Test 1 Fall 2008

Test 1 Fall 2008 - PHYS 2212, Test 1, September 17, 2008...

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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 : “<"””20-53,2-ox>mg~/¢ m: gamma) R A- J. — ~277> v“ m<i : “@549 , $0.127 20-729 (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» Swim-e. 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 . . . ’Ecapw-tm % 6: (+62 and —Q dlsks) lEfTZ-nge 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) Edi-1,0161% 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=|q|nA17 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 Epsilon-zero 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 Eiom-ze z 3 X 106 V/m BEarth (horizontal component) BEaTth z 2 X 10’5 T ...
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Test 1 Fall 2008 - PHYS 2212, Test 1, September 17, 2008...

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