Test3_fall2008 - PHYS 2212, Test 3, November 5, 2008 Name...

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Unformatted text preview: PHYS 2212, Test 3, November 5, 2008 Name (print) ____________ "liganuL __________ _; ___________________________________________________ __ Instructions Read all problems carefully before attempting to solve them. Your work must be legible, and the organization must be clear. You must show all work, including correct vector notation. Correct answers without adequate explanation will be counted wrong. Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anything you don’t want us to read! Make explanations correct but brief. You do not need to write a lot of prose. Include diagrams! —3 6 Show what goes into a calculation, not just the final number, e.g.: :43 = W = 5 X 104 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. i 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 1 (25 Points) A circuit is constructed from two identical 1.5 V batteries and two wires, as shown. The wires are made of the same material (tungsten), and have the same length (50 cm), but have different diameters: the left-hand wire has radius 1.0 cm and the right—hand wire has radius 0.5 cm. The electron mobility of tungsten is 1.8e—3 (m/s)/(V/m). Tungsten has 6e+28 mobile electrons per cubic meter. V B Battery #1 ‘G Bit #2 ,l l " Ill . 36'ny BGFEDCB Osiva’fc-twf’ E D E1 -3+AV,+AVZ :0 j v e m * " ' A - :: , £2) E 1,9Ll E1 (-19.1: A) 5: Ra.” QR\ ’> AV; “Al/i ' (a 5pts) Complete the graph for the potential vs. position along the circuit. :. AV, "‘ a Avg alga-y '-.===-. _. (b 6pts) On the circuit diagram, draw the electric field vectors at each of the 6 points indicated by X. Pa attention to the relative size f our vectors. . . '— __ " y i:\ MAO yy{ .3 T 18 Same. 35 E1 ' 1" Eu (c 7pts) Calculate the agnitude of the electric field at point C. Show all your work, and express your answer in N/C. (AV; .. AVL: "“ EI'AL’). 1% :. ~Ez.(0.5> CY :. M a I (d 7pts) How long does it take a given electron to travel from one end to the other of the thin wire? Show your work, and express your answer in seconds. Veloc‘da a} 7' Ué Qzélr; '35}; £1: at v etc“ 3 Problem 2 (25 Points) . (a lOpts) A particular capacitor has a separation between its plates of 0.03 mm. The area of one plate is 16 mg. The capacitor is initially uncharged. Draw a graph of the amount of charge on one plate of the capacitor vs. time, starting from the moment the capacitor is connected to a bulb and two ordinary 1.5 V batteries. Your graph should extend slightly past the time when the bulb goes out. charge (b 15pts) The circuit remains connected several minutes after the bulb has gone out. How much charge is now on the positive plate of the capacitor? Start from physics principles. Show your work clearly and express your final answer in Coulombs. C: 3Q2CV:6‘ _ . /I-———————-———.— Problem 3 (25 Points) . At a particular instant a proton is moving with velocity (5.5 X 107, 0,0) m/s, and an electron is moving with velocity (5 x 107, —5 X 107, 0) m/s. The electron is located 1.5 x 10—6 m below the proton (in the y direction). ' 5:: N a ‘9 0.. “1 \4 o , “r 5 (our ofpage (a 4pts) On the diagram draw an arrow representing the electric field at the location of the electron, due to the proton. Label it Ep. (b 4pts) On the diagram draw an arrow representing the magnetic field at the location of the electron, due to the proton. Label it Bp. (c 4pts) On the diagram draw an arrow representing the electric force on the electron at this instant. Label . it F51 . (d 4pts) On the diagram draw an arrow representing the magnetic force on the electron at this instant. Label it Fmag. (e 9pts) Calculate the magnetic force on the electron at this instant. Your answer should be a vector. Show all work. T: : Worm) Te: ~i.em“"‘c :; <5x\b7,—SMD7J '5). m/J V; ‘ f-W‘ i: <2: 4i WED Problem 4 (25 Points) An electron moving with a velocity <2 X 107, O, 0) m/s enters a region between two