21 The Electric Field Inside A Conductor 2m365b

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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 ed. 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 with end A of the rod. Part B After a great many s 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 s 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:


Part D How does end B of the rod react when the charged ball approaches it after a great many previous s 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:


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:

,

,

,


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 of any or all of the following variables: , ANSWER:

,

,

.


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


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 with end A of the rod Part B After several s 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 s 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 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 of , where

, and the unknown charge

. You may use

for

,

represents the permittivity of free space.

ANSWER:

=

Express your answer in 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 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 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 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 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 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:

.


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 of the given variables and the unit vectors

ANSWER: Part B What is

, .


, 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 of the given quantities.

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ANSWER:



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 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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with

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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 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 of quantities given in the problem introduction.

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ANSWER:


=

Hint B.5 The relationship between (angular) oscillation frequency and period Hint not displayed Express your answer in of ANSWER:

Summary

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and quantities given in the problem introduction.

=

9 of 14 problems complete (48.16% avg. score) 29.1 of 35 points

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21 The Electric Field Inside A Conductor 2m365b

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