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Wendy Alcius
commented on a video on YouTube.Did you go to UCF? I have exactly this project for my Midterm +donjuandj
Wendy Alcius
commented on a video on YouTube.so if delta dot V is equal to Zero that means it is incompressible, so the incompressible stream function for this flow field is zero right?
Wendy Alcius
commented on a video on YouTube.Wendy Alcius
Shared publicly VCC PHY2049 Fall 2012 OSC Logged in as Wendy Alcius Help Close
Chapter 21 Homework Assignment
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Chapter 21 Homework Assignment
Due: 11:59pm on Sunday, September 9, 2012
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy
The Electric Field inside a Conductor
Learning Goal:
To understand how the charges within a conductor respond to an externally applied electric field.
To illustrate the behavior of charge inside conductors, consider a long conducting rod that is suspended by insulating strings (see the figure). Assume that the rod is initially electrically neutral, and that it remains so for this discussion. The rod is positioned along the x axis, and an external electric field that points in the positive x direction (to the right) can be applied to the rod and the surrounding region. The atoms in the rod are composed of positive nuclei (indicated by plus signs) and negative electrons (indicated by minus signs). Before application of the electric field, these atoms were distributed evenly throughout the rod.
Part A
What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem introduction is applied? (Ignore internal fields in the rod for the moment.)
Hint 1. Formula for the force on a charge in an electric field
The force on a charge in an electric field is given by
.
ANSWER:
Both electrons and nuclei experience a force to the right.
The nuclei experience a force to the right and the electrons experience a force to the left.
The electrons experience a force to the left but the nuclei experience no force.
The electrons experience no force but the nuclei experience a force to the right.
Correct
Part B
What is the motion of the negative electrons and positive atomic nuclei caused by the external field?
Hint 1. How to approch this part
Newton's 2nd Law tells you that an object at rest will move in the direction of the force applied on it.
Hint 2. Masses and charges of nuclei and electrons
A nucleus contains as many protons as the atom does electrons. So if the atom has N electrons, the nucleus will contain N protons. This means that the force on the nucleus will be N times as much as that on the electron. This N is of the order of 10100.
However, the mass of a nucleus is roughly 2N times the mass of a proton, since it contains both protons and neutrons. Each proton itself weighs about 1836 times as much as an electron! So a typical nucleus really does weigh a lot more than an electron.
Given this information, how would the distance moved by a nucleus compare with that moved by an electron?
ANSWER:
Both electrons and nuclei move to the right.
The nuclei move to the right and the electrons move to the left through equal distances.
The electrons move to the left and the nuclei are almost stationary.
The electrons are almost stationary and the nuclei move to the right.
Correct
The nuclei of the atoms of a conducting solid remain almost in their places in the crystal lattice, while the electrons relatively move a lot. In an insulator, the electrons are constrained to stay with their atoms (or molecules), and at most, the charge distribution is displaced slightly.
The motion of the electrons due to the external electric field constitutes an electric current. Since the negatively charged electrons are moving to the left, the current, which is defined as the "flow" of positive charge, moves to the right.
Part C
Imagine that the rightward current flows in the rod for a short time. As a result, what will the net charge on the right and left ends of the rod become?
Hint 1. How to approach this part
Remember that the rod as a whole must remain electrically neutral even if the charges are redistributed. This is because applying an electric field does not change the charge on the rod, only redistributes it.
ANSWER:
left end negative and right end positive
left end negative and right end negative
left end negative and right end nearly neutral
left end nearly neutral and right end positive
both ends nearly neutral
Correct
Given that the positively charged nuclei do not move, why does the right end of the rod become positively charged? The reason is that some electrons have moved to the left end, leaving an excess of stationary nuclei at the right end.
Part D
The charge imbalance that results from this movement of charge will generate an additional electric field near the rod. In what direction will this field point?
Hint 1. Direction of the electric field
The electric field point away from positive charges and towards negative ones.
ANSWER:
It will point to the right and enhance the initial applied field.
It will point to the left and oppose the initial applied field.
Correct
An electric field that exists in an isolated conductor will cause a current flow. This flow sets up an electric field that opposes the original electric field, halting the motion of the charges on a nanosecond time scale for metersized conductors. For this reason, an isolated conductor will have no static electric field inside it, and will have a reduced electric field near it. This conclusion does not apply to a conductor whose ends are connected to an external circuit. In a circuit, a rod (or wire) can conduct current indefinitely.
Charge Distribution on a Conducting Shell  2
A positive charge is kept (fixed) offcenter inside a fixed spherical conducting shell that is electrically neutral, and the charges in the shell are allowed to reach electrostatic equilibrium. The large positive charge inside the shell is roughly 16 times that of the smaller charges shown on the inner and outer surfaces of the spherical shell.
Part A
Which of the following figures best represents the charge distribution on the inner and outer walls of the shell?
Hint 1. Symmetry inside shell
The field inside the conductor must be zero. The charge inside the shell is offcenter, and hence the charge on the inner surface of the shell will arrange itself asymmetrically to cancel the field of the large positive charge.
Hint 2. Symmetry outside shell
The field inside the conductor must be zero. To the charges on the outer surface, it is as if the inside of the conductor were completely neutral. Thus, the charges on the outer surface will feel no force other than their own mutual repulsion, and will therefore have no preferred direction.
ANSWER:
1
2
3
4
5
Correct
PhET Tutorial: Charges and Electric Fields
Learning Goal:
To understand the spatial distribution of the electric field for a variety of simple charge configurations.
For this problem, use the PhET simulation Charges and Fields. This simulation allows you to place multiple positive and negative pointcharges in any configuration and look at the resulting electric field.
Start the simulation. You can click and drag positive charges (red) or negative charges (blue) into the main screen. If you select Show Efield in the green menu, red arrows will appear, showing the direction of the electric field. Faint red arrows indicate that the electric field is weaker than at locations where the arrows are brighter (this simulation does not use arrow length as a measure of field magnitude).
