Unformatted text preview: AP® Physics C: Electricity and Magnetism
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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
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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 workenergy 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.
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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.
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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.
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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
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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 righthand rule, − (north × down) = east, the lefthand
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.
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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.
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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.
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 Physics, Electricity And Magnetism, Magnetism, Magnetic Field, College Entrance Examination Board, college entrance examination, Entrance Examination Board

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