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Unformatted text preview: PHYS 2212, Test 1, September 17, 2008 Name (print) _______________ __\_/_<_% _ ____ _ _. ______________________________________________________________ _ Instructions 0 Read all problems carefully before attempting to solve them. 0 Your work must be legible, and the organization must be clear. a You must show all work, including correct vector notation. 0 Correct answers Without adequate explanation will be counted wrong. 0 Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anything
you don’t want us to read! 0 Make explanations correct but brief. You do not need to write a lot of prose. 0 Include diagrams! . . . , we. _ —
0 Show what goes into a calculatlon, not Just the ﬁnal number, e.g.. c.— — (2x10“ )(4x10 ) _
"%L\',.~;;‘95 i r GiVe standard SI units with your results. Unless speciﬁcally asked to derive a result, you may start from the formulas given on the
formula sheet, including equations corresponding to the fundamental concepts. If a formula
you need is not given, you must derive it. If you‘cannot do some portion of a problem, invent ,a symbol for the quantity you can’t
calculate (explain that you are doing this), and use it to do the rest of the problem. Honor Pledge “In accordance with the Georgia Tech Honor Code, I have neither given
nor received unauthorized aid on this test.” Sign your name on the line above PHYS 2212
Do not write on this page! Problem
Problem 1 (25 pts
Problem 2 (25 pt Problem 3 (25 pts
Problem 4 (25 pts) Problem 1 (25 Points) A point charge of +22 MC (22 x 10—6 C) is located at (2, 7, 5) m. (a, 12 pts) At observation location (—3, 5, 2> In, what is the (vector) electric ﬁeld contributed by this
charge? You need to show all your work carefully in order to get credit, ﬁaL—l? H Chm)“ X 22 MD M ‘6 __'__
’i z .7
: “<"””2053,2ox>mg~/¢ m: gamma) R A J. — ~277>
v“ m<i : “@549 , $0.127 20729 (b, 13 pts) Next, a singly charged chlorine ion is placed at the location (—3, 5, 2) In. What is the (vector)
force on the chlorine? Show your work. ,__ — :”\.C7<\D‘HC*~
a; 31E 6; ‘16
:. is E r, (2.30 ,o.‘\2 ,3~92~>X‘0 N Wat Problem 2 (25 Points) A neutral metal cube is placed a distance d = 15 cm from a dipole consisting of two tiny plastic spheres
with charges +5 x 10‘9 C and —5 x 10—9 C, separated by a distance of s = 7 millimeters, as shown in the diagram. ++++ (a, 8 pts) On the diagram above, draw the approximate charge distribution in and / or on the metal cube. (b, 8 pts) Draw a vector showing the electric ﬁeld at the center of the cube due only to the charges in
—+
and/or on the cube, and label it Ecube. (c, 9 pts) Calculate the magnitude of the electric ﬁeld vector you drew in part Explain your
reasoning clearly; how do you know what the magnitude of this vector is? End 7/0 \Mséxe ” I ' EL "l’ Ecuga '1'» O
I M \écokel :léle " Hﬁﬁ l5} — 43
.. chMOl x'ESMO X’IMD
u.» M Q61 \0'2’)3 ’ : ﬁssgN/C Problem 3 (25 Points) A small glass ball is rubbed all over with a small silk cloth and acquires a charge of +5 nC. The silk cloth
and the glass ball are placed 30 cm apart. If any of the electric ﬁelds you are asked to draw are zero,
state this explicitly. (a, 3 pts) On the diagram below, at the location marked ”x,” draw and label two arrows, representing the
electric ﬁeld due to the silk cloth and the electric ﬁeld due to the glass ball. In your drawing, pay close
attention to the relative lengths of these arrows. —~ 5 9 E Esuu C D
‘1'“
E
silk cloth ﬁlm’s glass ball (b, 3 pts) Now a positively charged metal block is placed between the two objects. On the diagram below
draw the approximate charge distribution for the negatively charged metal block. silk cloth glass ball positively charged metal block (c7 16 pts) Using the same diagram from part (b), at the location marked ”x”, draw and label four arrows
using the same scale as in part (a): The electric ﬁeld due to the silk cloth Edam (4 pts), the electric ﬁeld
due to the glass ball ﬁlm” (4 pts), the electric ﬁeld due to the charges in and/or on the metal block
imam; (4 pts), The net electric ﬁeld End (4 pts). (g, 3 pts) Now the metal block is removed, and replaced by a neutral plastic block. On the diagram
below show the polarization of a molecule at each location marked ”x” below. silk cloth glass ball plastic block Problem 4 (25 Points) A metal sphere of radius 0.25 m carries an initial charge Q1 2 +1.4 X 10‘5 coulomb. (a, 6 pts) An electric ﬁeld Emma; = 3 X 106 N / C is required to ionize air. Is the electric ﬁeld from the
sphere, anywhere, sufﬁcient to ionize the surrounding air? Justify your answer. " _ J.“ S: A s ~5 a __ _____~‘ .—~ ‘ _ C Web I; Esézjxxb 7H Hlxm : Q'OWlDé—N
'er 6» Swime. O K grace. "62‘Roxéius >> E<§S . 1’. ﬁg EdoéN/c . (b, 6 pts) Your body is initially neutral, and the soles of your shoes are made of insulating material. You
move your ﬁnger close to the sphere but without touching the sphere. When your ﬁnger gets very close to
the sphere you see a spark in the air between your ﬁnger and the sphere. Explain why a spark appears,
despite Q1 being insufﬁcient to ionize the air before you brought your ﬁnger near. Remember to include appropriate diagrams! /
£3 E Homd \S \oolox‘xzeé clue 50
H + ﬁecﬂﬁtm’éfl‘m4 n p r
Fat; and “We can Lomzeo,
(c, 6 pts) After seeing the spark you move your ﬁnger away from the sphere. The charge on the sphere is
now Q2. Is Q2 less than, equal to, or greater than Q1? Explain brieﬂy. labia 3t ’5th (d, 7 pts) In terms of Q1 and Q2, what is your NJ
(ﬂame: so}:®l’@2 S‘vmglé WAD an “a This page is for extra work, if needed. Things you must know Relationship between electric ﬁeld and electric force Conservation of charge
Electric ﬁeld of a point charge The Superposition Principle
Relationship between magnetic ﬁeld and magnetic force Magnetic ﬁeld of a moving point charge Other Fundamental Concepts _, _ d1? d]? _ —’ d]? N _..
