test2_solution - Problem I {213 pts} A capacitor consists...

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Unformatted text preview: Problem I {213 pts} A capacitor consists cfttm parallel circular metal disks, both with a radius of IE cm and separated by ISL? mm. The disks carry trnil‘ortt'rlt.r distributed charges of equal magnitude and opposite sign. One of the disks has a sheet of plastic against it. The plastic sheet is {Lil mm thick and hes a dielectric oonalant'cf2.5. I|r'r"itlt the plastic in place, the potential difference ll m P}, — If... = 30' V. [Locations A and B are at the centers of the disks on the outside surfaces.) ta. 2 [315} At location C. in the center of the air gap. drew the direction of the j. : _:_ __ _ electric field. .. . .. {I}, 2 pts} Which plate, a. or a, has the positive charge? 53 (b. 12 pts) Calculate the magnitude of charge on one ofthe disks. Uscti= — 2543 = [Eta/Mr +{Eflfld‘q1m {figure not to scale} £- iffo if; Eat—“(HH— ‘: 191’— : 6’ 5-“?qu .. = “f 1"le a! 0. 9e“3 + 0 3e'3)h Y‘Gc “1 (-v—EK +£€,rc) ( 2'3- ' (e, 4 pls} Location D is 2 mm directlyr below C. What is VD — If}? Explain briefly. A I:- VDFV I because. E _l—. {5.6 film-'1; Flt}! CAD. W o 5 Problem 2 (30 pts)w '~ ‘ l ’ U.) 0 B --———--+ 4 -------------------------------------------- --L=70cm """"""""""""" --_---‘-> A thin, glass rod, length L = 70 cm, with a uniformly distributed positive charge of Q = 3e—8 C lies on the y-axis. Glued to the center of the rod is a small plastic sphere with a negative charge of q = —6e—10 C. Location A is 1.2 cm from the rod, and location B is 3 cm from the rod, both on a line perpendicular to the rod. (You may safely ignore polarization of either the rod or the sphere in your answers to the questions below.) (a, 2 pts) Draw the direction of the electric field due to the sphere at location B. Label the vector Esphere. (b, 2 pts) Draw the direction of the electric field due to the rod at location B. Label the vector Erod. (c, 20 pts) Calculate the potential difference VB — VA. B ...\ '5 A _ 4- a .Z’ VB’VA = " A Ell/ET '42“ djfESPHflE.’U — £5100 "£6 .. : -— 5"“ . A: .1— -'-/- —- "é —IO "L" fl; (VB V‘lrwrké A Emu“ aw Z]§é("b Q>-C e X???) 0,03 0-0” A re - 2(0/1.) I .20 f3 - e A _ __ _ I (VB“VA)Roo = "‘ Ekoo'flz “f ’1‘ Jr _ ‘1’?“ ‘- 1: r a” A “[7760 V Y‘ Q A : -4 £3 [a r] '3 = .4. BEE/um $0») Lyric. 1. VA 9m. L = -creo(2)c3e—€)(M_oa>www»: - 70M ’v’ 0 7 varvA: (VB—mwmé JrO/fmkon: .970 v- 7mm v = »— ‘lBéfiV (f, 6 pts) If an electron were released from rest at location B, would it reach location A? If so, calculate the kinetic energy of the electron when it reaches A. If it is not possible for the electron to reach A, briefly explain why. CquSe'fx’t/AT/m/ 0F EUERé)’; l NOTE: A K :0 VA—vvlah: "Q/G"VA): I<A~I<I3 4" U4 "43:0 ' M'UB=;(%~VB)=_3(VA-VB) KA + ("6)(\//+”Vi3)=0 / ufl—u8 <0) 50 ELEc'new SPoN 71!. u E at)ch “WE—S Kit: (QKVAfli/B): Qéefilq)(+q36'civ) \ 7‘0 Lac. 4 K4=é.