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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 & MAGNETISM
2004 SCORING GUIDELINES
General Notes about 2004 AP Physics Scoring Guidelines
1. The solutions contain the most common method(s) of solving the freeresponse questions, and the
allocation of points for these solutions. Other methods of solution also receive appropriate credit for
correct work. 2. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) is
correctly substituted into an otherwise correct solution to part (b), full credit will usually be awarded.
One exception to this may be cases when the numerical answer to a later part should be easily
recognized as wrong, e.g. a speed faster than the speed of light in vacuum.
3. Implicit statements of concepts normally receive credit. For example, if use of the equation expressing a
particular concept is worth one point, and a student’s solution contains the application of that equation to
the problem but the student does not write the basic equation, the point is still awarded.
4. The scoring guidelines typically show numerical results using the value g = 9.8 m s 2 , but use of 10 m s 2 is of course also acceptable.
5. Numerical answers that differ from the published answer due to differences in rounding throughout the
question typically receive full credit. The exception is usually when rounding makes a difference in
obtaining a reasonable answer. For example, suppose a solution requires subtracting two numbers that
should have five significant figures and that differ starting with the fourth digit (e.g. 20.295 and 20.278).
Rounding to three digits will lose the accuracy required to determine the difference in the numbers, and
some credit may be lost. Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 1
15 points total Distribution
of points (a) i. 3 points
For field lines inside the shell that point outward and are reasonably close to being
90∞ to both the line charge and the shell surface, with obviously more lines on
the right side than the left
For field lines outside the shell that are radial and noncontinuous with those inside
the shell, reasonably close to 90∞ to the shell surface, and approximately
evenly spaced
For no field lines inside the shell, given that there are field lines drawn elsewhere ii. 1 point
1 point
1 point 2 points
For only negative charges on the inside surface of the shell, with obviously more
charges on the right side
For only positive charges on the outside of the shell, approximately evenly spaced 1 point
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 1 (continued)
Distribution
of points (b) 4 points
4 Va 3 Vb 2 Vc 1 Vd 3 Ve For giving the lowest number to Vd (i.e. placing it at the highest potential)
For giving Vc a higher number than Vd (i.e. placing it at a lower potential) 1 point For giving Vb and Ve the same number 1 point For giving the highest number to Va (i.e. placing it at the lowest potential)
(c)
i. 1 point 1 point 2 points
Using Gauss’s law:
Q
Ú EidA = ε0 E Ú dA = Q ε0 For the correct expression for the value of the integral of the area
A = 2 p r (where is the length of the Gaussian surface)
For the correct expression for the total charge enclosed, which comes only from
the line charge
Q=l
Substituting and solving for the field:
l
E (2p r ) = 1 point
1 point ε0 E= l
2pε0 r Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 1 (continued) Distribution
of points
(c) (continued)
ii.
2 points
For any indication that the total field or the total charge enclosed is the sum of
contributions from the line charge and the shell
Qtot = Qline + Qshell or
Etot = Eline + Eshell
The field from the line charge is the same as calculated in part i.
l
Eline =
2pε0 r
For the shell, the charge enclosed is that between r and r1 ( Qshell = r p r 2  p r12 ) For substitution of the correct enclosed charge for the shell into Gauss’s law
r
Eshell ( 2p r ) =
p r 2  p r12 ε0 ( Eshell =
Etot =
iii. r
2ε0 r 1 point 1 point ) (r 2  r12 )
( r
l
+
r 2  r12
2pε0 r 2ε0 r ) 2 points
For any indication that the total field or the total charge enclosed is the sum of
contributions from the line charge and the shell
Qtot = Qline + Qshell or
Etot = Eline + Eshell
The field from the line charge is the same as calculated in part i.
l
Eline =
2pε0 r
For the shell, the charge enclosed is that between r2 and r1 ( Qshell = r p r22  p r12 ) For substitution of the correct enclosed charged for the shell into Gauss’s law
r
Eshell ( 2p r ) =
p r22  p r12 ε0 ( Etot ( 1 point ) ) r
r 2  r12
2ε0 r 2
r
l
=
+
r 2  r12
2pε0 r 2ε0 r 2 Eshell = 1 point ( ) Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 2
15 points total
(a) Distribution
of points 1 point
Once the switch is closed, VR1 = VC .
From the graph, one can see that at t = 0 , VC = 0 . Therefore, all the voltage drop
occurs across resistor R2 .
For the correct answer, with units
VR 2 = 20 V (b) 1 point 1 point
From the graph, the maximum voltage across the capacitor (and thus also R1 ), is 12 V.
The remaining voltage drop occurs across R2 .
VR 2 = 20 V  12 V
For the correct answer, with units
VR 2 = 8 V (c) 1 point 3 points
A long time after the switch is closed, I C = 0 .
