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Eðlisfræði 2, vor 2007
21. Electric Charge and Field
Assignment is due at 2:00am on Wednesday, January 24, 2007
Credit for problems submitted late will decrease to 0% after the deadline has passed.
The wrong answer penalty is 2% per part. Multiple choice questions are penalized as described in the online help.
The unopened hint bonus is 2% per part.
You are allowed 4 attempts per answer.
Charge, Conductors, and Insulators
The Electric Field inside a Conductor
Learning Goal: To understand how the charges within a conductor respond to an externally applied electric
field.
To illustrate the behavior of charge inside conductors, consider a long conducting rod that is suspended by
insulating strings (see the figure). Assume that the rod is initially electrically neutral, and that it remains so for
this discussion. The rod is positioned along the x axis, and an external electric field that points in the positive x
direction (to the right) can be applied to the rod and the surrounding region. The atoms in the rod are composed of
positive nuclei (indicated by plus signs) and negative electrons (indicated by minus signs). Before application of the
electric field, these atoms were distributed evenly throughout the rod.
Part A
What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem
introduction is applied? (Ignore internal fields in the rod for the moment.)
Hint A.1 Formula for the force on a charge in an electric field
Hint not displayed
ANSWER: Answer not displayed
Part B
What is the motion of the negative electrons and positive atomic nuclei caused by the external field?
Hint B.1 How to approch this part
Hint not displayed
Hint B.2 Masses and charges of nuclei and electrons
Hint not displayed
ANSWER: Answer not displayed
Part C
Part not displayed
Part D
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Part not displayed
Charging a Conducting Rod
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod
suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer
to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem,
"strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist
between two charged balls.
Part A
A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to
end A of the rod . What happens to end A of the rod when the ball approaches it closely this first time?
Hint A.1 The key property of conductors
The key property of a conductor is that the charges are free to move around inside in response to internal electric
fields; in a static situation, they will arrange so that the internal field is zero.
Hint A.2 How much charge moves to end A?
It is stated that the ball is much closer to the end of the rod than the length of the rod. Therefore, if points down
the rod several times the distance of approach (but still much closer to end A than end B) are to experience no
electric field, the charge on end A of the rod must be comparable in magnitude to the charge on the ball (so that
their fields will cancel).
ANSWER:
It is strongly repelled.
It is strongly attracted.
It is weakly attracted.
It is weakly repelled.
It is neither attracted nor repelled.
This charge is said to be "induced" by the presence of the electric field of the charged ball: It is not transferred
by the ball.
Now consider what happens when the small metal ball is repeatedly given a negative charge and then brought into
contact with end A of the rod.
Part B
After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is
far away)?
ANSWER:
There is positive charge on end B and negative charge on end A.
There is negative charge spread evenly on both ends.
There is negative charge on end A with end B remaining neutral.
There is positive charge on end A with end B remaining neutral.
Part C
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How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end
A? Assume that the phrase "a great many" means that the total charge on the rod dominates any charge movement
induced by the near presence of the charged ball.
ANSWER:
It is strongly repelled.
It is strongly attracted.
It is weakly attracted.
It is weakly repelled.
It is neither attracted nor repelled.
Part D
How does end B of the rod react when the charged ball approaches it after a great many previous contacts with end
A?
Hint D.1 The rod is a conductor
Because the rod is a conductor, the charge is free to distribute itself over the entire rod. It must be distributed so
that the internal electric field in the rod is zero, and there is only one distribution that achieves this. There is no
memory in this situation: The charge will always distribute itself into the same final result. The rod is
symmetric. Therefore, the final distribution of charge must also be symmetric, and hence the same charge must
be on end A as on end B.
ANSWER:
It is strongly repelled.
It is strongly attracted.
It is weakly attracted.
It is weakly repelled.
It is neither attracted nor repelled.
Charge Distribution on a Conducting Shell - 1
A positive charge is kept (fixed) at the center inside a fixed spherical neutral conducting shell.
Part A
The positive charge is equal to roughly 16 of the smaller charges shown on the surfaces of the spherical shell.
