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Homework 1Due: 11:59pm on Sunday, February 13, 2011

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Coulomb's Law Tutorial

Learning Goal: To understand how to calculate forces between charged particles, particularly thedependence 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 thesquare 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 ofcharge (particle 0) fixed at the origin, and anotherof 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 allof , , , , , , and .

ANSWER: =

Correct

Part B

Now add a third, negatively charged, particle, whosecharge is (particle 2). Particle 2 fixed on the

y-axis at position . 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 allof , , , , , , , , 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 terms of any or all of the following variables: , , , .

ANSWER: =

Correct

Part D

Now add a fourth charged particle, particle 3, withpositive charge , fixed in the yz-plane at

. What is the net force on particle 0 due solely to

this charge?

Hint D.1 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: = Answer not displayed

Hint D.2 Vector components

The force vector points from to . Because is symmetrically located between the y-axis and thez-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 vectorinto 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 oflength 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 lawIDENTIFY the relevant concepts:Coulomb’s law comes into play whenever you need to know the electric force acting between chargedparticles.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 coordinatesystem.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 toeach 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 fromparticle 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, isa 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 chargedistribution into infinitesimal pieces, use Coulomb’s law for each piece, and then integrate to find the vectorsum.In many situations, the charge distribution will be symmetrical. Whenever possible, exploit any symmetries tosimplify the problem-solving process.EVALUATE your answer:Check whether your numerical results are reasonable, and confirm that the direction of the net electric forceagrees 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 vectorsum of all the forces on that particle due to the presence of other charged particles. To do this, you will needto 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 particle3 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

To calculate the electric force acting on particle 3, you should begin by drawing a free-body diagramindicating 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 thecomponent forces in the and the directions. Then, sum the component forces for each direction:

.

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

Hint B.2 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 theirprecise 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: = 7.19×10−4

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 simpletrigonometry. The angle between particle 1 and particle 3 is 60 :

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

ANSWER: , = 3.60×10−4,6.23×10−4

Answer Requested ,

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

will provide you with the information needed to calculate the magnitude and direction of the net forceon 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 xcomponent of force will therefore be equal to the value calculated from Coulomb's law, and the ycomponent will be zero.

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

If a vector 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: , = 1.90×10−3,19.1Answer 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 whatdirection 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 notits 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 inthe 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 chargesattract. 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 fromeach 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 and positivecharge 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 is .

In the situation below, you should use Coulomb's law to find the contribution to the electric field at the

point from a piece of charge on the ring at a distance away. Then, you can integrate over

the ring to find the value of . Consider an infinitesimal piece of the ring with charge . Use Coulomb's

law to write the magnitude of the infinitesimal 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

.


Hint B.2 Simplifying with symmetry

By symmetry, the net field must point along the z axis, away from the ring, because the horizontalcomponent 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 of the electric field caused by the

charge on an infinitesimally small portion of the ring in the z direction?

Express your answer in terms of , 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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Hint B.3 Integrating around the ring

If you combine your results from the first two hints, you will have an expression for , the vertical

component of the field due to the infinitesimal charge . The total field is

.

If you are not comfortable integrating over the ring, change to a spatial variable. Since the total

charge 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 locatedon the positive and negative z axis. If the charge is positive, the electric field should point outward. Forpoints 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 fromthe origin. If the charge is negative, the electric field should point toward the origin. For points on thepositive z axis, the negative sign from the charge causes the electric field to point in the negative zdirection, which points toward the origin. For points on the negative z axis, the negative sign from the zcoordinate 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 and in simple harmonic motion. What will be

the angular frequency of these oscillations? Use the approximation to simplify your calculation;

that is, assume that .

Hint D.1 Simple harmonic motion

Hint not displayed

Hint D.2 Find the force on the charge

Hint not displayed

Express your answer in terms of given charges, dimensions, and constants.

ANSWER:

=

Correct

Dipole Motion in a Uniform FieldConsider an electric dipole located in a region with anelectric field of magnitude pointing in the positive y

direction. The positive and negative ends of the dipolehave charges and , respectively, and the two

charges are a distance apart. The dipole has

moment of inertia about its center of mass. The

dipole is released from angle , and it is allowed

to rotate freely.

