hw01

Manage this Assignment:

1. Electric Forces and Fields Due: 11:00pm on Thursday, January 14, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy.

Print Version with Answers

Charging a Conducting Rod

Description: This problem explores the behavior of charge on conductors

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.

Page 1 of 32MasteringPhysics: Assignment Print View

1/8/2010http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1323344

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.

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.

It is neither attracted nor repelled.

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

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.



A Test Charge Determines Charge on Insulating and Conducting Balls

Description: Determine the force between pairs of conducting and/or insulating balls of unknown charge by analyzing the forces between each ball and a positive test charge.

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?

Hint A.1 What is the net charge on ball A?

Since the test charge is positively charged, and there is a strongly attractive force between ball A and the test charge, what must be the nature of the net charge of ball A?

ANSWER: positive negative zero

Hint A.2 What is the net charge on ball B?

Since the test charge is positively charged, and there is a strongly repulsive force between ball B and the test charge, what must be the nature of the net charge of ball B?

ANSWER: positive negative



zero

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?

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.

Hint 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: strongly attractive strongly repulsive weakly attractive neither attractive nor repulsive

Part C

What is the nature of the force between balls A and D?

Hint C.1 What are the surface charges on ball D?

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?

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?

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.

ANSWER: attractive repulsive neither attractive nor repulsive

The Charge inside a Conductor

Description: A spherical conductor has a cavity containing a fixed charge. Conceptual questions about the charge arrangement on the surfaces of a conductor and changes if an additional external charge is brought near the conductor. (version for algebra-based courses)

A neutral conducting sphere contains a spherical cavity. A point charge is placed at the center of the cavity.

Part A

What is the total surface charge on the interior surface of the conductor?

Hint A.1 Gauss's law

Gauss's law states that the flux of an electric field through a Gaussian surface is given by



,

where is the electric field magnitude at that surface, is that surface's area, is the charge inside that

surface, and is a constant often called permittivity of free space.

Hint A.2 Gauss's law and properties of conductors

The electric field in the interior of the conducting material (but not necessarily the cavity) must be zero. Knowing this, you can use Gauss's law to find the net charge on the interior surface of the cavity. Use the following Gaussian surface: an imaginary sphere, centered at the cavity, that has a slightly larger radius than that of the cavity, so that it encompasses the surface of the cavity. This Gaussian surface lies within the conductor, so the flux of the field through the Gaussian surface must be zero. Thus, by Gauss's law, the net charge inside the Gaussian surface must be zero as well. But you know that there is a point charge within the Gaussian surface. If the net charge within the Gaussian surface must be zero, how much charge must be present on the surface of the cavity?

ANSWER: =

Part B

What is the total surface charge on the exterior surface of the conductor?

Hint B.1 Properties of a conductor

In the problem introduction you are told that the conducting sphere is neutral. Also, recall that the free charges within a conductor always accumulate on the conductor's surface or surfaces. You found the net charge on the conductor's interior surface in Part A. If the conductor is to have zero net charge (as it must, since it is neutral), how much charge must be present on its exterior surface?

ANSWER: =

Part C

What is the magnitude of the electric field inside the cavity as a function of the distance from the point

charge? Let , as usual, denote .

Hint C.1 How to approach the problem

The net electric field inside the conductor is created by three portions of charge:

1. charge ; 2. the charge on the cavity's walls, ; and 3. the charge on the outer surface of the conductor, .

However, the net electric field inside the material conductor must be zero. How must and be distributed for this to happen?



Here's a clue: The first two contributions listed above cancel each other out, outside the cavity. Then the electric field produced by inside the spherical conductor must be zero also. How must be distributed for this to happen?

After you have figured out how and are distributed, it will be easy to find the field in the cavity, either by adding field contributions from all charges, or using Gauss's law.

Hint C.2 Find the characteristics of the electric field

Both and are uniformly distributed. Unfortunately there is no easy way to justify this, which is why a clue was given in the previous hint. You might discover this result by assuming the simplest possible distribution (i.e., a uniform one) or by trial and error; you can then check that it works (gives no net electric field inside the conductor).

If is distributed uniformly over the surface of the conducting sphere, it does not produce an electric field inside the sphere. What kind of field does produce inside the cavity?

A spherically symmetrical distribution of charge produces no electric field inside itselfmuch like a uniform shell, whether thin or thick, produces no gravitational field inside itself.