Feel free to play around with the simulation. When you are done, click Clear All before beginning Part A.
Part A
Select Show Efield and grid in the green menu. Drag one positive charge and place it near the middle of the screen, right on top of two intersecting bold grid lines. You should see something similar to the figure below.
ANSWER:
The electric field produced by the positive charge
is directed radially away from the charge at all locations near the charge.
is directed radially toward the charge at all locations near the charge.
wraps circularly around the positive charge.
Correct
This means that another positive charge, if placed near the original charge, would experience a force directed radially away from the original charge.
Part B
Now, letâ™s look at how the distance from the charge affects the magnitude of the electric field. Select Show numbers on the green menu, and then click and drag one of the orange EField Sensors. You will see the magnitude of the electric field given in units of V/m (volts per meter, which is the same as newtons per coulomb).
Place the EField Sensor 1 away from the positive charge (1 is two bold grid lines away if going in a horizontal or vertical direction), and look at the resulting field strength.
Consider the locations to the right, left, above, and below the positive charge, all 1 away.
ANSWER:
For these four locations, the magnitude of the electric field is
greatest to the left of the charge.
the same.
greatest to the right of the charge.
greatest below the charge.
greatest above the charge.
Correct
This result implies that the strength of the electric field due to one point charge depends solely on the distance away from the charge. Mathematically, we say the electric field is spherically symmetric.
Part C
What is the magnitude of the electric field 1 m away from the positive charge compared to the magnitude of the electric field 2 m away?
Hint 1. How to approach the problem
Use an EField Sensor to determine the field strength both at 1 away and at 2 away from the charge. Then, take the ratio of the two field strengths.
ANSWER:
The magnitude of the electric field 1 away from the positive charge is
four times
two times
onehalf
equal to
onequarter
the magnitude of the electric field 2 away.
Correct
The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is proportional to the inverse of the distance squared (, where is the distance from the charge). You should verify this by looking at the field strength 3 or 4 meters away. This is consistent with Coulombâ™s law, which states that the magnitude of the force between two charged particles is .
Part D
If the field strength is = 9 a distance of 1 from the charge, what is the field strength a distance of 3 from the charge?
Hint 1. How to approach the problem
The magnitude of the electric field is inversely proportional to distance squared (). So if the distance is increased by a factor of three, the field strength must decrease by a factor of three squared. You could use the simulation to make a measurement (you might have to drag the charge away from the center so you have enough room to get 3 away).
ANSWER:
=
1
Correct
Correct. Since , if the distance is increased by a factor of three, the electric field is decreased by a factor of nine.
Part E
Remove the positive charge by dragging it back to the basket, and drag a negative charge (blue) toward the middle of the screen. Determine how the electric field is different from that of the positive charge.
Which statement best describes the differences in the electric field due to a negative charge as compared to a positive charge?
ANSWER:
Nothing changes; the electric field remains directed radially outward, and the electric field strength doesnâ™t change.
The electric field changes direction (now points radially inward), but the electric field strength does not change.
The electric field changes direction (now points radially inward), and the magnitude of the electric field decreases at all locations.
Correct
The electric field is now directed toward the negative charge, but the field strength doesnâ™t change. The electric field of a point charge is given by . Because of the sign of the charge, the field produced by a negative charge is directed opposite to that of a positive charge but the magnitude of the field is the same.
Part F
Now, remove the negative charge, and drag two positive charges, placing them 1 apart, as shown below.
Letâ™s look at the resulting electric field due to both charges. Recall that the electric field is a vector, so the net electric field is the vector sum of the electric fields due to each of the two charges.
Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)?
ANSWER:
The electric field is zero at any location along a vertical line going through the point directly between the two charges.
The electric field is roughly zero near the midpoint of the two charges.
The electric field is nonzero everywhere on the screen.
Correct
Directly between the two charges, the electric fields produced by each charge are equal in magnitude and point in opposite directions, so the two vectors add up to zero.
Part G
Consider a point 0.5 above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 . What is the magnitude of the total electric field due to both charges at this location?
ANSWER:
25
zero
36
Correct
Notice that this number is less than twice the magnitude of the field due to each charge. This occurs because the horizontal components of the electric field due to each charge exactly cancel out (add to zero). Only the vertical components of the fields add together.
Part H
Make an electric dipole by replacing one of the positive charges with a negative charge, so the final configuration looks like the figure shown below.
ANSWER:
The electric field at the midpoint is
directed to the right.
zero.
directed to the left.
Correct
The electric field due to the positive charge is directed to the right, as is the electric field due to the negative charge. So the net electric field, which is the sum of these two fields, is also to the right.
Part I
Make a small dipole by bringing the two charges very close to each other, where they are barely touching. The midpoint of the two charges should still be on one of the grid point intersections (see figure below).
Measure the strength of the electric field 0.5 directly above the midpoint as well as 1 directly above. Does the strength of the electric field decrease as 1 over distance squared ()?
Hint 1. How to approach the problem
If the strength of the field is decreasing as , then the ratio of the magnitudes of the electric field measured at two distances, say 0.5 away and 1 away, would be
.
Compare this value to the value you measure with an EField Sensor.
ANSWER:
No, it decreases less quickly with distance.
Yes, it does.
No, it decreases more quickly with distance.
Correct
In fact, it turns out that the strength of the electric field decreases roughly as ! So the field 1 above the midpoint is roughly eight times weaker than at 0.5 above the midpoint. The important lesson here is that, in general, a distribution of charges produces an electric field that is very different from that of a single charge.
Part J
Make a long line of positive charges, similar to that shown in the figure below. Try to place all of the charges centered along a horizontal grid line. Feel free to look at the electric field, as it is interesting.
Measure the strength of the electric field 1 directly above the middle as well as 2 directly above. Does the strength of the electric field decrease as 1 over distance squared ()?
ANSWER:
No, it decreases more quickly with distance.
No, it decreases less quickly with distance.
Yes, it does.
