pts) Rank the sections according to the magnitude of the current density, greatest first (circle one): A>&>% Q>%>h k>k>fi JA>JC>JB antic cumE/qr QENSI'T V 7* E: b/R l5 LQREEJT m N4MOLOGSY' Fife (c) (3 pts) Rank the sections according to the magnitude of the electric field, greatest first (circle one): EA>EB>EC EC>EB>EA EB>EC>EA EA > EC > EB all tie FD Ac . m. Wire C is madfifipper and has a onal area of 4.50x 10”6 m2. Wire D is made | of aluminum and has a resistivity of 2.75 X10"8 90m and a cross-sectional area f C 1.33 X 10“6 ml. The wires are joined as shown in the figure, and a current of /d€wn the two wires. muggy ARE IN JEKIE'S 1.0». __8. (Man) ( ? H _ =7ReZ : [CLGH [0 Am) “Jug-am?) 4' (2-?) sue—och) (Lump hat) ,__. O‘ZLpgfl r v = .L Reg 2 (memo-1934):: \o.L+o; v 2 4m, r-S'fi Leona-r (b) (6 pts) Calculate H1; rate at which ener is dissi ated between points 1 and 2: . . L. z c. ’1) z: 2’ R c. ‘2 8/ E P; Ac. I c 4N5 . #8 7. t . . 2 Cut) Cl~6qmo 41““)(510‘4‘3 :‘O'LS—W? WINE“ m L Wmlb “a (c) (6 pts) Calculate the (1 ~ r electrons in wire C, given that copper has a charge carrier density 0 8.49X1028 m" : é ~s an) : Ge 5 /4<. : (boa) Nero m m2) he ' 5‘ c : 4.5%on A/M’L AN-S’. Te. (%‘H%kIGSA/mt) __ (U6 : :: -‘ “ce K315“; meta/m3 XLCUX top-WC») Question 3 [8 points] In the figure below to the left, the switch is closed to point A at time t = 0. After closing the switch, there is a current 1' through resistor R. In the figure below (to the right) it shows this current i, as a function of time t, for four sets of values of resistance R and capacitance C. c; 53915 1 N or R C ‘Set(1): R=2.5§2andC=3.0tLF L CECi" can't/T) Moo—3W db ' - ' " AgariquI—HHEM! /‘ a 186M402 _ 0 and C: 6.0 ’ E— e c: ‘2‘. E/R 75) 9’} fi' 2:) 8 : U 5 COMET ’_________ (a) (4 pts) In set (1) above, the R and C values go with which curve (circle one): 5 C“ V 3 CM 5T 3&0“: H (0 curve (a) (ii) curve (b) 6033 ‘3 “Ra If E . - Rzz‘s-fl'>5.0—§l "‘5 (111) curve (c) (1v) curve ((1) a ‘ _> He LL LAME z. to) 3mm. Claim“ ?:RC= 7-545 =5 TIME 7 (b) (4 pts) —-—-—-———___. In set (4) above, the R and C vaiues go with which curve (circle one): (i) curve (a) (ii) curve (b) R C: E = '3 0A 5 (iii) curve (0) (iv) curve ((1) L-ON 6- T! M E LONG“ 7/“ IE < 5‘ , Rb Rs 1 SHALL 0(0) {3 504 4 2’34 A.) l Problem 3 [18 points] In the figure below, the resistors have resistance values of R 1 = 1.2 Q, R; : 2.5 Q, R3 : 10.09, and the ideal battery has an EMF of 5 =31) 111V. (a) (5 pts) Find the equivalent resistance for R2 and R3. PemLLz-‘L C-RL—tm—‘g :3 Rtgtw T. 1 R13 fizz-HES ._. 2. 4‘2, lQO—Q —* EL __ ‘10412 E13 ANS'. 154+ (Op-{L /—-"J (b) (6 pts) What is the current through the battery? \/n3 : 8 :2 R513 same; 562 _ {3.33 = Rn; ' R1+R23 : 1241+ 2,04). = 3-24 4/515! cl : \/r13 A 8 _ B-oxzofia‘v __ 938M0— Rng {2113 7.2.041, . A (c) (7 pts) What is the power being dissipated in the resistor R2? (-3 a . . 2. - * '2. ! ll-‘LLVL : C’Z._ R1. *\VL/RZ 56mm: 3,! z: a,” 2 LE“ : Mus/Rm : (if/ml :. =1.33MO“’A 2) v23 :43 RR :cq.sxxro“’4) ‘3 XCZfl-Q') Parana/l: VzZV3 :1 \/é_.5 : 1.88%th V' = l-SS’XIOHSV —_. - -6 K ‘35 P : val/K2- : CLQS’KIO 3V)1/C2.F.a,) :1 L‘Hxlo UWTS I4wa Question 4 [8 points] 55 55 v 3 F3 2? F; (a) (c) (1) (3) (iii) (C) (iv) (a) and (b) (V) (b) and (6) (vi) (a) and (c) (vii) (a) and (b) and (0) ONLY 8 co MéIJTENT (AJer R1614?” M4ij RULE é N59— chi/402651 ..,___————-' c/v’A/ZGE" @ “"489 CONSUTEM“ w/ POS’T’W‘? Lie) x (ix/AH I iélé Problem 4 [13 points] An ion source S is producing positively charged 6Li ions, which have a charge +3 and a mass of 9.99X 10—27 kg, as shown in the figure. E _.\ 6 + — O o O TrE-b- y LI —'"—‘_-—>V I- e e o 1/4 set—+9 Fe (a) (3 pts) The ions are accelerated by a potential difference of 10 kV. What is their final speed? . KE : (ie)x(1-okv) = 3 ~t<=t r: fox :0 x Léoxto d“ : l-GOXIOflISZF 17K ‘3- i awe-.- aE-wr => an: arm. Tb Y‘fl 2— CMWO-Ifi‘r) 5‘ 8 ‘. :1 Lei-‘1‘! x to”? lee) : AN W (b) (5 pts) The 6Li ions enter horizontally a region in which there is a uniform magnetic field withmagnitude B = 1.2 T, pointing out of the page as shown. What is the direction and magnitude of the force F; on the ions due to the magnetic field? (Draw the direction on the figure 01- use the coordinate system to specify it.)__' A A N 5 ( RHR i=er porn-sz =5; 17—}; :5 Down) 1‘ l“ J34 i _. . _ I -l'$ Fl5 _ 651V 8 = C l‘QOXlO me: >C5756xtogm/s‘ )(sL.zT )3 \ ls 07X”) N { Ari/b l, (c) (5 pts) What is the direction and magnitude of an electric field necessary to create a force, which balances F3 such that the 6Li ions pass Uri-deflected through region with the two fields? (Draw the direction on the figure or use the coordinate system to specify it.) I}; '2 gE l6 [quftliel +0 E” Ma " I [WUJT poi/ti” up 5:: E Poin‘l‘s U]? +1); A4515, '“ S‘ :- CEéexrobm/a 0.2.7): 6.7?er V/m $1 ...
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This note was uploaded on 05/20/2011 for the course PHYS 2102 taught by Professor Gimmnaco during the Fall '08 term at LSU.

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Sp09test2s - PRINT Your Name 50 L UT 0 N Instructor Louisiana State University Physics 2102 Exam 2 March 5th 2009 Please be sure to PRINT your name

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