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Version 025 – MT1 v3 – yao – (57465) This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 1 1 − = (2q)2pk 3 (2r − s) (2r + s)3 " # 4pkq 1 1 = 3 − 3 8r 3 1− s 1+ s 2r 001 10.0 points 1 2r Using the small ǫ approximation, where ǫ = s/2r, we obtain due to p ≈ Fon p1 pkq 3 p2 × 6ǫ = k . 2r 3 2 r4 This leads to: The figure shows the set up of problem 15.P.44 in the M&I text, which concerns the interaction of two identical permanent dipoles of moment p = qs separated by a distance r. Let us consider a new situation: replace the left dipole with a new dipole of moment p1 = Qs′ , where s′ = 2s and Q = 2q. Let the distance between the two dipoles be increased to r ′ = 2r. The right dipole, denoted p, remains unchanged. Find R, the ratio of the force on p1 due to p to the force on the original left dipole p0 due to p: R= due to p Fon p1 due to p Fon p0 R= = 002 due to p Fon p1 due to p Fon p0 2 3 p 2 k r4 2 6k pr4 = 1 . 4 10.0 points . due to p is given in the figure. Note that Fon p0 Choose the correct answer. 1. 1/4 correct 2. 1/2 3. 2 4. 1 5. 4 Explanation: p p due to p + (−Q)EA = QEB Fon p1 2pk 2pk =Q· −Q· ′ 3 ′ 3 s r′ − 2 r ′ + s2 A water molecule is a permanent dipole with a known dipole moment p = qs. There is a water molecule in the air a very small distance d from the midpoint of a charged rod of length L carrying a uniformly distributed negative charge −Q. The axis of the dipole lies along a line drawn through the center of the rod. Note that s ≪ d ≪ L. You may neglect the small change in the dipole moment of the water molecule induced by the rod. Version 025 – MT1 v3 – yao – (57465) Choose the answer that correctly expresses the magnitude and direction (along the xaxis) of the electric force on the water molecule, FHrod . Your f inal result must be 2O expressed only in terms of k, Q, q, s, p, d and/or L and any constant numerical factors. 1. −2 kQp correct Ld2 2. 0 kQp Ld2 kQp 4. 2 3 d kQp 5. − 3 d kQp 6. − 2 Ld kQq 7. 2 d kQp 8. 2 2 Ld kQp 9. 3 d kQp 10. −2 3 d Explanation: We are told that L ≫ d, so we may use the approximate field of the rod. From the figure, we see that the dipole moment points to the left, meaning that the positive dipole charge is closer to the rod. Combining these two facts, we obtain k(−Q) q q rod FH2 O = 2 − L d − 2s d + 2s 1 kQq 1 = −2 . s − s Ld 1 − 2d 1 + 2d 3. Letting ǫ = s/2d and applying the small ǫ approximation, kQq [(1 − ǫ)−1 − (1 + ǫ)−1 ] Ld kQq [1 + (−1)(−ǫ) − {1 + (−1)(ǫ)}] = −2 Ld Qq s kQq · 2ǫ = 2k · = −2 Ld Ld d FHrod = −2 2O = −2 2 kQp . Ld2 003 10.0 points Two large, parallel, insulating plates are charged uniformly throughout the plates. Each plate has the same amount and sign of charge per unit area +σ. What is the magnitude of the resultant electric field E? 1. σ between the plates, zero outside 2 ǫ0 2. zero everywhere 3. zero between the plates, 4. σ everywhere ǫ0 σ outside 2 ǫ0 2σ outside ǫ0 2σ 2σ 6. between the plates, outside ǫ0 ǫ0 2σ 7. between the plates, zero outside ǫ0 σ between the plates, zero outside 8. ǫ0 σ 9. zero between the plates, outside corǫ0 rect 5. zero between the plates, 10. σ σ between the plates, outside 2 ǫ0 2 ǫ0 Explanation: Each plate produces a constant electric field σ of E = directed away from the plate for 2 ǫ0 positive charge density, and toward