Correct
In fact, it turns out that the strength of the electric field decreases roughly as . So the field 1 m above the midpoint is roughly half the strength at 0.5 . This is another example showing that a distribution of charges produces an electric field that is very different from that of a single charge.
PhET Interactive Simulations
University of Colorado
http://phet.colorado.edu
Video Tutor: Charged Rod and Aluminum Can
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point.
Part A
Consider the situation in the figure below, where two charged rods are placed a distance on either side of an aluminum can. What does the can do?
Hint 1. How to approach the problem.
This problem asks you to think about induced charge on the surface of an object and the resulting polarization force.
To get started, draw a diagram. Draw the induced surface charges on the outside of the can. Next, draw a force diagram (freebody diagram) to show the forces exerted on the can. Aluminum is a conductor.
ANSWER:
Rolls to the left
Stays still
Rolls to the right
Correct
The positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can. However, the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod. The net force acting on the can is zero.
Part B
Now, consider the situation shown in the figure below. What does the can do?
ANSWER:
Rolls to the left
Rolls to the right
Stays still
Correct
The polarization force is always attractive, so the can does not move.
Part C
Using the setup from the first question, imagine that you briefly touch the negatively charged rod to the can. You then hold the two rods at equal distances on either side of the can. What does the can do?
Hint 1. How to approach the problem.
This problem asks you to consider what happens to a conductor after being touched by a charged object. What charge will the can have after being touched?
ANSWER:
Does not move
Rolls toward the positively charged rod
Rolls away from the positively charged rod
Correct
The can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.
A Test Charge Determines Charge on Insulating and Conducting Balls
Learning Goal:
To understand the electric force between charged and uncharged conductors and insulators.
When a test charge is brought near a charged object, we know from Coulomb's law that it will experience a net force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.
Consider three plastic balls (A, B, and C), each carrying a uniformly distributed charge equal to either +,  or zero, and an uncharged copper ball (D). A positive test charge (T) experiences the forces shown in the figure when brought very near to the individual balls. The test charge T is strongly attracted to A, strongly repelled from B, weakly attracted to C, and strongly attracted to D.
Assume throughout this problem that the balls are brought very close together.
Part A
What is the nature of the force between balls A and B?
Hint 1. What is the net charge on ball A?
Since the test charge is positively charged, and there is a strongly attractive force between ball A and the test charge, what must be the nature of the net charge of ball A?
ANSWER:
positive
negative
zero
Correct
Hint 2. What is the net charge on ball B?
Since the test charge is positively charged, and there is a strongly repulsive force between ball B and the test charge, what must be the nature of the net charge of ball B?
ANSWER:
positive
negative
zero
Correct
ANSWER:
strongly attractive
strongly repulsive
weakly attractive
neither attractive nor repulsive
Correct
Part B
What is the nature of the force between balls A and C?
Hint 1. What is the charge on ball C?
Recall that ball C is composed of insulating material, which means that it can be polarized, but the charges inside are otherwise not free to move around inside the ball. Since the test charge experiences only a weak force due to ball C, if we compare to ball A we conclude that the charge on ball C must be
ANSWER:
+

zero
Correct
ANSWER:
strongly attractive
strongly repulsive
weakly attractive
neither attractive nor repulsive
Correct
Recall that ball C is composed of insulating material, which can be polarized in the presence of an external charged object such as ball A. Once polarized, there will be a weak attraction between balls A and C, because the positive and negative charges in ball C are at slightly different average distances from ball A. If ball C had a very small negative charge the test charge would have the same response (weakly attractive) but it would have a weak repulsive interaction with ball A. However, a smaller negative charge is not one of the options.
Part C
What is the nature of the force between balls A and D?
Hint 1. What are the surface charges on ball D?
Recall that copper is a conductor, in which charges can freely flow. When ball D is brought close to ball A, what will be the nature of the surface charge density on the side of ball D that is closest to ball A?
ANSWER:
positive
negative
zero
Correct
The negatively charged ball A (see Part A) will exert an attractive force on the positive charges in ball D and a repulsive force on the negative charges (namely, the electrons). Since ball D is made of copper, which is a conductor, the electrons will be repelled from negatively charged ball A and will migrate to the side of ball D farthest from ball A. The deficit of electrons on the side of ball D that is closest to ball A results in a positive net surface charge density on that side of ball D. Because the positive charge on ball D is much closer to ball A than the negative charge, the attractive force that ball A experiences due to the positive charges on ball D is stronger than the repulsive force ball A experiences due to the negative charges on ball D.
ANSWER:
attractive
repulsive
neither attractive nor repulsive
Correct
Part D
What is the nature of the force between balls D and C?
ANSWER:
attractive
repulsive
neither attractive nor repulsive
Correct
Because the test charge T is neither strongly attracted to nor repelled from ball C, ball C must have zero net charge. Since ball D also has zero net charge, there will not be any force between the two balls.
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 yaxis 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 yaxis at position . What is the new net force on particle 0, from particle 1 and particle 2?
Express your answer (a vector) using any or all of , , , , , , , , and .
ANSWER:
=
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, with positive charge , fixed in the yzplane at . What is the net force on particle 0 due solely to this charge?
Express your answer (a vector) using , , , , , , and . Include only the force caused by particle 3.
Hint 1. Find the magnitude of force from particle 3
What is the magnitude of the force on particle 0 from particle 3, fixed at ?
Express your answer using , , , .
Hint 1. Distance to particle 3
Use the Pythagorean theorem to find the straight line distance between the origin and .
ANSWER:
=
Hint 2. Vector components
The force vector points from to . Because is symmetrically located between the yaxis and the zaxis, the angle between , the unit vector pointing from particle 3 to particle 0, and the yaxis is radians. You have already calculated the magnitude of the vector above. Now break up the force vector into its y and z components.
ANSWER:
=
Correct
The Trajectory of a Charge in an Electric Field
An charge with mass and charge is emitted from the origin, . A large, flat screen is located at . There is a target on the screen at y position , where . In this problem, you will examine two different ways that the charge might hit the target. Ignore gravity in this problem.