Which of the pictures best represents the charge distribution on the inner and outer walls of the shell?
Hint A.1 Effects of symmetry
The charge is centered, and the shell is of uniform thickness in all directions. Hence the charge distribution must
also be the same in all directions.
ANSWER:
1
2 3
4 5
Charge Distribution on a Conductor with a Cavity
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A positive charge is brought close to a fixed neutral conductor that has a cavity. The cavity is neutral; that is, there
is no net charge inside the cavity.
Part A
Which of the figures best represents the charge distribution on the inner and outer walls of the conductor?
Hint A.1 Conductors have no internal field
At steady state, conductors have no internal electric field (otherwise, charge would flow). Therefore, the
arrangement of charges on the surfaces of the conductor must exactly cancel out any external electric field to
ensure that the internal field is zero.
Hint A.2 Charges on the cavity walls
Think about what the answer would be for a conductor without a cavity. Would there be net charges on the
surface of some imaginary sphere drawn inside of the conductor? Would this change if you removed all of the
material inside of that sphere?
ANSWER:
1
2 3
Charge Distribution on a Conducting Shell - 2
A positive charge is kept (fixed) off-center inside a fixed spherical conducting shell that is electrically neutral, and
the charges in the shell are allowed to reach electrostatic equilibrium. The large positive charge inside the shell is
roughly 16 times that of the smaller charges shown on the inner and outer surfaces of the spherical shell.
Part A
Which of the following figures best represents the charge distribution on the inner and outer walls of the shell?
Hint A.1 Symmetry inside shell
Hint not displayed
Hint A.2 Symmetry outside shell
Hint not displayed
ANSWER:
1
2 3
4 5
A Test Charge Determines Charge on Insulating and Conducting Balls
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Learning Goal: To understand the electric force between charged and uncharged conductors and insulators.
When a test charge is brought near a charged object, we know from Coulomb's law that it will experience a net
force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also
experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral
conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their
molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak
net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will
accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force
if the test charge is placed very close to the conductor's surface.
Consider three plastic balls (A, B, and C), each carrying a uniformly
distributed unknown charge (which may be zero), and an uncharged copper ball
(D). A positive test charge (T) experiences the forces shown in the figure when
brought very near to the individual balls. The test charge T is strongly
attracted to A, strongly repelled from B, weakly attracted to C, and strongly
attracted to D.
Assume throughout this problem that the balls are brought very close
together.
Part A
What is the nature of the force between balls A and B?
Part A.1 What is the net charge on ball A?
Part not displayed
Part A.2 What is the net charge on ball B?
Part not displayed
ANSWER:
strongly attractive strongly repulsive weakly attractive neither attractive nor repulsive
Part B
What is the nature of the force between balls A and C?
Part B.1 What is the charge on ball C?
Recall that ball C is composed of insulating material, which means that it can be polarized, but the charges
inside are otherwise not free to move around inside the ball. Since the test charge experiences only a weak force
due to ball C, what must be the nature of the net charge on ball C?
ANSWER:
positive negative zero
ANSWER:
strongly attractive strongly repulsive weakly attractive neither attractive nor repulsive
Recall that ball C is composed of insulating material, which can be polarized in the presence of an external
charged object such as ball A. Once polarized, there will be a weak attraction between balls A and C, because
the positive and negative charges in ball C are at slightly different average distances from ball A.
Part C
What is the nature of the force between balls A and D?
Part C.1 What are the surface charges on ball D?
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Recall that copper is a conductor, in which charges can freely flow. When ball D is brought close to ball A, what
will be the nature of the surface charge density on the side of ball D that is closest to ball A?
ANSWER:
positive negative zero
The negatively charged ball A (see Part A) will exert an attractive force on the positive charges in ball D and a
repulsive force on the negative charges (namely, the electrons). Since ball D is made of copper, which is a
conductor, the electrons will be repelled from negatively charged ball A and will migrate to the side of ball D
farthest from ball A. The deficit of electrons on the side of ball D that is closest to ball A results in a positive
net surface charge density on that side of ball D. Because the positive charge on ball D is much closer to ball
A than the negative charge, the attractive force that ball A experiences due to the positive charges on ball D is
stronger than the repulsive force ball A experiences due to the negative charges on ball D.
