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Phys1120 Sample Midterm1 solution - 91131511205me NAME ‘3...

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Unformatted text preview: 91131511205me NAME: ‘3 ' K Student II) C Section number and TA Name: __ _ _ _ Exam Version 0001 Please do not omn the exam until you are asked to. 0 Your exam should have l0 pages. numbered El .l thru El . l0. 0 This exam consists of 15 multiple choice questions; worth 5 points each. and a long-answer section worth 25 points. for a total of IOU points. 0 Write your answers in the space provided on each mge for the written question. Use the hack of the page if needed. Circle the correct letter for the multiple choice questions. SHOW YOUR WORK ON THE WRITTEN QUESTIONS. Full credit will he given only [or the correct ansWer mmmpanied by your reaming. PLEASE! I. Print your name on each page in the space provided. 2. Print your student Identification Number on the page above. 3. Bubble your name. ID. and exam tension number on the bubble sheet provided. 4. Write the W neatly in the big blank space at the bottom of the bubble. slwet. Out 'l‘A's names are Charles Bailey. Jeremy lirmwr. Cmr‘g Magic. Daniel MrKinmm. and i'mmgwoa H mm: Don't waste time erasing on the written problems. hm put a line through defectne reasoning. Ask for more paper if needed. At the end of the exam. check that you have completed all the questions By handing in this exam. you agree to the following statement: "On my honor. as a University of Colorado Student. I have neither given nor received unauthorized moistunce on this work" Signature- Possibly mm information: ‘ . ' k I: 9.0x l0“ N.mle‘. basic charge unit 1.6x l0 w (7. n. = 8.85s l0"' C’I(N.m'l 6 . 6.7x to " N.m’/kg= Good luck! Spring 2% El .l First Midtcmt PHYS I120 Exam I NAME:_3¢[email protected]____ Circle the time“ answer and Bubble in questions l-lS ON YOUR BUBBLE SHEET! WW that a spherical body of mm M. feels from a second spherical body of mass M; when their centers are separated by a distance. r. is attractive. and has a nugnitude given by: GMJI, Ft.) z I" This force is very similar in form to Coulomhk Law for the force between static point charges. I. We can ignore static electric forces when calculating the motion of planets around the Sun became: A) electric threes drop to nearly rem much more quickly with distance than gravity forcex B) gravity forces are intrinsically larger than electric forces. no. electric forces are actually part of the gravity force. anets and the Sun are nearly electrically neutral. so the electric forces are very small. :) planctsand Sunarenot ' tparticles. so the el tric forces donutl'olkiw Coulomb‘s! w. Gr 42: £— MgbjmlmszfiJrA "ed”! 2. A plastic rod has an initial charge of +3 111C. The rod isruhbed with a neutral cloth. After the rubbing. the rod hm a final charge ot‘ +6 inC. + 301C A) Conservation of charge requires that the cloth has a +6 mC net charge. c: Conservation of charge requires that the cloth has a +3 mC net charge. onscrvation of charge requires that the cloth has" a -3 mC net charge. g ) Conservation of charge requires that protons moved to the cloth. 4- F.) Conservation ot'charge requires that both object-e remain neutral. QM” 3. A charge oi" +7 microC is heated 3 meters from a clause of -3 milliC. The electric force on the 7 niicrriC charge from the -3 niilli(‘ charge is: ‘_ ' 5 A) A mptllsivc force 0le N. ( , X c attractive force uf63 N. F .. k a 0 (3 Via rt attractive force of 2| N ’ 3 2» ) An attractive force of 2| NIC. inc Fm“ W E) Thefurcciszem. "3MC - q [07 abd0,7) :- X 1 Spring 2008 El .2 FIN Midterm PHYS ”20 Exam 1 NAME: : 21AM WW two identical negated) chats“! panic!“ each have mass. m. and eharge.- -.Q The two particles are placed a distance R apart in empty space and are released from rest. Each particle feels only the Coulomb force due to the other particle. (There Is no gravity in this problem.) F: ma ‘Q Q . . ’ R2. 4. The initial magnitude or the aeeel «r ‘ . . each particle. immediately after relezme. is: ’2 I Alla-Ll; mic Cle R R’ El none of the above at 5. As the particles continue to move alter then release. the magnitude of the amelemtton of each i ‘ ‘lt: ‘reascs Btincreases (‘i stays the same DlIIIitially decreases. lmttheninereascs PM EA So W A: A?“ R 0. "‘ E; 6. The electric field 4 cm from a point charge of +3 C Is: A) l6. 9; l0' NIC pointing towards the charge 8) 6. 8x IO" NK‘ pointing away from the charge 6. 8300" NC pointing towards the charge D) 6 91; It)" NIC pointing U from the charge Thel'rcldisze I E : kg; </ " Spring 2008 EL} FIN Midterm PHYS 1120 Exam l NAME: 321% W No Mick‘s. one with positive charge +Q and the other with negative charge -20 are limited on the x-axis as shown below. The three questions below ask about the electric field on the x-axls In the three regions: I. some finite distance to the left ol‘the two charges (anywhere but infinity). ll. somewhere in between the two charges. and Ill. some finite distance to the right of the two charges (anywhere but inf )9!) “416 l :32, $6.?