Part A
Assume that the charge is emitted with velocity in the positive x direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive y direction. What should the magnitude of the electric field be if the charge is to hit the target on the screen?
Express your answer in terms of , , , , and .
Hint 1. How to approach the problem
Once you determine the force on the charge due to the electric field, this becomes a standard twodimensional kinematics problem. To solve the problem, first determine the equations of motion in both the x and y directions. Then use the fact that at some final time you know that the position of the charge is to obtain two equations in terms of the two unknowns and . Eliminate and solve for .
Hint 2. Find the equation of motion in the x direction
Find an expression for , the charge's x position as a function of time.
Express your answer in terms of as well as any of the given variables and constants.
Hint 1. Find the force in the x direction
What net force does the charge experience in the x direction?
Hint 1. Formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
ANSWER:
=
0
Hint 2. A helpful kinematic equation
Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is
.
ANSWER:
=
Hint 3. Find the equation of motion in the y direction
Find an expression for , the charge's y position as a function of time.
Express your answer in terms of as well as any of the given variables and constants.
Hint 1. Find the force in the y direction
What is the net force acting on the charge in the y direction?
Express your answer in terms of the given variables and constants.
Hint 1. Formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
ANSWER:
=
Hint 2. A helpful kinematic equation
Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is
.
ANSWER:
=
Hint 4. Combine Your Results
At some final time , you have and . Starting with these two equations, eliminate and solve for .
Hint 5. Find
Use the equation for the motion of the charge in the x direction to find .
Express your answer in terms of the variables and .
ANSWER:
=
ANSWER:
=
Correct
Part B
Now assume that the charge is emitted with velocity in the positive y direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive x direction. What should the magnitude of the electric field be if the charge is to hit the target on the screen?
Express your answer in terms of , , , , and .
Hint 1. How to approach the problem
Just as in the previous part, once you determine the force on the charge due to the electric field, this becomes a standard twodimensional kinematics problem. To solve the problem, first determine the equations of motion in both the x and y directions. Then use the fact that at some final time you know that the position of the charge is to obtain two equations in terms of the two unknowns and . Eliminate and solve for .
Hint 2. Find the equation of motion in the y direction
Find an expression for , the charge's y position as a function of time.
Express your answer in terms of as well as any of the given variables and constants.
Hint 1. Find the force in the y direction
What net force does the charge experience in the y direction?
Hint 1. Formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
ANSWER:
=
0
Hint 2. A helpful kinematic equation
Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is
.
ANSWER:
=
Hint 3. Find the equation of motion in the x direction
Find an expression for , the charge's x position as a function of time.
Express your answer in terms of as well as any of the given variables and constants.
Hint 1. Find the force in the x direction
What is the net force acting on the charge in the x direction?
Express your answer in terms of the given variables and constants.
Hint 1. Formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
ANSWER:
=
Hint 2. A helpful kinematic equation
Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is
.
ANSWER:
=
Hint 4. Combine your results
At some final time , you have and . Starting with these two equations, eliminate and solve for .
Hint 5. Find
Use the equation for the motion of the charge in the y direction to find .
Express your answer in terms of the variables and .
ANSWER:
=
ANSWER:
=
Correct
The equations of motion for this part are identical to the equations of motion for the previous part, with and interchanged. Thus it is no surprise that the answers to the two parts are also identical, with and interchanged.
Charged Ring
Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius and positive charge distributed evenly along its circumference.
Part A
What is the direction of the electric field at any point on the z axis?
Hint 1. How to approach the problem
Approach 1
In what direction is the field due to a point on the ring? Add to this the field from a point on the opposite side of the ring. In what direction is the net field? What if you did this for every pair of points on opposite sides of the ring?
Approach 2
Consider a general electric field at a point on the z axis, i.e., one that has a z component as well as a component in the xy plane. Now imagine that you make a copy of the ring and rotate this copy about its axis. As a result of the rotation, the component of the electric field in the xy plane will rotate also. Now you ask a friend to look at both rings. Your friend wouldn't be able to tell them apart, because the ring that is rotated looks just like the one that isn't. However, they have the component of the electric field in the xy plane pointing in different directions! This apparent contradiction can be resolved if this component of the field has a particular value. What is this value?
Does a similar argument hold for the z component of the field?
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?
Use in your answer, where .
Hint 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.
Use in your answer, where . You may also use some or all of the variables , , and .
ANSWER:
=
Hint 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 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.
ANSWER:
=
Hint 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
.
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
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 .
Express your answer in terms of given charges, dimensions, and constants.
Hint 1. Simple harmonic motion
Recall the nature of simple harmonic motion of an object attached to a spring. Newton's second law for the system states that
, leading to oscillation at a frequency of
(here, the prime on the symbol representing the spring constant is to distinguish it from ). The solution to this differential equation is a sinusoidal function of time with angular frequency . Write an analogous equation for the ball near the charged ring in order to find the term.
Hint 2. Find the force on the charge
What is , the z component of the force on the ball on the ball at the point ? Use the approximation .
Express your answer in terms of , , , , and .
Hint 1. A formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
Therfore, in particular,
.
You have already found in Part B. Use that expression in the equation above to find an expression for the z component of the force on the ball at the point . Don't forget to use the approximation given.
ANSWER:
=
ANSWER:
=
Correct
The Electric Field Produced by a Finite Charged Wire
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 1. Consider opposite ends of the wire
Consider the sum of the electric fields produced by the two equivalent charge elements at opposite ends of the wire. What is the direction of the resulting electric field?
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 .
Express your answer in terms of , , , and .
Hint 1. How to approach the problem
To compute the field at point , divide the rod into infinitesimal segments and find the electric field produced by each. Then integrate to find the total electric field. The electric field due to an arbitrary infinitesimal segment of wire has both x and y components. You could integrate these two components separately, but because of symmetry, you already know that the total electric field points only in the y direction. So you need to work only with the y component of the field.
