2003 answers Elec

2003 answers Elec - AP® Physics C: Electricity and...

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Unformatted text preview: AP® Physics C: Electricity and Magnetism 2003 Scoring Guidelines The materials included in these files are intended for use by AP teachers for course and exam preparation; permission for any other use must be sought from the Advanced Placement Program®. Teachers may reproduce them, in whole or in part, in limited quantities for noncommercial, face-to-face teaching purposes. This permission does not apply to any third-party copyrights contained herein. This material may not be mass distributed, electronically or otherwise. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here. These materials were produced by Educational Testing Service® (ETS®), which develops and administers the examinations of the Advanced Placement Program for the College Board. 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The College Board is committed to the principles of equity and excellence, and that commitment is embodied in all of its programs, services, activities, and concerns. For further information, visit www.collegeboard.com Copyright © 2003 College Entrance Examination Board. All rights reserved. College Board, Advanced Placement Program, AP, AP Vertical Teams, APCD, Pacesetter, Pre-AP, SAT, Student Search Service, and the acorn logo are registered trademarks of the College Entrance Examination Board. AP Central is a trademark owned by the College Entrance Examination Board. PSAT/NMSQT is a registered trademark jointly owned by the College Entrance Examination Board and the National Merit Scholarship Corporation. Educational Testing Service and ETS are registered trademarks of Educational Testing Service. Other products and services may be trademarks of their respective owners. For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. AP PHYSICS C: ELECTRICITY AND MAGNETISM 2003 SCORING GUIDELINES Question 1 15 points total Distribution of points Answers shown here are expressed in terms of Coulomb’s law constant k, but equivalent answers in terms of 1 4F ε 0 were acceptable. (a) 3 points The sphere of charge can be treated as a point charge located at the sphere’s center. i. (2 points) kq E= 2 r For indicating that the total charge is Q For a correct answer kQ E= 2 r 1 point 1 point ii. (1 point) For a correct answer kQ V= r 1 point Credit was also awarded for integrating E to obtain V (b) 3 points For indicating that the proton will move away from the charged sphere For indicating that the velocity of the proton will increase or reach a finite value For indicating that the acceleration of the proton will decrease (c) 1 point 1 point 1 point 3 points For a correct statement of conservation of energy K = Ur − U R For the substitution of an electrical potential energy with the correct form − keQ − keQ K= − r R For a correct answer æ 1 1ö K = keQ ç − ÷ è R rø Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 2 1 point 1 point 1 point AP PHYSICS C: ELECTRICITY AND MAGNETISM 2003 SCORING GUIDELINES Question 1 (cont’d.) Distribution of points (c) (continued) Alternate solution For showing correct use of the work-energy theorem K = W = ò F • dr For setting up the correct integration R − kQe K =ò dr r2 r æ 1ö K = − kQe ç − ÷ è rø R r 1 point æ 1 æ 1ö ö æ1 1ö = − kQe ç − − ç − ÷ ÷ = − kQe ç − ÷ è r Rø è R è r øø 1 point For a correct answer æ 1 1ö K = keQ ç − ÷ è R rø (d) Alternate points 1 point 3 points H0 can be determined by integrating the volume distribution and setting it equal to the total charge Q For indicating that an integration is necessary 1 point R Q = ò H ( r ) dV 0 For showing a correct volume element dV = 4π r 2 dr For substitution of H ( r ) and dV 1 point 1 point R rö æ Q = ò H0 ç1 − ÷ 4F r 2 dr è Rø 0 R R æ æ r3 r 4 ö æ R3 R3 ö r3 ö R3 Q = 4 FH0 ò ç r 2 − ÷ dr = 4FH0 ç − = 4FH0 ç − ÷ = 4FH0 4ø 12 Rø è 3 4R ÷ 0 ø è3 0è H0 = 3Q F R3 Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 3 AP PHYSICS C: ELECTRICITY AND MAGNETISM 2003 SCORING GUIDELINES Question 1 (cont’d.) Distribution of points (e) 3 points For writing Gauss’s law with a charge element dq between E and OR showing the relationship 1 point ò dq 1 k ò dq OR dE = r 2 dq ε0 1 k E 4F r 2 = ò H ( r ) dV OR E = 2 ò H ( r ) dq A0 r ÑE ò • dA = For correct substitution of H0 , dV, and correct limits r r 0 0 1 point 3Q æ rö 1 − ÷ 4F r 2 dr 3ç è Rø ò H ( r ) dV = ò F R r r r 12Q æ r3 ö 12Q æ r 3 r 4 ö Qæ 3r 4 ö H ( r ) dV = 3 ò ç r 2 − ÷ dr = 3 ç − = 3 ç 4r 3 − ò R 0è Rø R è 3 4R ÷ 0 R è R÷ ø ø 0 1 Q æ 3 3r 4 ö 4r − 4πε 0 r 2 R 3 ç R÷ è ø For a correct answer kQr æ 3r ö E = 3 ç4 − ÷ Rè Rø E= OR E= k Q æ 3 3r 4 ö 4r − r 2 R3 ç R÷ è ø 1 point Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 4 AP PHYSICS C: ELECTRICITY AND MAGNETISM 2003 SCORING GUIDELINES Question 2 15 points total (a) Distribution of points 4 points For including the resistor, capacitor, battery and switch elements with common symbols and clear labels in the diagram One point was deducted for using one uncommon symbol or mislabeling one element No credit was given for a circuit missing any elements or with two or more uncommon symbols or mislabeled elements For correctly showing the resistor, capacitor, and battery connected in series For placing the ammeter in series with the resistor (b) 2 points 1 point 1 point 3 points For any statement of Ohm’s law R =V I One point each for substituting the values of the initial conditions consistent with the graph into the above equation R = (12 V ) ( 0.010 A ) R = 1200 Ω Alternate