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Physics B
AP Review: Electricity and Magnetism
Charge
Comes in + and –
The proton has a charge of e.
The electron has a charge of –e.
e = 1.602  10-19 Coulombs.
The Coulomb is the SI unit of charge
Name:________________
IMPORTANT NOTE:
The electric field inside a conductor is always zero,
whether or not the conductor is charged or near some
external charges.
Problem: Draw field around a + charge
Charge distribution
Positively charged objects have too few electrons.
Negatively charged objects have too many electrons.
If the charged object is an insulator, the excess charge is
usually distributed evenly throughout the object.
If the charged object is a conductor, the excess charge will
accumulate on the surface of the conductor. There will
be no excess charge in the center.
+
Charge is conserved
In any nuclear reaction (or any process whatsoever) total
charge remains constant.
Problem: Draw field around a - charge
Charges apply force to each other
Like charges repel each other; unlike charges attract each
other
Coulomb’s Law
Calculates force between two charges
F = kq1q2/r2
Applies only to spherically symmetric charges
-
Problem: Coulomb’s Law (1988)
54. Two isolated charges, + q and - 2q, are 2
centimeters apart. If F is the magnitude of
the force acting on charge -2q, what are the
magnitude and direction of the force acting
on charge + q ?
Magnitude
Direction
(A) 1/2 F
Toward charge —2q
(B) 1/2 F
Away from charge -2q
(C) F
Toward charge -2q
(D) F
Away from charge -2q
(E) 2F
Toward charge -2q
Show your work
Problem: Draw field around dipole
-
+
Problem: Draw field between capacitor plates and
describe the field in words
+++++++++++++++++++++++++
-----------------------
IMPORTANT NOTE:
The constant k in Coulomb’s Law is equivalent to the
following.
k = 1/4o
Both appear on AP Exams.
++
Magnitude
+ + of the electric field
E = kq/r2
Applies+ only
+ to spherically symmetric charges
Electric Fields
Exist even when just one charge is present.
Start on + charges and terminate on – charges.
Electric field lines indicate direction force would be on a tiny +
charge put in the field.
Electric field lines are not vectors themselves. The field vectors
are tangent to the field lines.
The resulting electric field vector gives the direction of the
electric force on a positive charge placed in the field.
Principle of Superposition
The electric field at a given point in space is the vector sum
of the electric fields due to all charges in the vicinity.
The resulting vector gives the direction of the electric force
on a positive charge placed in the field.
1
(A)
Problem: Field strength analysis (1988)
20. A hollow metal sphere of radius R is
positively charged. Of the following distances
from the center of the sphere, which location will
have the greatest electric field strength?
(A) O (center of the sphere)
(B) 3R/4
(C) 5R/4
(D) 2R
(E) None of the above because the field is of
constant strength
Explain your reasoning
(B)
(C)
(D)
(E)
Explain your reasoning:
Electric Polarization
Electric fields cause polarization (redistribution of charge) on
neutral objects
Conductors are especially vulnerable to this effect. When placed
in an electric field, the charges redistribute themselves so that
the electric field inside the conductor is zero.
Remember our electroscope experiments? The electroscope
is a conductor. When a charged rod is brought near, the
charges on the electroscope move. That makes the vanes
separate, since they assume the same charge.
The electric field inside the electroscope’s metal parts will be
zero.
Problem: Electric Field (1988)
57. Charges +Q and -4Q are situated as shown
above. The net electric field is zero nearest
which point?
(A) A
(B) B
(C) C
(D) D
(E) E
Show your work:
Problem: Electrical polarization (1998)
13. Which of the following is true about the net
force on an uncharged conducting sphere in
a uniform electric field?
(A) It is zero.
(B) It is in the direction of the field.
(C) It is in the direction opposite to the field.
(D) It produces a torque on the sphere about the
direction of the field.
(E) It causes the sphere to oscillate about an
equilibrium position.
Explain your reasoning:
(Hint: Draw a picture of the sphere and how it
polarizes!)
Problem: Electric Field (1998)
The figure above shows two particles, each with
a charge of +Q, that are located at the opposite
corners of a square of side d.
17. What is the direction of the net electric field
at point P ?