ANSWER: It produces no electric field. It produces a field that is the same as that produced by a point charge located at the center of the sphere It produces a field that is the same as that produced by a point charge located at the position of the charge in the cavity

ANSWER: 0

Part D

What is the electric field outside the conductor?

Hint D.1 How to approach the problem

The net electric field inside the conductor has three contributions:

1. from the charge ; 2. from the charge on the cavity's walls, ; and 3. from the charge on the outer surface of the spherical conductor, .

However, the net electric field inside the conductor must be zero. How must and be distributed for this to happen?

Here's a helpful clue: The first two contributions above cancel each other outside the cavity. Therefore, the electric field produced by inside the spherical conductor must be be zero also. How must be distributed for this to happen? What sort of field would such a distribution produce outside the conductor?



Hint D.2 Electric field due to

If is distributed uniformly over the surface of the conducting sphere, it does not produce a net electric field inside the sphere. What are the characteristics of the field it produces outside the sphere?

ANSWER: 0 The same as the field produced by a point charge located at the center of the sphere The same as the field produced by a point charge located at the position of the charge in the cavity

Placing Charges Conceptual Question

Description: Simple conceptual question about placing various charges around a central charge such that the free body diagram given is correct.

Below are free-body diagrams for three electric charges that lie in the same plane. Their relative positions are unknown.

Part A

Along which of the lines (A to H) in the figure should charge 2 be placed so that the free-body diagrams of charge 1 and charge 2 are consistent?

Hint A.1 How to approach the problem

Newtons 3rd law states that the forces exerted by a pair of objects on each other are always equal in magnitude and opposite in direction. Identifying the forces that correspond to 3rd-law pairs in the free-body diagrams will enable you to place the particles in their proper relative position.

Hint A.2 Placing charge 2

The two forces acting on charge 2 correspond to the forces exerted on it by charge 1 and charge 3. This means that one of these forces must pair with a force on charge 1 of equal magnitude and opposite direction and the other must pair with a force on charge 3 of equal magnitude and opposite direction. Also note that charge 2 should be repelled by charge 1, since both are negative. Therefore, the vector that represents the force of charge 1 on charge 2 must point away from charge 1. This information is all you need to place charge 2 in its correct position.



ANSWER:

Part B

Along which of the lines (A to H) in the figure should charge 3 be placed so that the free-body diagrams of charge 1, charge 2, and charge 3 are consistent?

ANSWER: D

Part C

Along which of the lines (A to H) in the figure should charge 2 be placed so that the free-body diagrams of charge 1 and charge 2 are consistent?

ANSWER: H

Part D

Along which lines (A to H) in the figure should charge 3 be placed so that the free-body diagrams of charge 1,



charge 2, and charge 3 are consistent?

ANSWER: F

Electric Force of Three Collinear Points Ranking Task

Description: Short conceptual problem related to the Coulomb force between three collinear point charges. (ranking task)

In the diagram below, there are three collinear point charges: , , and . The distance between and is the same as that between and . You will be asked to rank the Coulomb force on due to and .

Part A

Rank the six combinations of electric charges on the basis of the electric force acting on . Define forces pointing to the right as positive and forces pointing to the left as negative. Rank positive forces as larger than negative forces.

Hint A.1 Definition of electric force



The electric force between a pair of charges is proportional to the product of the charge magnitudes ( and ) and inversely proportional to the square of the distance ( ) between them. This result is summarized mathematically by Coulombs law:

.

The direction of the force is such that opposite charges attract and like charges repel each other.

Hint A.2 Determine the net force for one combination of charges

For combination of charges ( , , ), what is the direction of the net electric force

on due to the other charges?

Hint A.2.1 Find the direction of the force on due to

For combination of charges ( , , ), what is the direction of the electric force

on due to ? Remember that like charges repel each other and opposite charges attract each other.

ANSWER: to the right to the left There is no force in any direction.

Hint A.2.2 Determine the direction of the force on charge due to

For combination of charges ( , , ), what is the direction of the electric force

on due to ?

ANSWER: to the right. to the left There is no force.

Hint A.2.3 Find the magnitude of the net force on

In combination of charges ( , , ), which of the two forces on , that from

or that from , is larger in magnitude?

ANSWER: the force from the force from Neither; they are equal in magnitude.

ANSWER: to the right to the left There is no net force.



Rank from largest to smallest, placing the largest on the left and the smallest on the right. To rank items as equivalent, overlap them.

ANSWER:

View

Forces in a Three-Charge System

Description: Find the total force, magnitude and direction, on a charge exerted by two other charges.