the plate for negative charge density. Between the two plates, the two fields cancel each other so that Enet = 0. Outside of the two plates, the fields add together, so that Enet = σ . ǫ0 004 (part 1 of 3) 10.0 points A solid conducting sphere of radius R1 and Version 025 – MT1 v3 – yao – (57465) total charge q1 is enclosed by a conducting shell with an inner radius R2 and outer radius R3 and total charge q2 . OA = a and OC = c. 1. ΦS = 2. ΦS = q2 R1 q1 3. ΦS = S 4. ΦS = O R2 5. ΦS = A C R3 Find the electric field at C. 1. EP = 2. EP = 3. EP = 4. EP = 5. EP = 6. EP = k (q1 + q2 ) c k q1 c2 k q2 c2 k q1 k q2 + 2 R12 R3 k (q1 + q2 ) R32 k q2 c 7. EP = 0 k (q1 + q2 ) correct c2 Explanation: Consider a spherical Gaussian surface through the point C. C is outside of the shell, so the enclosed charge qencl = q1 + q2 can be treated as a point charge and qencl q1 + q2 EA (4 π c2 ) = = ǫ0 ǫ0 (q1 + q2 ) 1 q1 + q2 =k . EA = 2 4 π ǫ0 c c2 8. EP = 005 (part 2 of 3) 10.0 points Find the total flux emanating through the Gaussian surface S drawn through the point A. 6. ΦS = 7. ΦS = 8. ΦS = 3 −q2 ǫ0 q1 correct ǫ0 q2 ǫ0 q2 − q1 ǫ0 −(q1 + q2 ) ǫ0 q1 − q2 ǫ0 q1 + q2 ǫ0 −q1 ǫ0 Explanation: The enclosed charge is q1 , so the flux is ΦS = q1 . ǫ0 006 (part 3 of 3) 10.0 points Find the surface charge density on the inner surface of the shell (i.e, at R2 ). q2 4 π R22 q2 − q1 = 4 π R22 q1 correct =− 4 π R22 q1 =− 4 π R12 q1 = 4 π R12 q2 =− 4 π R22 q1 = 4 π R22 q1 − q2 = 4 π R22 1. σ = 2. σ 3. σ 4. σ 5. σ 6. σ 7. σ 8. σ Explanation: Draw a gaussian surface inside the shell — since this surface is inside a conductor, we know immediately that ΦE = 0. From this Version 025 – MT1 v3 – yao – (57465) we conclude that the charge enclosed by this surface is zero; the surface encloses q1 and the charge on the inner surface of the shell, which must be −q1 . Since the inner surface of the shell is at R2 , the charge density there is σ=− q1 . 4 π R22 007 10.0 points A lithium-7 nucleus (3 protons and 4 neutrons) and a neutral carbon atom of polarizability α are at a distance r apart (r is much greater than the diameter d of the atom). Due to polarization of the atom by the lithium nucleus, there is a force F between the 7 Li nucleus and the carbon atom. If we change r to F′ 5.3r, what will be the ratio of where F ′ is F the new force between the two? 1. 0.0029802 2. 0.024485 3. 0.00021779 4. 0.00054192 5. 0.00019869 6. 0.00015236 7. 0.0012621 8. 0.00043602 9. 0.00013987 10. 0.00023912 4 axis of this dipole. The magnitude of the electric field due to this dipole at the location of the 7 Li nucleus will be given by Ep = 1 2p 4πǫ0 r 3 Here, we have used the fact that r ≫ d. The magnitude of force on the nucleus due to the dipole (and vice versa) can be obtained as F = Ep · 3e. Combining all the equations, we get F = Ep · 3e = = 1 4πǫ0 2 1 2(αEe )3e 4πǫ0 r3 18αe2 . r5 This implies that when r to 5.3r, the electric field field, and consequently the force, is decreased by a factor of (5.3)−5 = 0.00023912, and hence the answer is F′ = (5.3)−5 = 0.00023912 F 008 10.0 points Correct answer: 0.00023912. Explanation: The force will be proportional to r −5 and F′ = hence the correct answer will be F 0.00023912 . The calculation is as follows. The magnitude of electric field due to the 7 Li nucleus at the location of the carbon