ANSWER:
attractive repulsive neither attractive nor repulsive
Part D
What is the nature of the force between balls D and C?
ANSWER:
attractive repulsive neither attractive nor repulsive
Because the test charge T is neither strongly attracted to nor repelled from ball C, ball C must have zero net
charge. Since ball D also has zero net charge, there will not be any force between the two balls.
Coloumb's Law
Coulomb's Law Tutorial
Learning Goal: To understand how to calculate forces between charged particles, particularly the dependence on
the sign of the charges and the distance between them.
Coulomb's law describes the force that two charged particles exert on each other (by Newton's third law, those two
forces must be equal and opposite). The force exerted by particle 2 (with charge ) on particle 1 (with charge )
is proportional to the charge of each particle and inversely proportional to the square of the distance between
them:
,
where
and
is the unit vector pointing from particle 2 to particle 1. The force vector will be parallel or
antiparallel to the direction of , parallel if the product
and antiparallel if
charges are of opposite sign and repulsive if the charges are of the same sign.
; the force is attractive if the
Part A
Consider two positively charged particles, one of charge (particle 0) fixed at
the origin, and another of charge (particle 1) fixed on the y-axis at
.
What is the net force on particle 0 due to particle 1?
Express your answer (a vector) using any or all of ,
, and .
ANSWER:
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,
,
,
,
= Answer not displayed
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Part B
Now add a third, negatively charged, particle, whose charge is
(particle 2).
Particle 2 fixed on the y-axis at position
. What is the new net force
on particle 0, from particle 1 and particle 2?
Express your answer (a vector) using any or all of ,
, , , , and .
ANSWER:
,
,
,
= Answer not displayed
Part C
Particle 0 experiences a repulsion from particle 1 and an attraction toward particle 2. For certain values of and ,
the repulsion and attraction should balance each other, resulting in no net force. For what ratio
is there no net
force on particle 0?
Express your answer in terms of any or all of the following variables: ,
ANSWER:
,
,
.
= Answer not displayed
Part D
Now add a fourth charged particle, particle 3, with positive charge , fixed in
the yz-plane at
. What is the net force on particle 0 due solely to this
charge?
Part D.1 Find the magnitude of force from particle 3
Part not displayed
Hint D.2 Vector components
Hint not displayed
Express your answer (a vector) using ,
particle 3.
ANSWER:
,
,
,
, , and . Include only the force caused by
= Answer not displayed
Charging an Insulator
This problem explores the behavior of charge on realistic (i.e. non-ideal) insulators. We take as an example a long
insulating rod suspended by insulating wires. Assume that the rod is initially electrically neutral. For convenience,
we will refer to the left end of the rod as end A, and the right end of the rod as end B . In the answer options for this
problem, "weakly attracted/repelled" means "attracted/repelled with a force of
magnitude similar to that which would exist between two balls, one of which
is charged, and the other acquires a small induced charge". An
attractive/repulsive force greater than this should be classified as "strongly
attracted/repelled".
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Part A
A small metal ball is given a negative charge, then brought near (i.e., within a few millimeters) to end A of the
rod. What happens to end A of the rod when the ball approaches it closely this first time?
Hint A.1 What is an insulator?
An insulator is a material which does not allow charge/current to flow easily through it.
Part A.2 Charge at end A
Keeping in mind that like charges repel each other, and opposite charges attract each other, what sort of charge is
induced at end A of the (non-ideal) insulating rod?
ANSWER:
A small positive charge
A small negative charge
Select the expected behavior.
ANSWER:
strongly repelled strongly attracted weakly attracted weakly repelled neither attracted nor
repelled
Currently, you can think of this in the following way: When the sphere is brought near the rod, a positive
charge is induced at end A (and correspondingly, end B acquires a negative induced charge). This means that
some charge must have flowed from A to B. Since charge flow is inhibited in an insulator, the induced charges
are typically small. Later you will learn how to model insulators more accurately and formulate a slightly more
accurate argument.