“ D'?‘ 51': +0017“; 5W 4___ :41 ‘3}. ' 0 aroma «2M3 .52 .20 +62 b 7. lnregion .Ihe electric field .. Mm is every“ here non zero and pointing right '5 everywhere non-zero and pointing left It be rem at some location. 8. l regionfl. the electric field .. *gwrywhcnc nonvzero and pointing right = ) is everywhere non-zen» and pointing left C) can be rem at some location. 9. ln «gloom. the electric field .. - is ewrywhere non-zero and pointing right mu eterywhere non-zero and pointing left D can be new at some location. ID A flat rectangular surface has an area of- " 5 in" An elecmc field of 6 NIC is directed at tut angle of 63 degrees to the surface normal sector as shown The electric flux through the surfaces is: $65 N'm’lC SW [(7 8N'm‘1C l3 4 N'm‘K E) Hux' Ls zero. -A= iEHAtiQ i if =(eé’cY2900’543 Spring 2008 15 L4 First Midterm PHYS ”20 Exam 1 NAME: SQ igAQflé ow' - s ‘es‘ ' :Apointmsitivecluirgem=2Cissurroundedhy a charge neutral spherical metallic shell as shown in the figure. The shell has an inner mdius'ol‘ l.5 meters and mi outer radius of 2.5 meters. The punitive point charge causes a rearrangement of the charges in the metal. leading to some separation of negative and positive charge as shown. l I. At point A. Lugs; from the point charge. the electric field .. sq“ {'QC, E: is zero due to Gauss's Law. 6177-0 my @ I Barlow NK‘ in magnitude I0 I I'I tern because [3-0 In static equilihrium inside metals. : ‘. ilk/0 I.) DI is 3 .6x l0'° NIC due to superposition of the point charge and neg surtace charge. '6 :2. At point B. mrmmmcpoimhugc. theelectnc field .. 6‘0 or 61‘3“.“V‘y A) is NOT mo due Io Gauss's Law. / Move . : ls4 mo" NC in magnitude. we ' be 5' ‘ , zero obecause 5:0 In static equilibrium insrde metals. I is 9x [0° NIC due to superposition of the point charge and neg. sud'aee charge. l3. At point c. 3mm from the point charge. the elecrric field” ‘3 wt 84454144 = 's zero due to Gwss’s Law 5e Q a 2x l0° NIC in magnitude. H ( :6) $ 2C ' . Is tern because [3:0 in static equilibrium inside metals 5'; %—L) 2!: 0) Is 4; l0" NK‘ due to summition of the point charge and outer surface charge. 3 3 M H. A closed irregularl) shaped surface of 700 m encloses a total charge of 40an C. The surface area is then increased by 2:: to I400 III. while the enclosed charge drops by” -x t I 2000 C. Therefore. the total electric flux through the new surface; 66b; masses. is not rem. but remains unchanged d950'i 5;, DI Is zero both before and after the changes in area and charge. OS. { J’Cfmfifi 6' Spring 2008 E [.5 First Midterm PHYS 1120 Exam l NAME: g0 'JW‘A I5. Two point charges an: positioned as shown on the y-axis. one change with +2 nanoC is located 2 m below the on'gin. and one charge with -2 nanoC Lg located 2 m above lhc origin. The ciccuic potcnlial at the origin (assuming that zcm volts is at infinity) is: ..c g: 2.23:“ 3W M“ M ) -9 v0 Ls $6! gfl (5 . D) -18\'Oll$ k ,2 L x ‘0 volts V 2 kg :. __<é_va. ) \/ :ggzgéfl Swing 2008 Ella Finn Midlcml PHYS I120 Exam l NAME: SOL/Lama WW Two point clutt‘gcs. ~Q. an: shown on a square grid in the diagram below. Also shown an thtcc points. labeled as A. B. and C: 3) Draw arrows on the figutc below to indicate the direction (Jtheforre that would he felt by a negative test charge placed at the points A. B. and C. If any of the forces are um. state that explicitly. — [.5 5 m ‘2 c. . Ditcction of force l'clt .eed at A. B. and (I M Draw mm on the figutc below to indicate the dimw‘m «Wine electric/field at the points A. B. and (T. If any of the electric fields an: 1cm. state that explicitly. Direction of electric field at A. B. and C Spting 2008 EL? Fitst Midtcttn PHYS 1 120 Exam I NAME:_§2]A’£L_ c) An uncharged metal ball is now placed at location C. Sketch the distn'bulion of chargw on the metal hall to indicate what. if anything. llamas in this situation. (I) Will the clccuic field magnitude at B (circle ( - a mtoc at B to show the direction ol'clcctt'ic fic ~ ‘ i create. or rcmln the some? Sketch Stating 2008 E I .8 First Midtcmt PHYS ”20 Exam l NAME:__S21£0]M— W: A charge of +0 is placed at the origin of a Cartesian criorilinutc system ac shown in the diagram. An imaginary 'Gaussian surface' of cubic shape is centered on the charge. 2 u) is the total elect -' ~ ' A new spherical Gaussian surface is drawn that completely encloses the charge. thc original cubic surface. and has. larger surface area than the cube. b) Is the clcctric flux thmugh l ’ ' ace (citric 2 one) greater than. less than. «$n flux through the cubic surface. Exnlain m; rcgsmigg W: E'5 Spring 2008 El .9 First Midterm PHYS “20 Exam l NAME:.$_LJZMM_— Now an additional positive charge. +Q. is placed to the light of and outside the cubic surface a.» dump. c) Docslhc electric flux thmugh the .. .ul lace of the cube (citclc one) than". nr mnm'n Ilu' some? E81191!!- mm. Spring 2008 El . In First Midterm ...
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