2
Problem: Electrical field calculation from
point charges (1993)
Show your work
68. The diagram above shows an isolated,
positive charge Q. Point B is twice as far away
from Q as point A. The ratio of the electric
field strength at point A to the electric field
strength at point B is
(A) 8 to 1
(B) 4 to 1
(C) 2 to 1
(D) 1 to 1
(E) 1 to 2
Show your work:
Electrical Potential
The electric potential is related to potential energy.
Potential gets more positive as you near positive charges and more
negative as you near negative charges.
Potential change is usually more important than the absolute potential.
Potential is a scalar, and requires no vector analysis.
Positive charges like to DECREASE their potential, and
move to lower potentials.
Negative charges like to INCREASE their potential, and
move to higher potentials.
If current can flow, it will flow, as long as there is a
difference in potential to drive current flow.
Problem: Electric Potential (1993)
70. Two conducting spheres of different radii, as
shown above, each have charge - Q. Which of
the following occurs when the two spheres are
connected with a conducting wire?
Calculating Force from Field
F = Eq
Problem: Electric Force from Field (1988)
Questions 17- l8
An electron is accelerated from rest for a time of
10-9 second by a uniform electric field that exerts
a force of 8.0 x 10-15 newton on the electron.
17. What is the magnitude of the electric field
(A) 8.0 x 10-24N/C
(B) 9.1 x 10-22 N/C
(C) 8.0 x 10-6N/C
(D) 2.0 x 10-5 N/C
(E) 5.0 x 104 N/C
Show your work
(A) No charge flows.
(B) Negative charge flows from the larger sphere to the
smaller sphere until the electric field at the surface of
each sphere is the same.
(C) Negative charge flows from the larger sphere to the
smaller sphere until the electric potential of each sphere
is the same.
(D) Negative charge flows from the smaller sphere to the
larger sphere until the electric field at the surface of
each sphere is the same.
(E) Negative charge flows from the smaller sphere to the
larger sphere until the electric potential of each sphere
is the same.
Explain your reasoning
Electrical Potential: spherical calculation
Calculates potential a given distance from charge.
V = kq/r
Applies only to spherically symmetric charges
18. The speed of the electron after it has
accelerated for the 10-9 second is most
nearly
(A) 101 m/s
(B) 103 m/s
(C) 105 m/s
(D) 107 m/s
(E) 109 m/s
Electrical Potential: uniform field calculation
Electrical Potential in a uniform electric field (that is, and
electric field that is like the one you drew in the capacitor
above)
V = -Ed
3
Problem: Electric Potential, Field, and Force
(1993)
Problem: Electric Potential and Potential
Energy (1998)
17. Two large parallel conducting plates P and Q
are connected to a battery of emf , as shown
above. A test charge is placed successively at
points I, II, and III. If edge effects are
negligible, the force on the charge when it is at
point III is
The figure above shows two particles, each with
a charge of +Q, that are located at the
opposite corners of a square of side d.
18. What is the potential energy of a particle of
charge +q that is held at point P ?
A) zero
(A) of equal magnitude and in the same direction as the force
on the charge when it is at point I
(B) of equal magnitude and in the same direction as the force
on the charge when it is at point II
(C) equal in magnitude to the force on the charge when it is
at point I, but in the opposite direction
(D) much greater in magnitude than the force on the charge
when it is at point II, but in the same direction
(E) much less in magnitude than the force on the charge
when it is at point II, but in the same direction
B) 2 qQ/(4od)
C) qQ/(4od)
D) 2 qQ/(4od)
E) 22 qQ/(4od)
Explain your reasoning
Show your work
Electrical Potential Energy
For absolute potential energy
U = qV
For potential energy change
U = qV
Problem: Charged conductor (1988)
59. A positive charge Of 10-6 coulomb is placed
on an insulated solid conducting sphere.
Which of the following is true?
Problem: Electric Potential and Potential
Energy (1988)
16. An electron volt is a measure of
(A) energy
(B) electric field
(C) electric potential due to one electron
(D) force per unit electron charge
(E) electric charge
Explain your choice (Hint: relate this to an
equation)
(A) The charge resides uniformly throughout the sphere.
(B) The electric field inside the sphere is constant in
magnitude, but not zero.
(C) The electric field in the region surrounding the sphere
increases with increasing distance
from the sphere.
(D) An insulated metal object acquires a net positive charge
when brought near to, but not in
contact with, the sphere.