Coulomb's law for the magnitude of the force between two particles with charges and separated by a

distance is

,

where . is the permittivity of free space.

Consider two point charges located on the x axis: one charge, = -11.5 , is located at = -1.720 ; the second

charge, = 33.0 , is at the origin .

Part A

What is the net force exerted by these two charges on a third charge = 46.0 placed between and at

= -1.055 ?

Your answer may be positive or negative, depending on the direction of the force.


First, draw a diagram of the system. Next, find the magnitudes of the forces exerted on the third charge by each of the other charges. Then determine the direction of each of these forces. Finally, use vector addition to find the net force on the third charge.

Hint A.2 Calculate the force on the third charge by the first charge

Calculate the magnitude of the force that the first charge exerts on the third charge.

Hint A.2.1 Coulomb's law



Express your answer numerically in newtons to three significant figures.


distance is

,

where is the permittivity of free space.

ANSWER: =

Hint A.3 Calculate the force on the third charge by the second charge

Calculate the magnitude of the force that the second charge exerts on the third charge.


Hint A.3.1 Coulomb's law


distance is

,

where is the permittivity of free space.

ANSWER: =

Hint A.4 What are the directions of the forces?

In what directions do the forces on the third charge point in our system?

Note that since all the forces are in the x direction, you can drop the vector notation. Let be the force on

due to , and let be the force on due to .

ANSWER: and both point along the x direction. and both point along the x direction.




points along the x direction and points along the x direction. points along the x direction and points along the x direction.

Hint A.5 Relating the net force and the forces between pairs of charges

Since all the forces are in the x direction, you can drop the vector notation. Let be the net force on .

Similarly, let be the force on due to , and be the force on due to . Then .

ANSWER:

Force on =

Electric Fields and Forces

Description: Electric forces and electric fields.

Learning Goal: To understand Coulomb's law, electric fields, and the connection between the electric field and the electric force.

Coulomb's law gives the electrostatic force acting between two charges. The magnitude of the force between

two charges and depends on the product of the charges and the square of the distance between the charges:

,

where . The direction of the force is along the line connecting the two

charges. If the charges have the same sign, the force will be repulsive. If the charges have opposite signs, the force will be attractive. In other words, opposite charges attract and like charges repel.

Because the charges are not in contact with each other, there must be an intermediate mechanism to cause the force. This mechanism is the electric field. The electric field at any location is equal to the force per unit charge experienced by a charge placed at that location. In other words, if a charge experiences a force , the electric field at that

point is

.

The electric field vector has the same direction as the force vector on a positive charge and the opposite direction to that of the force vector on a negative charge.

An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a



point charge has magnitude

.

The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by , will not significantly affect the electric field if it is

small compared to .

Imagine an isolated positive point charge with a charge (many times larger than the charge on a single electron).

Part A

There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?

Check all that apply.

According to Coulomb's law, the force between two particles depends on the charge on each of them and the distance between them.

ANSWER: the distance between the positive charge and the electron the charge on the electron the mass of the electron the charge of the positive charge the mass of the positive charge the radius of the positive charge the radius of the electron

Part B

For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?


The electrostatic force cannot exist unless two charges are present. The electric field, on the other hand, can be created by only one charge. The value of the electric field depends only on the charge producing the electric field and the distance from that charge.

ANSWER: the distance between the positive charge and the electron the charge on the electron the mass of the electron the charge of the positive charge the mass of the positive charge the radius of the positive charge the radius of the electron



Part C

Recall that the positive charge is much greater than the charge on the electron.

Because the charge on the electron is so much smaller than the positive charge, the electron does not significantly affect the electric field. However, if the electron were replaced by a negative charge comparable to the positive charge, it would significantly affect the electric field.

ANSWER: The presence of the electron will not

significantly affect the value of the electric field near the electron.

Part D

If the total positive charge is = 1.62106 , what is the magntidue of the electric field caused by this charge at point P, a distance = 1.53 from the charge?

Enter your answer numerically in newtons per coulomb.

ANSWER: =

Part E

What is the direction of the electric field at point P?

Enter the letter of the vector that represents the direction of .



ANSWER: G

Part F

Now find the magnitude of the force on an electron placed at point P. Recall that the charge on an electron has magnitude .

Enter your answer numerically in newtons.

Hint F.1 Determine how to approach the problem

What strategy can you use to calculate the force between the positive charge and the electron?

ANSWER: Use Coulomb's law.Multiply the electric field due to the positive charge by the charge on the electron. Do either of the above. Do neither of the above.

