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MasteringPhysics: Assignment Print View

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Homework 1 Due: 11:59pm on Sunday, February 13, 2011 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy [Switch to Standard Assignment View]

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

; the

force is attractive if the charges are of opposite sign and repulsive if the charges are of the same sign. 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: = Correct

Part B Now add a third, negatively charged, particle, whose charge is (particle 2). Particle 2 fixed on the y-axis at position

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. What is the new net

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force on particle 0, from particle 1 and particle 2? Express your answer (a vector) using any or all of , , , , , , , , and .

ANSWER: = Correct

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: = Correct

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?

Hint D.1

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Find the magnitude of force from particle 3

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What is the magnitude of the force on particle 0 from particle 3, fixed at

?

Hint D.1.1 Distance to particle 3 Hint not displayed Express your answer using , ANSWER:

Hint D.2

,

,

.

= Answer not displayed

Vector components

The force vector points from to . Because is symmetrically located between the y-axis and the z-axis, the angle between , the unit vector pointing from particle 3 to particle 0, and the y-axis is radians. You have already calculated the magnitude of the vector above. Now break up the force vector into its y and z components. Express your answer (a vector) using ,

,

,

,

, , and . Include only the force caused by

particle 3. ANSWER: = Correct

± PSS 21.1 Coulomb's Law 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 3.0 . Two of the particles have a negative charge: = -9.0 and = -18.0 . 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: 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

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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:

Correct

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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 answer as a magnitude Hint B.1

acting on particle 3 due to the presence of the other two particles. Report you and a direction

measured from the positive x axis.

How to approach the problem

To calculate the electric force acting on particle 3, you should begin by drawing a free-body diagram indicating the forces acting on particle 3 due to particle 1 and particle 2. You know that . Use Coulomb's law to calculate the magnitude of each of these forces. Apply vector algebra to find the component forces in the and the directions. Then, sum the component forces for each direction: . From

Hint B.2

and

you can find the magnitude and direction of the resulting electric force vector.

Draw a free-body diagram

Identify the forces on the positively charged particle 3. Draw your vectors starting at the origin. The orientation of your vectors will be graded but their precise length will not. ANSWER:

View Answer Requested

Hint B.3

Calculate the force on particle 3 due to particle 1

Using the equation for Coulomb's law, calculate the magnitude of the force on particle 3 due to particle 1. Keep in mind that . Express your answer in newtons using three significant figures.

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ANSWER:


−4 = 7.19×10 Answer Requested

Use the same method to calculate the force on particle 3 due to particle 2. Hint B.4

Calculate the component forces on particle 3 due to particle 1

Calculate the x component and the y component forces acting on particle 3 due to particle 1, using simple trigonometry. The angle between particle 1 and particle 3 is 60 :

Enter the components of the force in newtons separated by a comma. ANSWER:

,

−4 −4 = 3.60×10 ,6.23×10 Answer Requested

,

Use the same method to calculate the component force on particle 3 due to particle 2. The sum of all the components in each direction

will provide you with the information needed to calculate the magnitude and direction of the net force on particle 3. Hint B.5

How to calculate the component forces on particle 3 due to particle 2

Because particles 2 and 3 both lie on the x axis, there will be no y component to calculate. The x component of force will therefore be equal to the value calculated from Coulomb's law, and the y component will be zero. Hint B.6 If a vector

How to determine the magnitude and direction of a vector from its components has components

and

, the magnitude

and direction

are given by

, where

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

,

−3 = 1.90×10 ,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 All attempts used; correct answer displayed

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 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

.

Charged Ring

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Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius charge

and positive

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:

parallel to the x axis parallel to the y axis parallel to the z axis in a circle parallel to the xy plane Correct

Part B What is the magnitude of the electric field along the positive z axis? Hint B.1

Formula for the electric field

You can always use Coulomb's law,

, to find the electric field (the Coulomb force per unit

charge) due to a point charge. Given the force, the electric field say at

due to

In the situation below, you should use Coulomb's law to find the contribution point

from a piece of charge

the ring to find the value of

on the ring at a distance

is

to the electric field at the

away. Then, you can integrate over

. Consider an infinitesimal piece of the ring with charge

law to write the magnitude of the infinitesimal

.

