This preview has intentionally blurred parts. Sign up to view the full document

View Full Document

Unformatted Document Excerpt

066 Version Exam 1 chiu (57460) This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points A lightweight metal ball hangs from a thread, to the right of an plastic rod. (See Figure 1.) Both are initially uncharged. 1 Statement III : IIIa. The induced dipole moment for a conductor occurs over a macroscopic length scale. IIIb. The induced dipole moment for a conductor occurs over an atomic length scale. 1. Ib, IIa, IIIb 2. Ia, IIb, IIIa correct 3. Ia, IIa, IIIb 4. Ib, IIa, IIIa 5. Ia, IIa, IIIa 6. Ia, IIb, IIIb 7. Ib, IIb, IIIa 8. Ib, IIb, IIIb Explanation: Answer to statement I: The excess negative charge at the left end leads to the alignment of induced dipoles in the interior of the insulator rod. For the metal rod case, free electrons are on the surface of the rod. Ia is correct. Answer to statement II: Since electrons are free to move along the surface of a conductor, the average location of the free electrons in the conductor is closer to the average location of the induced negative charges in in the ball than is the case for the insulator. So the former gives a stronger attraction. IIb is correct. Answer to statement III: The induced dipole moment for the conductor is distributed throughout the conductor. Hence, the dipole moment induced is across a macroscopic length scale. IIIa is correct. 002 (part 1 of 2) 10.0 points Suppose you want to create an electric eld = 2, 2, 0 N /C at the origin located at 0, 0, 0 . Where would you place an electron to create this eld? Denote the coordinate vector of the electron by r = r r Determine r The plastic (insulator) case: You rub the left end of the plastic rod with wool, depositing charged molecular fragments whose total negative charge is that of 1 109 electrons. The metal (conductor) case: You perform a similar experiment with a conducting metal rod. You touch the left end of the rod with a charged metal object, depositing the same excess electrons to the left end. You then remove the object. Statement I : Consider two cases below: (i). There are polarized dipoles inside of the rod. (ii). There are electrons at the surface. Ia. (i) is applicable for the insulator. (ii) is applicable for the conductor. Ib. (ii) is applicable for the insulator. (i) is applicable for the conductor. Statement II : Compare the attraction between the rod and the ball IIa. The insulator rod has more attraction. IIb. The conductor rod has more attraction. Version 066 Exam 1 chiu (57460) 1. 1, 1, 0 1, 1, 0 2 1, 1, 0 3. 2 1, 1, 0 correct 4. 2 2. 5. 1, 1, 0 6. 2 1, 1, 0 7. 2 1, 1, 0 1, 1, 0 2 Explanation: The electron has a negative charge. The direction of the vector should along the direction of the electric eld. 8. E =E= r E 1 = 2, 2, 0 r 8 1 = 1, 1, 0 r 2 003 (part 2 of 2) 10.0 points Find the magnitude r (in units of m). 1. 2.26 107 m 2. 0.000452 m 3. 2.26 105 m correct 4. 0.000226 m 5. 2.26 106 m 6. 0.00226 m 7. 0.000678 m 8. 1.13 106 m 2 The magnitude of E is given by |E | = E = 2 2. But, we also know that E= r= ke r2 ke = 2.26 105 m E 004 10.0 points A wire 3 m long lying on the X-axis has a nearly uniform charge. The electric eld at location 0.02, 0.01, 0 m is 0, 66000, 0 N /C . What is the total charge on the wire? 1. -1.57e-07 2. -1.27e-07 3. -1.75e-07 4. -1.68e-07 5. -2.5e-07 6. -2.28e-07 7. -1.01e-07 8. -1.6e-07 9. -2.35e-07 10. -2.46e-07 Correct answer: 2.46 107 C. Explanation: Since the E-eld is negative, the charge generating it must also be negative. The distance r is given by r= x2 + y 2 + z 2 = 0.02236 m The magnitude of the electric eld is given by the formula E 1 2(Q/L) 4 0 r Hence the charge is given by Q = 2 0E L r = 2.46 107 C Explanation: 005 10.0 points A large, thin plastic disk with radius R = 1.6 m meter carries a uniformly distributed negative charge of Q = 7.7 105 C as shown in the Figure below. A circular piece of aluminum foil is placed d = 3 mm from the Version 066 Exam 1 chiu (57460) disk, parallel to the disk. The foil has a radius of r = 4 cm and a thickness t = 1 mm. The foil is neutral. Plastic disk R = 1. 