Now consider what happens when the small metal ball is repeatedly given a negative charge and then brought into
contact with end A of the rod
Part B
After several contacts with the charged ball, how is the charge on the rod arranged?
Hint B.1 What is an insulator?
An insulator is a material which does not allow charge/current to flow easily through it.
Select the best description.
ANSWER:
positive charge on end B and negative charge on end A
negative charge spread evenly on both ends
negative charge on end A with end B remaining almost neutral
positive charge on end A with end B remaining almost neutral
none of the above
When the sphere is touched to end A, some of its negative charge will be deposited there. However, since
charge cannot flow easily through an insulator, most of this charge will just sit at end A and will not distribute
itself over the rod, as it would if the rod was a conductor.
Part C
How does end A of the rod react when the ball approaches it after it has already made several contacts with the rod,
such that a fairly large charge has been deposited at end A?
Select the expected behavior.
ANSWER:
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strongly repelled strongly attracted weakly attracted weakly repelled neither attracted nor
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repelled
More on insulators
You may have learnt that any material is made of atoms, which in turn consist of a nucleus and electrons. In
the atoms of some materials, some of the electrons are "bound" to the nucleus very weakly, which leaves them
free to move around the volume of the material. Such electrons are called "free" electrons, and such materials
are called conductors, because the charge (i.e. electrons) can move around easily. In insulators, all the electrons
in the atom are bound quite tightly to the nucleus, i.e. there are no free electrons available to move through the
insulator.
Mystery Charge
Consider the following configuration of fixed, uniformly charged spheres (see figure):
a blue sphere fixed at the origin with positive charge .
a red sphere fixed at the point
with unknown charge ,
a yellow sphere fixed at the point
with unknown charge
The net electric force on the blue sphere is observed to be
, where
.
.
Part A
What is the sign of the charge on the yellow sphere?
ANSWER:
positive
negative
Part B
What is the sign of the charge on the red sphere?
ANSWER:
positive
negative
Part C
Suppose that the magnitude of the charge on the yellow sphere is determined to be . Calculate the charge
the red sphere.
on
Hint C.1 How to approach the problem
From the problem statement, you know that the x component of the net force acting on the blue sphere must
vanish. The red sphere and the yellow sphere each exert a force on the blue sphere. You know the charge of the
yellow sphere. This allows you to calculate the x component of the force that the yellow sphere exerts on the
blue sphere. You need to find the appropriate charge
for the red sphere such that the x components of the two
forces sum to zero.
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Part C.2 Find the force due to the yellow sphere
Find
the x component of the force that the yellow sphere exerts on the blue sphere.
Hint C.2.a How to approach this part
Use Coulomb's Law to find the force due to the yellow charge on the blue charge. Then resolve this force into
components.
Note that Coulomb's Law gives the total force between charges. To find a component, you must first apply
Coulomb's Law to find the total force and then resolve the force. It is incorrect to first resolve the relative
position vector and then use Coulomb's Law separately for each component.
Express your answer in terms of ,
, and . You may use
for
, where
represents the
permittivity of free space.
ANSWER:
=
Part C.3 Find the force due to the red sphere
Find
, the x component of the force that the red sphere exerts on the blue sphere.
Express your answer in terms of ,
where
, and the unknown charge
. You may use
for
,
represents the permittivity of free space.
ANSWER:
=
Express your answer in terms of ,
ANSWER:
,
, and .
=
The Trajectory of a Charge in an Electric Field
An charge with mass and charge is emitted from the origin,
. A large, flat screen is located at
.
There is a target on the screen at y position , where
. In this problem, you will examine two different ways
that the charge might hit the target. Ignore gravity in this problem.
Part A
Assume that the charge is emitted with velocity in the positive x direction. Between the origin and the screen,
the charge travels through a constant electric field pointing in the positive y direction. What should the magnitude
of the electric field be if the charge is to hit the target on the screen?