. Use Coulomb's

at a point on the positive z axis due to the charge

shown in the figure.

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Use


in your answer, where

. You may also use some or all of the variables

, , and

. ANSWER:

Hint B.2


Simplifying with symmetry

By symmetry, the net field must point along the z axis, away from the ring, because the horizontal component of each contribution of magnitude is exactly canceled by the horizontal component of a similar contribution of magnitude

from the other side of the ring. Therefore, all we care about is the z

component of each such contribution. What is the component charge on an infinitesimally small portion of the ring

Express your answer in of

of the electric field caused by the

in the z direction?

, the infinitesimally small contribution to the electric field; ,

the coordinate of the point on the z axis; and , the radius of the ring.

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ANSWER:

Hint B.3



Integrating around the ring

If you combine your results from the first two hints, you will have an expression for component of the field due to the infinitesimal charge

, the vertical

. The total field is .

If you are not comfortable integrating charge

over the ring, change to a spatial variable. Since the total

is distributed evenly about the ring, convince yourself that .

Use

in your answer, where

.

ANSWER: = Correct

Notice that this expression is valid for both positive and negative charges as well as for points located on the positive and negative z axis. If the charge is positive, the electric field should point outward. For points on the positive z axis, the field points in the positive z direction, which is outward from the origin. For points on the negative z axis, the field points in the negative z direction, which is also outward from the origin. If the charge is negative, the electric field should point toward the origin. For points on the positive z axis, the negative sign from the charge causes the electric field to point in the negative z direction, which points toward the origin. For points on the negative z axis, the negative sign from the z coordinate and the negative sign from the charge cancel, and the field points in the positive z direction, which also points toward the origin. Therefore, even though we obtained the above result for postive and , the algebraic expression is valid for any signs of the parameters. As a check, it is good to see that if

is much greater than

the magnitude of

is approximately

, independent of the size

of the ring: The field due to the ring is almost the same as that due to a point charge

at the origin.

Part C Imagine a small metal ball of mass

and negative charge

. The ball is released from rest at the point

and constrained to move along the z axis, with no damping. If

, what will be the ball's

subsequent trajectory? ANSWER:

repelled from the origin attracted toward the origin and coming to rest

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oscillating along the z axis between

and

circling around the z axis at Correct

Part D The ball will oscillate along the z axis between the angular frequency

in simple harmonic motion. What will be

of these oscillations? Use the approximation

that is, assume that Hint D.1

and

to simplify your calculation;

.

Simple harmonic motion Hint not displayed

Hint D.2

Find the force on the charge Hint not displayed

Express your answer in of given charges, dimensions, and constants. ANSWER: = Correct

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 moment of inertia

apart. The dipole has

about its center of mass. The

dipole is released from angle

, and it is allowed

to rotate freely.

Part A

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What is Hint A.1


, the magnitude of the dipole's angular velocity when it is pointing along the y axis? How to approach the problem Hint not displayed

Hint A.2

Find the potential energy Hint not displayed

Hint A.3

Find the total energy at the moment of release Hint not displayed

Hint A.4

Find the total energy when Hint not displayed

Express your answer in of quantities given in the problem introduction. ANSWER: = Correct

Thus

increases with increasing

, as you would expect. An easier way to see this is to use the

trigonometric identity

to write

as

.

Part B If

is small, the dipole will exhibit simple harmonic motion after it is released. What is the period

of the

dipole's oscillations in this case? Hint B.1


Hint B.2

Compute the torque Hint not displayed

Hint B.3

The small-angle approximation Hint not displayed

Hint B.4

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Find the oscillation frequency

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Hint not displayed Hint B.5

The relationship between (angular) oscillation frequency and period Hint not displayed


and quantities given in the problem introduction.

ANSWER: = Correct

Electric Field due to Multiple Point Charges 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 Hint A.1

, 12.0

).


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

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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 Find the magnitude and sign of Hint B.1

is now placed at point B, located at coordinates (0

, 15.0

).

needed to make the total electric field at point A equal to zero.


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 All attempts used; correct answer displayed

Electric Force of Three Collinear Points 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 .

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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 Coulomb’s 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 ( force on

,

,

), what is the direction of the net electric

due to the other charges?