6 m t = 1 mm r= d = 3 mm Close-up of foil 3 point indicated by the x-symbol, Enet Edisk + Ef oil Q 1q 1 + 2 0 Adisk 0 Af oil Q Af oil | q |= 2 Adisk =0 =0 =0 Q r2 = 2 R2 7.7 105 C = 2 = 2.41 108 C . 4 cm (0.04 m)2 (1.6 m)2 Calculate the magnitude q of the induced charge on the left circular face of the foil. 1. 6.94e-09 2. 2.44e-09 3. 2.23e-08 4. 1.13e-08 5. 3.84e-08 6. 2.41e-08 7. 9.94e-09 8. 1.4e-08 9. 3.25e-08 10. 1.44e-08 Your answer must be within 2.0% Correct answer: 2.41 108 C. Explanation: 006 10.0 points Consider the following two objects. A: a uniformly charged disk. B: a uniformly charged ring. Choose the correct pair of statements. When z R, the E eld is approximately constant for Ia. A only Ib. B only Ic. A and B When z R, the E eld is approximately proportional to z 2 for IIa. A only IIb. B only IIc. A and B 1. Ic, IIa 2. Ic, IIb 3. Ia, IIb 4. Ib, IIb 5. Ic, IIc 6. Ib, IIa Let : R = 1.6 m , r = 4 cm = 0.04 m , d = 3 mm , Q = 7.7 105 C t = 1 mm . and 7. Ia, IIc correct 8. Ia, IIa 9. Ib, IIc Explanation: In the gure, within the enlarged foil, at the Version 066 Exam 1 chiu (57460) Ia. For z R, the E eld of the ring rises linearly but the E eld of the disk is approximately constant. Ering = 1 qz 4 0 (R2 + z 2 )3/2 Q/A 20 4 Edisk = IIc. All nite charge distributions resemble a point charge when viewed from afar. 007 10.0 points A positive and a negative particle of equal magnitude of charge e are equidistant from a point C. The two particles and C are all collinear, and the two particles produce an E eld pointing north at C. If the particles are replaced by new particles 11 times farther away from C, what magnitude of charge must these new particles have in order to produce the same E eld at C? (Express your answer in units of Coulombs.) 1. 7.056e-17 2. 4.624e-17 3. 4.096e-17 4. 7.744e-17 5. 3.136e-17 6. 2.56e-18 7. 1.936e-17 8. 1.6e-17 9. 5.76e-18 10. 7.84e-18 Correct answer: 1.936 1017 C. Explanation: The eld due to the two particles is Ec = 1 2q 4 0 r 2 Two dipoles are oriented as shown in the gure above. Each dipole consists of two charges +q and -q, held apart by a rod of length s, and the center of each dipole is a distance d from location A. The dipole moment is given by p = qs. The magnitude and the direction of the dipole eld vector contributed by the lower-left dipole only is given by kp , up correct d3 kp 2. 3 , down d kp 3. 2 , up d 2kp 4. 2 , up d kp 5. 2 , down d 2kp 6. 3 , down d 2kp 7. 3 , up d 2kp 8. 2 , down d Explanation: At A the bottom one has a contribution of kp in the upwards direction. d3 1. 009 (part 2 of 2) 10.0 points Determine the magnitude of the resultant eld vector contributed by both dipoles. 1. 008 (part 1 of 2) 10.0 points 4kp d3 From this formula, observe that that if E is held xed, q is proportional to r 2 . Hence the value of q is q = e (11)2 = 1.936 1017 C Version 066 Exam 1 chiu (57460) 2. 3. 4. 5. 6. 7. 8. 9. 5kp d3 3kp correct d3 kp d3 kp 5d3 kp 3d3 kp 4d3 2kp d3 kp 2d3 5 the same direction and a negative ratio implies that they are pointing in the opposite direction. Determine this ratio. 1. 2 2. 2 3. - 3 4. - 2 correct 5. 3 6. 1 7. - 8 8. -2 9. -1 10. 8 Explanation: Notice that at A, E1 is perpendicular to E3 . The magnitude of E13 is given by E13 = where E1 = 2 2 E1 + E3 = 2kp At A the top dipole contributes 3 in the d upwards direction. Hence, the resulting total 3kp eld is 3 . d 010 10.0 points Explanation: 2E1 kq , and by symmetry E3 = E1 . a2 By inspection the magnitude of E2 is k (2q ) kq E2 = = 2 = E1 2 a ( 2a ) Consider the setup shown in the gure, where charges Q1, Q2 and Q3 and the point A occupy four corners of a square with the length a at each side. Given that Q1 = Q3 = q > 0 and Q2 = 2q . The resultant electric eld at A contributed by charges Q1 and Q3 is labeled as E13 . The electric eld at A contributed by Q2 is labeled as E2 . Verify that the vectors E13 and E2 are aligned along a same line. Such an E13 alignment implies that the ratio may be E2 represented by a number. Here, a positive ratio implies two that vectors are pointing in E13 and