HT1_Solution

HT1_Solution - Department of Physics, Lehigh University...

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Unformatted text preview: Department of Physics, Lehigh University Physics 21 — Introductory Physics II Spring 2011 Hour Exam 1 February 16, 2011 Closed Notes 4: 10 — 5:10 PM Student’s Name Recitation Section Number Recitation Leader’s Name The test is a multiple-choice examination. First, check to be sure that you have a six-page examination, including the equation sheet. Please answer each question by writing the letter corresponding to your choice in the answer area. Answers marked outside the answer areas will not be scored. Problem 1: You have two charges. One charge is located at x= - 3.0m. and has a value of 4.0 Coulombs. The other charge is located at x = +3.0 m and has a charge value of - 2.0 Coulombs. 1) What is the magnitude of the force between the two charges? (a) 7.0 N (b) 6.0 x 10 9 N (c) 57 N (d) 2.0 x 109N Answer: 2) What is the force on a charge, q, of charge value 1.0 C placed at the center (x = 0.0 In) between them? (a) 6.0 x 109N (b) 84 x 109 N (c) 2.0 xio9N 0\ (d) 6700 N Answer: 3) Where on the x axis will the force on a 1.0 C charge be zero due to the two original charges? (a) 35.673 m (b) 4.6 in (c) 89.21 m (d) 17.484 In ' Answer: 4) What is the x component of electric field at x I 0.0 m, y = 3.0 In due to the two original charges? (a) 89 x109 N/C (b) 43.25 x 109 N/C (c)2.12X109N/C C (d) — 76 N/C Answer: 5) What is the y component of electric field at x = 0.0 m, y = 3.0 In due to the two original -.harges? (a) 5.21 x 104N/C 7 5 (b) 650 x 109 MC (0) -37 NC (d) 7.07 X 108 MC Answer: Problem 2: 1) Consider the following statements: I. If an enclosing surface has zero total electric flux through it, then E = O everywhere on the surface. II. If an enclosing surface has E z 0 everywhere on it, then the total electric flux through it is zero. (a) Only statement I is correct (b) Only statement His correct (0) Neither statement is correct ((1) Both statements are correct Answer: 2) A point charge Q is at the center of two Gaussian spheres, sphere l of radius R and sphere 2 of radius 2R. How does the total electric flux through the two spheres compare? (a) It depends on the sign of the charge (b)eR=e2R=0 (C) @R = 9523 i 0 C (d) @523 < @523 Answer: 3) A point charge Q is at the center of two Gaussian spheres, sphere 1 of radius R and sphere 2 of radius 2R. How does the strength of the electric field at the surfaces of the two spheres compare? (5) ER = E2R 3 0 (b) ER 2 Ezra 5’5 0 (C) ER > Em C (d) E R < E2}; Answer: ' 4) A point charge 9! is at the center a Gaussian sphere of radius r0. Later, the charge is moved to the right a distance ro/Z, but the sphere remains where it was. How do the total electric flu}: through the sphere and electric field at the surface of the sphere change? (a) the flux doesn't change but the field changes (b) the flux changes but the field doesn't change (c) neither the flux nor the field change Gk (d) both the flux and the field change Answer: 5) A charge of 6e is at the center of a cube of side length 1.00 X 10'6 n1. What is the electric flux through one face of the cube? (a) 1.81X10'8N'm2/C (b)3.02><10'9N-m2/C ‘ . (c)1.00 X10“12N-m2/C Gk (d) 0 (due to symmetry) ' Answer: 1 . J i i ;x 1 e ~._ Pk ,. 3 mfimgmiff in; to (we. (ET, I P F -' D {we V‘VWL/{fiffi can C31 E H l . f {h i_ :1, . z ,5; .: A i fiiflf’}{€‘4?fi ’11 HEEL”; :‘qu' '1‘. ’ trey-:23 0W a i I r; ,rJ‘ i ‘ amol :Lx (Tl/ms) “FL: +o+¢f 71:qu :3 afrgl i a . Chara; ; 1 ‘ ; 1 O. .. 