solution For using the given equation for the current i (t ) = εe 1 point 2 points Alternate points 1 point −t 4 R For substituting consistent current and time values from the graph into the above equation For a value of resistance consistent with the substituted values Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 5 1 point 1 point AP PHYSICS C: ELECTRICITY AND MAGNETISM 2003 SCORING GUIDELINES Question 2 (cont’d.) Distribution of points (c) 4 points For the equation for the time constant τ = RC For substituting the value of the time constant from the given equation for current, i.e. 4 s For substituting the resistance from part (b) C = ( 4 s ) (1200 Ω ) For a value of capacitance consistent with the substituted values, including units C = 3.3 × 10−3 F Alternate solution For any statement of the equation relating capacitance, charge, and voltage C =Q V For an equation relating current and charge i = dQ dt OR Q = ò i dt 1 point 1 point 1 point 1 point Alternate points 1 point ò For evaluating the integral Q = i dt , including clearly identifying the limits of integration ε For example, Q = 6 e Rò 0 −t 4 6 ( 1 point ) dt = −4 ( 0.010) e − t 4 0 = − ( 0.040) e − 6 4 − e 0 = 0.031 C The value of Q and the potential difference associated with the time interval used in the integral are then substituted into the first equation. Using the above example, the associated voltage would be 12 V − ( 0.0025 A )(1200 Ω ) = 9 V , and the capacitance would be 3.4 × 10−3 F . For a value of capacitance consistent with the result of the integration and the correct voltage difference for that time interval, including units (d) 1 point 1 point 4 points For correctly labeling the capacitor, resistor, battery, and the A and B switch elements For a circuit that has the resistor, capacitor, and battery connected in series when the switch is closed at A, and will charge the capacitor to 12 V For a circuit that has the resistor and capacitor connected in series when the switch is closed at B, and will discharge the capacitor from 12 V to zero For a voltmeter connected in parallel across the capacitor Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 6 1 point 1 point 1 point 1 point AP PHYSICS C: ELECTRICITY AND MAGNETISM 2003 SCORING GUIDELINES Question 3 15 points total Distribution of points (a) 2 points For an arrow downward on or near antenna For one correct justification statement Examples: F = qv × B , the right-hand rule, − (north × down) = east, the left-hand rule for electrons (b) 1 point 1 point 5 points An equilibrium is reached when the electric force due to the separation of charge in the antenna balances the magnetic force on an electron For a correct expression for the magnetic force, showing a cross product or an angular dependence For a correct expression for the electric force Equating the electric and magnetic forces: qE = qv × B E = K B sin G For substitution of sin 55° For substitution of other given values E = K B sin G = ( 75 m s) 6 × 10−5 T sin 55° ( For the correct answer E = 0.0037 V m (or N C ) ) 1 point 1 point 1 point 1 point 1 point (c) 2 points Using the relationship between electric field and potential difference for the linear situation: V = Ed For substitution of the value of E from part (b) For substitution of the correct length, 15 m V = ( 0.0037 V m )(15 m ) V = 0.0553 V An alternate method is to do part (c) first, using principles noted in part (b), then substituting V into the relationship V = Ed to obtain E. Points equivalent to those above were awarded for this method. Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 7 1 point 1 point AP PHYSICS C: ELECTRICITY AND MAGNETISM 2003 SCORING GUIDELINES Question 3 (cont’d.) Distribution of points (d) 1 point For indicating that the top of the wing is at the higher potential 1 point Note: To earn credit, this answer must be consistent with the student's answer to part (a). (e) 5 points i. (3 points) For indicating that there must be a change in the magnetic flux through the closed loop For specifying a correct change in the plane's orientation or the connected wire that could result in a change in flux through the closed loop. 2 points 1 point ii. (2 points) For any indication of the position of the connected wire in the closed loop For showing the correct direction of the resulting induced current on the diagram, based upon the conditions indicated in part (e) i. Examples Ex. 1: The connecting wire trails behind the antenna, over the body of the plane. The plane goes into a forward dive, increasing its angle with respect to horizontal. The forward dive would decrease magnetic flux downward through the loop, so a current will be induced to create an increased downward flux. Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 8 1 point 1 point AP PHYSICS C: ELECTRICITY AND MAGNETISM 2003 SCORING GUIDELINES Question 3 (cont’d.) Distribution of points (e) (continued) Ex. 2: The connecting wire loops under the wing, directly under the antenna. The plane makes a level right turn. The turn will decrease the northward component of magnetic flux through the loop, so a current will be induced to create an increased flux to the north. Other acceptable concepts were arranging the loop so its area can be changed, or shielding a part of the circuit from the magnetic field. Copyright 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 9 ...
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This note was uploaded on 08/05/2009 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.

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