Hint 2. Find the field due to an infinitesimal segment
Find , the y component of the electric field produced at point by the infinitesimal segment of the wire between and .
Express your answer in terms of , , , , and .
Hint 1. Find the total electric field
Find the magnitude of the electric field at point produced by the infinitesimal wire segment between and .
Express your answer in terms of , , , , and the constant .
Hint 1. Formula for electric field due to a point charge
The electric field at position away from an infinitesimal charge element is given by .
Hint 2. Find
What is the distance from an infinitesimal segment of wire at position to the point ?
Express your answer in terms of and .
ANSWER:
=
Hint 3. Find an expression for
The wire has charge per unit length , and the length of an infinitesimal charge element is . Write an expression for , the total charge on the infinitesimal element .
Express your answer in terms of and .
ANSWER:
=
ANSWER:
=
Hint 2. Find the y component of
What is , the y component of the electric field at point due to an infinitesimal segment of wire located at position ? Assume that the magnitude of the electric field at point due to the infinitesimal wire segment is .
Express your answer in terms of , , and .
ANSWER:
=
ANSWER:
=
Hint 3. A necessary integral
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 apart. The dipole has moment of inertia about its center of mass. The dipole is released from angle , and it is allowed to rotate freely.
Part A
What is , the magnitude of the dipole's angular velocity when it is pointing along the y axis?
Express your answer in terms of quantities given in the problem introduction.
Hint 1. How to approach the problem
Because there is no dissipation (friction, air resistance, etc.), you can solve this problem using conservation of energy. When the dipole is released from rest, it has potential energy but no kinetic energy. When the dipole is aligned with the y axis, it is rotating, and therefore has both kinetic and potential energy. The sum of potential and kinetic energy will remain constant.
Hint 2. Find the potential energy
Find the dipole's potential energy due to its interaction with the electric field as a function of the angle that the dipole's positive end makes with the positive y axis. Define the potential energy to be zero when the dipole is oriented perpendicular to the field: .
Express your answer in terms of , , , and .
Hint 1. The formula for the potential energy of a dipole
The general formula for the potential energy of an electric dipole with dipole moment in the presence of a uniform electric field is .
Hint 2. The dipole moment
The dipole moment of the electric dipole , when it makes an angle with the positive y axis can be written as
.
ANSWER:
=
Hint 3. Find the total energy at the moment of release
Find , the total energy (kinetic plus potential) at the moment the dipole is released from rest at angle with respect to the y axis. Use the convention that the potential energy is zero when the dipole is oriented perpendicular to the field: .
Express your answer in terms of some or all of the variables , , , and .
ANSWER:
=
Hint 4. Find the total energy when
Find an expression for , the total energy (kinetic plus potential) at the moment when the dipole is aligned with the y axis. Use the convention that the potential energy is zero when the dipole is oriented perpendicular to the field: .
Express your answer in terms of quantities given in the problem introduction and .
Hint 1. What is kinetic energy as a function of angular velocity?
What is the kinetic energy of a body rotating with angular velocity around an axis about which the moment of inertia is ?
ANSWER:
=
ANSWER:
=
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?
Express your answer in terms of and quantities given in the problem introduction.
Hint 1. How to approach the problem
The equation of motion for a simple harmonic oscillator can always be written in the standard form . To solve this problem, you need to write the equation of motion for the dipole in the standard form with replaced by the angular variable . This will allow you to read off the expression for , which has a simple relationship to the period of oscillation. (Note: Here, the variable does not represent the angular velocity of the dipole; rather, it denotes the frequency of the dipole's oscillation.) Start with the angular analogue of Newton's second law: . Recall that , the angular acceleration, is equal to the second derivative of , just as linear acceleration is equal to the second derivative of position.
Hint 2. Compute the torque
What is the magnitude of the torque that the electric field exerts about the center of mass of the dipole when the dipole is oriented at an angle with respect to the electric field?
Express the magnitude of the torque in terms of quantities given in the problem introduction and .
Hint 1. Formula for torque on a dipole
The torque on a dipole with dipole moment in an electric field is given by . Alternatively, the torque can be related to the potential energy by .
Hint 2. The dipole moment
When it makes an angle with the positive y axis, the dipole moment of the electric dipole can be written as
.
ANSWER:
=
Hint 3. The smallangle approximation
Because is small, you can apply the smallangle approximation to the expression for torque, and take the torque to be .
Up to this point we have been interested only in the magnitude of the torque. Now let's think about the direction. After all, torque is a vector quantity. For a system to oscillate, the torque must be a restoring torque; that is, the torque and the (small) angular displacement must be in opposite directions. (Recall that small angular displacements can be treated as vectors, since they obey vector addition, while large angles do not.) If you did the vector algebra carefully, you would find that the correct vector equation is
.
For future purposes we will write this as , keeping in mind that now represents the component of in the direction, rather than the magnitude of .
Hint 4. Find the oscillation frequency
Putting together what you have so far yields
.
Compare this to the standard form for a simple harmonic oscillator to obtain the oscillation frequency for the motion of the dipole.
Express your answer in terms of quantities given in the problem introduction.
ANSWER:
=
Hint 5. The relationship between (angular) oscillation frequency and period
The relationship between , the angular oscillation frequency of the dipole, and the period of oscillation is given by
.
ANSWER:
=
Correct
Charging an Insulator
This problem explores the behavior of charge on realistic (i.e. nonideal) insulators. We take as an example a long insulating rod suspended by insulating wires. Assume that the rod is initially electrically neutral. For convenience, we will refer to the left end of the rod as end A, and the right end of the rod as end B . In the answer options for this problem, "weakly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two balls, one of which is charged, and the other acquires a small induced charge". An attractive/repulsive force greater than this should be classified as "strongly attracted/repelled".
Part A
A small metal ball is given a negative charge, then brought near (i.e., within a few millimeters) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Select the expected behavior.
Hint 1. What is an insulator?
An insulator is a material which does not allow charge/current to flow easily through it.
