mastering physics_ assignment 1 print view

Electric Forces and Fields

Due: 7:00pm on Monday, January 30, 2012

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

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For Items 7, 8, 12, 14, and 18, please submit your answers online, and also submit written work showing

how you solved these problems. Put your written solutions in the homework box on the second floor of the

Physics Building by the same due date and time as for the online assignment.

Item 1

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?

Hint not displayed

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

Hint not displayed

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ANSWER:strongly attractive

strongly repulsive

weakly attractive

neither attractive nor repulsive

Correct

Part B

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

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:positive

negative

zero

Correct

ANSWER:strongly attractive

strongly repulsive

weakly attractive


Correct

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?

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

Hint not displayed

ANSWER:attractive


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repulsive


Correct

Part D

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

ANSWER:attractive

repulsive


Correct

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.

Item 2

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".

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?

Hint not displayed


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Hint A.2 Charge at end A

Hint not displayed

Select the expected behavior.

ANSWER:strongly repelled

strongly attracted

weakly attracted

weakly repelled

neither attracted nor repelled

Correct

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

Correct

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


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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?



strongly attracted

weakly attracted

weakly repelled


Correct

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.

Item 3

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


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Hint not displayed

Hint A.2 How much charge moves to end A?

Hint not displayed

ANSWER:It is strongly repelled.

It is strongly attracted.

It is weakly attracted.

It is weakly repelled.

It is neither attracted nor repelled.

Correct

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.


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.

Correct

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.







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Correct

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

Hint not displayed






Correct

Item 4

Learning Goal: To understand interactions between a grounded conductor and a charged ball that

is repeatedly brought into contact with it.

This problem explores the behavior of charge on grounded conductors. We take as an example a

long conducting rod suspended by conducting wires that are connected to ground. 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?


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Hint A.1 The key property of conductors

Hint not displayed

Hint A.2 How much charge moves to end A?

Hint not displayed



strongly attracted

weakly attracted

weakly repelled


Correct


then brought into contact with end A of the rod.

Part B

After a great many contacts with the charged ball, how is any charge on the rod arranged (when

the charged ball is far away)?

Hint B.1 The rod is grounded

Hint not displayed

Select the best description.

ANSWER:positive charge on end B and negative charge on end A

negative charge spread evenly on both

negative charge on end A with end B remaining neutral

both ends neutral

positive charge spread evenly on both

Correct

Part C

How does end A of the rod react when the (re)charged ball approaches it after a great many

previous contacts with end A?


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Hint C.1 The rod is a grounded conductor

Hint not displayed



strongly attracted

weakly attracted

weakly repelled


Correct

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 grounded conductor

Hint not displayed



strongly attracted

weakly attracted

weakly repelled


Correct

Item 5

Learning Goal: To practice Problem-Solving Strategy 21.1 Coulomb's Law.

Three charged particles are placed at each of three corners of an equilateral triangle whose sides

are of length 2.1 . Two of the particles have a negative charge: = -8.8 and = -17.6 .

The remaining particle has a positive charge, = 8.0 . What is the net electric force acting on

particle 3 due to particle 1 and particle 2?

Problem-Solving Strategy: Coulomb's law

IDENTIFY the relevant concepts:


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Coulomb’s law comes into play whenever you need to know the electric force acting between

charged particles.

SET UP the problem using the following steps:

Make a drawing showing the locations of the charged particles, and label each particle with its

charge.

If three or more particles are present and they do not all lie on the same line, set up an xy

coordinate system.

Often you will need to find the electric force on just one particle. If so, identify that particle.

EXECUTE the solution as follows:

For each particle that exerts a force on the particle of interest, calculate the magnitude of that force

using .

Sketch a free-body diagram showing the electric force vectors acting on the particle(s) of interest

due to each of the other particles. Recall that the force exerted by particle 1 on particle 2 points

from particle 2 toward particle 1 if the two charges have opposite signs, but points from particle 2

directly away from particle 1 if the charges have the same sign.

Calculate the total electric force on the particle(s) of interest. Recall that the electric force, like any

force, is a vector.

As always, using consistent units is essential. If you are given non-SI units, don’t forget to convert!

If there is a continuous distribution of charge along a line or over a surface, divide the total charge

distribution into infinitesimal pieces, use Coulomb’s law for each piece, and then integrate to find the

vector sum.

In many situations, the charge distribution will be symmetrical. Whenever possible, exploit any

symmetries to simplify the problem-solving process.

EVALUATE your answer:

Check whether your numerical results are reasonable, and confirm that the direction of the net electric force

agrees with the principle that like charges repel and opposite charges attract.

IDENTIFY the relevant concepts

To determine the angle of the force vector on a single charged particle, you will need to calculate

the vector sum of all the forces on that particle due to the presence of other charged particles. To

do this, you will need to use Coulomb's law.

SET UP the problem using the following steps

Part A

Identify the most appropriate xy coordinate system.

ANSWER:


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Correct

You are asked to find the net force acting on particle 3. Centering the xy coordinate system on

particle 3 will make this easier.