Hint A.1 How to approach the problem
Hint not displayed
Part A.2 Find the equation of motion in the x direction
Part not displayed
Part A.3 Find the equation of motion in the y direction
Part not displayed
Hint A.4 Combine Your Results
Hint not displayed
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Part A.5 Find
Part not displayed
Express your answer in terms of
ANSWER:
, ,
,
, and .
=
Part B
Now assume that the charge is emitted with velocity in the positive y direction. Between the origin and the
screen, the charge travels through a constant electric field pointing in the positive x direction. What should the
magnitude of the electric field be if the charge is to hit the target on the screen?
Hint B.1 How to approach the problem
Just as in the previous part, once you determine the force on the charge due to the electric field, this becomes a
standard two-dimensional kinematics problem. To solve the problem, first determine the equations of motion in
both the x and y directions. Then use the fact that at some final time
you know that the position of the
charge is
to obtain two equations in terms of the two unknowns and . Eliminate
and solve for .
Part B.2 Find the equation of motion in the y direction
Find an expression for
, the charge's y position as a function of time.
Part B.2.a Find the force in the y direction
Part not displayed
Hint B.2.b A helpful kinematic equation
Hint not displayed
Express your answer in terms of
ANSWER:
as well as any of the given variables and constants.
=
Part B.3 Find the equation of motion in the x direction
Find an expression for
, the charge's x position as a function of time.
Part B.3.a Find the force in the x direction
Part not displayed
Hint B.3.b A helpful kinematic equation
Hint not displayed
Express your answer in terms of
ANSWER:
as well as any of the given variables and constants.
=
Hint B.4 Combine your results
Hint not displayed
Part B.5 Find
Part not displayed
Express your answer in terms of
ANSWER:
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, ,
,
, and .
=
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The equations of motion for this part are identical to the equations of motion for the previous part, with and
interchanged. Thus it is no surprise that the answers to the two parts are also identical, with and
interchanged.
Finding Electric Field using Coulomb's Law
Charged Ring
Consider a uniformly charged ring in the xy plane, centered at the origin. The
ring has radius and positive charge distributed evenly along its
circumference.
Part A
What is the direction of the electric field at any point on the z axis?
Hint A.1 How to approach the problem
Hint not displayed
ANSWER: Answer not displayed
Part B
What is the magnitude of the electric field along the positive z axis?
Part B.1 Formula for the electric field
Part not displayed
Part B.2 Simplifying with symmetry
Part not displayed
Hint B.3 Integrating around the ring
Hint not displayed
Use
in your answer, where
ANSWER:
.
= Answer not displayed
Part C
Imagine a small metal ball of mass and negative charge
constrained to move along the z axis, with no damping. If
. The ball is released from rest at the point
and
, what will be the ball's subsequent trajectory?
ANSWER: Answer not displayed
Part D
Part not displayed
The Electric Field Produced by a Finite Charged Wire
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A charged wire of negligible thickness has length units and has a linear
charge density . Consider the electric field at the point , a distance above
the midpoint of the wire.
Part A
The field
points along one of the primary axes. Which one?
Hint A.1 Consider opposite ends of the wire
Hint not displayed
ANSWER: Answer not displayed
Part B
Part not displayed
Electric Dipoles
Torque on a Dipole in a Uniform Field
Consider an electric dipole whose dipole moment (a vector pointing from the
negitive charge to the positive charge) is oriented at angle with respect to the
y axis. There is an external electric field of magnitude (independent of the
field produced by the dipole) pointing in the positive y direction. The positive
and negative ends of the dipole have charges
and , respectively, and the
two charges are a distance apart. The dipole has a moment of inertia about
its center of mass. It will help you to imagine that the dipole is free to rotate
about a pivot through its center.
Part A
What is the net force
that the dipole experiences due to the electric field?
Part A.1 What is the force on the positive charge?
Part not displayed
Express
in terms of the given variables and the unit vectors
ANSWER:
Part B
What is
, .
= Answer not displayed
, the magnitude of the torque that the electric field exerts about the center of mass of the dipole?
