Physics 2212 Test 2

Physics 2212 Test 2 - Physics 2212 G/H Test form Name...

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Unformatted text preview: Physics 2212 G/H Test form Name Solul-rovts Spring 2008 Quiz 2 Instructions: seat: 1) Select a seat with. a RED stenciled letter matching the seat assignment at right. . A 2) Print your name, test form number (above) , and nine-digit student number in the section of the answer card labeled "STUDENT IDENTIFICATION". 3) Bubble your test form number in columns 1-3, skip column 4, then bubble in your student number in columns 5-13. 4) For each multiple-choice question, select the answer most nearly correct, circle this answer on your test, and bubble itin on your answer card. Show all relevant work on your quiz.. Do not put extra marks on the card. 5) Turn in your test and answer card and leave. Your score will be posted under "Course Standing" when it has been graded. Quiz grades become final when the next quiz is given. 6) You may use a calculator that cannot store letters, but no other aids. k = 8.99x109 N-mz/Cz . a, = 8.85x10‘12 CZ/N-m2 e = 1.60x10-19 c Blank space — Feel free to use this space for scratch Standard Integrals: work. If you have any hand-graded work here, please . . n+1 label It clearly, and make a note of this fact on the u” du = u + C (n # _1) relevant page. n + 1 [undu = lnlu! +C (n = -—1) du = ——co.9(u) + C [c0500 do = sin(u) + 0 f8“ du = e“ + C /u67‘du= (u—-1)e“+C page 1 of 6 221 T he following problem will be hand-graded. Show all your work (for this problem. Make no marks and leave no space on your answer card for it. » . A - Ltat ==3th [I] An unknown charge distribution produces the electric field: 2 .. _ _, A 2 . ‘ firs «‘s the. E(x>yaz)- k, wM‘c where f and g are positive constants. M box “at. (a) (15 points) Compute the net flux through the surface of a box £9931; ‘ Ma s of dimensions (L x 2L x 3L) with one corner located at the origin, as shown in the diagram. Use the standard convention that outward flux is positive. A: @ch‘tim'i‘ EXT-:0. MIGKMMi-makgug’ “A”! fig.th 41:91:30 (9 NA: Mt E, =cgis£d «A mgmtmlydmeizcl "mes, fi‘c'bkm lei-igcciavd 0394‘: on NBL+£¢¢¢ will be 2224 GAE: jerk M1119. mimmlfikvkas «NEH-j ’=b [email protected] @ (boii‘DW‘ che Q."— Z-=o: Eiugg'lwi'vO 40 Ii“ 50! (9 on 429%.- a=sg ant, hams! yaw ' § =9 if: T565. 2 E-[aL‘i‘] a (93L‘Y2L‘)=lnr3L“i Ngi Rm T5 i7: ib'fé‘E—i’quf @Bi‘i. = m w (b) (5 points) What is the total charge contained in the above box? gram Gwss‘s Law; qm : 3‘3 Qm L/b Q?“ i: Page 2 of 6 221 The following problem will be hand-graded. Show all your work &r this problem. Make no marks and leave no space on your answer card for it. [[1] ' A very 10mg, hollOw cylinder has an inner cavity radius a and outer radius b. The cylinder is an insulator, carrying a charge density: p(r) = C r2 , where r is the distance from the cylinder axis and C is constant. ' For each of the following regions; use Gauss’ Law to determine an expression for the electric field. In each case, express your answer in terms of the permittivity constant, the parameters a, b, and C, and the distance r from the axis of the cylinder. Be sure to specifil the direction of the field vector! (a) (5 points) Find E(r) within the cavity (r < a). Nair 1 gr 9-} EL 1““ “a; m" {or as 'P : Q». :30 ‘Y"“&" “W? (l ‘33 )° “ ‘ .L AL) Flux éyll » "b-B(°3'9‘F‘AL=‘0(0) / ~ egmcidvsr‘TY-t IBM 550' Q = E(r).a‘u‘p-AL_ (b) (10points) Find E(r) within the insulating cylinder (a < r < b). New .h L at!“ (“Nee \ujjlg Lilhéfir.‘ [4' ‘P‘ =NA¥cs cl; 6-5 "9: FLM Egg Jae-Anus 1,} ‘s’ =AJM garrqu a<s<r~ J : - 5w at,“ WQRS 99 f )hn We Sefq,rJ-k ‘9 Q“ = flow = (csfifimrsoouls) in = awcaLSZ 5335 = Eat-LOW“ 'l'xwh Cities; (-4“! ecp3.am-4( =i %Cr4-a“fl ~b E(r) 5%!) 