parallel plates that are 6 mm apart and have a potential difference of 200 V. The electron is deflected toward the top plate. Elufi'muo Fm. avatar «an! A g r is at». *C'k + Pm“ y g g P‘DOe's»®Pl’W x Z (a 5pts) At the location of the electron, draw the the electric field vector due to the parallel plates and label it E. (b 5pts) At the location of the electron, draw the electric force vector acting on the electron and label it Felec- (c 5pts) At the location of the electron, draw the direction of the magnetic field that has to be applied if the electron is to travel between the plates without any deflection, and label it B. (d 10pts) What is the magnitude of the magnetic field for which the electron travels between the plates without any deflection? NO 6F e.“ é m mag ‘ C. PM: ' QWC— 95%. AL ’ 5.22%" m :. 229‘s CND ’q of \ex -:- \E-xm“T ; team T I /\ b'wed‘lm ‘5 $1 "HZ‘ . This page is for extra work, if needed. Things you must know Relationship between electric field and electric force Conservation of charge Electric field of a point charge The Superposition Principle Relationship between magnetic field and magnetic force Magnetic field of a moving point charge Other Fundamental Concepts _, d1? d" ~ d_' _. . 02E 31%: net fiandd—fzmaifv<<c AUez = qAV AV = —fiondl z —Z(E$Ax+EyAy+EzAz) @elszlfidA @mangggfid/l ' ffiofidAzw ffiofidAz E0 - a do - - lemfl : fENC . : drgag . : .U'O ZL‘nside path _. _. (I) —» d -v A |emf| I: fENC O = d dimly f B . = M0 [inside path + 60% . ndA Specific Results —» 1 2 s . —» 1 qs . ’Edipole,a$is ” 47(60 Ti?) (0n ax1s, 7“ >> 8) lEdl-pole z 47TH) E (on J. ams, 7° >> 3) c 1 Q . . - a E = —— J. f t l t d l t 2 , = E - rod 4WD T T2 + (L/2)2 (7‘ rom cen er) e ec I‘lC IpO e momen p qs p (1 applied —‘ 1 Emd = ——Q———— (r J. from center) 47T60 “H”? + (L/2)2 ~ 1 2Q/L . ~ 1 qz . z . 2 __ 1 Brad 471.60 T (If 7" << ‘Emng 47760 (Z2 + R2)3/2 (Z a 011g 8X18) ~ Q/A z . {a Q/A[ z] Q/A , ' = — —— - z — — x f R ‘Edzsk 260 1 (Z2 + R2)1/2 (2 along ems) Edzsk 260 R 260 (1 2 << ) - A a A . . . Ecapacitor Q Q/ (+Q and ——Q disks) Efm-nge x Q/ Just out51de capac1tor 60 60 AB = flIAl X T (short Wire) A}? = IATX E 471’ T2 ~‘ fill/0 ~,LL02I a. _1—o l I wzre — ~ (7’ < L) ‘Bwn‘e — Bearth tan6‘ a _ M0 217TR2 ~ M0 21'7rR2 , _ _ 2 {Bloop —EW~E 23 (on ax15,z>>R) M—IA—IWR ~ 2 . ~ #0 u . leipoleyaxis “ fit—7:7? (011 3X13: 7” >> 8) ‘Bdipoleyi’ ~ 55 (on J. ax1s, 7" >> 5) - E d -' _ 1 —qaL A _ A A - _ m rad — 471,60 C2T 1} — rad X Brad Brad — C 1=nA77 I=|q|nA17 UzuE I L 0=|q|nu Jzing R_a ‘ * Eu '6 1 1 t Edielectric = ppm d AV = q — — — due to a point charge K 47TEO Tf T‘i . IAVI . . . I = R for an ohmlc res1stor (R Independent of AV); power 2 I AV Q = C |AV| |AV| . . Q = C IAVI Power = I AV I = (ohmic res1stor) . d-' ~ 2 K % émv2 1f 1} << 0 circular motion: E—i = % z 17% Math Help {ix 5: (awawz) >< (bz,by,bz) = (ay bz — a; by)§: — (ax bz # a2 1275)?) + (aan by —' (1y bw)2 / dav 1 ( + ) + / d3: 1 + / dm 1 + C : n ——.._— 2 _ C —_ = A.__.___ ac+a a m c (a:+a)2 a+ac (a+x)3 2(a+ar)2 ‘1 2 2 a 3 /adav=ax+c /a$dm=§x +c /a$ daz=§x +c Constant Symbol Approximate Value Speed of light c 3 X 108 m/s Gravitational constant G 6.7 x 10'11 N - m2/kg2 Approx. grav field near Earths surface 9 9.8 N/ kg Electron mass mi3 9 x 10’31 kg Proton mass mp 1.7 X 10’27 kg Neutron mass mm 1.7 x 10‘27 kg 1 Electric constant 4 9 X 109 N - m2/C2 7760 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 no 471' 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 R, x 1 X 10‘10 m Proton radius Rp % 1 x 10—15 m E to ionize air Eicmize z 3 x 106 V/m BEarth (horizontal component) B Earth % 2 X 10‘5 T ...
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This note was uploaded on 04/10/2011 for the course PHYS 2212 taught by Professor Kindermann during the Spring '09 term at Georgia Institute of Technology.

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Test3_fall2008 - PHYS 2212, Test 3, November 5, 2008 Name...

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