AP® Physics C: Electricity and Magnetism
2006 Free-Response Questions
The College Board: Connecting Students to College Success
The College Board is a not-for-profit membership association whose mission is to connect students to college success and
opportunity. Founded in 1900, the association is composed of more than 5,000 schools, colleges, universities, and other
educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and
3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and
teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement
Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied
in all of its programs, services, activities, and concerns.
© 2006 The College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP
Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Board. Admitted Class Evaluation
Service, CollegeEd, connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the
Cornerstones are trademarks owned by the College Board. PSAT/NMSQT is a registered trademark of the College Board and
National Merit Scholarship Corporation. All other products and services may be trademarks of their respective owners.
Permission to use copyrighted College Board materials may be requested online at:
www.collegeboard.com/inquiry/cbpermit.html.
Visit the College Board on the Web: www.collegeboard.com.
AP Central is the official online home for the AP Program: apcentral.collegeboard.com.
TABLE OF INFORMATION FOR 2006 and 2007
UNITS
CONSTANTS AND CONVERSION FACTORS
1 unified atomic mass unit,
1 u = 1.66 × 10−27 kg
Name
= 931 MeV c 2
Proton mass,
m p = 1.67 × 10
−27
kg
m
kilogram
kg
mn = 1.67 × 10−27 kg
Electron mass,
me = 9.11 × 10−31 kg
second
e = 1.60 × 10−19 C
ampere
Avogadro’s number,
Universal gas constant,
Boltzmann’s constant,
Speed of light,
Planck’s constant,
N 0 = 6.02 × 1023 mol−1
kelvin
R = 8.31 J (moli K)
k B = 1.38 × 10
−23
J/K
c = 3.00 × 10 m/s
8
h = 6.63 × 10−34 J is
= 4.14 × 10−15 eVis
hc = 1.99 × 10−25 J i m
= 1.24 × 10
⑀0 = 8.85 × 10−12 C 2 N i m 2
3
Vacuum permittivity,
eV i nm
Coulomb’s law constant, k = 1 4π ⑀0 = 9.0 × 109 Nim 2 C2
Vacuum permeability,
µ0 = 4π × 10−7 (T i m) A
Magnetic constant, k ′ = µ0 /4π = 10−7 (T ⋅ m) A
Universal gravitational constant,
Acceleration due to gravity
at Earth’s surface,
1 atmosphere pressure,
G = 6.67 × 10
−11
m
3
g = 9.8 m s 2
1 atm = 1.0 × 10 N m
5
2
= 1.0 × 10 Pa
1 eV = 1.60 × 10−19 J
5
1 electron volt,
kg is
2
Symbol
meter
Neutron mass,
Electron charge magnitude,
PREFIXES
10
10
10
10
A
10
K
mol
hertz
Hz
newton
N
pascal
Pa
9
106
s
mole
Prefix Symbol
giga
G
Factor
10
10
3
-2
-3
-6
-9
-12
mega
M
kilo
k
centi
c
milli
m
micro
m
nano
n
pico
p
VALUES OF
TRIGONOMETRIC
FUNCTIONS FOR COMMON
ANGLES
joule
J
watt
W
θ
sin θ
cos θ
coulomb
C
0
0
1
0
volt
V
30
1/2
3 /2
3 /3
ohm
W
henry
H
37
3/5
4/5
3/4
farad
F
45
2 /2
2 /2
1
tesla
T
53
4/5
3/5
4/3
degree
Celsius
C
60
3 /2
1/2
3
90
1
0
•
electronvolt
eV
tan θ
The following conventions are used in this examination.
I. Unless otherwise stated, the frame of reference of any problem is assumed to be inertial.
II. The direction of any electric current is the direction of flow of positive charge (conventional current).
III. For any isolated electric charge, the electric potential is defined as zero at an infinite distance from the charge.