Part A


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What is , the magnitude of the dipole's angular velocity when it is pointing along the y axis?


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 terms 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 not displayed

Hint B.2 Compute the torque

Hint not displayed

Hint B.3 The small-angle approximation

Hint not displayed

Hint B.4 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

Express your answer in terms of and quantities given in the problem introduction.

ANSWER:

=

Correct

Electric Field due to Multiple Point ChargesTwo point charges are placed on the x axis. The firstcharge, = 8.00 , is placed a distance 16.0

from the origin along the positive x axis; the secondcharge, = 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 ).


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


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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 innewtons per coulomb to three significant figures.

ANSWER: = 0,0.300Correct

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

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 TaskIn 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 forcespointing to the right as positive and forces pointing to the left as negative. Rank positive forces as largerthan 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 issummarized 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 ( , , ), 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 eachother.

ANSWER: 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 ( , , ), what is the direction of the electric

force on due to ?

ANSWER: 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 ( , , ), 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.

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 rankitems as equivalent, overlap them.

ANSWER:

View Correct


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Finding the Zero-Field PointTwo 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 is the total electric fieldfrom the two charges zero?


Hint not displayed

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 terms 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:


Your answer can be reduced to

.

In order to arrive at this expression from its initial form,

,

the assumption must be made that (otherwise there's a zero in the denominator). Therefore, to

solve the special (or singular) case when , it would be best to use a symmetry argument tosolve this problem.

The Electric Field Produced by a Finite Charged Wire


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A charged wire of negligible thickness has length

units and has a linear charge density . Consider the

electric field at the point , a distance above the

midpoint of the wire.

Part A

The field points along one of the primary axes. Which one?

Hint A.1 Consider opposite ends of the wire

Hint not displayed

ANSWER:

Correct

Part B

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


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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 ofinfinite charged sheets. This approximation is useful when a problem deals with points whose distance from afinite charged sheet is small compared to the size of the sheet. In this problem, you will look at the electricfield from two finite sheets and compare it to the results for infinite sheets to get a better idea of when thisapproximation 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. Thedisk centered at has positive charge density

, and the disk centered at has negative chargedensity , 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 ?


When calculating the electric field from more than one charge or a continuous charge distribution, onemakes use of the superposition principle, which states that the electric field from multiple charges equalsthe vector sum of the fields from each individual charge. To find the electric field at a point, add the fielddue to each disk at that point. Be careful of whether the field magnitudes should add or subtract. Theeasiest way to be sure is to draw a figure with the two disks and arrows for the electric field direction oneach side of each disk. If the arrows for the two disks point the same way, then the magnitudes add. Ifthey 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 is the radius of the disk, is the charge density on the disk, is the permittivity of free space,

and 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 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 greatestdeviation 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?


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Hint B.1 Percent difference

Recall that the percent difference between two numbers, in this case the electric field from Part A and

the electric field due to two infinite sheets, is given by

.

Express your answer to two significant figures.

ANSWER: = 50Answer Requested

As mentioned above, this is the point on the x axis where the deviation from the result for two infinitesheets is greatest. A common component of electrical circuits called a capacitor is usually made fromtwo 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 quitegood for a capacitor.This applet shows the electric field lines from a pair of finite plates (viewed edge-on). You can adjustthe surface charge density. You can also move the test charge around and increase or decrease itscharge to see what sort of force it would experience. Notice that the deviation from uniform electricfield only becomes noticeable near the edges of the capacitor plates.

A Charged Sphere with a CavityAn insulating sphere of radius , centered at the origin, has a uniform volume charge density .

Part A

Find the electric field inside the sphere (for < ) in terms of the position vector .


Hint not displayed

Hint A.2 Determine the enclosed charge

Hint not displayed

Hint A.3 Calculate the integral over the Gaussian surface

Hint not displayed

Express your answer in terms of , , and .