atom will be given by E7 Li = 1 3e 4πǫ0 r 2 From this, the magnitude of the induced dipole moment of the carbon atom can be written down as p = αE7 Li . This dipole moment will be pointing away from the 7 Li nucleus, and hence the 7 Li nucleus will lie on the The center of the spherical metal ball of radius R, carrying a negative charge −Q, is located a distance r from the center of a short, thin, neutral copper wire of length L. Assume that L , R ≪ r. The induced charges at the two ends of the wire is ±q. Determine the polarizability of the wire segment which has a length L = 0.0115 m. The units of polarizability are C2 m/N. 1. 1.013e-17 2. 3.811e-17 3. 1.389e-17 4. 1.736e-18 Version 025 – MT1 v3 – yao – (57465) 5. 3.814e-18 6. 2.112e-17 7. 3.417e-17 8. 4.764e-18 9. 4.234e-17 10. 2.4e-17 Your answer must be within ± 2.0% Correct answer: 2.112 × 10−17 . Explanation: First, we must determine the magnitude of the induced charge. Since the wire is a conductor, we must have E = 0 inside; using this fact, we can calculate q: EQ (A) = Edip (A) 1 Q 1 q =2· 2 2 4πǫ0 r 4πǫ0 L 2 Q 8q = 2 2 r L 2 Q L . q= 8 r The polarizability is given by α = p qL = kQ E r2 Using the value of q given above (Q/8)(L/r)2 L α = kQ r2 α = C2 m L3 = 2.112 × 10−17 8k N 009 (part 1 of 2) 10.0 points Two plastic balls are equally but oppositely charged and held in place by insulating threads, as shown in the following figure. + 5 Neutral Metal ~ F − ~ F Before metal ball They attract each other with an electric force of magnitude F . Then, an uncharged metal ball is held in place by insulating threads between the balls, closer to the left ball. What happens to the magnitudes of the net electric forces on the left and right charged balls? 1. The magnitudes of the net electric forces on both balls are increased. correct 2. The magnitude of the net electric force on the right ball is reduced, and that on the left ball is increased. 3. The magnitudes net electric forces on both balls are reduced. 4. The neutral metal ball has no effect on the magnitudes of the net electric forces. 5. The magnitude of the net electric force on the left ball is reduced, and that on the right ball is increased. Explanation: When placed between the two oppositely charged balls, the metal sphere polarizes. The electric field of both the positive and negative charge point to the right at the position of the neutral ball, polarizing it such that negative charge collects on the left and positive charge collects on the right. The negative charge on the left side of the metal sphere exerts a (small) rightward force on the left positively charged ball, adding to the pull from the negatively charged ball on the right, meaning the net force increases. Similarly, the excess positive charge on the Version 025 – MT1 v3 – yao – (57465) 6 right side of the metal sphere exerts a (small) leftward force on the negatively charged ball on the right, increasing the net force on that ball as well. See the explanation for part 1. Since both balls generate a field that points to the right at the location of the neutral sphere, it must polarize like option VII. 010 (part 2 of 2) 10.0 points In the following diagram, choose the image (IIX) that best represents what happens to the neutral metal ball when it is placed between the two oppositely charged balls. (Option IX means there is no change.) 