Part B.1 Find the torque on the positive charge
Part not displayed
Hint B.2 Now consider the negative charge
Hint not displayed
Express the magnitude of the total torque in terms of the given quantities.
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= Answer not displayed
Part C
Part not displayed
Transition will be visible after you complete previous item(s).
Part D
Part not displayed
Part E
Part not displayed
Dipole Motion in a Uniform Field
Consider an electric dipole located in a region with an electric field of
magnitude pointing in the positive y direction. The positive and negative
ends of the dipole have charges
and , respectively, and the two charges
are a distance apart. The dipole has moment of inertia about its center of
mass. The dipole is released from angle
, and it is allowed to rotate freely.
Part A
What is
, the magnitude of the dipole's angular velocity when it is pointing along the y axis?
Hint A.1 How to approach the problem
Because there is no dissipation (friction, air resistance, etc.), you can solve this problem using conservation of
energy. When the dipole is released from rest, it has potential energy but no kinetic energy. When the dipole is
aligned with the y axis, it is rotating, and therefore has both kinetic and potential energy. The sum of potential
and kinetic energy will remain constant.
Part A.2 Find the potential energy
Find the dipole's potential energy
due to its interaction with the electric field as a function of the angle that
the dipole's positive end makes with the positive y axis. Define the potential energy to be zero when the dipole is
oriented perpendicular to the field:
.
Hint A.2.a The formula for the potential energy of a dipole
Hint not displayed
Hint A.2.b The dipole moment
Hint not displayed
Express your answer in terms of
ANSWER:
, ,
, and .
=
Part A.3 Find the total energy at the moment of release
Find
, the total energy (kinetic plus potential) at the moment the dipole is released from rest at angle
respect to the y axis. Use the convention that the potential energy is zero when the dipole is oriented
perpendicular to the field:
.
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Part B
If is small, the dipole will exhibit simple harmonic motion after it is released. What is the period
dipole's oscillations in this case?
of the
Hint B.1 How to approach the problem
The equation of motion for a simple harmonic oscillator can always be written in the standard form
. To
solve this problem, you need to write the equation of motion for the dipole in the standard form with replaced
by the angular variable . This will allow you to read off the expression for , which has a simple relationship to
the period of oscillation. (Note: Here, the variable does not represent the angular velocity of the dipole; rather,
it denotes the frequency of the dipole's oscillation.) Start with the angular analogue of Newton's second law:
. Recall that , the angular acceleration, is equal to the second derivative of , just as linear acceleration is equal
to the second derivative of position.
Part B.2 Compute the torque
What is the magnitude of the torque
that the electric field exerts about the center of mass of the dipole when
the dipole is oriented at an angle with respect to the electric field?
Hint B.2.a Formula for torque on a dipole
Hint not displayed
Hint B.2.b The dipole moment
Hint not displayed
Express the magnitude of the torque in terms of quantities given in the problem
introduction and .
ANSWER:
=
Hint B.3 The small-angle approximation
Because is small, you can apply the small-angle approximation to the expression
take the torque to be
.
for torque, and
Up to this point we have been interested only in the magnitude of the torque. Now let's think about the direction.
After all, torque is a vector quantity. For a system to oscillate, the torque must be a restoring torque; that is, the
torque and the (small) angular displacement must be in opposite directions. (Recall that small angular
displacements can be treated as vectors, since they obey vector addition, while large angles do not.) If you did the
vector algebra carefully, you would find that the correct vector equation is
.
For future purposes we will write this as
in the direction, rather than the magnitude of .
, keeping in mind that now represents the component of
Part B.4 Find the oscillation frequency
Putting together what you have so far yields
.
Compare this to the standard form
for a simple harmonic oscillator to obtain the oscillation frequency
for the motion of the dipole.
Express your answer in terms of quantities given in the problem introduction.
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Hint B.5 The relationship between (angular) oscillation frequency and period
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Express your answer in terms of
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and quantities given in the problem introduction.
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9 of 14 problems complete (48.16% avg. score)
29.1 of 35 points
17/4/07 12:25