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Unformatted text preview: Physics 2212 G/H Test form Name Solulrovts 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 ninedigit student number in the
section of the answer card labeled "STUDENT IDENTIFICATION". 3) Bubble your test form number in columns 13, skip column 4, then bubble in your student number
in columns 513. 4) For each multiplechoice 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 Nmz/Cz . a, = 8.85x10‘12 CZ/Nm2 e = 1.60x1019 c
Blank space — Feel free to use this space for scratch Standard Integrals:
work. If you have any handgraded 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 handgraded. 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. MIGKMMimakgug’ “A”! ﬁg.th 41:91:30 (9 NA: Mt E, =cgis£d «A mgmtmlydmeizcl
"mes, ﬁ‘c'bkm leiigcciavd 0394‘: on NBL+£¢¢¢ will be 2224 GAE: jerk M1119. mimmlﬁkvkas «NEHj
’=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 handgraded. 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 »
"bB(°3'9‘F‘AL=‘0(0) / ~ egmcidvsr‘TYt
IBM 550' Q = E(r).a‘u‘pAL_
(b) (10points) Find E(r) within the insulating cylinder (a < r < b). New .h L at!“ (“Nee
\ujjlg Lilhéﬁr.‘ [4' ‘P‘ =NA¥cs cl; 65 "9: FLM Egg JaeAnus 1,} ‘s’ =AJM garrqu a<s<r~ J :  5w at,“ WQRS 99 f )hn We Sefq,rJk
‘9 Q“ = ﬂow = (csﬁﬁmrsoouls)
in = awcaLSZ 5335 = EatLOW“ 'l'xwh Cities; (4“! ecp3.am4( =i %Cr4a“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 handgraded. 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 AC/Nmz.
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. nillaw um:
"rs—=51? —> l3l=al§l ——> 3255 mam new '3' =60wa ﬁt.) (Lentil g) : (b) (5 points) What is the magnitude of the current I? 5»?an 75 org gvac'l’kzcméésaibaoHkwm’
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 clasha ciang C“) Stan‘58 "WA é‘rgsion; o‘FIavAi are 03% as L=n41v4 LDI=EDA¢VA [wise>1 1'. I ..___.._.. .. 3'
ﬁvd ‘4. he  Eﬁ [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 crosssectional areas are also 120°
identical,A1 = A 2 = A. Wire‘3 has twice the crosssectional 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
Indy15 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 ideadoehkzl,\ (inc: 444:4?
WMB‘ AMCL‘OA 'cen ‘l‘lwe am Amok13 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¢ 12‘ rs ST“! «0 Md: 7% ijyl
6"“ l" Ware A ,th «9 July Nan H“? “M” '2 MvméorﬁwB
march! (b, an with) MAWe   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 (75 ‘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)
BtuH «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 crosssections 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 ‘
{chad1‘0 3 2 4 1 143m “gamma «as
(6) E1=E2=E3=E4=0 gh‘kﬁm’TLJZW‘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.
 Spring '08
 N/A
 Physics

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