011 10.0 points A positive charge q and a dipole of moment ~p contribute to a net field at location A that is zero, as shown in the following figure. Both ~p and q are a distance a from point A, with a ≫ dipole separation. I + + II + + III + ++ + + + IV − − + + − − + − − − − + + ++ + + a + q a + + VI − −− − VII A + V − − + + ++ + + b + + − − VIII − − −− − − − − − − − IX − − − ~p If ~p points to the right, we denote it +p; if ~p points to the left, we denote it −p. Choose the relationship between p and q that results in a net field of 0 at A. 1. p = qa 2. p = −qa correct qa 2 qa 4. p = 2 Explanation: At A, the field due to the point charge points to the left. In order to get cancellation of E at A, the field due to the dipole must point to the right. Recall that, on the axis perpendicular to the dipole moment, the dipole field points in the direction opposite the dipole moment, so the dipole moment must point to the left: ~p = −p. To determine the magnitude of the relationship between p and q, we simply set the point charge field at A and the dipole field at A equal to each other: 3. p = − 1. VIII 2. VI 3. I 4. V 5. II 6. IV 7. VII correct 8. III 9. IX Explanation: |Edip⊥ | = |Eq | kq k|~p| = 2 3 a a Version 025 – MT1 v3 – yao – (57465) |~p| = qa . Combining the two results, our answer must be: p = −qa. 7 Two vectors can define a plane. When these two vectors are plotted in this plane, we have 012 10.0 points ~ has components Vector A Ax = −2.1, Ay = 6.2, Az = 6.7, B Bz = 5. What is the angle θAB between these vectors? (Answer between 0◦ and 180◦ .) 1. 121.865 2. 105.72 3. 113.517 4. 140.101 5. 102.412 6. 118.162 7. 111.323 8. 142.246 9. 157.02 10. 125.71 5. 7 By = −7.7, 10 Bx = 7.7, 2◦ ~ has components while vector B A 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. Correct answer: 105.72◦. Explanation: ~ = kAk q A2x + A2y + A2z q = (−2.1)2 + 6.22 + 6.72 = 9.36696 and q q 2 2 2 ~ kBk = Bx + By + Bz = 7.72 + (−7.7)2 + 52 = 11.9825 , so using ~ ·B ~ = Ax B x + Ay B y + Az B z A = (−2.1) 7.7 + 6.2 (−7.7) + 6.7 (5) = −30.41 , cos θAB θAB ~ ·B ~ A −30.41 = = ~ kBk ~ (9.36696) (11.9825) kAk = −0.270938 = arccos(−0.270938) = 105.72◦ . (a) You touch the ball briefly with your fingers then release it. (b) A block of metal that is known to be charged is now moved near the ball. Version 025 – MT1 v3 – yao – (57465) (c) The ball briefly 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 positively 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 6 of Figure 1? 8 sion implies that either the ball is positively charged (this rules out I-L) or the effective positive charge in the ball is closer to the rod which produces a net repulsion (this rules out M). Since the positive charge of the rod attracts electrons in the positively charged ball, the far side of the ball should appear more positively charged. This condition rules out E,F,N. G is the correct answer. Positive charge on the ball leads to repulsion. It has more positive charge on the far side, i.e. on the side away from the charged rod. This is expected since the negative charge on the surface of the balls repositions to be as close to the positively charged rod as possible. 