ANSWER: =

Part G

What is the direction of the force on an electron placed at point P?

Enter the letter of the vector that represents the direction of .

ANSWER: C

Visualizing Electric Fields



Description: Select the correct drawing of electric field lines for several situations and answer questions about why other choices are incorrect. Then, these ideas are demonstrated with an applet.

Learning Goal: To understand the nature of electric fields and how to draw field lines.

Electric field lines are a tool used to visualize electric fields. A field line is drawn beginning at a positive charge and ending at a negative charge. Field lines may also appear from the edge of a picture or disappear at the edge of the picture. Such lines are said to begin or end at infinity. The field lines are directed so that the electric field at any point is tangent to the field line at that point.

The figure shows two different ways to visualize an electric field. On the left, vectors are drawn at various points to show the direction and magnitude of the electric field. On the right, electric field lines depict the same situation. Notice that, as stated above, the electric field lines are drawn such that their tangents point in the same direction as the electric field vectors on the left. Because of the nature of electric fields, field lines never cross. Also, the vectors shrink as you move away from the charge, and the electric field lines spread out as you move away from the charge. The spacing between electric field lines indicates the strength of the electric field, just as the length of vectors indicates the strength of the electric field. The greater the spacing between field lines, the weaker the electric field. Although the advantage of field lines over field vectors may not be apparent in the case of a single charge, electric field lines present a much less cluttered and more intuitive picture of more complicated charge arrangements.

Part A

Which of the following figures correctly depicts the field lines from an infinite uniformly negatively charged sheet? Note that the sheet is being viewed edge-on in all pictures.

Hint A.1 Description of the field

Recall that the field around an infinite charged sheet is always perpendicular to the sheet and that the field strength does not change, regardless of distance from the sheet.

ANSWER: A B C



D

Part B

In the diagram from part A , what is wrong with figure B? (Pick only those statements that apply to figure B.)


ANSWER: Field lines cannot cross each other. The field lines should be parallel because of the sheet's symmetry. The field lines should spread apart as they leave the sheet to indicate the weakening of the field with distance. The field lines should always end on negative charges or at infinity.

Part C

Which of the following figures shows the correct electric field lines for an electric dipole?

ANSWER: A B C D



This applet shows two charges. You can alter the charge on each independently or alter the distance between them. You should try to get a feeling for how altering the charges or the distance affects the field lines.

Part D

In the diagram from part C , what is wrong with figure D? (Pick only those statements that apply to figure D.)


In even relatively simple setups as in the figure, electric field lines are quite helpful for understanding the field qualitatively (understanding the general direction in which a certain charge will move from a specific position, identifying locations where the field is roughly zero or where the field points a specific direction, etc.). A good figure with electric field lines can help you to organize your thoughts as well as check your calculations to see whether they make sense.

ANSWER: Field lines cannot cross each other. The field lines should turn sharply as you move from one charge to the other. The field lines should be smooth curves. The field lines should always end on negative charges or at infinity.

Part E

In the figure , the electric field lines are shown for a system of two point charges, and . Which of the

following could represent the magnitudes and signs of and ?

In the following, take to be a positive quantity.



Very far from the two charges, the system looks like a single charge with value . At large

enough distances, the field lines will be indistinguishable from the field lines due to a single point charge .

ANSWER: , , , , ,

Electric Field Conceptual Question

Description: Simple conceptual question about identifying where on the x axis the electric field would be zero, given two charges.

Part A

For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is zero.

Hint A.1 Zeros of the electric field

The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude.



If no such region exists on the horizontal axis choose the last option (nowhere).

If there is such a point, then select that region.

ANSWER: A B C D E nowhere

Part B


Hint B.1 Zeros of the electric field

The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region.

Hint B.2 Determine the regions where the electric fields could cancel

In which region(s) do the electric fields from the two source charges point in opposite directions?

List all the correct answers in alphabetical order.

Since the two charges produce fields that point in opposite directions in these regions, if the magnitude of the fields are equal, the net electric field will be zero.

ANSWER: BCD

Hint B.3 Consider the magnitude of the electric field




For each of the three regions found in the previous hint, determine whether it is possible for the magnitudes to be equal. As an example, consider the point directly between the two charges. Which charge produces the largest magnitude field directly between the two charges?

Therefore, the point directly between the two charges is not the correct answer since the right charge dominates at this point. Check the other two possible regions.

