Solutions to Conceptual Practice Problems PHYS 1112 In-Class Exam #2A+2B

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1 Solutions to Conceptual ractice roblems HYS 1112 In-Class xam #2A+2B Thu. Mar. 19, 2009, 11:00am-12:15pm and 2:00pm-3:15pm C 2.01: In a two-source interference experiment two sources are oscillating in phase with the same period. The first intensity minimum, closest to the central intensity maximum M and to the left of M, is located at point, as shown in Fig A wave crest A from source 1 and a wave trough B from source 2 arrive simultaneously at. Therefore, A and B must have departed from their respective sources as follows: Fig M l 1 l 2 Source 1 Source 2 (A) A departed 1/2 period before B. (B) B departed 1/2 period before A. (C) A departed 1 period before B. (D) B departed 1 period before A. () A departed 3/2 period before B. C 2.01 Answer: (B) At, the pathlength difference l l 2 l 1 (see Fig. 2.01) must be half a wavelength: l = 1 λ, since it s the first interference minimum to the left of center M (with l = 0 at 2 M). Hence, B has 1 λ further to travel than A, which takes an extra travel time of 1period. 2 2 So B must leave 1 period earlier than A to arrive simultaneously at. [Recall: v = λ/τ; so 2 it takes 1 period τ for a crest or trough to travel 1 wavelength λ.] C 2.02: A diffaction grating is illuminated with coherent (laser) light with a wavelength λ < d where d is the spacing between adjacent slits in the grating. The first intensity maximum, to the right of and closest to the central intensity maximum M, is located at point Q, as shown in Fig A wave crest A from slit R and a wave crest B from the neighboring slit S to the right of R, arrive simultaneously at Q. Therefore, A and B must 1

2 have departed from their respective slits as follows: Fig M Q l R l S R S Diff. Grating Laser Beam (A) A departed 1/2 period before B. (B) B departed 1/2 period before A. (C) A departed 1 period before B. (D) B departed 1 period before A. () A departed 3/2 period before B. C 2.02 Answer: (C) At Q, the pathlength difference l l R l S (see Fig. 2.02) must be one full wavelength: l = λ, since it s the first interference maximum to the right of center M (with l = 0 at M). Hence, A has 1 λ further to travel than B, which takes an extra travel time of 1 period. So A must leave 1 period earlier than B to arrive simultaneously at Q. [Recall: v = λ/τ; so it takes 1 period τ for a crest or trough to travel 1 wavelength λ.] C 2.03: A beam of coherent (laser) light of wavelength λ is incident upon a diffraction grating with line spacing d, with λ < d, as shown in Fig Assume y is the distance (in cm) between the two 1st-order intensity maxima, observed on a screen at a distance L on the other side of the grating. This distance y will 2

3 1st order maxima Δy Fig Screen L θ Diff. Grating Laser Beam (A) decrease if we increase L (at fixed λ and d) (B) decrease if we increase λ (at fixed L and d) (C) increase if we decrease λ (at fixed L and d) (D) increase if we increase d (at fixed λ and L) () decrease if we increase d (at fixed λ and L) C 2.03 Answer: () At 1st-order maximum: sin θ = λ/d. Also, y = 2L tan θ, by trigonometry (see Fig. 2.03). So sin θ, hence θ, hence tan θ, hence y, will decrease if d increases or if λ decreases. Therefore, () is correct; and (B), (C) and (D) are wrong. Because of y = 2L tan θ, y increases if L increases: hence (A) is wrong. C 2.04: A beam of coherent (laser) light of wavelength λ is incident upon two parallel slits with a spacing d and λ < d, as shown in Fig Assume y is the distance (in cm) between the two adjacent dark fringes (interference intensity minima), observed closest to the central maximum on a screen at a distance L on the other side of the slits. This distance y will 1st dark fringes Δy Fig Screen L θ Double Slit Laser Beam 3