ANSWER: =

Correct


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Part B

A spherical cavity is excised from the inside of thesphere. The cavity has radius and is centered at

position , where , so that the entire cavity

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


Hint not displayed

Hint B.2 Find the field due to the imaginary sphere

Hint not displayed

Express your answer as a vector in terms 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 RodAn infinitely long conducting cylindrical rod with apositive 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 from the axis of the cylindrical rod?

Hint A.1 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

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

ANSWER: =

Correct

Part B

What is , the surface charge density (charge per unit area) on the inner surface of the conductingshell?

Hint B.1 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 problemstatement 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, , outside the shell?

Hint D.1 How to approach the problem

Hint not displayed

Hint D.2 Find the charge within the Gaussian surface

Hint not displayed

Hint D.3 Find the flux in terms of the electric field

Hint not displayed

ANSWER: =

Correct

The Electric Field of a Ball of Uniform Charge DensityA solid ball of radius has a uniform charge density .

Part A

What is the magnitude of the electric field at a distance from the center of the ball?

Hint A.1 Gauss's law

Hint not displayed

Hint A.2 Find

Hint not displayed



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ANSWER:

=

Correct

Notice that this result is identical to that reached by applying Coulomb's law to a point charge centeredat 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 thecharge were concentrated at its center.

Part B

What is the magnitude of the electric field at a distance from the center of the ball?

Hint B.1 How does this situation compare to that of the field outside the ball?

Hint not displayed


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 SpheresTwo small insulating spheres with radius 3.00×10−2 are separated by a large center-to-center distance of0.540 . One sphere is negatively charged, with net charge -1.30 , and the other sphere is positively

charged, with net charge 3.40 . The charge is uniformly distributed within the volume of each sphere.

Part A

What is the magnitude of the electric field midway between the spheres?


Hint not displayed

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 = 8.85×10−12 .

ANSWER: = 5.80×105

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, theelectric field from each sphere separately will point toward the negatively charged sphere, and so thetotal field will also point in that direction.

Back to Square OneFour 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 not displayed

Hint A.2 Find the distance to the center

Hint not displayed

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

ANSWER:

=

Correct


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Part B

What is the contribution to the electric potential energy of the system, due to interactions involving the

charge ?

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 of this system of charges?

Hint C.1 How to approach the problem

Hint not displayed

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 of charge at a very large distance from the other charges?

Hint D.1 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 thepotential 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 intokinetic 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 thethree terms corresponding to the three remaining pairs of charges, or you could have subtracted theinitial from the total energy of the system before charge was removed.

Bouncing ElectronsTwo electrons, each with mass and charge , are released from positions very far from each other. Withrespect to a certain reference frame, electron A has initial nonzero speed toward electron B in the positivex direction, and electron B has initial speed 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 electronsapproach each other, they slow due to their electric repulsion. This repulsion eventually pushes them awayfrom each other.

Part A

Which of the following statements about the motion of the electrons in the given reference frame will be trueat 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 thenext instant. If at a given moment the electrons are moving away from each other, then they werecloser in the previous instant. The electrons will be traveling in the same direction at the same speed atthe moment they reach their minimum separation. Only in a reference frame in which the totalmomentum is zero (the center of momentum frame) would the electrons be stationary at their minimumseparation.

Part B

What is the minimum separation that the electrons reach?


Hint not displayed

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 areference 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 TaskIn 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 rankchanges in the electric potential along paths betweenpairs 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 largest-magnitude 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 DropletsA uniformly charged spherical droplet of mercury with electric potential breaks into identical spherical

droplets, each with electric potential . The 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 not displayed

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 TetrahedraTwo conductors, A and B, are each in the shape of a tetrahedron, but of different sizes. They are charged inthe 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 not displayed

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 terms 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 massof this particle is , what was its initial speed ?


Hint not displayed


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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

Express your answer in terms of , , , 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 not displayed

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 sits in the xy plane, with its center at the origin of thecoordinate axes. The annulus has an inner radius and outer radius .

Part A

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

the annulus a distance from its center.

Hint B.1 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 terms of some or all of the variables , , , and . Use .