(E) When a second conducting sphere is connected by a
conducting wire to the first sphere, charge is transferred
until the electric potentials of the two spheres are equal.
Explain your reasoning:
4
<< ADVANCED TOPIC >>
Capacitor
Consists of two “plates” in close proximity.
When “charged”, there is a voltage across the plates, and they
bear equal and opposite charges.
Energy in a Capacitor
UE = ½ C (V)2
UE: electrical potential energy (J)
C: capacitance in (F)
V: potential difference between plates (V)
Calculating capacitance
C = q / V
C: capacitance in Farads (F)
q: charge (on positive plate) in Coulombs (C)
V: potential difference between plates in Volts (V)
Problem: Energy in a Capacitor (1998)
70. A 4 F capacitor is charged to a potential
difference of 100 V. The electrical energy stored
in the capacitor is
Capacitors in Circuits
Circuit drawing
(A)
(B)
(C)
(D)
Equivalent capacitance
If you have capacitors in series, you add the reciprocal of the
capacitances and then take the reciprocal of the result
1/Ceq = Ci)
If you have capacitors in parallel, you add the capacitances
Ceq = Ci
Equivalent capacitance equations are the opposite of
equivalent resistance equations.
2 x 10-10 J
2 x 10-8 J
2 x 10-6 J
2 x 10-4 J
2 x 10-2 J
(E)
Show your work:
Problem: Equivalent Capacitances (1993)
Questions 15-16 refer to the circuit below.
Capacitance of parallel plate capacitor
A fancy equation, but basically it says that capacitance is
related linearly with plate area, and inversely with
spacing between the plates
C = e0A/d
C: capacitance (F)
e: dielectric constant of filling
0 : permittivity (8.85 x 10-12 F/m)
A: plate area (m2)
d: distance between plates(m)
15. The equivalent capacitance for this network
is most nearly
(A) 10/7 uF
(B) 3/2 uF
(C) 7/3 uF
(D) 7 uF
(E) 14 uF
Show your work:
Problem: Parallel Plate Capacitor (1998)
14. Two parallel conducting plates are
connected to a constant voltage source. The
magnitude of the electric field between the
plates is 2,000 N/C. If the voltage is doubled
and the distance between the plates is reduced
to 1/5 the original distance, the magnitude of
the new electric field is
(A)
800 N/C
(B) 1,600 N/C
(C) 2,400 N/C
(D) 5,000 N/C
(E) 20,000 N/C
Show your work:
16. The charge stored in the 5-microfarad
capacitor is most nearly
(A) 360 uC
(B) 500 uC
(C) 710 uC
(D) 1,100 uC
(E) 1,800 uC
Show your work:
5
Conductors
Conduct electricity easily.
Have high “conductivity”.
Have low “resistivity”.
Metals are examples.
Problem: Parallel Plate Capacitor (1998)
64. Two parallel conducting plates, separated by
a distance d, are connected to a battery of emf
 . Which of the following is correct if the plate
separation is doubled while the battery remains
connected?
(A) The electric charge on the plates is doubled.
(B) The electric charge on the plates is halved.
(C) The potential difference between the plates is doubled.
(D) The potential difference between the plates is halved.
(E) The capacitance is unchanged.
Explain your reasoning
Insulators
Don’t conduct electricity easily.
Have low “conductivity”.
Have high “resistivity”.
Rubber is an example.
Resistivity and Conductivity
Depend on the identity of the material, not its shape, size, or
configuration.
Available in tables of data.
Resistors
Devices put in circuits to reduce the current flow.
Represented by this symbol:
Calculating resistance from resistivity
R = L/A
Problem: Parallel Plate Capacitor (1988)
14. The capacitance of a parallel-plate capacitor
can be increased by increasing which of the
following?
Problem: Calculating Resistance (1998)
(A) The distance between the plates
(B) The charge on each plate
(C) The area of the plates
(D) The potential difference across the plates
(E) None of the above
Explain your reasoning
Two concentric circular loops of radii b and 2b, made
of the same type of wire, lie in the plane of the page, as
shown above.
65. The total resistance of the wire loop of radius
b is R. What is the resistance of the wire loop of
radius 2b ?