4 (A) increase if we increase L (at fixed λ and d) (B) increase if we decrease λ (at fixed L and d) (C) decrease if we increase λ (at fixed L and d) (D) increase if we increase d (at fixed λ and L) () decrease if we decrease d (at fixed λ and L) C 2.04 Answer: (A) At the dark fringe (interference minimum) nearest to center: sin θ = 1 λ/d. Also, y = 2 2L tan θ, by trigonometry (see Fig. 2.04). So sin θ, hence θ, hence tan θ, hence y, will increase if d decreases or if λ increases. Therefore, (B), (C), (D) and () are wrong. Because of y = 2L tan θ, y increases if L increases: hence (A) is correct. C 2.05: If two point charges q 1 and q 2 at some distance r repel each other with a force of 160N, what force would they exert on each other if r is quadrupled ( 4); and q 1 and q 2 are unchanged? The two charges will (A) repel each other with a force of 22µN (B) attract each other with a force of 10N (C) repel each other with a force of 10N (D) attract each other with a force of 65000MN () repel each other with a force of 65000MN C 2.05 Answer: (C) By Coulomb s law, the force F = k q 1 q 2 /r 2 1/r 2. Hence, changing r r = 4r will change F F = F/4 2 = 160N/16 = 10N. Since q 1 and q 2 initially repel each other, they both have the same sign. Since the signs of q 1 and q 2 are unchanged, the two charges will continue to repel each other after r is increased. C 2.06: If two point charges q 1 and q 2 at some distance r attract each other with a force of 10N, what force would they exert on each other if q 1 is quadrupled ( 4) and its sign is reversed; and q 2 and r are unchanged? (A) repel each other with a force of 40N (B) attract each other with a force of 2.1mN (C) repel each other with a force of 2.1mN (D) attract each other with a force of 56000MN () repel each other with a force of 56000MN C 2.06 Answer: (A) 4

5 By Coulomb s law, the force F = k q 1 q 2 /r 2 q 1. Hence, changing q 1 q 1 = 4 q 1 will change F F = 4 F = 4 10N = 40N. Since q 1 and q 2 initially attract each other, they initially have opposite sign. Since the sign of q 1 is reversed and q 2 is unchanged, the two charges will both have the same sign, after q 1 is changed, and therefore repel each other. C 2.07: If two point charges q 1 and q 2 at some distance r attract each other with a force of 6N, what force would they exert on each other if q 1 is doubled ( 2) without changing its sign; q 2 is tripled ( 3) and its sign is reversed; and r is unchanged? The two charges will (A) repel each other with a force of 1N (B) attract each other with a force of 1N (C) repel each other with a force of (1/6)N (D) attract each other with a force of (1/6)N () repel each other with a force of 36N C 2.07 Answer: () By Coulomb s law, the force F = k q 1 q 2 /r 2 q 1 q 2. Hence, changing q 1 q 1 = 2 q 1 and q 2 q 2 = 3 q 2 will change F F = 2 3 F = 6 6N = 36N. Since q 1 and q 2 initially attract each other, they initially have opposite sign. Since the sign of q 2 is reversed and the sign of q 1 is unchanged, the two charges will both have the same sign, after q 1 is changed, and therefore repel each other. C 2.08: If two point charges q 1 and q 2 at some distance r repel each other with a force of 4.5N, what force would they exert on each other if q 1 is unchanged; q 2 is doubled ( 2) and its sign is reversed; and r is tripled ( 3)? The two charges will (A) repel each other with a force of 1N (B) attract each other with a force of 1N (C) repel each other with a force of (1/6)N (D) attract each other with a force of (1/6)N () repel each other with a force of 36N C 2.08 Answer: (B) By Coulomb s law, the force F = k q 1 q 2 /r 2 q 2 /r 2. Hence, changing q 2 q 2 = 2 q 2 and r r = 3r will change F F = 2 F/3 2 = 2 4.5N/9 = 1N. Since q 1 and q 2 initially repel each other, they initially have the same sign. Since the sign of q 2 is reversed and q 1 is unchanged, the two charges will have opposite sign, after q 2 is changed, and therefore attract each other. 5

6 C 2.09: Q in Fig is a positive point charge. Which arrow drawn at could correctly represent the electric field vector generated by Q at? Fig (A) Q (A) (B) (C) (D) () C 2.09 Answer: (A) For a single positive point charge Q, at any observation point, the electric field vector, must (a) point along the straight line connecting Q and and (b) away from Q. Hence (A) is the correct answer. C 2.10: Q in Fig is a negative point charge. Which arrow drawn at could correctly represent the electric field vector generated by Q at? Fig Q (C) (A) (B) (C) (D) () C 2.10 Answer: (C) For a single negative point charge Q, at any observation point, the electric field vector 6