EXECUTE the solution as follows

Part B

Find the net force acting on particle 3 due to the presence of the other two particles. Report

you answer as a magnitude and a direction measured from the positive x axis.

Hint B.1 How to approach the problem

Hint not displayed

Hint B.2 Draw a free-body diagram

Hint not displayed

Hint B.3 Calculate the force on particle 3 due to particle 1

Hint not displayed

Hint B.4 Calculate the component forces on particle 3 due to particle 1

Hint not displayed

Hint B.5 How to calculate the component forces on particle 3 due to particle 2


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Hint not displayed

Hint B.6 How to determine the magnitude and direction of a vector from its components

Hint not displayed

Express the magnitude in newtons and the direction in degrees to three significant figures.

ANSWER: , = 3.80×10−3,19.1

Answer Requested ,

EVALUATE your answer

Part C

Assume that particle 3 is no longer fixed to a corner of the triangle and is now allowed to move. In

what direction would particle 3 move the instant after being released?

Draw the velocity vector for particle 3 below. The orientation of your vector will be graded,

but not its length.

ANSWER:

View

Correct

Specifically, from Newton's 2nd law, , you know that a mass accelerates in the same direction

as the net force acting upon it. Therefore, at the instant after being released, particle 3 accelerates in

the same direction as . Moreover, since particle 3 starts from rest, its velocity at that instant will be

. In other words, the initial direction of particle 3 is the same direction as its acceleration, and

therefore the same direction as the applied net force.

Let us interpret this result in terms of electric forces. In general, like charges repel and unlike

charges attract. If particle 3 were free to move, it would move toward the negative charges

and . If and were the same size, particle 3 would start to move toward them along a

direction equidistant from each charge, that is, at an angle of from the positive x axis.

Instead, , so particle 3 will be more strongly attracted toward particle 2 and will

move off in a direction less than .

Item 6


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Two small aluminum spheres, each of mass 0.0250 kilograms, are separated by 80.0 centimeters.

Part A

How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 grams

per mole, and its atomic number is 13.)

Hint A.1 The definition of mole and atomic number

Hint not displayed

Hint A.2 How many electrons per mole of aluminum?

Hint not displayed

Hint A.3 How many electrons per kilogram of aluminum?

Hint not displayed

Express your answer numerically.

ANSWER: 7.25×1024

Correct

Part B

How many electrons would have to be removed from one sphere and added to the other to cause

an attractive force between the spheres of magnitude (roughly one ton)? Assume

that the spheres may be treated as point charges.


Hint not displayed

Hint B.2 Find the relationship between the charges of the spheres

Hint not displayed


ANSWER: 5.27×1015

Correct

Part C

What fraction of all the electrons in one of the spheres does this represent?



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ANSWER: 7.26×10−10

Correct

Submit written work for the following problem.

Item 7

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.5

apart, the repulsive force between them has a magnitude of 0.250 .

Part A

What is the charge on each sphere if the two charges are equal?

ANSWER: = 7.65×10−7

Correct

Part B

What is the charge on first sphere if it has four times the charge of the other?

ANSWER: = 1.53×10−6

Correct

Part C

What is the charge on the second sphere?

ANSWER: = 3.82×10−7

Correct


Item 8

Four identical charges are placed at the corners of a square of side .

Part A

Find the magnitude total force exerted on one charge by the other three charges.

Express your answer in terms of the variables Q, L and appropriate constants.

ANSWER:

=

Answer Requested


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Item 9

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.

ANSWER:the distance between the positive charge and the electron


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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

Correct

According to Coulomb's law, the force between two particles depends on the charge on each

of them and the distance between them.

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?


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

Correct

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.

Part C

If the total positive charge is = 1.62×10−6 , what is the magnitude 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.


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ANSWER: = 6220

Correct

Part D

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

Enter the letter of the vector that

represents the direction of .

ANSWER: G

Correct

Part E

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

electron has magnitude .

Hint E.1 Determine how to approach the problem

Hint not displayed

Enter your answer numerically in newtons.

ANSWER: = 9.95×10−16

Correct

Part F

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 .


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ANSWER: C

Correct

Item 10

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.


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Hint A.1 Description of the field

Hint not displayed

ANSWER:A

B

C

D

Correct

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.

Correct

Part C

Which of the following figures shows the

correct electric field lines for an electric

dipole?


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ANSWER:A

B

C

D

Correct

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.)


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.

Correct

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


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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.

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.

ANSWER:,

,

,

,

,

Correct

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 .

Item 11

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.


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Part A

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

Hint A.1 How to approach the problem

Hint not displayed

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

Hint not displayed

Hint A.3 Determine the directions of the electric fields

Hint not displayed

Hint A.4 Calculate the components of

Hint not displayed

Hint A.5 Calculate the components of

Hint not displayed

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.

ANSWER: = 0,0.300

Correct

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 not displayed


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Hint B.2 Determine the sign of the charge

Hint not displayed

Hint B.3 Calculating the magnitude of the new charge

Hint not displayed

Express your answer in nanocoulombs to three significant figures.