ANSWER: =


It is interestering to note that the potential at any point on the axis of a disk of radius can be

obtained from the expression above by setting and . 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 ofthe annulus. Moreover, by adding the potentials due to these two disks, using the formula above, youwould 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 thecharge 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 terms, of which

one is subtracted from the other. If you are far from the disk, the disk looks like a point, and thepotential should drop off, just like the potential due to a point charge. Indeed, on account of thenegative second term in the expressions, this is the case. Try some values or check that the derivativeof 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 CylinderA 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 at the origin?


Hint not displayed

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 terms of , , and .

ANSWER: =


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

Potential of a Charged DiskA 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 on the z axis as a function of , for ?


Hint not displayed

Hint A.2 Find the potential due to a ring

Hint not displayed

Hint A.3 A useful antiderivative

Hint not displayed

Express your answer in terms of , , and . You may use instead of .


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ANSWER: =

Correct

Part B

What is the magnitude of the electric field on the axis, as a function of , for ?

Hint B.1 Direction of the electric field

Hint not displayed

Hint B.2 Electric field from potential

Hint not displayed

Express your answer in terms 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 plane, the

general expression for the magnitude of the electric field on the z axis for all is

.

Note the use of instead of .

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


Hint not displayed

Hint A.2 The potential due to a point charge

Hint not displayed

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

ANSWER: =

Correct

Part B

What is the magnitude of the electric field on the z axis as a function of , for ?

Hint B.1 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 terms 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 functionof , the vector cancellation from points on opposite sides of the ring becomes very strong for small . on account of these vector cancellations. On the other hand , even

though all the individual '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 RodA 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 not displayed

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


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 likethat due to a point charge.

Part B

What is , the electric potential at point , located at distance from one end of the rod (on the x axis)?


Hint not displayed

Hint B.2 Find the distance from point B to a segment of the rod

Hint not displayed

Give your answer in terms of , , , and .

ANSWER: =

Correct

This result can be written as

.

As before, for , . Thus, for , the logarithm approaches , in which case

the result 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.


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± Potential of Two Charged SpheresA metal sphere with radius is supported on an insulating stand at the center of a hollow, metal sphericalshell 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 not displayed

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 terms of some or all of thevariables , , , , and any appropriate constants.

ANSWER: = 0Correct

Note that if were not defined in the introduction, the exact value for the potential could

not be calculated (since potential is always in terms of some reference potential), only the potentialdifference . 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




Hint not displayed

Hint B.2 Find an expression for the electric field

Hint not displayed

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 terms of some or all of thevariables , , , , and any appropriate constants.

ANSWER: =

Correct

Part C



Hint C.1 How to approach the problem

Hint not displayed

Hint C.2 Find an expression for the electric field

Hint not displayed

Hint C.3 Relationship between potential and electric field

Hint not displayed

Use for the permittivity of free space and express your answer in terms of some or all of thevariables , , , , and any appropriate constants.

ANSWER: =

Correct

Note that even though there is no electric field inside the innermost charged sphere, there is still anonzero potential difference. However, since this potential does not depend on the variable , but onlyon 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 . The only region this

would affect, however, is , since the enclosed charge would be the same in the middle and innerregions ( and , respectively).

Moving a Charge

Part A


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A point charge with charge = 2.40 is held stationary at the origin. A second point charge with charge

= -4.30 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 not displayed

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 = 8.99×109 for Coulomb's constant: .

ANSWER: = -0.643Correct

Speed of an Electron in an Electric FieldTwo stationary positive point charges, charge 1 of magnitude 3.10 and charge 2 of magnitude 1.65 ,

are separated by a distance of 32.0 . An electron is released from rest at the point midway between thetwo charges, and it moves along the line connecting the two charges.

Part A

What is the speed of the electron when it is 10.0 from charge 1?


Hint not displayed

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: = 5.28×106

Correct

Note that the electric field between the two charges is not constant, so the easiest way to do thesecalculations 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 todo and not necessary for the problem.

Stopping the ProtonAn infinitely long line of charge has a linear charge density of 8.00×10−12 . A proton is at distance 19.0

from the line and is moving directly toward the line with speed 1200 .

Part A

How close does the proton get to the line of charge?


Hint not displayed

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: 0.180Correct

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 theproblem, 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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