7 , must (a) point along the straight line connecting Q and and (b) towards Q. Hence (C) is the correct answer. C 2.11: Q 1 and Q 2 in Fig are positive point charges. Which arrow drawn at could correctly represent the total electric field vector generated by Q 1 and Q 2 at? Fig Q 2 1 () 2 Q 1 (A) (B) (C) (D) () C 2.11 Answer: () At observation point, the positive point charge Q 1 produces an electric field vector 1 pointing away from Q 1, i.e., upward; the positive point charge Q 2 produces an electric field vector 2 pointing away from Q 2, i.e., rightward; as shown in Fig The total electric field is the vector sum of 1 and 2, i.e., = Therefore, must have a vertical component pointing upward and a horizontal component pointing rightward. Hence, () is the only possible correct answer. C 2.12: Q 1 in Fig is a positive point charge, and Q 2 is a negative point charge. Which arrow drawn at could correctly represent the total electric field vector generated by Q 1 and Q 2 at? 7

8 Fig Q 2 (D) 1 2 Q 1 (A) (B) (C) (D) () C 2.12 Answer: (D) At observation point, the positive point charge Q 1 produces an electric field vector 1 pointing away from Q 1, i.e., upward; the negative point charge Q 2 produces an electric field vector 2 pointing towards Q 2, i.e., leftward; as shown in Fig The total electric field is the vector sum of 1 and 2, i.e., = Therefore, must have a vertical component pointing upward and a horizontal component pointing leftward. Hence, (D) is the only possible correct answer. C 2.13: In xperiment 1, a proton travels from point A to point B, in the electric field generated by/between two oppositely charged capacitor plates, as shown in Fig In xperiment 2, an electron travels from point A to point B, in the same electric field between the same two charged capacitor plates. A proton has a charge +e and an electron has a charge e. Fig B A Assume the proton experiences a drop in electric potential, V = 250V, and a loss of potential energy, U = 40aJ (where 1aJ J), in traveling from A to B. What will the electron experience? (A) V = 500V and U = 80aJ (B) V = 1000V and U = 160aJ (C) V = +1000V and U = +160aJ (D) V = 1000V and U = +160aJ 8

9 () V = 250V and U = +40aJ C 2.13 Answer: () Both proton and electron are traveling in the same -field; V is completely determined by the -field; and it is independent of the test charge traveling through this field. So, V is the same for both: V (electron) = V (proton) = 250V. However, U = q V ; and q = +e for the proton, but q = e for the electron. Hence, U for the electron has the opposite sign as for the proton: U(electron) = U(proton) = +40aJ. So, the answer is (). C 2.14: In xperiment 1, an electron travels from point A to point B, in the electric field generated by/between two oppositely charged capacitor plates, as shown in Fig In xperiment 2, a proton travels from point A to point B, in the same electric field between the same two charged capacitor plates. A proton has a charge +e and an electron has a charge e. Fig A B Assume the electron experiences a rise in electric potential, V = +250V, and a loss of potential energy, U = 40aJ (where 1aJ J), in traveling from A to B. What will the proton experience? (A) V = +250V and U = +40aJ (B) V = 100V and U = 16aJ (C) V = +100V and U = +16aJ (D) V = 100V and U = +16aJ () V = 50V and U = 8aJ C 2.14 Answer: (A) Both proton and electron are traveling in the same -field; V is completely determined by the -field; and it is independent of the test charge traveling through this field. So, V is the same for both: V (electron) = V (proton) = +250V. However, U = q V ; and q = +e for the proton, but q = e for the electron. Hence, U for the proton has the opposite sign as for the electron: U(proton) = U(electron) = +40aJ. So, the answer is (A). 9

Conceptual Practice Problems for PHYS 1112 In-Class Exam #2A+2B

Conceptual Practice Problems for PHYS 1112 In-Class Exam #2A+2B Conceptual ractice roblems for HYS 1112 In-Class xam #2A+2B Thu. Mar. 19, 2009, 11:00am-12:15pm and 2:00pm-3:15pm C 2.01: In a two-source interference experiment two sources are oscillating in phase with