Hint A.2.1 Find the direction of the force on For combination of charges ( force on other. ANSWER:

due to

,

due to ,

), what is the direction of the electric

? that like charges repel each other and opposite charges attract each

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

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

due to

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For combination of charges ( force on

due to

ANSWER:

,

,

), what is the direction of the electric

?

to the right. to the left There is no force. Correct

Hint A.2.3 Find the magnitude of the net force on In combination of charges ( from

or that from

ANSWER:

,

,

), which of the two forces on

, that

, is larger in magnitude?

the force from the force from Neither; they are equal in magnitude. Correct

ANSWER:

to the right to the left There is no net force. Correct

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 Correct

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Finding the Zero-Field Point Two particles with positive charges

and

are separated by a distance .

Part A Along the line connecting the two charges, at what distance from the charge from the two charges zero? Hint A.1

is the total electric field


Hint A.2

Find the distance from the second charge to the zero-field point Hint not displayed

Hint A.3

Find the electric field from the first charge Hint not displayed

Hint A.4

Find the electric field from the second charge Hint not displayed

Hint A.5

Solving the quadratic equation and choosing the correct answer Hint not displayed

Express your answer in of some or all of the variables ,

,

and

. If your answer

is difficult to enter, consider simplifying it, as it can be made relatively simple with some work. ANSWER: All attempts used; correct answer displayed

Your answer can be reduced to . In order to arrive at this expression from its initial form, , the assumption must be made that solve the special (or singular) case when solve this problem.

(otherwise there's a zero in the denominator). Therefore, to , it would be best to use a symmetry argument to

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 electric field

at the point

. Consider the

, a distance

above the

midpoint of the wire.

Part A The field Hint A.1

points along one of the primary axes. Which one? Consider opposite ends of the wire Hint not displayed

ANSWER:

Correct

Part B What is the magnitude

of the electric field at point

of the constant , defined by

Hint B.1

? Throughout this part, express your answers in

.


Hint B.2

Find the field due to an infinitesimal segment Hint not displayed

Hint B.3

A necessary integral Hint not displayed


,

,

, and .

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ANSWER: = Correct

What About Finite Sheets? 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 density

has negative charge

, where the charge density is charge per

unit area.

Part A What is the magnitude Hint A.1

of the electric field at the point on the x axis with x coordinate

?

How to approach the problem

When calculating the electric field from more than one charge or a continuous charge distribution, one makes use of the superposition principle, which states that the electric field from multiple charges equals the vector sum of the fields from each individual charge. To find the electric field at a point, add the field due to each disk at that point. Be careful of whether the field magnitudes should add or subtract. The easiest way to be sure is to draw a figure with the two disks and arrows for the electric field direction on each side of each disk. If the arrows for the two disks point the same way, then the magnitudes add. If they point in opposite directions, the magnitudes subtract. Hint A.2

The magnitude of the electric field due to a single disk

The magnitude

of the electric field along the x axis for a charged disk centered at

is

, where and

is the radius of the disk,

is the charge density on the disk,

is the permittivity of free space,

is the x coordinate. Be careful in determining the direction in which the electric field due to each

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disk points. Hint A.3

Determine the general form of the electric field between the disks

Which of the following equations represents the magnitude

of the electric field between the two disks?

Hint A.3.1 Determine whether the magnitudes should add or subtract Hint not displayed

ANSWER:

Correct

Express your answer in 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 the point

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of plate radius to separation between the plates does the electric field at

on the x axis differ by 1 percent from the result

for infinite sheets?

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Hint B.1


Percent difference

Recall that the percent difference between two numbers, in this case the electric field the electric field

from Part A and

due to two infinite sheets, is given by .

Express your answer to two significant figures. ANSWER:

= 50 Answer Requested

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.

A Charged Sphere with a Cavity An insulating sphere of radius , centered at the origin, has a uniform volume charge density . Part A Find the electric field Hint A.1

inside the sphere (for < ) in of the position vector

.


Hint A.2

Determine the enclosed charge Hint not displayed

Hint A.3

Calculate the integral over the Gaussian surface Hint not displayed


, , and

.

