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08C 2220 Exam 1a Key

08C 2220 Exam 1a Key - Physics 2220 Spring 2008 Examination...

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Unformatted text preview: Physics 2220, Spring 2008 Examination 1, Name: Key A Student ID: Instructor (circle one): Littler Matteson ' ' Weathers _ This test consists of 7 multiple-choice questions and 5 free-response problems for a tetal of 110 ’ points (so that 10 points of extra credit are possible). To receive credit for the free-response problems, you must Show all of your work on the pages provided. Don t hand 1n any extra sheets or other paper. You may also earn partial credit for the some of the multiple choice problems if you show your work. Suggested procedure for solving the problems: 1. 95":“9’!” Read each problem carefully and make sure you know what is being asked before , starting the problem. Draw a figure for the problem. List the parameters given. _ . Write down the equations to be used. Solve for the answer symbolically. ' , Substitute numbers into you final equation and circle your answer. , : WORK THE EASY PROBLEMS FIRST” 1; (4 points) Consider two tennis balls sitting side by side on a table, separated by about 0.1 m. Given that each ball contains approximately 3 >.<1025 protons and electrons, each of which corresponds to a charge of magnitude 5 ><106 C, why is the force between the two balls so small? ‘ a. The elementary unit of charge e is very small. b. The halls are not conducting. c. The halls are too far apart. (1. The gravitational force cancels the electrical force. 63) Each hall is neutral or nearly so. 'Wmdhévm; between like. adumlilw. 9044394 Wang. 2. (4 points) Two charged conducting spheres of different radii are connected by a long, straight conducting wire. When the system is in equilibrium, the greater charge density resideseon’ a. the sphere with the greater charge. @ the sphere with the smaller radius. 0. the sphere with the greater radius. This TWWM W TTWJ ‘M We fat/Feud d. the sphere at the higher electric potential. ‘m d4“ ', Ouéwfl‘wllaeeflupwblw: ' e. neither; the chargeis evenly distributed. , ' ' , " Each splm is at it» Saw. palatial: ‘R'. , E; Q, - Ti: Q’s i 3. (4 points) You can remain safe inside a car (essentially a metal box) during a lightning strike because a. the electric potential is zero inside the car. ® the eleCtric field is zero inside the car. c. the tires insulate the car from the ground. (1. the eliectricwfliiitw near the caris zer6;V V ' ' V' V e. the car remains electrically neutral. 112‘!"le «(AW ls studded 4. (5 points) A point charge q is located at the, center of a cubic Gaussiansurface of side a. If the electric flux outward through one face of the cube, dueto the charge is +5.0 ><105 N'mZ/C, whatisq? , , I, _ . , _ a. 4.4uC ' ' It mwdmfiq‘f'wgkmlmlstlmfifilflw d.-:;~4,..4 MC , : l Ml‘i e? ‘V “’4’ c. q‘cannot be deterrnined’Without’k’knowing a. 5. (6 points) The electric potential in some'region of space is represented by the equipotential contour plot shown at right. The contours are , , drawn at intervals of 10 V potential difference. In the folloWing statements, the subscripts A L and B refer to the corresponding points in the figure. Identify all correct statements: _ a. EA = EB (13) lEAl > IE3! (3- WA! < lEBI ., d. EA and EB aredirected to theright ' EA and EB are directed to the left a my. a MmldtflWmc/UWW) . E ls'direchélé’ you/aid": he, lit/WWW ‘Iniw‘ticla V0104? 