02 05,9 _ k2") Q“): a,“ M? Di} cyifivcjkr = S 'QTI'S‘AL'AS I 221 The following problem will be hand-graded. Show all your work tor this problem. Make no marks and leave no space on your answer card for it. [111] A cylindrical copper wire has radius R = 1.0 mm and conductivity o= 6.0x 107 A-C/N-mz. A uniform electric field of magnitude E = 0.010 N/C driVes a charge current I through the wire. (a) (5 points) What is the current density J in the rod? _ _ 31$ w R quCom) ‘Hm Cami éeusrl rs a tailed. nil-law um: "rs—=51? —> l3l=al§l ——> 325-5 mam new '3' =60wa fit.) (Lentil g) : (b) (5 points) What is the magnitude of the current I? 5»?an 75 org gvac'l’kzcméé-sai-baoH-kwm’ I =WS 34A -—> I :SAL ~— TOW)» :[ = (c.0xlos’fir‘3'60 "0"6301 2 (c) (5 points) The density of conduction electrons in copper is n = 8.5x1028 e‘lm3. Compute the drift speed vd of the mobile electrons in the rod. (5 points) gé‘ “’f ‘4“4‘30J‘VN3 win—£5 amoué, «04»! I -'-’- C') Le [i=fidknh cmfil’ e=Mg§ELAReL clash-a ciang C“) Stan-‘58 "WA- é‘rgsion; o‘FIavA-i are 03% as L=n41v4 LDI=EDA¢VA [wise->1 1'. I ..___.._.. .. 3' fivd- ‘4. he -- E-fi- [QAvedgeuj ~mmvcl~vs<m~3§ol kaMCL)] Soilv q (“axles .. 4 4x105 d (“mugfi5kwu %2 i . 5 Page 4 of 6 221 Question value] 0 points (1) Consider the junction of three wires with identical conductivities depicted at right. The currents through two of the wires are equal, 11 = 12 = 1. Their cross-sectional areas are also 120° identical,A1 = A 2 = A. Wire‘3 has twice the cross-sectional of the other two wires: A3 = 2A. What is the current 13 through the third wire? (a) I3=0 ' [dapth lag) 13=-21[ _, 21?": 2‘1,” (0) 13: *1 Indy-15 7— *(d) 1 =21 % (e) 1; =1 I; = — (11+ch =l “all ’6 - POP'L‘Ql finial: - Mam $13» mm m Mattie AOAo—iaqu who clz fight MW ‘skeaith * “Hal-‘5 Odom, noi- idea-doe-hkzl,\ (inc: 444:4? WMB‘ AMCL‘OA 'cen ‘l‘lwe am Amok-13 3’3 ,Ulh‘cin 5m flow 99;" Question value 10p0ints 9"“ P (2) An uncharged conductor has two internal cavities. A point charge q is placed inside one of the cavities, and it polarizes the conductor. LetA denote the surface of the cavity that contains q, let B denote the surface of the cavity that does not contain any charge, and let C denote the outer surface of the conductor. 11 Which of the statements below best describes the polarization charges induced by q? (a) There will be no polarization charge within the , conductor but induced surface chares onA and C. p ‘ MN“ wilt: ’q gig mdudor: an 91 mg“, 50 0:“ =0 -——-> aiding: ~z Inga} read: an MC.“ A (b) There will be an induced volume charge density pp within the material of the conductor. W‘ *(c) There will be no polarization charge within the ("53+ conductor, but induced surface charges on A. a 6'5 (A, wTi'hm modular, ((1) There will be an induced volume charge density pp “WWL‘K: 530 5° Q-‘nso within the conductor and induced surface charges on C. _' .4: 39 um Amy; “’41,”: (c) There will be no polarization charge inside the ( , ‘ conductor, but surface charges on A, B, and C. ° (>5 b , Wily. . 520 40 Q», 30 CGKAOC‘i'uF ‘fs Ought;ng 2 7‘; Linag¢ 1-2‘ rs ST“! «0 Md: 7% ijyl 6"“ l" Ware A ,th «9 July Nan H“? “M” '2 MvméorfiwB march! (b, an with) MAW-e - - Page5of6 ( “8‘ 9129? (a: an were? C [paler $044.16?) 221 Question value 10p0ints V (3) ‘You are given a distribution of three point charges and four Gaussian Surfaces A, B, C, and D. Rank the flux values dJA, $3, CDC, and (DD through the four (closed) surfaces, from greatest to least. a. ’A‘ : a + #(a) > q>B>q>C 5 Q45» ? (b) ¢D>¢B>¢C>¢A (7-5 ‘3 t 0.5“: O (C) ¢A=¢D>¢B=¢C ' ' *(d) «DA >> (DD 6‘5 'C’ 3 0:3“ 1 O (c) ¢A>¢B>¢C>¢D GS (DI : Qfiw=+2 ‘fiut L =2"; QM we 5:2 ‘flw‘F J , - - a > Q =42) Btu-H «A71- Qpélleo (Ga 00) C 0 $415., $ or task <i=<fo3>fi=® Question value] 0 points (4) Four cylindrical wires all have the same conductivity, 1 —I> ’A but carry different currents Ii. The wires have different cross-sections A,- and lengths Li. Assume that the currents are caused by uniform internal 52/31,... electric fields E,-. How do the electric field magnitudes compare? (a) E3>E2>E1>E4 " E4>E1 =E2>E d E >E =E >E ‘ {chad-1‘0 3 2 4 1 143m “gamma «as (6) E1=E2=E3=E4=0 gh-‘kfim’TLJ-ZW‘A.“ Page 6 of 6 ...
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This note was uploaded on 04/13/2008 for the course PHYSICS 2212 taught by Professor N/a during the Spring '08 term at Georgia Tech.

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Physics 2212 Test 2 - Physics 2212 G/H Test form Name...

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