ANSWER: = Correct

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Part B A spherical cavity is excised from the inside of the sphere. The cavity has radius and is centered at position , where

, so that the entire cavity

is contained within the larger sphere. Find the electric field inside the cavity.

Hint B.1


Hint B.2

Find the field due to the imaginary sphere Hint not displayed

Express your answer as a vector in of any or all of ,

,

, and

.

ANSWER: = Correct

Notice that the electric field inside the hole is uniform: Both its magnitude and direction are constant.

A Conducting Shell around a Conducting Rod An infinitely long conducting cylindrical rod with a positive charge per unit length is surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge per unit length of and radius , as shown in the figure.

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Part A What is

, the radial component of the electric field between the rod and cylindrical shell as a function

of the distance Hint A.1

from the axis of the cylindrical rod?

The implications of symmetry Hint not displayed

Hint A.2

Apply Gauss' law Hint not displayed

Hint A.3

Find the charge inside the Gaussian surface Hint not displayed

Hint A.4

Find the flux Hint not displayed


, , and

, the permittivity of free space.

ANSWER: = Correct

Part B What is shell? Hint B.1

, the surface charge density (charge per unit area) on the inner surface of the conducting

Apply Gauss's law Hint not displayed

Hint B.2

Find the charge contribution from the surface Hint not displayed

ANSWER: = Correct

Part C What is , the surface charge density on the outside of the conducting shell? (Recall from the problem statement that the conducting shell has a total charge per unit length given by .)

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Hint C.1


What is the charge on the cylindrical shell? Hint not displayed

ANSWER: = Correct

Part D What is the radial component of the electric field, Hint D.1

, outside the shell?


Hint D.2

Find the charge within the Gaussian surface Hint not displayed

Hint D.3

Find the flux in of the electric field Hint not displayed

ANSWER: = Correct

The Electric Field of a Ball of Uniform Charge Density A solid ball of radius

has a uniform charge density .

Part A What is the magnitude of the electric field Hint A.1

at a distance

from the center of the ball?

Gauss's law Hint not displayed

Hint A.2

Find Hint not displayed


, , and

.

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ANSWER: = Correct

Notice that this result is identical to that reached by applying Coulomb's law to a point charge centered at the origin with . The field outside of a uniformly charged sphere does not depend on the size of the sphere, only on its charge. A uniformly charged sphere generates an electric field as if all the charge were concentrated at its center. Part B What is the magnitude of the electric field

at a distance

from the center of the ball?

How does this situation compare to that of the field outside the ball?

Hint B.1

Hint not displayed Express your answer in of , ,

, and

.

ANSWER: = Correct

Part C Let

represent the electric field due to the charged ball throughout all of space. Which of the following

statements about the electric field are true? Hint C.1

Plot the electric field Hint not displayed

Check all that apply. ANSWER:

. . . The maximum electric field occurs when The maximum electric field occurs when

. .

The maximum electric field occurs as Correct

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Charged Insulating Spheres Two small insulating spheres with radius 3.00×10−2 0.540

are separated by a large center-to-center distance of

. One sphere is negatively charged, with net charge -1.30

charged, with net charge 3.40

, and the other sphere is positively

. The charge is uniformly distributed within the volume of each sphere.

Part A What is the magnitude Hint A.1

of the electric field midway between the spheres?


Hint A.2

Using Gauss's law Hint not displayed

Hint A.3

Calculate the field due to the negatively charged sphere Hint not displayed

Hint A.4

Determine the direction of the electric field from the first sphere Hint not displayed

Hint A.5

Calculate the field due to the positively charged sphere Hint not displayed

Hint A.6

Determine the direction of the electric field from the positively charged sphere Hint not displayed

Hint A.7

Vector addition Hint not displayed

Take the permittivity of free space to be ANSWER:

= 8.85×10−12

.

5 = 5.80×10 Correct

Part B What is the direction of the electric field midway between the spheres? ANSWER:

toward the positively charged sphere toward the negatively charged sphere

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upward perpendicular to the line connecting the centers of the spheres downward perpendicular to the line connecting the centers of the spheres Correct Since the electric field will point toward a negative charge and away from a positive charge, the electric field from each sphere separately will point toward the negatively charged sphere, and so the total field will also point in that direction.

