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Unformatted text preview: AP® Physics C: Electricity and Magnetism
2006 FreeResponse Questions The College Board: Connecting Students to College Success
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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 kilogram kg Electron mass, me = 9.11 × 10−31 kg
e = 1.60 × 10−19 C ampere 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 c = 3.00 × 10 m/s Vacuum permittivity, Coulomb’s law constant, k = 1 4π
Vacuum permeability, 0 1 atmosphere pressure, m 1 atm = 1.0 × 10 N m 2 = 1.0 × 10 Pa
1 eV = 1.60 × 10−19 J
5 1 electron volt, Pa micro n pico 1 2 m nano 9 m p VALUES OF
TRIGONOMETRIC
FUNCTIONS FOR COMMON
ANGLES kg is 2 J watt W θ sin θ cos θ coulomb C 0 0 1 0 V 30 1/2 3 /2 3 /3 W henry H 37 3/5 4/5 3/4 F 45 2 /2 2 /2 1 tesla 3 g = 9.8 m s 2
5 N 10 milli 6 c farad µ0 = 4π × 10−7 (T i m) A
G = 6.67 × 10 Hz centi 3 k ohm = 9.0 × 109 Nim 2 C 2 Magnetic constant, k ′ = µ0 / 4π = 10−7 (T ⋅ m) A
Universal gravitational constant,
Acceleration due to gravity
at Earth’s surface, 10 kilo 2 M volt eV i nm −11 10 mega 3 joule = 1.24 × 10
−12 2
C Nim2
0 = 8.85 × 10
3 10 mol pascal = 4.14 × 10−15 eVis
hc = 1.99 × 10−25 J i m 10 K newton h = 6.63 × 10−34 J is 10 A hertz J/K 9 106 s mole 8 10 kg second Prefix Symbol
giga
G Factor m mn = 1.67 × 10−27 kg Avogadro’s number,
Universal gas constant, Symbol meter Neutron mass,
Electron charge magnitude, PREFIXES 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 + 12
at
2 u 2 = u0 2 + 2a ( x  x0 ) Â F = Fnet = ma
dp
dt F= J = Ú F dt = Dp p = mv
F fric £ m N
W= ÚF K= 12
mu
2 P= dW
dt ∑ dr P=Fv 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 = Â t = t net = I a Ú r dm = Â mr
2 2 rcm = Â mr Â m L = r ¥ p = Iw
12
Iw
2 T= 12
kx
2 E= F
q ÚE Tp = 2p UG = 12
at
2 r2 Gm1m2
r q Â rii 1
4p 0 V= i UE = qV = 1 q1q2
4p 0 r Q
V C= k C= 0A d Â Ci Cp = i 1
1
=Â
Cs
Ci
i dQ
dt r
A V = IR
Rs =
ˆ
r Â Ri 1
=
Rp 1 ÂR i P = IV FM = qv ¥ B 3 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 = m0 I dB = m0 I d ¥ r
4p r3 ÚI d ¥B Bs = m0 nI
fm = Ú B ∑ d A d fm
dt e i i n =
=
=
=
=
=
=
=
=
=
=
= F= I = Neud A g Gm1m2 dV
dr E = rJ m
k Ts = 2 p F
I
J
L 0 E= A
B
C
d
E e Q dA= ∑ R= 2p
1
=
w
f FG =  w = w0 + at
q = q0 + w0 t + 1 q1q2
4p 0 r 2 1
1
Uc = QV = CV 2
2
2 u = rw K= F= I= 2 t=r¥F I= ELECTRICITY AND MAGNETISM = e = L UL = dI
dt 12
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
dn
( x ) = nxn 1
dx
dx
(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= 43
pr
3 S = 4p r 2 c Ú a
90° q a
c
b
c tan q = x dx = e x dx
= ln x
x Ú sin x dx =  cos x a 2 + b2 = c2 cos q = dx = Ú cos x dx = sin x b Right Triangle sin q = 1
x n + 1 , n π 1
n +1 n a
b 4 2006 AP® PHYSICS C: ELECTRICITY AND MAGNETISM
FREERESPONSE 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.
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5 2006 AP® PHYSICS C: ELECTRICITY AND MAGNETISM
FREERESPONSE 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.
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6 2006 AP® PHYSICS C: ELECTRICITY AND MAGNETISM
FREERESPONSE 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.
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7 2006 AP® PHYSICS C: ELECTRICITY AND MAGNETISM
FREERESPONSE 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.
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This note was uploaded on 02/25/2010 for the course PHY 15166 taught by Professor Mr.donald during the Spring '10 term at 4.1.
 Spring '10
 mr.donald
 Physics, Magnetism

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