(A) R/4
(B) R/2
(C) R
(D) 2R
(E) 4R
Show your work or explain your reasoning
Current
Flow of positive charge
I = Q/t
Direct Current (DC)
Uniform current that flows in one direction in a circuit.
Cell
What produces the current in a circuit.
Represented by this symbol:
Battery
Multiple cells in series.
Represented by this symbol (for a 2-cell battery)
Problem: Calculating Resistance (1988)
40. The five resistors shown below have the lengths
and cross-sectional areas indicated and are
made of material with the same resistivity.
Which resistor has the least resistance?
Electromotive force (
The potential, or voltage, that can theoretically be produced
by the cell.
Depends on the chemistry of cell.
The emf is the maximum voltage the cell can produce.
Usually internal resistance of the cell causes the actual
voltage to be lower.
6
(A) 8 coulombs
(B) 8 newtons
(C) 8 joules
(D) 8 calories
(E) 8 newton-amperes
Show your work
Explain your reasoning
Problem: Power and Energy (1988)
19. An immersion heater of resistance R
converts electrical energy into thermal energy
that is transferred to the liquid in which the
heater is immersed. If the current in the heater is
I, the thermal energy transferred to the liquid in
time t is
(A) IRt
(B) I2Rt
(C) IR2t
(D) IRt2
(E) IR/t
Show your work
Ohm’s Law
V = IR
Ohmmeter
Placed across resistor or other circuit element to measure
resistance when no current is flowing.

Voltmeter
Placed across resistor or other circuit element to measure
potential change when current is flowing.
V
Problem: Power and Energy (1998)
20. A certain coffeepot draws 4.0 A of current
when it is operated on 120 V household lines. If
electrical energy costs 10 cents per
kilowatt-hour, how much does it cost to operate
the coffeepot for 2 hours?
(A)
2.4 cents
(B)
4.8 cents
(C)
8.0 cents
(D)
9.6 cents
(E) 16 cents
Show your work
Ammeter
Placed in a circuit in place of a wire to measure the current
flowing in that part of the circuit.
A
Power in Electrical Circuits
P = I V
Energy in Electrical Circuits
Energy is power times time
E = (P)(t)
Note: the kilowatt hour is a unit of energy, not a unit of
power.
Problem: Power and Energy (1993)
5l. The product
2 amperes x 2 volts x 2 seconds
is equal to
Resistors in series
Req = Ri
7
Problem: Equivalent Resistance (1998)
Resistors in parallel
1/Req = Ri)
Problem: Equivalent Resistance (1984)
15. The electrical resistance of the part of the
circuit shown between point X and point Y is
(A)
(B)
(C)
(D)
18. Which two arrangements of resistors shown
above have the same resistance between the
terminals?
(A) I and II
(B) I and IV
(C) II and III
(D) II and IV
(E) III and IV
Show your work
1
1 
3
2
3
2 
4
4
(E)
6
Show your work
Problem: Ohm’s Law (1998)
Problem: Equivalent Resistance (1988)
16. When there is a steady current in the circuit,
the amount of charge passing a point per unit of
time is
I5. The total equivalent resistance between points
X and Y in the circuit shown above is
(A) 3 
(B) 4 
(C) 5 
(D) 6 
(E) 7 
Show your work
(A)
(B)
(C)
(D)
(E)
8
the same everywhere in the circuit
greater at point X than at point Y
greater in the 1 resistor than in the 2 resistor
greater in the 1 resistor than in the 3 resistor
greater in the 2 resistor than in the 3 resistor
Show your work or state your reasoning
Problem: General Circuit Problems (1993)
Questions 20-22 relate to the following circuit
diagram, which shows a battery with an internal
resistance of 4.0 ohms connected to a 16-ohm
and a 20-ohm resistor in series. The current in
the 20-ohm resistor is 0.3 amperes.
Problem: Circuit Problem (1993)
50. In the diagrams above, resistors R1. and R2
are shown in two different connections to the
same source of emf  that has no internal
resistance. How does the power dissipated by
the resistors in these two cases compare?
20. What is the emf of the battery?
(A) 1.2 V
(B) 6.0 V
(C) 10.8 V
(D) 12.0 V
(E) 13.2 V
Show your work
(A) It is greater for the series connection.
(B) It is greater for the parallel connection.
(C) It is the same for both connections.