Back to Square One Four point charges form a square with sides of length , as shown in the figure. In the questions that follow, use the constant

in place of

.

Part A What is the electric potential

at the center of the square?

Make the usual assumption that the potential tends to zero far away from a charge. Hint A.1


Hint A.2

Find the distance to the center Hint not displayed


, and appropriate constants.

ANSWER: = Correct

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Part B What is the contribution charge

to the electric potential energy of the system, due to interactions involving the

?

Hint B.1

Find the electric potential at the point with charge Hint not displayed



ANSWER: = Correct

Part C What is the total electric potential energy Hint C.1

of this system of charges?


Hint C.2

How many pairs? Hint not displayed



ANSWER: = Correct

Imagine now that charge

is released, and it drifts away from the rest of the charges, which remain fixed

in place. Part D What would be the kinetic energy Hint D.1

of charge

at a very large distance from the other charges?

What happens to energy? Hint not displayed


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ANSWER: = Correct

It should not come as a surprise that the answer to Part D is equal to the intial contribution to the potential energy of the system due to the presence of charge . Initially, the kinetic energy of charge is zero. Because the electric potential between two charges is inversely proportional to the distance between them, after charge

has drifted far away from the others,

(see B) is (very close to)

zero. Since total energy is conserved, the change in potential energy must have been converted into kinetic energy. Part E What will be the potential energy

of the system of charges when charge

is at a very large distance

from the other charges? Hint E.1

What happens to energy? Hint not displayed



ANSWER: = Correct

There are two ways you could have approached this question. You could have found the sum of the three corresponding to the three remaining pairs of charges, or you could have subtracted the initial from the total energy of the system before charge was removed.

Bouncing Electrons Two electrons, each with mass

and charge , are released from positions very far from each other. With

respect to a certain reference frame, electron A has initial nonzero speed x direction, and electron B has initial speed

toward electron B in the positive

toward electron A in the negative x direction. The electrons

move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other. Part A Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two electrons reach their minimum separation?

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ANSWER:


Electron A is moving faster than electron B. Electron B is moving faster than electron A. Both electrons are moving at the same (nonzero) speed in opposite directions. Both electrons are moving at the same (nonzero) speed in the same direction. Both electrons are momentarily stationary. Correct

If at a given moment the electrons are still moving toward each other, then they will be closer in the next instant. If at a given moment the electrons are moving away from each other, then they were closer in the previous instant. The electrons will be traveling in the same direction at the same speed at the moment they reach their minimum separation. Only in a reference frame in which the total momentum is zero (the center of momentum frame) would the electrons be stationary at their minimum separation. Part B What is the minimum separation Hint B.1

that the electrons reach?


Hint B.2

Find the initial energy Hint not displayed

Hint B.3

Find the final energy Hint not displayed

Hint B.4

Find the initial momentum Hint not displayed

Hint B.5

Find the final momentum Hint not displayed

Hint B.6

Some math help Hint not displayed

Express your answer in term of ,

, , and

(where

).

ANSWER: =

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Correct An experienced physicist might approach this problem by considering the system of electrons in a reference frame in which the initial momentum is zero. In this frame the initial speed of each electron is . Try solving the problem this way. Make sure that you obtain the same result for , and decide for yourself which approach is easier.

Change in Electric Potential Ranking Task In the diagram below, there are two charges of

and

and six points (a through f) at various distances

from the two charges. You will be asked to rank changes in the electric potential along paths between pairs of points.

Part A Using the diagram to the left, rank each of the given paths on the basis of the change in electric potential. Rank the largest-magnitude positive change (increase in electric potential) as largest and the largestmagnitude negative change (decrease in electric potential) as smallest. Hint A.1

Change in electric potential Hint not displayed

Hint A.2

Determine the algebraic sign of the change in potential Hint not displayed

Hint A.3

Conceptualizing changes in electric potential Hint not displayed

Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER:

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View Correct

Charged Mercury Droplets A uniformly charged spherical droplet of mercury with electric potential droplets, each with electric potential