014 10.0 points Consider a thin glass rod of length L lying along the x-axis with its left end at the origin. The rod carries a uniformly distributed positive charge Q. 2. M L O 1. H −a +Q 3. L 4. J 5. I 6. G correct 7. F What is the electric field due to the rod at position −a on the x-axis, a > L, as shown in the figure above? Let the negative direction point to the left. {Hint: If you are unsure of your result, evaluate your integral to see if the result makes physical sense.} 8. E 9. N 10. K Explanation: For slide 6 of Figure1, it is stated that the rod is positively charged. The observed repul- 1 Q 1. E(−a) = 4 π ǫ0 L ZL dx (x − a)2 0 1 Q 2. E(−a) = − 4 π ǫ0 L ZL 0 dx x2 Version 025 – MT1 v3 – yao – (57465) 1 Q 3. E(−a) = 4 π ǫ0 L ZL 9 ~ E dx x2 0 ZL 1 Q 4. E(−a) = − 4 π ǫ0 L B F ~ dx cor(x + a)2 0 rect A α 5. E(−a) = 1 Q 4 π ǫ0 L ZL dx (x + a)2 0 1 Q 6. E(−a) = − 4 π ǫ0 L ZL dx (x − a)2 0 Explanation: The charges of the rod are distributed between 0 and L. The distance from −a to a point on the rod is given by x − (−a) = x + a and the integration limits must be 0 and L. Looking at the figure, since the charge on the rod is positive, the field at −a must point to the left and our answer must be negative. 1 ∆Q ∆E = − 4 π ǫ0 (x + a)2 Q L ∆x 1 =− 4 π ǫ0 (x + a)2 ZL dx 1 Q E=− 4 π ǫ0 L (x + a)2 Now imagine that a lithium nucleus, containing three protons and four neutrons, is placed at point B. What is the value of the electric field at location B due to the alpha particle? (Remember that e represents the fundamental unit of charge, 1.6 × 10−19 C. Also, r BA is the position vector pointing from point A to point B.) ~ 2. E ~ 3. E ~ 4. E ~ 5. E ~ 6. E 0 Digression: to use the hint, you must realize that the electric field of the rod must be less than that of a point charge Q at the origin but greater than that of a point charge Q at L: |EQL (−a)| < |Erod (−a)| < |EQ0 (−a)| . Determining bounds on the possible values of your answer is a useful way to check the validity of your results. 015 (part 1 of 4) 10.0 points The following figure shows an alpha particle (two protons and two neutrons) at location ~ at point B. A creating an electric field E Any charged particle placed at location B ~. experiences a force F 1 2e BA r̂ correct 4πǫ0 |~r BA |2 3e BA 1 = r̂ 4πǫ0 |~r BA | 1 2e BA =− r̂ 4πǫ0 |~r BA | 1 11e BA r̂ = 4πǫ0 |~r BA |2 5e BA 1 =− r̂ 4πǫ0 |~r BA |2 e 1 = r̂ BA 4πǫ0 |~r BA |2 e2 1 =− r̂ BA 4πǫ0 |~r BA |2 5e2 BA 1 r̂ = 4πǫ0 |~r BA |2 ~ = 1. E ~ 7. E ~ 8. E Explanation: The electric field at point B is not related to the charge at point B, only the one at point A, so the answers with multiples of e2 are incorrect. Recalling the form of the electric field and remembering that the alpha particle has charge 2e, ~ = E 1 q r̂, 4πǫ0 |~r |2 we see that it has the distance squared in the denominator. Therefore the correct choice is Version 025 – MT1 v3 – yao – (57465) ~ = E 1 2e BA r̂ . 4πǫ0 |~r BA |2 016 (part 2 of 4) 10.0 points What is the force on the lithium nucleus? 11e2 ~ = −1 r̂ BA 1. F 2 BA 4πǫ0 |~r | 3e BA ~ = 1 r̂ 2. F 4πǫ0 |~r BA | −1 2e2 BA ~ 3. F = r̂ 4πǫ0 |~r BA | 6e2 BA ~ = 1 4. F r̂ correct 4πǫ0 |~r BA |2 ~ = −1 5e r̂ BA 5. F 4πǫ0 |~r BA |2 2e2 BA ~ = 1 6. F r̂ 4πǫ0 |~r BA |2 5e2 BA ~ = 1 7. F r̂ 4πǫ0 |~r BA |2 e2 ~ = 1 r̂ BA 8. F 4πǫ0 |~r BA |2 Explanation: Force involves both particles, and has the general form 1 Q1 Q2 r̂, 4πǫ0 |~r |2 where ~r is the vector pointing between the two charges. In this case, Q1 = 2e and Q2 = 3e, so we end up with ~ = F ~ = F 6e2 BA 1 r̂ . 