2
ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2006 and 2007
MECHANICS
u = u0 + at
x = x0 + u0 t +
1 2
at
2
u 2 = u0 2 + 2a ( x - x0 )
 F = Fnet = ma
dp
dt
F=
J = Ú F dt = Dp
p = mv
F fric £ m N
W =
ÚF
K =
1 2
mu
2
P=
dW
dt
∑
dr
P = Fⴢv
DUg = mgh
ac =
a
F
f
h
I
J
K
k
=
=
=
=
=
=
=
=
=
L =
m=
N =
P =
p =
r =
r =
T =
t =
U=
u =
W=
x =
m =
q =
t =
w =
a =
acceleration
force
frequency
height
rotational inertia
impulse
kinetic energy
spring constant
length
angular momentum
mass
normal force
power
momentum
radius or distance
position vector
period
time
potential energy
velocity or speed
work done on a system
position
coefficient of friction
angle
torque
angular speed
angular acceleration
u
= w2 r
r
Fs = - k x
Us =
 t = t net = I a
Ú r dm = Â mr
2
2
rcm =  mr  m
L = r ¥ p = Iw
1 2
Iw
2
T =
1 2
kx
2
E=
F
q
ÚE
Tp = 2p
UE = qV =
C =
Q
V
C =
k ⑀0 A
d
UG = 1 2
at
2
r2
Gm1m2
r
1 q1q2
4 p⑀0 r
 Ci
Cp =
i
1
1
=Â
Cs
C
i i
dQ
dt
r
A
V = IR
Rs =
rˆ
 Ri
1
=
Rp
i
1
ÂR
i
i
P = IV
FM = qv ¥ B
3
n
=
=
=
=
=
=
=
=
=
=
=
=
area
magnetic field
capacitance
distance
electric field
emf
force
current
current density
inductance
length
number of loops of wire
per unit length
number of charge carriers
per unit volume
power
charge
point charge
resistance
distance
time
potential or stored energy
electric potential
velocity or speed
resistivity
magnetic flux
dielectric constant
N =
P =
Q=
q =
R =
r =
t =
U=
V =
u =
r =
fm =
k =
ÚB
∑
d ᐉ = m0 I
dB =
F=
I = Neud A
g
Gm1m2
i
E = rJ
m
k
Ts = 2 p
q
 rii
1
4 p⑀0
V =
F
I
J
L
⑀0
dV
dr
E =-
A
B
C
d
E
e
Q
dA=
∑
R=
2p
1
=
w
f
FG = -
w = w0 + at
q = q0 + w0 t +
1 q1q2
4 p⑀0 r 2
1
1
Uc = QV = CV 2
2
2
u = rw
K =
F =
I =
2
t=r¥F
I =
ELECTRICITY AND MAGNETISM
m0 I d ᐉ ¥ r
4p r3
Ú I dᐉ ¥ B
Bs = m0 nI
fm = Ú B ∑ d A
d fm
dt
e
=-
e
= -L
UL =
dI
dt
1 2
LI
2
ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 2006 and 2007
GEOMETRY AND TRIGONOMETRY
Rectangle
A = bh
Triangle
A=
1
bh
2
Circle
A = pr2
C = 2p r
Parallelepiped
V = wh
Cylinder
A=
C=
V=
S =
b =
h =
=
w=
r =
CALCULUS
area
circumference
volume
surface area
base
height
length
width
radius
df
d f du
=
dx
du dx
d n
( x ) = nxn -1
dx
d x
(e ) = e x
dx
d
(1n x ) = 1
dx
x
d
(sin x ) = cos x
dx
d
(cos x ) = - sin x
dx
V = pr2
Úx
S = 2p r + 2p r 2
Úe
Sphere
V =
4 3
pr
3
S = 4p r
2
c
Ú
a
90°
q
cos q =
b
c
tan q =
a
b
x
dx = e x
dx
= ln x
x
Ú sin x dx = - cos x
a 2 + b2 = c2
a
c
dx =
Ú cos x dx = sin x
b
Right Triangle
sin q =
1
x n + 1 , n π -1
n +1
n
4
2006 AP® PHYSICS C: ELECTRICITY AND MAGNETISM
FREE-RESPONSE QUESTIONS
PHYSICS C: ELECTRICITY AND MAGNETISM
SECTION II
Time— 45 minutes
3 Questions
Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions,
which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in the
pink booklet in the spaces provided after each part, NOT in this green insert.