ANSWER: the charge on the right the charge on the left neither, because they have the same magnitude

ANSWER: A B C D E nowhere

Part C



Hint C.1 Zeros of the electric field


ANSWER: A



B C D E nowhere

Part D


Hint D.1 Zeros of the electric field


ANSWER: A B C D E Nowhere along the finite x axis

Electric Field Vector Drawing

Description: Simple conceptual question about determining the electric field vector in a region of space from the change in the path of a charged particle. (vector applet)

Each of the four parts of this problem depicts a motion diagram for a charged particle moving through a region of uniform electric field. For each part, draw a vector representing the direction of the electric field.

Part A



Draw a vector representing the direction of the electric field. The orientation of the vector will be graded. The location and length of the vector will not be graded.

The motion diagram shows that the particle's acceleration points to the right. Because the particle has positive charge, the electric field should point to the right.

Hint A.1 Relationship between electric field and electric force

The relationship between the electric force that acts on a particle and the electric field at the location of the particle is

.

This formula indicates that the force and the electric field point in the same direction for a positively charged particle, and in opposite directions for a negatively charged particle.

Hint A.2 Determining the direction of the electric field

The acceleration of the particle can be determined from the change in its velocity. By Newtons 2nd law, the force acting on the particle is parallel to its acceleration. Finally, since this is a positively charged particle, the electric field is parallel to the force. Putting this all together results in an electric field that is parallel to the particles acceleration.

ANSWER:

View

Part B

Draw a vector representing the direction of the electric field. The orientation of the vector will be graded.

Hint B.1 Relationship between electric field and electric force


.


Hint B.2 Determining the direction of the electric field

The acceleration of the particle can be determined from the change in its velocity. By Newtons 2nd law, the force acting on the particle is parallel to its acceleration. Finally, since this is a negatively charged particle, the electric field is directed opposite to the force. Putting this all together results in an electric field that is directed opposite to the particles acceleration.



The location and length of the vector will not be graded.

The motion diagram shows that the particle's acceleration points to the right. Because the particle has negative charge, the electric field should point to the left.

ANSWER:

View

Part C


Hint C.1 Relationship between electric field and electric force


.


Hint C.2 Determining the direction of the electric field

The acceleration of the particle can be determined from the change in its velocity. By Newtons 2nd law, the force acting on the particle is parallel to its acceleration. Finally, since this is a positively charged particle, the electric field is parallel to the force. Putting this all together results in an electric field that is parallel to the particles acceleration.

Because the electric field is uniform, you can find the direction of the particle's acceleration by subtracting any two consecutive velocity vectors graphically. If and are any two consecutive velocities, you can subtract

from by placing at the tip of . is the vector that starts at the tail of and ends at the tip of .

To find the direction of the particle's acceleration graphically, use two unlabeled vectors to represent and

. Pick any two vectors and that would make your subtraction easier; you can verify your result by

subtracting any other pair of consecutive vectors.

ANSWER:



View

Part D


Hint D.1 Relationship between electric field and electric force


.


Hint D.2 Determining the direction of the electric field

The acceleration of the particle can be determined from the change in the illustrated velocity vectors. By Newtons 2nd law, the force acting on the particle is parallel to its acceleration. Finally, since this is a negatively charged particle, the electric field is directed opposite to the electric force. Putting this all together results in an electric field that is directed opposite to the particles acceleration.

Because the electric field is uniform, you can find the direction of the particle's acceleration by subtracting any two consecutive velocity vectors graphically. If and are any two consecutive velocities, you can subtract

from by placing at the tip of . is the vector that starts at the tail of and ends at the tip of .

To find the direction of the particle's acceleration graphically, use two unlabeled vectors to represent and

. Pick any two vectors and that would make your subtraction easier; you can verify your result by

subtracting any other pair of consecutive vectors.

ANSWER:

View



Electric Field due to Multiple Point Charges

Description: Calculate the electric field due to two charges of the same sign. Then add a third charge at a new point and find its charge (including sign) to cancel the field of the first two charges at a particular point.

Two point charges are placed on the x axis. The first charge, = 8.00 , is placed a distance 16.0 from the

origin along the positive x axis; the second charge, = 6.00 , is placed a distance 9.00 from the

origin along the negative x axis.

Part A

Calculate the electric field at point A, located at coordinates (0 , 12.0 ).


Find the contributions to the electric field at point A separately for and , then add them together (using vector addition) to find the total electric field at that point. You will need to use the Pythagorean theorem to find the distance of each charge from point A.

Hint A.2 Calculate the distance from each charge to point A

Calculate the distance from each charge to point A.