4 _, F1 .— ~ In \ I s: ‘ ‘9‘??? r5”? ' r e i: Ss‘vxcg Splmzrgg EmCiOSC fha Sarfli J‘ 1 1 ' “Cu: :7 1 I f g 1 {3i , A Chargfi> é—‘wflv‘ imam “HQ age-ML ‘Rfiw 3TW‘X} fi'fla’!‘ Es'mf‘fi ’Ma ‘= I . 3.! m‘ ‘ ' Charfifl. QWCIBSQOE Jg Ada/{€312}; {EA/e: JroJFal TIN/4X, a; may} if;{"§_ t Cine ' £31 — I . i E2 r‘fii‘ny baud; Law 7L0 C. rPna'itRE, “A: racirug 2’" Emelogmfl ghom‘e a J t E CD A k 2 (-7 Q ~ m 1 j a «4;, M $> Efifir ;»29 _9 a“ QL o“ @fieal a , Problem 3: Aplastic disk of radius R, charged on its top surface to a uniform surface charge density (7. raw 1) What is the total charge on the surface of the disk? (a) 0(4mr22) (b) 0(3 HR3) (c) can?) (d) a 3 Answer: L I 2) What integral is used to derive the potential at point P? (a) V = ff Cb) V = I: evzfi;%%fi @V2fififl ! 3) The potential at point P is V = a? (V22 + R2 — 2) What is the Electric field? 0 __C’_ 2 _ Ema. Z “DE—250W 1) (W3 mum—2) __=L _ z *1 z (GE—2500 m2“) (d) Iii—250927432) Answer: 4. A point charge(charge q) is placed directly above point P a distance 2 away (22 fiom the disk), What is the new electric potential at point P? (a)V=2-::;(\/mmz)—k—q (b)V=2—:;(VZZ+R2—Z)+% z k (c)V=—g—2—:;(m—z) (aw: (5%(W—2D2H? z V b Answer: ——W__ 5. What is the electric potential energy of the point charge (U = qV)? (a) U = 571(sz + R2 — z) (b) U = Egg/(22y + R2 — 22) @U=E @ U:E 2 Answer: Problem 4: As showu in figure, a potential difference V= 15.0 V is applied across a capacitor arrangement with capacitances C1 = 2.0 ,uF, C2 = 6.0 ,uF, and C3 = 4.0 ,uF. After a while these capacitors are fully charged. 1) What is the charge on C3? (3) 40 so (b) 60 ac (c) 180 pC \g (d) 82.5 [4C Answer: 2) What is the energy stored in C3? (3.) 450 to (b) 30 M (c) 120 [.LJ ((1) 900 pl Answer: OK 3) What is the effective capacitance C535: of C1 and C2 combined? (a) 8.0 ,uF (c) 0.67 JuF C& (d) 1.5 ,uF Answer: 4) What is the charge on C1? (a) VC1 (b) VCgff \E (C) VCZ “D ((1) None of the above Answer: 5) Insert in C3 a material with dielectric constant K = 3.0. What is the energy stored in C3? (21) 150 [.LJ (b) 120 a] (c) 1350 “J C (d) 300 pl Answer: i xi :2 3V 31 (fiaMFjfls‘gygz _ * 1 Cut—(1 9. .n W J. C»! + C2. l .' anzefimg 245:} ‘3' .5}ng ,5 f, " .a‘r Emilia/:1 0/; W ; Problem 5: The circuit below has values V = 30 V, R1= 10 Q, R2 = 4 Q, R3 = 10 9., R4 = 2.5 Q. Calculate the following currents. R: 1) What is the current through resistor R1? (3) 1 Amp ([3) 3 Amps (c) 6 Amps L (d) 10 Amps Answer: 2) What is the current through resistor R2? (a) 5 Amps (b) 6 Amps (c) 8 Amps (d) 2 Amps Answor: 3) What is the current through resistor R3? (a) 6 Amps 00 2 Amps . (c) 0.5 Amps ‘ (d) 1Amp Answer: V 4) What is the current through resistor R4? (a) 4 Amps (b) 2 Amps (c) 10 Amps (‘31) 193-11113 Answerzw 5) What is the current from the power supply? (a) 30 Amps (13) 8 Amps ~ (c) 12 Amps (c1) 15 Amps Answer: Es D CMfr6/\* "H/‘Jaij (UH-hr R! "an, "E..._ i3. ‘J E. _ {-3 Z) {614+ +£h’5farjim EL Can-L in $1 C“ an raLKJ-ar U5! 6% Aa'lfermlnc_ 44m VON—c, L down Q (he? owes"! (21 ‘Cfoh. V‘: 114;; 30~§v‘-{ to lo w R [a r Y2 Lt +611“ Samar ‘ o 2. s“ " A" 5) (M! 4’ an} gozx {I s: 0:0 £617.; {hwy arm; H04 "I “Cram :u‘ ’23 19.14;? £1} {13 +127 3 LL'H—u Ufii-‘Lfidt GD ...
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HT1_Solution - Department of Physics, Lehigh University...

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