ANSWER: =

0.300

Correct


Item 12

A point charge is placed at each corner of a square with side length . The charges all have the

same magnitude . Two of the charges are positive and two are negative, as shown in the following

figure.

Part A

What is the direction of the net electric field at the center of the square?

ANSWER:rightward direction

leftward direction

downward direction

upward direction

Correct


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Part B

What is the magnitude of the net electric field at the center of the square due to the four charges in

terms of and ?

Express your answer in terms of the variables , , and appropriate constants.

ANSWER: = Answer not displayed

Item 13

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:

Correct

Part B

What is the magnitude of the electric field at point ? Throughout this part, express your


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answers in terms of the constant , defined by .


Hint not displayed

Hint B.2 Find the field due to an infinitesimal segment

Hint not displayed

Hint B.3 A necessary integral

Hint not displayed

Express your answer in terms of , , , and .

ANSWER:

=

Correct


Item 14

Positive charge is uniformly distributed around a semicircle of radius .

Part A

Find the magnitude of the electric field at the center of curvature P.

Express your answer in terms of the given quantities and appropriate constants.


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ANSWER:

=

Correct

Part B

What is the direction of the electric field at the center of curvature P.

ANSWER:downward

upward

Correct

Item 15

Frequently in physics, one makes simplifying approximations. A common one in electricity is the notion

of infinite charged sheets. This approximation is useful when a problem deals with points whose

distance from a finite charged sheet is small compared to the size of the sheet. In this problem, you

will look at the electric field from two finite sheets and compare it to the results for infinite sheets to

get a better idea of when this approximation is valid.

Consider two thin disks, of negligible thickness, of radius oriented perpendicular to the x axis

such that the x axis runs through the center of each disk. The disk centered at has positive

charge density , and the disk centered at

has negative charge density ,

where the charge density is charge per unit

area.

Part A

What is the magnitude of the electric field at the point on the x axis with x coordinate ?


Hint not displayed

Hint A.2 The magnitude of the electric field due to a single disk


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Hint not displayed

Hint A.3 Determine the general form of the electric field between the disks

Hint not displayed

Express your answer in terms of , , , and the permittivity of free space .

ANSWER:

=

Correct

Notice that as approaches , this expression approaches , the result for two infinite

sheets. Also, note that the minimum value of the electric field, which corresponds in this case

to the greatest deviation from the result for two infinite sheets, occurs halfway between the

disks (i.e., at ).

Part B

For what value of the ratio of plate radius to separation between the plates does the electric

field at the point on the x axis differ by 1 percent from the result for infinite sheets?

Hint B.1 Percent difference

Hint not displayed

Express your answer to two significant figures.

ANSWER: = 50

Correct

As mentioned above, this is the point on the x axis where the deviation from the result for two infinite

sheets is greatest. A common component of electrical circuits called a capacitor is usually made from

two thin charged sheets that are separated by a small distance. In such a capacitor, the ratio is far

greater than 50. Based on your result, you can see that the infinite sheet approximation is quite good for

a capacitor.

This applet shows the electric field lines from a pair of finite plates (viewed edge-on). You can adjust the

surface charge density. You can also move the test charge around and increase or decrease its charge

to see what sort of force it would experience. Notice that the deviation from uniform electric field only

becomes noticeable near the edges of the capacitor plates.

Item 16


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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

Hint A.1 Relationship between electric field and electric force

Hint not displayed

Hint A.2 Determining the direction of the electric field

Hint not displayed

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.

ANSWER:

View

Correct

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.

Part B

Hint B.1 Relationship between electric field and electric force

Hint not displayed

Hint B.2 Determining the direction of the electric field

Hint not displayed



ANSWER:


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View

Correct

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.

Part C

Hint C.1 Relationship between electric field and electric force

Hint not displayed

Hint C.2 Determining the direction of the electric field

Hint not displayed



ANSWER:

View

Correct

Part D

Hint D.1 Relationship between electric field and electric force

Hint not displayed

Hint D.2 Determining the direction of the electric field

Hint not displayed



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ANSWER:

View

Correct

Item 17

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 not displayed

Hint A.2 Find the equation of motion in the x direction

Hint not displayed

Hint A.3 Find the equation of motion in the y direction

Hint not displayed

Hint A.4 Combine Your Results

Hint not displayed

Hint A.5 Find

Hint not displayed


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Express your answer in terms of , , , , and .

ANSWER:

=

Correct

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 not displayed

Hint B.2 Find the equation of motion in the y direction

Hint not displayed

Hint B.3 Find the equation of motion in the x direction

Hint not displayed

Hint B.4 Combine your results

Hint not displayed

Hint B.5 Find

Hint not displayed

Express your answer in terms of , , , , and .

ANSWER:

=

Correct

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.


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Item 18

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A

proton is released from rest at the surface of the positively charged plate and strikes the surface of

the opposite plate, 1.54 distant from the first, in a time interval of 1.46×10−6 .

Part A

Find the magnitude of the electric field.


Part B

Find the speed of the proton when it strikes the negatively charged plate.


Score Summary:

Your score on this assignment is 71%.

You received 127.73 out of a possible total of 180 points.


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