Hint 2. Charge at end A
Keeping in mind that like charges repel each other, and opposite charges attract each other, what sort of charge is induced at end A of the (nonideal) insulating rod?
ANSWER:
A small positive charge
A small negative charge
ANSWER:
strongly repelled
strongly attracted
weakly attracted
weakly repelled
neither attracted nor repelled
Correct
Currently, you can think of this in the following way: When the sphere is brought near the rod, a positive charge is induced at end A (and correspondingly, end B acquires a negative induced charge). This means that some charge must have flowed from A to B. Since charge flow is inhibited in an insulator, the induced charges are typically small. Later you will learn how to model insulators more accurately and formulate a slightly more accurate argument.
Now consider what happens when the small metal ball is repeatedly given a negative charge and then brought into contact with end A of the rod
Part B
After several contacts with the charged ball, how is the charge on the rod arranged?
Select the best description.
Hint 1. What is an insulator?
An insulator is a material which does not allow charge/current to flow easily through it.
ANSWER:
positive charge on end B and negative charge on end A
negative charge spread evenly on both ends
negative charge on end A with end B remaining almost neutral
positive charge on end A with end B remaining almost neutral
none of the above
Correct
When the sphere is touched to end A, some of its negative charge will be deposited there. However, since charge cannot flow easily through an insulator, most of this charge will just sit at end A and will not distribute itself over the rod, as it would if the rod was a conductor.
Part C
How does end A of the rod react when the ball approaches it after it has already made several contacts with the rod, such that a fairly large charge has been deposited at end A?
Select the expected behavior.
ANSWER:
strongly repelled
strongly attracted
weakly attracted
weakly repelled
neither attracted nor repelled
Correct
More on insulators
You may have learnt that any material is made of atoms, which in turn consist of a nucleus and electrons. In the atoms of some materials, some of the electrons are "bound" to the nucleus very weakly, which leaves them free to move around the volume of the material. Such electrons are called "free" electrons, and such materials are called conductors, because the charge (i.e. electrons) can move around easily. In insulators, all the electrons in the atom are bound quite tightly to the nucleus, i.e. there are no free electrons available to move through the insulator.
Exercise 21.19
Three point charges are arranged along the xaxis. Charge = +3.00 is at the origin, and charge = 5.00 is at = 0.200 . Charge = 8.00 .
Part A
Where is located if the net force on is 7.00 in the direction?
ANSWER:
=
0.144
Correct
Exercise 21.38
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.56distant from the first, in a time interval of 1.43×10−6 .
Part A
Find the magnitude of the electric field.
ANSWER:
=
159
Correct
Part B
Find the speed of the proton when it strikes the negatively charged plate.
ANSWER:
=
2.18×104
Correct
Exercise 21.53
Positive electric charge is distributed along the yaxis with charge per unit length .
Part A
Consider the case where charge is distributed only between the points and . For points on the axis, find the component of the electric field as a function of .
Express your answer in terms of the variables , , and appropriate constants.
ANSWER:
=
Part B
Consider instead the case where charge is distributed along the entire yaxis with the same charge per unit length . Find the component of the electric field as a function of for values.
ANSWER:
=
Exercise 21.57
Point charges 4.00 and 4.00 are separated by a distance of 3.80 , forming an electric dipole.
Part A
Find the magnitude of the electric dipole moment.
ANSWER:
=
1.52×10−11
Correct
Part B
Find the direction of the electric dipole moment?
ANSWER:
from to
from to
Correct
Part C
The charges are in a uniform electric field whose direction makes an angle of 36.6 with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.50×10−9 ?
ANSWER:
=
828
Correct
Problem 21.82
Two tiny spheres of mass = 9.00 carry charges of equal magnitude, 72.0 , but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 . When a horizontal uniform electric field that is directed to the left is turned on, the spheres hang at rest with the angle between the strings equal to in the following figure.
Part A
Which ball (the one on the right or the one on the left) has positive charge?
ANSWER:
Which ball (the one on the right or the one on the left) has positive charge?
The one on the right
The one on the left
Part B
What is the magnitude of the field?
Express your answer with the appropriate units.
ANSWER:
=
Try Again; 5 attempts remaining
Score Summary:
Your score on this assignment is 85.4%.
You received 13.67 out of a possible total of 16 points.
Chapter 21 Homework Assignment
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Chapter 21 Homework Assignment
Due: 11:59pm on Sunday, September 9, 2012
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy
The Electric Field inside a Conductor
Learning Goal:
To understand how the charges within a conductor respond to an externally applied electric field.
To illustrate the behavior of charge inside conductors, consider a long conducting rod that is suspended by insulating strings (see the figure). Assume that the rod is initially electrically neutral, and that it remains so for this discussion. The rod is positioned along the x axis, and an external electric field that points in the positive x direction (to the right) can be applied to the rod and the surrounding region. The atoms in the rod are composed of positive nuclei (indicated by plus signs) and negative electrons (indicated by minus signs). Before application of the electric field, these atoms were distributed evenly throughout the rod.
Part A
What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem introduction is applied? (Ignore internal fields in the rod for the moment.)
Hint 1. Formula for the force on a charge in an electric field
The force on a charge in an electric field is given by
.
ANSWER:
Both electrons and nuclei experience a force to the right.
The nuclei experience a force to the right and the electrons experience a force to the left.
The electrons experience a force to the left but the nuclei experience no force.
The electrons experience no force but the nuclei experience a force to the right.
Correct
Part B
What is the motion of the negative electrons and positive atomic nuclei caused by the external field?
Hint 1. How to approch this part
Newton's 2nd Law tells you that an object at rest will move in the direction of the force applied on it.
Hint 2. Masses and charges of nuclei and electrons
A nucleus contains as many protons as the atom does electrons. So if the atom has N electrons, the nucleus will contain N protons. This means that the force on the nucleus will be N times as much as that on the electron. This N is of the order of 10100.