. The

breaks into

identical spherical

small droplets are far enough apart form one another that

they do not interact significantly. Part A Find

, the ratio of

, the electric potential of the initial drop, to

, the electric potential of one

of the smaller drops. Hint A.1


Hint A.2

Find the charge on the small droplets Hint not displayed

Hint A.3

Find the radius of a small droplet Hint not displayed

The ratio should be dimensionless and should depend only on ANSWER: = Correct

Conducting Tetrahedra Two conductors, A and B, are each in the shape of a tetrahedron, but of different sizes. They are charged in the following manner: Tetrahedron A is charged from an electrostatic generator to charge . Tetrahedron A is briefly touched to tetrahedron B. Steps 1 and 2 are repeated until the charge on tetrahedron B reaches a maximum value.

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Part A If the charge on tetrahedron B was

after the first time it touched tetrahedron A, what is the final charge

on tetrahdedron B? Hint A.1


Hint A.2

Find the ratio of the conductors' charges Hint not displayed

Hint A.3

Find the maximum charge on A Hint not displayed

Express your answer in of . ANSWER: = Correct

Not So Fast! Four point charges, fixed in place, form a square with side length .

Part A The particle with charge

is now released and given a quick push; as a result, it acquires speed .

Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is , what was its initial speed ? Hint A.1


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Hint A.2


Finding the potential energy Hint not displayed

Hint A.3

Find the initial potential energy Hint not displayed

Hint A.4

Find the final potential energy Hint not displayed

Hint A.5

Find the kinetic energy Hint not displayed

Hint A.6

Formula for kinetic energy Hint not displayed


,

, and appropriate constants. Use

instead of

. The

numeric coefficient should be a decimal with three significant figures. ANSWER: = Correct

Part B When the particle with charge reaches the center of the original square, it is, as stated in the problem, momentarily at rest. Is the particle at equilibrium at that moment? Hint B.1


ANSWER:

yes no Correct

The exact value of the net force can be found by a calculation.

Potential of a Charged Annulus

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An annular ring with a uniform surface charge density coordinate axes. The annulus has an inner radius

sits in the xy plane, with its center at the origin of the

and outer radius

.

Part A If you can find symmetries in a physical situation, you can often greatly simplify your calculations. In this part you will find a symmetry in the annular ring before calculating the potential along the axis through the ring's center in Part B. Consider three sets of points: points lying on the vertical line A; those on circle B; and those on the horizontal line C, as shown in the figure. Which set of points makes the same contribution toward the potential calculated at any point along the axis of the annulus?

Hint A.1

Definition of the potential due to a point charge Hint not displayed

ANSWER:

points on line A points on circle B points on line C Correct

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Part B By exploiting the above symmetry, or otherwise, calculate the electric potential the annulus a distance Hint B.1

at a point on the axis of

from its center.

How to exploit the angular symmetry of the problem Hint not displayed

Hint B.2

Find the area of an annular slice Hint not displayed

Hint B.3

Doing the integral Hint not displayed

Hint B.4

A formula for the integral Hint not displayed

Express your answer in of some or all of the variables

,

,

, and

. Use

.

ANSWER: = All attempts used; correct answer displayed

It is interestering to note that the potential at any point on the axis of a disk of radius obtained from the expression above by setting

and

can be

. Doing so, one obtains .

Conversely, the annulus can be thought of as the superposition of two disks, one with charge density and radius , and the other with charge density and radius . In the region from the center to , the opposite charge densities cancel out, so the net charge distribution would be just like that of the annulus. Moreover, by adding the potentials due to these two disks, using the formula above, you would recover the potential of the annulus. It is also instructive to look at the general behavior of these potentials as a function of the parameters. Clearly, the potential increases with increasing charge densities, as well as with increasing areas (if the charge density is held constant), which intuitively seems reasonable. However, if the distance increases, it is not clear whether the potential should grow, since

appears in both , of which

one is subtracted from the other. If you are far from the disk, the disk looks like a point, and the potential should drop off, just like the potential due to a point charge. Indeed, on of the negative second term in the expressions, this is the case. Try some values or check that the derivative of is indeed negative. You can also check that the above expression actually reduces to the potential due to a point charge for

.

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Potential of a Charged Cylinder A hollow cylinder of radius

and height

has a total charge

uniformly distributed over its surface. The axis

of the cylinder coincides with the z axis, and the cylinder is centered at the origin, as shown in the figure.