For any indication that I R1 = I R 2 1 point Using Ohm’s law to calculate the current in R1 :
I R1 = VR1 R1 = (12 V ) (15 kW ) = 0.8 ¥ 103 A
Using Ohm’s law to calculate R2 : ( R2 = VR 2 I R 2 = (8 V ) 0.8 ¥ 10 3 A ) For the correct numerical answer
For the correct units
R2 = 10 kW 1 point
1 point NOTE: There was a discrepancy in this question between information in the graph and
the circuit diagram. The value of the time constant as shown in the graph was about
a factor of ten too large when compared to the value determined from the information
in the diagram. Determination of R2 by equating the expression for the time constant
in terms of resistance and capacitance to the time constant from the graph yields a
negative resistance. Since it was not the fault of the students that the correct analysis
yielded a negative value of resistance, full credit could be earned for this method. Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 2 (continued)
Distribution
of points (d) 3 points
Using the relationship for the energy stored in a capacitor:
1
U = CV 2
2
For substituting the correct value of voltage
For substituting the correct value of capacitance
1
2
U = (20 mF ) (12 V )
2
For the correct answer, with correct units
U = 1.44 ¥ 10 3 J (e) 1 point
1 point 1 point 4 points For a correct value for the initial current
From part (a), VR 2 = 20 V immediately after the switch is closed 1 point I max R 2 = ( 20 V ) (10 kW) = 2 ¥ 10 3 A
For a correct value for the final current
From part (c), I R 2 = 0.8 ¥ 103 A a long time after the switch is closed 1 point For a correctly curving graph
For labeling the vertical axis (including units)
Only the general shape of the graph and the endpoints were graded. The time constant
of the graph was not evaluated. 1 point
1 point Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 2 (continued)
Distribution
of points (f) 3 points
For indicating that the energy would be greater
For a correct and complete explanation
For example: Consider the circuit a long time after the switch is closed, when there
is no current in the capacitor. If R2 is replaced with a smaller resistance, then
the total resistance decreases. This results in a larger current through the resistors.
Therefore, the voltage across R1 , and thus across the capacitor, increases. Since
1
U = CV 2 , energy increases.
2
One of the two explanation points could be earned for an incomplete but correct
explanation. 1 point
2 points Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 3
15 points total
(a) Distribution
of points 4 points
The magnetic field from the wire with current I is given by:
mI
B= 0
(this equation can be remembered or derived from Ampere’s law)
2p r
Using the equation for the flux:
fm = Bid A Ú For correctly including the dimension 4 of the loop when reexpressing the area
integral as an integral over r
dA = 4 dr
For the correct limits on the integral
4 fm = mI 0
Ú 2p r 4 dr = 2 m0 I
p 4 Ú fm = 2 m0 I
ln r
p = 2 m0 I
( l n 4  ln
p 1 point ) For the correct answer
2 m0 I
fm =
ln 4
p
(b) 1 point dr
r For correctly performing the integration
4 1 point 1 point 3 points
For correctly indicating that the induced current is counterclockwise
For a complete correct explanation
For example: The decreasing current means that the magnetic field, which points out
of the plane of the page, is decreasing in magnitude. Using Lenz’s law, the induced
current thus needs to create a field to counteract this change. The current must
therefore be counterclockwise to create a field directed out of the page 1 point
2 points Only one point was awarded for incomplete explanations, such as merely saying
that Lenz’s law or the righthand rule justifies the answer. Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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2004 SCORING GUIDELINES
Question 3 (continued)
Distribution
of points (c) 4 points
For a correct relationship between the current, resistance, and derivative of the flux
I loop = e R = (1 R ) (  d f dt )
For using the expression for flux from part (a)
2 m0 I
fm =
ln 4
p
For substituting I (t ) in the flux equation
2 m0 I 0
fm =
(ln 4) e  kt
p
For correctly differentiating this expression to determine the emf
e =  d f =  (  k ) 2 m0 I0 (ln 4) e kt
dt
p
m
e = 2 kp 0 I0 (ln 4) e kt
Substituting into the expression for current:
2 k m0 I 0
I loop =
(ln 4) e  kt
pR (d) 1 point
1 point 1 point 1 point 4 points
For using a correct expression for power
P = I 2R
For substituting the expression for current from part (c)
2
2
2 k m0 I 0
Ê 2 k m0 I 0
P=Á
(ln 4) e  kt ˆ R = Ê
(ln 4)ˆ 1 e 2kt
˜
Á
˜R
Ë pR
¯
Ë
¯
p
For indicating that the dissipated energy is the integral of the power 1 point
1 point 1 point • U= Ú P dt
0 • 2
Ê 2 k m0 I 0
U =Á
(ln 4 )ˆ 1 Ú e 2 kt dt
˜R
Ë
¯
p
0 For correctly integrating the expression
2 ()
1
( 21k )(0  1)
R Ê 2 k m0 I 0
U =Á
(ln 4)ˆ 1  1 e 2 kt
˜ R 2k
Ë
¯
p
2
Ê 2 k m0 I 0
U =Á
(ln 4)ˆ
˜
Ë
¯
p 1 point • 0 2
Ê 2 m0 I 0
ˆk
U =Á
(ln 4)˜
Ëp
¯ 2R Copyright © 2004 by College Entrance Examination Board. All rights reserved.
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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.
 Spring '08
 Reich
 Physics, Electricity And Magnetism, Magnetism

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