a—E E—Fnet ﬁanda~ma1fv<<c
AUel = qAV AV = — ff E  dl z — Z (ExAx + EyAy + EzAz)
(Delszofsz @mangéoﬁdA
fﬁoﬁdA = qud6 f3 . MA =
60
s s (1(1) a s
lemfl = f ENC . dl I dngag f B . dl : H0 Elmside path
—. d<I> ~ d ~ A
emf : fENC 0 dl : dV’tmg f B 0 dl : lug Imside path + 60% f E O ndA
Speciﬁc Results a 1 2qs . —» 1 qs _
lEdipole,aacis z 47mg T—3 (on 3X15, 7' >> 3) lEdipole,1‘ '3 47r€O T—3 (011 J 3‘Xlsa T >> 3) s 1 a
Emd = 4WD W (7" __ from center) electric dipole moment 19 = qs, 15' = a Eapplied _. 1 Q E = ————— 7" __ from center rod 47r€0 Ir r————T2 + ( ) a 1 2Q/L _ —. 1 qz .
Emd ~ 47m) T (if T << L) ’Ering = 4WD W (2 along ams) ~ _ Q/A Z . ~ Q/A Z Q/A .
‘Edisk — 260 1 — m (2 along aXlS) Edisk z 260 [ — ~ 260 (1fz < R) ’ Q A . ~ Q A s . . .
’Ecapwtm % 6: (+62 and —Q dlsks) lEfTZnge z 6/0 Just outsxde capaCItor 4 IA? A ~ ~ ~ AB = 'u—g T: T (short Wire) AF 2 IAl X B “ #0 LI ,uo 2I _. ..
13m = z 1;? <7" < D leim = W a _ p0 2I71'R2 N ,uo 2I7rR2 ‘ 2 Bloop ~EW§~E Z3 (on ax1s, Z>>R) p,=IA:I7rR ’ H0 2 . ~ .
‘Bdipolemis z ET—g (on anus, 7' >> 3) Edi1,0161% z fit—2% (on J. ax1s, T‘ >> 3) * 1 —q5:_[_ A A A 4 ‘Erad rad : 4mm cgr 1’ : rad X Brad Brad : c
i=nA17 I=qnA17 17=uE I L U—lqlnu J—Z_aE R—a E  1 1
' Edielectm'c = applied AV = q — — — due to a point charge
K 471’60 rf m
IAV . . .
I = R for an ohmic reelstor (R Independent of AV); power = I AV
Q = C IAVI
AV . .
Q = C IAVI Power = IAV I = R (ohmic reSIStor)
d“ ‘ 2
K m %mv2 ifv << 0 circular motion: 125i 2 % I17] % ﬂ};—
Math Help
a x 5: (ax,ay,az) >< (bm,by,bz)
= (ay bz — az by)§7 — (ax bz — az 1%)?) + (aac by — .ay bx)2
da: dzc 1 d1: 1
Z 1 —— Z — ———— : ————————
/a:+a “(am)” /(m+a)2 a+x+c /(a+x)3 2mm)?“
‘1 2 2 a 3
/adx=a:r+c /awdx=§a: +c /ax dngx +c
Constant Symbol Approximate Value
—T——__—————8_——#_—
Speed of hght c 3 X 10 m/s
Gravitational constant G 6.7 x 10'11 N  m2/kg2
Approx. grav ﬁeld near Earth’s surface g 9.8 N/kg
Electron mass me 9 x 10'31 kg
Proton mass mp 1.7 x 10“27 kg
Neutron mass m” 1.7 x 10‘27 kg
1
Electric constant 4 9 X 109 N  m2/C2
7T60
Epsilonzero 60 8.85 x 10’12 (N  m2/C2)"1
Magnetic constant 5—0 1 x 10—7 T  m/ A
7r
Mu—zero [1,0 47r X 10‘7 T  m/A
Proton charge 6 1.6 x 10“19 C
Electron volt 1 eV 1.6 x 10'19 J
Avogadro’s number N A 6.02 X 1023 molecules /mole
Atomic radius Ra z 1 x 10‘10 m
Proton radius RI, z 1 X 10—15 m
E to ionize air Eiomze z 3 X 106 V/m BEarth (horizontal component) BEaTth z 2 X 10’5 T ...
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 Spring '09
 Kindermann
 Physics

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