Part A What is the electric potential Hint A.1

at the origin?


Hint A.2

Potential due to a thin ring Hint not displayed

ANSWER:

Correct

Part B What is the potential

in the limit as

goes to zero?

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Hint B.1


How to take the limit Hint not displayed

Express your answer in of , , and

.

ANSWER: = All attempts used; correct answer displayed

Note that this expression is the same as that for the potential at the center of a charged a ring! The reason for this is that if the radius of the cylinder is much larger than the length of the cylinder, the cylinder looks and behaves much like a ring.

Potential of a Charged Disk A disk of radius

has a total charge

uniformly distributed over its surface. The disk has negligible

thickness and lies in the xy plane. Throughout this problem, you may use the variable

in place of

.

Part A What is the electric potential Hint A.1

on the z axis as a function of , for

?


Hint A.2

Find the potential due to a ring Hint not displayed

Hint A.3

A useful antiderivative Hint not displayed


, , and . You may use

instead of

.

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ANSWER: = Correct

Part B What is the magnitude Hint B.1

of the electric field on the

axis, as a function of , for

?

Direction of the electric field Hint not displayed

Hint B.2

Electric field from potential Hint not displayed

Express your answer in of some or all of the variables

, , and . You may use

instead of

. ANSWER: = Correct

Since the magnitude of the electric field (and potential) must be symmetric about the general expression for the magnitude of the electric field on the z axis for all

plane, the

is .

Note the use of

instead of .

Potential of a Charged Ring A ring with radius

and a uniformly distributed total charge

lies in the xy plane, centered at the origin.

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Part A What is the potential Hint A.1

due to the ring on the z axis as a function of ?


Hint A.2

The potential due to a point charge Hint not displayed


, ,

, and

or

.

ANSWER: = Correct

Part B What is the magnitude of the electric field Hint B.1

on the z axis as a function of , for

?

Determine the direction of the field Hint not displayed

Hint B.2

The relationship between electric field and potential Hint not displayed

Express your answer in of some or all of the quantities

, ,

, and

or

.

ANSWER: |

|= Correct

Notice that while the potential is a strictly decreasing function of , the electric field first increases till and then starts to decrease. Why does the electric field exhibit such a behavior? Though the contribution to the electric field from each point on the ring strictly decreases as a function of , the vector cancellation from points on opposite sides of the ring becomes very strong for small .

on of these vector cancellations. On the other hand

though all the individual

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, even

's point in (almost) the same direction there, because the contribution to the

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electric field, per unit length of the ring

as

.

Potential of a Finite Rod A finite rod of length

has total charge , distributed uniformly along its length. The rod lies on the x -axis

and is centered at the origin. Thus one endpoint is located at

, and the other is located at

.

Define the electric potential to be zero at an infinite distance away from the rod. Throughout this problem, you may use the constant

in place of the expression

.

Part A What is

, the electric potential at point A (see the

figure), located a distance

above the midpoint of

the rod on the y axis?

Hint A.1


Hint A.2

Find the electric potential of a section of the rod Hint not displayed

Hint A.3

A helpful integral Hint not displayed


,

, , and .

ANSWER: = Correct

If

, this answer can be approximated as

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. For

,

. For this problem, this means that the logarithm can be further

approximated as

, and the expression for potential reduces to

. This is

what we expect, because it means that from far away, the potential due to the charged rod looks like that due to a point charge. Part B What is

Hint B.1

, the electric potential at point

, located at distance

from one end of the rod (on the x axis)?


Hint B.2

Find the distance from point B to a segment of the rod Hint not displayed

Give your answer in of ,

,

, and .

ANSWER: = Correct

This result can be written as . As before, for the result reduces to

,

. Thus, for

, the logarithm approaches

, in which case

. This is what we expect, because it means that from far away, the potential

due to the charged rod looks like that due to a point charge.

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± Potential of Two Charged Spheres A metal sphere with radius is ed on an insulating stand at the center of a hollow, metal spherical shell with radius . There is charge on the inner sphere and charge on the outer spherical shell. Take the potential

to be zero when the distance

from the center of the spheres is infinite.