‘i‘?e’/77 ( I » I'rohlcm s {211 pts} ' .M a particular instant in timo,_al1 alpha particle {helium nucleus, consists oftwo protons and two neutrons] is at location < —I .Sa—fi, I). t]:- m and has a trclocit}r of <1], 21:5, 013. ms. An clcclron is at < Chic—IS, 43666-6, I]? m and has a 1I.I'cln|:ncil'_t,r of €335.13, I}? Ms. Calculstc thtnct magnetic field ductu tilts: two particles at the origin {location 0). Show all work. Express you answer :13 a 3-WD1PDI1EI'II fi'flflfll‘. 3. £23th ’1?! “9!: _. (L 5‘3“: )4- 1—3.3‘2’13'771'! {u *1 FIR. o’er (-0.53"‘} 0,366.; -C,o> ;Eiflf‘ sma- M V I'E‘C; m if] :F‘f sin a d. $=ch If“! "‘ s 2% WK r WY”) = -? 1r. 4‘? Ban, <0 I o! if? _E%——— > <0 011012;) 1: )(lrefxa an» Problem 4 (30 pts) I. E L FCTRO‘V (,lt/ll'i/IVZXSS 1’ Q U131? WI— (1011111353; ‘3 l but u mimith H l 0“ I A long wire is connected to a battery, and a steady current runs through the wire. The wire passes once over Compass #1, and twice over Compass #2, as shown in the diagram. While current runs in the wire, Compass #1 deflects 16 degrees to the West, as shown. When the wire lies on top of the compass, it is about 5 mm above the compass needle. Location B is at the center of the loop. (a: 4 pts) Draw an arrow on the diagram showing the direction of electron current in the segment of the wire that passes over Compass #1. Label the arrow “electron current”. (b: 4 pts) Draw an arrow on the diagram showing the direction in which the needle of Compass #2 will point. (c: 4 pts) What is the magnitude of the deflection of the needle of Compass #2? Explain your reasoning. AGO“. 32°. Twmg THE (WC/CWT FLOWS {mm 460W? COMPASS .2 COM/.4165?) 7'0 COM/ASS 1, so 8 15 0? 775455 ARI/9 7H6" pff—‘(gcnoy ,5 App/ear. ,2 TIMES “£61572 ((1: 4 pts) Draw (or state in words) the direction of the magnetic field at location B, due only to the current in the wire. [N79 P4 6 6' (e: 12 pts) What is the magnitude of the magnetic field at location B, due only to the current in the wire? Clearly show all steps in your work. ,_ __ CUKKEWT m wuee': J. ~— 0. /‘/3 A Bee-cmcs-i: 65mm tan/6 o - Loun'lo'v 8 IS 5AM€ DIST4A/cg FROM £65, QOSCM) .53e~$ tan/69 TOP & BoTTOM w; ( :5‘73 ~G7— ’8'! DUE TO ALL val/(6’5 mar MI J‘AME ' :1: DIRECT/aw AT (06, B, 8cm?” : :19 ’” Jo. c417 r‘ o? (31-) ,/ -7)(2)(3) {0 [LI/3) B *‘ 3m * ——° reg/,2. (cams: :I tr”? r 0 0 25) £10) (a) a , “In — -4: (5.73e'é)(fe'3) 8%,, 3,419 x10 7— J/ I _. (/ e «7) ( a ) (f: 2 pts) If you made any approximations or simplifying assumptions in your calculation in part (e), state them here: -A5sume’o wmes we’fze' Ve’w/ mug COMPARED 'TO Dar/mu: To LOC. 3 .Aswmw my .2 wines ou comp/ts #2 maze" ESSF‘VWLLY THE §AME DISTANCE PM”! 6, . Agfomé’O EAsTbru/EW‘ 51/05 Do Ua‘r CauTl?(l3d/‘Z{ Sléwltlotflfl‘z ...
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test2_solution - Problem I {213 pts} A capacitor consists...

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