(D) It is different for each connection, but one must know the
values of . R1 and R2 to know which is greater.
(E) It is different for each connection, but one must know the
value of  to know which is greater.
Explain your reasoning
21. What is the potential difference across the
terminals X and Y of the battery?
(A) 1.2 V
(B) 6.0 V
(C) 10.8 V
(D) 12.0 V
(E) 13.2 V
Show your work
Problem: Ohm’s Law (1988)
68. In the circuit shown above, the value of r for
which the current I is 0.5 ampere is
(A) 0 
(B) 1 
(c) 5 
(D) 10 
(E) 20 
Explain your reasoning
22. What power is dissipated by the 4-ohm
internal resistance of the battery?
A) 0.36 W
(B) 1.2 W
(C) 3.2 W
(D) 3.6 W
(E) 4.8 W
Show your work
9
Kirchoff’s 1st Rule (Junction rule)
The sum of the currents entering a junction equals the sum of
the currents leaving the junction.
Conservation of charge.
22. An electron is in a uniform magnetic field B
that is directed out of the plane of the page, as
shown above. When the electron is moving in
the plane of the page in the direction indicated by
the arrow, the force on the electron is directed
(A) toward the right
(B) out of the page
(C) into the page
(D) toward the top of the page
(E) toward the bottom of the page
State your reasoning
Kirchoff’s 2nd Rule (Loop rule)
The net change in electrical potential in going around one
complete loop in a circuit is equal to zero.
Conservation of energy.
Problem: Kirchoff’s Rules (1993)
14. Kirchhoff's loop rule for circuit analysis is an
expression of which of the following?
(A) Conservation of charge
(B) Conservation of energy
(C) Ampere's law
(D) Faraday's law
(E) Ohm's law
Explain your reasoning
Magnetic Fields
Formed by moving charge
Affect moving charge
Magnetic Forces can...
accelerate charged particles by changing their direction
cause charged particles to move in circular or helical paths
Magnetic Forces cannot...
change the speed or kinetic energy of charged particles
do work on charged particles
Magnetic Dipole
The magnetic force is centripetal
qvBsin = mv2/r
qB = mv/r
q/m = v/(rB)
Problem: Magnetic Force as a Centripetal
Force (1998)
Questions 46-47 A magnetic field of 0.1 T forces
a proton beam of 1.5 mA to move in a circle of
radius 0.1 m. The plane of the circle is
perpendicular to the magnetic field.
46.
Of the following, which is the best
estimate of the work done by the magnetic field
on the protons during one complete orbit of the
circle?
(A) 0 J
(B) 10-22 J
(C) 10-5 J
(D) 102 J
(E) 1020 J
Show your work
The magnetic field lines are complete loops. They exit the
magnet at the north pole and re-enter the south pole.
Magnetic Field (B-field) Units
Tesla (SI)
Gauss (1 T = 104 gauss)
Magnetic Monopoles
Do not exist!
Magnetic Force on Charged Particle
F = qvBsin
direction: Right Hand Rule
Problem: Magnetic Force (1998)
10
47. Of the following, which is the best estimate
of the speed of a proton in the beam as it moves
in the circle?
(A) 10-2 m/s
(B) 103 m/s
(C) 106 m/s
(D) 108 m/S
(E) 1015 m/s
Show your work
Problem: Magnetic Force (1984)
63. Two long, parallel wires, fixed in space,
carry currents I1 and 12. The force of
attraction has magnitude F. What currents
will give an attractive force of magnitude
4F?
1
(A) 2 I 1 and I 2
2
1
(B) I 1 and I 2
4
1
1
(C) I 1 and I 2
2
2
(D) 2 I 1 and 2 I 2
(E) 4 I 1 and 4 I 2
Velocity filter
Electric and magnetic fields can be used together to precisely
select the velocity of a charged particle.
Show your work
B directed into paper
E directed down
Velocity of charged particle directed right
q
Magnetic Field for Long Straight Wire
B = oI/(2r)
Problem: Magnetic Field (1993)
Magnetic Force on Current-carrying Wire
F = I L B sin
Problem: Magnetic Force (1988)
19. Two long, parallel wires are separated by a
distance a as shown above. One wire carries a
steady current I into the plane of the page
while the other wire carries a steady current I
out of the page. At what points in the plane of
the page and outside the wires, besides points
at infinity, is the magnetic field due to the
currents zero?