However, the mass of a nucleus is roughly 2N times the mass of a proton, since it contains both protons and neutrons. Each proton itself weighs about 1836 times as much as an electron! So a typical nucleus really does weigh a lot more than an electron.
Given this information, how would the distance moved by a nucleus compare with that moved by an electron?
ANSWER:
Both electrons and nuclei move to the right.
The nuclei move to the right and the electrons move to the left through equal distances.
The electrons move to the left and the nuclei are almost stationary.
The electrons are almost stationary and the nuclei move to the right.
Correct
The nuclei of the atoms of a conducting solid remain almost in their places in the crystal lattice, while the electrons relatively move a lot. In an insulator, the electrons are constrained to stay with their atoms (or molecules), and at most, the charge distribution is displaced slightly.
The motion of the electrons due to the external electric field constitutes an electric current. Since the negatively charged electrons are moving to the left, the current, which is defined as the "flow" of positive charge, moves to the right.
Part C
Imagine that the rightward current flows in the rod for a short time. As a result, what will the net charge on the right and left ends of the rod become?
Hint 1. How to approach this part
Remember that the rod as a whole must remain electrically neutral even if the charges are redistributed. This is because applying an electric field does not change the charge on the rod, only redistributes it.
ANSWER:
left end negative and right end positive
left end negative and right end negative
left end negative and right end nearly neutral
left end nearly neutral and right end positive
both ends nearly neutral
Correct
Given that the positively charged nuclei do not move, why does the right end of the rod become positively charged? The reason is that some electrons have moved to the left end, leaving an excess of stationary nuclei at the right end.
Part D
The charge imbalance that results from this movement of charge will generate an additional electric field near the rod. In what direction will this field point?
Hint 1. Direction of the electric field
The electric field point away from positive charges and towards negative ones.
ANSWER:
It will point to the right and enhance the initial applied field.
It will point to the left and oppose the initial applied field.
Correct
An electric field that exists in an isolated conductor will cause a current flow. This flow sets up an electric field that opposes the original electric field, halting the motion of the charges on a nanosecond time scale for metersized conductors. For this reason, an isolated conductor will have no static electric field inside it, and will have a reduced electric field near it. This conclusion does not apply to a conductor whose ends are connected to an external circuit. In a circuit, a rod (or wire) can conduct current indefinitely.
Charge Distribution on a Conducting Shell  2
A positive charge is kept (fixed) offcenter inside a fixed spherical conducting shell that is electrically neutral, and the charges in the shell are allowed to reach electrostatic equilibrium. The large positive charge inside the shell is roughly 16 times that of the smaller charges shown on the inner and outer surfaces of the spherical shell.
Part A
Which of the following figures best represents the charge distribution on the inner and outer walls of the shell?
Hint 1. Symmetry inside shell
The field inside the conductor must be zero. The charge inside the shell is offcenter, and hence the charge on the inner surface of the shell will arrange itself asymmetrically to cancel the field of the large positive charge.
Hint 2. Symmetry outside shell
The field inside the conductor must be zero. To the charges on the outer surface, it is as if the inside of the conductor were completely neutral. Thus, the charges on the outer surface will feel no force other than their own mutual repulsion, and will therefore have no preferred direction.
ANSWER:
1
2
3
4
5
Correct
PhET Tutorial: Charges and Electric Fields
Learning Goal:
To understand the spatial distribution of the electric field for a variety of simple charge configurations.
For this problem, use the PhET simulation Charges and Fields. This simulation allows you to place multiple positive and negative pointcharges in any configuration and look at the resulting electric field.
Start the simulation. You can click and drag positive charges (red) or negative charges (blue) into the main screen. If you select Show Efield in the green menu, red arrows will appear, showing the direction of the electric field. Faint red arrows indicate that the electric field is weaker than at locations where the arrows are brighter (this simulation does not use arrow length as a measure of field magnitude).
Feel free to play around with the simulation. When you are done, click Clear All before beginning Part A.
Part A
Select Show Efield and grid in the green menu. Drag one positive charge and place it near the middle of the screen, right on top of two intersecting bold grid lines. You should see something similar to the figure below.
ANSWER:
The electric field produced by the positive charge
is directed radially away from the charge at all locations near the charge.
is directed radially toward the charge at all locations near the charge.
wraps circularly around the positive charge.
Correct
This means that another positive charge, if placed near the original charge, would experience a force directed radially away from the original charge.
Part B
Now, letâ™s look at how the distance from the charge affects the magnitude of the electric field. Select Show numbers on the green menu, and then click and drag one of the orange EField Sensors. You will see the magnitude of the electric field given in units of V/m (volts per meter, which is the same as newtons per coulomb).
Place the EField Sensor 1 away from the positive charge (1 is two bold grid lines away if going in a horizontal or vertical direction), and look at the resulting field strength.
Consider the locations to the right, left, above, and below the positive charge, all 1 away.
ANSWER:
For these four locations, the magnitude of the electric field is
greatest to the left of the charge.
the same.
greatest to the right of the charge.
greatest below the charge.
greatest above the charge.
Correct
This result implies that the strength of the electric field due to one point charge depends solely on the distance away from the charge. Mathematically, we say the electric field is spherically symmetric.
Part C
What is the magnitude of the electric field 1 m away from the positive charge compared to the magnitude of the electric field 2 m away?
Hint 1. How to approach the problem
Use an EField Sensor to determine the field strength both at 1 away and at 2 away from the charge. Then, take the ratio of the two field strengths.
ANSWER:
The magnitude of the electric field 1 away from the positive charge is
four times
two times
onehalf
equal to
onequarter
the magnitude of the electric field 2 away.
Correct
The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is proportional to the inverse of the distance squared (, where is the distance from the charge). You should verify this by looking at the field strength 3 or 4 meters away. This is consistent with Coulombâ™s law, which states that the magnitude of the force between two charged particles is .
Part D
If the field strength is = 9 a distance of 1 from the charge, what is the field strength a distance of 3 from the charge?