Part A Calculate the potential

for

. (Hint: The net potential is the sum of the potentials due to the

individual spheres.) Hint A.1


Hint A.2

Find an expression for the electric field Hint not displayed

Hint A.3

The relationship between potential and electric field Hint not displayed

Use as the permittivity of free space and express your answer in of some or all of the variables , , , , and any appropriate constants. ANSWER:

= 0

Note that if

Correct

were not defined in the introduction, the exact value for the potential

could

not be calculated (since potential is always in of some reference potential), only the potential difference . Moreover, the value for that you calculate for the region is valid only for the region outside of both spheres, as you will see in the following parts. Part B Calculate the potential

for


individual spheres.) Hint B.1


Hint B.2


Hint B.3

Relationship between potential and electric field Hint not displayed

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Use for the permittivity of free space and express your answer in of some or all of the variables , , , , and any appropriate constants. ANSWER: = Correct

Part C Calculate the potential

for


individual spheres.) Hint C.1


Hint C.2


Hint C.3

Relationship between potential and electric field Hint not displayed

Use for the permittivity of free space and express your answer in of some or all of the variables , , , , and any appropriate constants. ANSWER: = Correct

Note that even though there is no electric field inside the innermost charged sphere, there is still a nonzero potential difference. However, since this potential does not depend on the variable , but only on the radii of the spheres and , this potential will be constant throughout the inner sphere. Furthermore, if the two spheres did not have the same magnitude of charge (e.g., and with ), the same method could still be used to calculate the potential, but in that case, the electric field in the outer region would not be zero, since we would have would affect, however, is regions (

. The only region this

, since the enclosed charge would be the same in the middle and inner

and , respectively).

Moving a Charge Part A

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A point charge with charge = -4.30

= 2.40

is held stationary at the origin. A second point charge with charge

moves from the point ( 0.105

, 0) to the point ( 0.285

, 0.260

). How much work

is

done by the electric force on the moving point charge? Hint A.1


Hint A.2

Calculate the initial electric potential energy Hint not displayed

Hint A.3

Calculate the final electric potential energy Hint not displayed

Express your answer in joules. Use ANSWER:

= 8.99×109

for Coulomb's constant:

.

= -0.643 Correct

Speed of an Electron in an Electric Field Two stationary positive point charges, charge 1 of magnitude 3.10 are separated by a distance of 32.0

and charge 2 of magnitude 1.65

,

. An electron is released from rest at the point midway between the

two charges, and it moves along the line connecting the two charges. Part A What is the speed Hint A.1

of the electron when it is 10.0

from charge 1?


Hint A.2

Calculate the potential at the midpoint Hint not displayed

Hint A.3

Calculate the initial potential energy Hint not displayed

Hint A.4

Calculate the initial kinetic energy Hint not displayed

Hint A.5

Calculate the final potential energy

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Hint not displayed Hint A.6

Putting it all together Hint not displayed

Express your answer in meters per second. ANSWER:

6 = 5.28×10 Correct

Note that the electric field between the two charges is not constant, so the easiest way to do these calculations is to use conservation of energy. It is possible to integrate along the path of the electron, using the electric field as a function of the distance from each charge, but this is much more difficult to do and not necessary for the problem.

Stopping the Proton An infinitely long line of charge has a linear charge density of 8.00×10−12 from the line and is moving directly toward the line with speed 1200

. A proton is at distance 19.0 .

Part A How close does the proton get to the line of charge? Hint A.1


Hint A.2

Calculate the final kinetic energy Hint not displayed

Hint A.3

Calculate the initial kinetic energy Hint not displayed

Hint A.4

Electric field around a line of charge Hint not displayed

Hint A.5

Potential difference over the path of the proton Hint not displayed

Hint A.6

Putting it all together Hint not displayed

Express your answer in meters.

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ANSWER:


0.180 Correct

Another way you could have solved this problem is by defining the point where

to be at either

the initial or final location of the proton, and calculating the corresponding potential at the other point. However, since only the potential difference is needed (or potential energy difference) to solve the problem, this would be one more unnecessary step to worry about.

Score Summary: Your score on this assignment is 85.1%. You received 85.07 out of a possible total of 100 points.

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