47. A wire in the plane of the page carries a
current I directed toward the top of the page, as
shown above. If the wire is located in a uniform
magnetic field B directed out of the page, the
force on the wire resulting from the magnetic
field is
(A) directed into the page
(B) directed out of the page
(C) directed to the right
(D) directed to the left
(E) zero
State your reasoning
(A) Only at point P
(B) At all points on the line SS'
(C) At all points on the line connecting the two wires
(D) At all points on a circle of radius 2d centered on point P
(E) At no points
Explain your reasoning
11
Hand Rule for magnetic force on moving positive charge
Place your fingers in direction of velocity. Then rotate your
wrist so that your fingers can bend into the direction of the
field. Your thumb will be pointing in the direction of the
force.
Induced Current
A system will respond to oppose changes in magnetic flux.
Changing the magnetic flux can generate electrical current.
Faraday’s Law of Induction
 = -NB/t
Hand Rule for magnetic force on moving negative charge
Use the method described above, then flip your thumb 180 o.
Alternately, you may use your left hand.
To generate voltage
 = -B/t
 = -(BAcos)/t
Change B
Hand Rule for magnetic force on current in wire
Place your fingers in direction of current. Then rotate your
wrist so that your fingers can bend into the direction of the
field. Your thumb will be pointing in the direction of the
force.
Change A
Change 
Problem: Faraday’s Law (1998)
Hand Rule for fields where current is straight
Curve your fingers.
Place your thumb in direction of current. Then your curved
fingers point in direction of curved magnetic field.
Hand Rule for fields where current is circular
Curve your fingers.
Place your curved fingers in direction of current. Then your
thumb points in direction of magnetic field in center of
circular current.
66.
A uniform magnetic field B that is
perpendicular to the plane of the page now
passes through the loops, as shown above. The
field is confined to a region of radius a, where a
< b, and is changing at a constant rate. The
induced emf in the wire loop of radius b is  .
What is the induced emf in the wire loop of
radius 2b ?
(A) Zero
 2
(B)
(C)

(D) 2
Magnetic Field Inside a Solenoid
B = onI
Magnetic Flux
The product of magnetic field and area.
B = BAcos
B: magnetic flux in Webers (Tesla meters2)
B: magnetic field in Tesla
A: area in meters2.
: the angle between the area and the magnetic field.
Problem: Magnetic Flux (1998)
(E)
4
Show your work
19. A rectangular wire loop is at rest in a uniform
magnetic field B of magnitude 2 T that is directed
out of the page. The loop measures 5 cm by 8 cm,
and the plane of the loop is perpendicular to the
field, as shown above. The total magnetic flux
through the loop is
(A) zero
(B) 2 x 10-3 Tm
(C) 8 x 10-3 Tm
(D) 2 x 10-1 Tm
(E) 8 x 10-1 Tm
Show your work
Problem: Faraday’s Law (1993)
67. A square loop of wire of resistance R and
side a is oriented with its plane perpendicular to
a magnetic field B. as shown above. What must
be the rate of change of the magnetic field in
order to produce a current I in the loop?
12
Explain your reasoning
(A) IR/a2
(B) la2/R
(C) la /R
(D) Ra/l
(E) IRa
Show your work
Motional emf
Faraday’s law can be used to show that a wire moving in a
magnetic field generates a potential equivalent to
 = BLv
Lenz’s Law
Induced current will flow in a direction so as to oppose the
change in flux.
Use in combination with hand rule to predict current
direction.
Problem: Motional emf (1993)
Problem: Lenz’s Law (1998)
4l. A wire of constant length is moving in a
constant magnetic field, as shown above. The
wire and the velocity vector are perpendicular
to each other and are both perpendicular to the
field. Which of the following graphs best
represents the potential difference  between
the ends of the wire as a function of the speed
v of the wire?
48.
A single circular loop of wire in the
plane of the page is perpendicular to a uniform
magnetic field B directed out of the page, as
shown above. If the magnitude of the magnetic
field is decreasing, then the induced current in
the wire is
(A) directed upward out of the paper
(B) directed downward into the paper
(C) clockwise around the loop
(D) counterclockwise around the loop
(E) zero (no current is induced)
Explain your reasoning
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