E2 are aligned along the same line and are pointing in the opposite direction. E13 The ratio is a negative number. So, we E2 have E13 2E 1 = E1 E2 E13 E2 011 = 2 10.0 points Version 066 Exam 1 chiu (57460) 6 A neutral copper block is polarized as shown in the above gure, due to an electric eld made by external charges (not shown). Consider the following statements: By inspection of the gure the correct answer is Ib, IIa, IIIa. The applied eld is responsible for having the positive surface charge at the bottom. So the applied eld is downward (Ib). The dipole eld direction runs from positive charge (the source) to the negative charge (the sink). So the eld goes up (IIa). Within the conducting medium, there are abundant free electrons. At equilibrium Enet at B must be zero. Otherwise free electrons will be moving. It will no longer in equilibrium.(IIIa) 012 10.0 points Ia. The applied external eld is pointing upward. Ib. The applied external eld is pointing downward. IIa. The dipole eld due to the surface charges is pointing upward. IIb. The dipole eld due to the surface charges is pointing downward. IIIa. At equilibrium, the magnitude of the net eld at B is zero. IIIb. The magnitude of the net eld at B is not zero even at equilibrium. 1. Ib, IIb, IIIa 2. Ib, IIa, IIIb 3. Ia, IIa, IIIb 4. Ia, IIb, IIIb 5. Ia, IIa, IIIa 6. Ib, IIa, IIIa correct 7. Ia, IIb, IIIa 8. Ib, IIb, IIIb Explanation: This problem concerns the two cases shown in the gures above. The top gure shows a positively charged glass sphere surrounded by a plastic shell while the bottom gure shows a positively charged glass sphere surrounded by a metal shell. Choose the correct pair of statements. If the magnitude of the E eld at the points Version 066 Exam 1 chiu (57460) is designated EP,p and EP,m for the plastic and metal cases, respectively, then Ia. EP 1,p > EP 1,m Ib. EP 1,p < EP 1,m Ic. EP 1,p = EP 1,m and IIa. EP 2,p > EP 2,m IIb. EP 2,p < EP 2,m IIc. EP 2,p = EP 2,m 1. Ib, IIb 2. Ic, IIb 3. Ib, IIa 4. Ia, IIa 5. Ia, IIb 6. Ia, IIc correct 7. Ib, IIc 8. Ic, IIa 9. Ic, IIc 7 Explanation: Ia. When exposed to an external E eld, the E eld inside of a conductor is zero, but the E eld inside of an insulator is not. IIc. Neither shell aects the E eld at P2 . For the conductor, the inner and outer surface charges produce canceling E elds at P2 . Likewise, for an insulator the radially symmetric distribution of dipoles produces a canceling E eld at P2 . 013 10.0 points The diagrams in Figure 1 show a sequence of events described below, which involves a small lightweight solid metal ball that is suspended from a cotton thread. (a) You touch the ball briey with your ngers then release it. (b) A block of metal that is known to be charged is now moved near the ball. (c) The ball briey touches the charged metal block. Then the ball swings away from the block and hangs motionless at an angle as shown in diagram 5 of Figure 1. (d) Finally the block is moved far away. A negatively charged rod is brought near the ball. The ball is repelled by the charged rod shown in diagram 6 of Figure 1. Among distributions E through N shown in Figure 2, which of the diagrams best shows the distribution of charge in the ball in slide 3 of Figure 1? Version 066 Exam 1 chiu (57460) 8 A plastic hollow sphere, which is uniformly charged with negative charge on its surface, is placed near the center of a horizontal plastic rod, which is uniformly charged with negative charge. What is the direction of the electric eld vector at P due to the charges on the plastic sphere alone? 1. L 2. K 3. E 4. F 5. G 6. J 7. H 8. M 9. N correct 10. I 1. direction 2 2. Zero magnitude correct 3. direction 4 4. direction 7 5. direction 5 Explanation: From the nal slide, we know the sphere is negatively charged. The only way it could have obtained this negative charge is from contact with the box, which therefore also must have been negatively charged. In slide 3, the sphere has a net charge of zero, but polarizes such that the positive side is closer to the negatively charged box. 014 (part 1 of 2) 10.0 points 6. direction 3 7. direction 1 8. direction 6 9. direction 8 Explanation: Since the charge distribution on the surface is uniform, the eld at P due to the charges on the sphere alone is 0. Version 066 Exam 1 chiu (57460) 015 (part 2 of 2) 10.0 points What is direction of the resultant eld vector contributed by the rod plus the sphere? 