2 in Was Cw to 1%,, Mt a! 4M . ' A and BM mmsw WWIMW 6. (5 points) How much electrical charge q would a dust particle of mass of m = l ><10‘13 kg have to carry in order for it to levitate in Earth’s gravitational field (g = 9.8 m/sz) if it is in an electric field E = 1 ><105 V/m directed downward? a. 1 ><1o-18 C W ~17 ‘ ' lg. i :13-7 CC_ " ' ‘Lmfl 2E, : ”'ij s-OLE : ma: 0 ‘FW [eviMm (Why) d. ~~1><10"18 c “‘5 ' _ W} @ ~1x10"”c _ » 3, or a. f. —1 x107 0 E 7. (4 points) The figure shows a paired set of charges on the left, and either a single charge or O—® Q paired set of Charges on the right. In which cases will the objects (i) +q ~q W on the left and right be attracted to each Fother? , (ii) % 3%, b (ii) only , I... In;{i)f,8Nmsitec4a/[email protected] 90 WWW +3 ’3‘ _ . “1, c, (iii) only 1‘ tsW ’ " L ' 37d " (iV)0n1Y n is b: lb cine/win (iii) 53-? 8—3; (i) and (iii) , E“ ( )Y‘WW «gnu W cm I an ' ' f. (iii) and (iv) (1 ('71)) [Mia [flu o. (ml'fia‘iahajfii, Midi!“ - 0% a0 g 0) (iii) an (iv) WWW. {Le W333... am... (1V) +4 ~61 -q +q ({w} IS Mae (ml bvlrwr‘ll» lint Maw CW claw 5? 8. (13 points) In a certain region of space, the electric potential is given by the function V(x, y, z)= axy2 + bxz/z + cy3, where a= 4.0 V/m3, b —— —2. O V/m, and c —- 5.0 V/m3. Find the electric field at (x, y, z)= (2.0 m, 2.0 m, 2.0 m). _ L ’2Vm m.- J I lax-“3i ‘43“? = —4%s-<23)‘-3-‘-2~M{.2* ~ 32% .,E?.,f,‘5‘.' Wt 393 = ~24 .0332“) 333.42..) .423 9V 2:- "ZV/m (2M) - V E? 9?, Z . p (231)” 17‘ . (15 points) A point charge q1 = +4.00 ”C is located at (x, y) = (0.00 m, 2.00 m), a second point charge q; = ~2.00 “C is located at (x, y) = (—1.00 m, 0.00 m), and a third point charge q3 = +2.00 “C is located at (x, y) = (1.00 m, 0.00 m). Find the force on 413 due to q] and q2. é a: I.., L. > as “'3 9 J. x Pu: '5 d g Ea A "e IQ " F:F23TE3 : _y...i§fz+ $130,531 . T25 43 , Le A f Wcl'jfls‘mg‘] m X “X: when. Gos6= 3 ~ 9'03 , J L smB; Yévy' T13?" (2....)2 (my = "2.56, “0““ N3 - l. 2? «IO'ZNJ ’ J , ~ '9 '6 . ~c . . => F: 3-"9V‘°°'N*%Z-2x'0'°cg 2110 c + 4on c M); + Wk] 10. (14 points) Find the total electrostatic potential energy for the same system of charges described 1n the previous problem. » _ c 2 (I141: Y Axis O w rm 1&2 Wm? ‘ =f(-l~o)‘+(o-2)z p. v' Em $113619 «109 Ant/ca “WW“ ““40 i ‘WLZ“’°““C)‘2"°"-9 + (effl-“fl um] 11. (18) A charge q = 20 11C is uniformly distributed throughout an insulating sphere of radius a= 5 cm. This sphere 1s located at the center of a hollowspherical conductor of inner radius [1— — 10 cm and outer radius c—— = 20 cm; this conductor carries a net charge of Q = —30 71C. Use Gauss s law to find the electric field a. forr=7cm ." : 'ke—i L x8%!b9N- 7E; . Zono‘iiéc _ _ 4N E—i HF (71104141 M) 3.641140 k. b. for r=15cm 57‘" (Mid: W) o. forr = 25 cm s’ l: gm)‘ 8-99110‘3Nmé (20 v3ozx/o‘ 96. 1 E= ‘ rt r .— (gs‘xm m)” = -l 441% ”/c 12. (18) For the same system described in the previous problem (i. e. q _=_20 nC, a = 5 cm, b = 10 cm, 6 #720 cm5 and Q,= ~30 nC), , ‘ 7 Find the eleefi'ie potential a. tow: 7cm b L ”Ax/=4- £2149» Vebisi‘vzez-§gd,zg gal , :. ice?( I, ~=’-> VG).- V(b)+ head.) 9 _ 2. w o—go)xlo"’c _ zom'qc 2ozzo"’c :kQ/m)- keq’ fa — 8'99)!” I’m/é I 02M. OJ». 4' 007m c. for r=15cm k, ta) 8.99xl09N'M%=(20‘~3o)m't V: "'%—’ z = -- f0 quflm 4 V c. forr = 25 cm 9 a. '95 v'i-éfl) = Weefiwl/ _ r' - v tawdm ...
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