4πǫ0 |~r BA |2 017 (part 3 of 4) 10.0 points The lithium nucleus is removed and an electron is placed at location B. Now what is the value of the electric field at location B? 1 11e BA r̂ 4πǫ0 |~r BA |2 1 5e BA =− r̂ 4πǫ0 |~r BA |2 1 e2 r̂ BA =− 4πǫ0 |~r BA |2 ~ = 1. E ~ 2. E ~ 3. E 10 1 e r̂ BA 2 BA 4πǫ0 |~r | 3e BA 1 r̂ = 4πǫ0 |~r BA | 1 2e BA = r̂ correct 4πǫ0 |~r BA |2 5e2 BA 1 r̂ = 4πǫ0 |~r BA |2 1 2e BA =− r̂ 4πǫ0 |~r BA | ~ = 4. E ~ 5. E ~ 6. E ~ 7. E ~ 8. E Explanation: See the explanation for part 1. The electric field does not depend on the particle at point B, so in this case the answer is the same as for part 1. 018 (part 4 of 4) 10.0 points What is the the force on the electron? ~ = 1. F ~ = 2. F ~ = 3. F ~ = 4. F ~ = 5. F ~ = 6. F ~ = 7. F ~ = 8. F 1 2e BA r̂ 4πǫ0 |~r BA | −1 2e2 BA r̂ 4πǫ0 |~r BA | −1 e2 r̂ BA 4πǫ0 |~r BA |2 −1 3e2 BA r̂ 4πǫ0 |~r BA |2 3e2 BA 1 r̂ 4πǫ0 |~r BA |2 1 e2 r̂ BA 4πǫ0 |~r BA |2 −1 2e2 BA r̂ correct 4πǫ0 |~r BA |2 1 2e2 BA r̂ 4πǫ0 |~r BA |2 Explanation: Since the electric field of the alpha particle is now acting on a negative charge, the force will be attractive, so we need an answer proportional to −r̂ BA . Since the force is proportional to Q1 Q2 , we need an answer with 2e2 . As always, distance dependence goes like 1/r 2 . The only correct option is Version 025 – MT1 v3 – yao – (57465) 11 VIII 2e2 ~ = −1 F r̂ BA . 2 BA 4πǫ0 |~r | I VII II III VI IV V 019 (part 1 of 2) 10.0 points Two dipoles consisting of two oppositely charged balls connected by a wooden stick are located as shown in the following figure. A block of plastic is located between them, as shown. Assume each of the molecules throughout the block has the same polarizability. Locations B, C, and D all lie on a line perpendicular to the axis of the dipole, passing through the midpoint of the dipoles. Plastic block + − D b C b 8. VII + + + − I − − + H − − + G D − + F 4. I b C − + 3. V 7. VI − + B 2. III correct 6. IV Before selecting answers to the following questions, you might want to draw your own diagram of this situation, showing all the fields and charge distributions requested. Answer the following questions by selecting either a direction (I-VIII) or an orientation of a polarized molecule (A-J) from the following two diagrams. A 1. II 5. VIII − + B Which of the arrows (I-VIII) best indicates the direction of the electric field at location D due only to the dipoles? E J Other Explanation: The electric field along the perpendicular bisector of a dipole is opposite its dipole moment. Since the upper dipole is closer to point D, it’s field dominates that of the other dipole. Since the dipole moment of the upper dipole points left, its field at D points right. Direction III is the correct answer. 020 (part 2 of 2) 10.0 points Which of the diagrams (A-J) best indicates the polarization of a molecule of plastic at location D? 1. Diagram D 2. Diagram B correct 3. Diagram I Version 025 – MT1 v3 – yao – (57465) 4. Diagram E 5. Diagram F 6. Diagram C 7. Diagram H 8. Diagram A 9. Diagram G 10. Diagram J Explanation: Since the field at D points to the right, the dipole moment of a molecule at D must also point to the right. Choice B shows the correct polarization for a rightward dipole moment. 12