E&M 1.
The square of side a above contains a positive point charge +Q fixed at the lower left corner and negative point
charges -Q fixed at the other three corners of the square. Point P is located at the center of the square.
(a) On the diagram, indicate with an arrow the direction of the net electric field at point P.
(b) Derive expressions for each of the following in terms of the given quantities and fundamental constants.
i. The magnitude of the electric field at point P
ii. The electric potential at point P
(c) A positive charge is placed at point P. It is then moved from point P to point R, which is at the midpoint of
the bottom side of the square. As the charge is moved, is the work done on it by the electric field positive,
negative, or zero?
____ Positive
____ Negative
____ Zero
Explain your reasoning.
(d)
i. Describe one way to replace a single charge in this configuration that would make the electric field at
the center of the square equal to zero. Justify your answer.
ii. Describe one way to replace a single charge in this configuration such that the electric potential at the
center of the square is zero but the electric field is not zero. Justify your answer.
© 2006 The College Board. All rights reserved.
Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).
GO ON TO THE NEXT PAGE.
5
2006 AP® PHYSICS C: ELECTRICITY AND MAGNETISM
FREE-RESPONSE QUESTIONS
E&M 2.
The circuit above contains a capacitor of capacitance C, a power supply of emf e , two resistors of resistances R1
and R2 , and two switches, S1 and S2 . Initially, the capacitor is uncharged and both switches are open. Switch S1
then gets closed at time t = 0.
(a) Write a differential equation that can be solved to obtain the charge on the capacitor as a function of time t.
(b) Solve the differential equation in part (a) to determine the charge on the capacitor as a function of time t.
Numerical values for the components are given as follows:
e = 12 V
C = 0.060 F
R1 = R2 = 4700 W
(c) Determine the time at which the capacitor has a voltage 4.0 V across it.
After switch S1 has been closed for a long time, switch S2 gets closed at a new time t = 0.
(d) On the axes below, sketch graphs of the current I1 in R1 versus time and of the current I 2 in R2 versus time,
beginning when switch S2 is closed at new time t = 0. Clearly label which graph is I1 and which is I 2 .
© 2006 The College Board. All rights reserved.
Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).
GO ON TO THE NEXT PAGE.
6
2006 AP® PHYSICS C: ELECTRICITY AND MAGNETISM
FREE-RESPONSE QUESTIONS
E&M 3.
A loop of wire of width w and height h contains a switch and a battery and is connected to a spring of force
constant k, as shown above. The loop carries a current I in a clockwise direction, and its bottom is in a constant,
uniform magnetic field directed into the plane of the page.
(a) On the diagram of the loop below, indicate the directions of the magnetic forces, if any, that act on each side of
the loop.
(b) The switch S is opened, and the loop eventually comes to rest at a new equilibrium position that is a distance x
from its former position. Derive an expression for the magnitude B0 of the uniform magnetic field in terms of
the given quantities and fundamental constants.
© 2006 The College Board. All rights reserved.
Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).
GO ON TO THE NEXT PAGE.
7
2006 AP® PHYSICS C: ELECTRICITY AND MAGNETISM
FREE-RESPONSE QUESTIONS
The spring and loop are replaced with a loop of the same dimensions and resistance R but without the battery and
switch. The new loop is pulled upward, out of the magnetic field, at constant speed u0 . Express algebraic answers to
the following questions in terms of B0 , u0 , R, and the dimensions of the loop.
(c)
i.
On the diagram of the new loop below, indicate the direction of the induced current in the loop as the loop
moves upward.
ii. Derive an expression for the magnitude of this current.
(d) Derive an expression for the power dissipated in the loop as the loop is pulled at constant speed out of the field.
(e) Suppose the magnitude of the magnetic field is increased. Does the external force required to pull the loop at
speed u0 increase, decrease, or remain the same?
_____ Increases
_____ Decreases
_____ Remains the same
Justify your answer.
END OF EXAM
© 2006 The College Board. All rights reserved.
Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).
8