9 1. direction 5 correct 2. direction 6 3. direction 7 4. Zero magnitude 5. direction 1 6. direction 4 7. direction 8 8. direction 2 9. direction 3 The center of the spherical metal ball of radius R, carrying a negative charge -Q is located at a distance r from A. The point A is the center of a short, thin neutral copper wire of length L. The induced charges at the two ends of the wire is q . Using the condition that the eld at A due to -Q cancels the eld at A due to the induced dipole in the copper wire, one nds that: q= QL 8r 2 Determine the polarizability of the wire segment which has a length L = 0.0075 m. 1. 1.389e-17 2. 3.814e-18 3. 5.859e-18 4. 2.4e-17 5. 4.234e-17 6. 2.112e-17 7. 1.736e-18 8. 1.608e-17 9. 1.849e-17 10. 3.417e-17 Your answer must be within 2.0% Correct answer: 5.859 1018 . Explanation: The polarizability is given by = Explanation: The E eld due to the charged shell is 0 and the E eld due to the rod points downward. By the principle of superposition the correct answer is that the E-eld points in direction 5. 016 10.0 points qL p = kQ E r2 Using the value of q given above = (Q/8)(L/r )2 L kQ r2 L3 = 5.859 1018 = 8k Version 066 Exam 1 chiu (57460) 017 10.0 points 10 Two small, negatively charged plastic spheres are placed near a neutral iron block, as shown in the above gure. Which choice (a to j) in the gure best indicates the direction of the net electric eld at location A? 1. d 2. e 3. g 4. a 5. j 6. f correct 7. c 8. i 9. b 10. h Explanation: At A there is no net eld due to the negative charge pair. The elds contributed by the pair are equal and opposite and they cancel each other. There are induced charges at the left and lower surfaces of the iron cube which are symmetrically distributed by the 45 line and they are positive. There are also negative charges symmetrically distribution about the 45 line far away. The nearside charges dominate. By inspection f is the correct direction. 018 10.0 points A negatively charged iron block is placed in a region where there is an electric eld downward (in the y direction due to charges not shown). Which of the diagram in the gure above best describes the charge distribution in and/or on the iron block? (Caution: You should bubble your numerical choice, NOT the letter label of the diagram) 1. d 2. f 3. b 4. e 5. c correct 6. a Explanation: For present problem, the following 3 conditions must be satised: (i) Conductor medium: Net charges can only be at the surface. (ii) Net charge is negative. (iii) E is downward, negative charge moves upward. More negative charge in the upper surface is expected. Only diagram c satises all three conditions. a violates (ii) b violates (ii) d violates (iii) e violates (i) f violates (iii) 019 10.0 points Version 066 Exam 1 chiu (57460) Which is the correct integral formulation for the electric eld on the axis of a uniformly charged disk of total charge Q and radius R? 2 11 1. Ez = 0 1 4 0 Q R2 (R 2 z d + z 2 )3/2 2. Ez = 0 3. Ez = correct 4. Ez = Q z 1 d 4 0 2 (R2 + z 2 )3/2 R 1 Q z r dr 2 2 + z 2 )3/2 R (r 0 20 Q z 1 r dr R2 (R2 + z 2 )3/2 0 20 R Q z 1 5. Ez = r dr 2 + z 2 )3/2 2 ( r 0 20 2 Q z 1 d 6. Ez = 4 0 2 (R2 + z 2 )3/2 0 R Q z 1 r dr 7. Ez = 2 + z 2 )3/2 2 ( R 0 20 Q z 1 d 8. Ez = 2 2 + z 2 )3/2 R (R 0 4 0 Explanation: The electric eld of a uniformly charged thin ring is given by 1 qz E= 4 0 (R2 + z 2 ) 3/2 A disk can be divided into a sequence of concentric rings. The only non-vanishing component of the electric eld will be in the z direction. 1 q z E z = 4 0 (r 2 + z 2 ) 3/2 Now, we evaluate q and it is given by 2r r q = Q R2 For an innitesimal ring, we have 1Q z r dr dEz = 22 2 0 R (r + z 2 ) 3/2 Thus, the value of Ez is R R Ez = 0 1 20 Q R2 z r dr (r 2 + z 2 )3/2 ... View Full Document

End of Preview

Sign up now to access the rest of the document