Enter the two distances, separated by a comma, in meters to three significant figures.

ANSWER:

, =

Hint A.3 Determine the directions of the electric fields

Which of the following describes the directions of the electric fields and created by charges and

at point A?

ANSWER: points up and left and points up and right.

points up and left and points down and left.



In this case, the electric fields due to the two charges have both x and y components that are nonzero. To find the total field, add these two components separately.

points down and right and points up and right.

points down and right and points down and left.

Hint A.4 Calculate the components of

Calculate the x and y components of the electric field at point A due to charge .

Express your answers in newtons per coulomb, separated by a comma, to three significant figures.

Hint A.4.1 Calculate the magnitude of the total field

Calculate the magnitude of the field at point A due to charge only.

Express your answer in newtons per coulomb to three significant figures.

ANSWER: =

Hint A.4.2 How to find the components of the total field

Once you have found the magnitude of the field, use trigonometry to determine the x and y components of the field. The electric field of a positive point charge points directly away from the charge, so the direction of the electric field at point A due to charge will be along the line joining the two. Use the position coordinates of

and point A to find the angle that the line joining the two makes with the x or y axis. Then use this angle to resolve the electric field vector into components.

ANSWER:

=

Hint A.5 Calculate the components of

Calculate the x and y components of the electric field at point A due to charge .

Hint A.5.1 Calculate the magnitude of the total field

Calculate the magnitude of the field at point A due to charge only.

Express your answer in newtons per coulomb to three significant figures.

ANSWER:



Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.

Express your answers in newtons per coulomb, separated by a comma, to three significant figures.

=

Hint A.5.2 How to find the components of the total field

Once you have found the magnitude of the field, use trigonometry to determine the x and y components of the field. The electric field of a positive point charge points directly away from the charge, so the direction of the electric field at point A due to charge will be along the line joining the two. Use the position coordinates of

and point A to find the angle that the line joining the two makes with the x or y axis. Then use this angle to resolve the electric field vector into components.

ANSWER:

=

ANSWER:

=

Part B

An unknown additional charge is now placed at point B, located at coordinates (0 , 15.0 ).

Find the magnitude and sign of needed to make the total electric field at point A equal to zero.

Hint B.1 How to approach the problem

You have already calculated the electric field at point A due to and . Now find the charge needed to make an opposite field at point A, so when the two are added together the total field is zero.

Hint B.2 Determine the sign of the charge

Which sign of charge is needed to create an electric field that points in the opposite direction of the total

field due to the other two charges, and ?

ANSWER: positive negative



Express your answer in nanocoulombs to three significant figures.

Hint B.3 Calculating the magnitude of the new charge

Keep in mind that the magnitude of the field due to is , and the field must be equal in

magnitude to the field due to charges and .

ANSWER:

=

Flux out of a Cube

Description: If a point charge is in the center of a cube, calculate the flux through one of the faces, and then do the same for a smaller cube.

A point charge of magnitude is at the center of a cube with sides of length .

Part A

What is the electric flux through each of the six faces of the cube?


Since the magnitude and direction of changes over the entire surface, you cannot use the expression

to find the electric flux. Integration is possible, of coursebut who likes to integrate if there is an easier way?

In this case, there is a way indeed. Just use Gauss's law to find the total electric flux through the cube and then exploit the symmetry of the problem.

Hint A.2 Calculate the total electric flux

Calculate the total electric flux coming out of the cube.

Hint A.2.1 Calculate the flux for a sphere

If the charge were in the center of a sphere of radius instead of a cube with sides of length , what would be

the total flux out of the sphere?

Use for the permittivity of free space.

ANSWER: =

Hint A.2.2 Flux and surface shape

The total flux through any closed surface depends solely on the total amount of charge enclosed by that surface, not on its shape.



Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 12 points.


The shape of the surface enclosing a charge, in this case a cube, does not affect the total electric flux through the surface. The flux depends only on the total enclosed charge.


ANSWER: =

Hint A.3 Flux through a face

Since the charge is at the center of the cube, by symmetry the flux through each face is the same.

ANSWER: =

Part B

What would be the flux through a face of the cube if its sides were of length ?


Just as the shape of the surface does not affect the total electric flux coming out of that surface, its size does not make any difference in the total electric flux either. The only relevant quantity is the total enclosed charge.

Hint B.1 How to approach the problem

Gauss's law states that the flux through a closed surface depends only on the charge enclosed by that surface. Does changing the size of the cube affect the total charge enclosed by the cube?

ANSWER: =