Hint 1. How to approach the problem
The magnitude of the electric field is inversely proportional to distance squared (). So if the distance is increased by a factor of three, the field strength must decrease by a factor of three squared. You could use the simulation to make a measurement (you might have to drag the charge away from the center so you have enough room to get 3 away).
ANSWER:
=
1
Correct
Correct. Since , if the distance is increased by a factor of three, the electric field is decreased by a factor of nine.
Part E
Remove the positive charge by dragging it back to the basket, and drag a negative charge (blue) toward the middle of the screen. Determine how the electric field is different from that of the positive charge.
Which statement best describes the differences in the electric field due to a negative charge as compared to a positive charge?
ANSWER:
Nothing changes; the electric field remains directed radially outward, and the electric field strength doesnâ™t change.
The electric field changes direction (now points radially inward), but the electric field strength does not change.
The electric field changes direction (now points radially inward), and the magnitude of the electric field decreases at all locations.
Correct
The electric field is now directed toward the negative charge, but the field strength doesnâ™t change. The electric field of a point charge is given by . Because of the sign of the charge, the field produced by a negative charge is directed opposite to that of a positive charge but the magnitude of the field is the same.
Part F
Now, remove the negative charge, and drag two positive charges, placing them 1 apart, as shown below.
Letâ™s look at the resulting electric field due to both charges. Recall that the electric field is a vector, so the net electric field is the vector sum of the electric fields due to each of the two charges.
Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)?
ANSWER:
The electric field is zero at any location along a vertical line going through the point directly between the two charges.
The electric field is roughly zero near the midpoint of the two charges.
The electric field is nonzero everywhere on the screen.
Correct
Directly between the two charges, the electric fields produced by each charge are equal in magnitude and point in opposite directions, so the two vectors add up to zero.
Part G
Consider a point 0.5 above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 . What is the magnitude of the total electric field due to both charges at this location?
ANSWER:
25
zero
36
Correct
Notice that this number is less than twice the magnitude of the field due to each charge. This occurs because the horizontal components of the electric field due to each charge exactly cancel out (add to zero). Only the vertical components of the fields add together.
Part H
Make an electric dipole by replacing one of the positive charges with a negative charge, so the final configuration looks like the figure shown below.
ANSWER:
The electric field at the midpoint is
directed to the right.
zero.
directed to the left.
Correct
The electric field due to the positive charge is directed to the right, as is the electric field due to the negative charge. So the net electric field, which is the sum of these two fields, is also to the right.
Part I
Make a small dipole by bringing the two charges very close to each other, where they are barely touching. The midpoint of the two charges should still be on one of the grid point intersections (see figure below).
Measure the strength of the electric field 0.5 directly above the midpoint as well as 1 directly above. Does the strength of the electric field decrease as 1 over distance squared ()?
Hint 1. How to approach the problem
If the strength of the field is decreasing as , then the ratio of the magnitudes of the electric field measured at two distances, say 0.5 away and 1 away, would be
.
Compare this value to the value you measure with an EField Sensor.
ANSWER:
No, it decreases less quickly with distance.
Yes, it does.
No, it decreases more quickly with distance.
Correct
In fact, it turns out that the strength of the electric field decreases roughly as ! So the field 1 above the midpoint is roughly eight times weaker than at 0.5 above the midpoint. The important lesson here is that, in general, a distribution of charges produces an electric field that is very different from that of a single charge.
Part J
Make a long line of positive charges, similar to that shown in the figure below. Try to place all of the charges centered along a horizontal grid line. Feel free to look at the electric field, as it is interesting.
Measure the strength of the electric field 1 directly above the middle as well as 2 directly above. Does the strength of the electric field decrease as 1 over distance squared ()?
ANSWER:
No, it decreases more quickly with distance.
No, it decreases less quickly with distance.
Yes, it does.
Correct
In fact, it turns out that the strength of the electric field decreases roughly as . So the field 1 m above the midpoint is roughly half the strength at 0.5 . This is another example showing that a distribution of charges produces an electric field that is very different from that of a single charge.
PhET Interactive Simulations
University of Colorado
http://phet.colorado.edu
Video Tutor: Charged Rod and Aluminum Can
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point.
Part A
Consider the situation in the figure below, where two charged rods are placed a distance on either side of an aluminum can. What does the can do?
Hint 1. How to approach the problem.
This problem asks you to think about induced charge on the surface of an object and the resulting polarization force.
To get started, draw a diagram. Draw the induced surface charges on the outside of the can. Next, draw a force diagram (freebody diagram) to show the forces exerted on the can. Aluminum is a conductor.
ANSWER:
Rolls to the left
Stays still
Rolls to the right
Correct
The positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can. However, the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod. The net force acting on the can is zero.
Part B
Now, consider the situation shown in the figure below. What does the can do?
ANSWER:
Rolls to the left
Rolls to the right
Stays still
Correct
The polarization force is always attractive, so the can does not move.
Part C
Using the setup from the first question, imagine that you briefly touch the negatively charged rod to the can. You then hold the two rods at equal distances on either side of the can. What does the can do?
Hint 1. How to approach the problem.
This problem asks you to consider what happens to a conductor after being touched by a charged object. What charge will the can have after being touched?
ANSWER:
Does not move
Rolls toward the positively charged rod
Rolls away from the positively charged rod
Correct
The can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.
A Test Charge Determines Charge on Insulating and Conducting Balls
Learning Goal:
To understand the electric force between charged and uncharged conductors and insulators.
When a test charge is brought near a charged object, we know from Coulomb's law that it will experience a net force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.
Consider three plastic balls (A, B, and C), each carrying a uniformly distributed charge equal to either +,  or zero, and an uncharged copper ball (D). A positive test charge (T) experiences the forces shown in the figure when brought very near to the individual balls. The test charge T is strongly attracted to A, strongly repelled from B, weakly attracted to C, and strongly attracted to D.
Assume throughout this problem that the balls are brought very close together.
Part A
What is the nature of the force between balls A and B?
Hint 1. What is the net charge on ball A?
Since the test charge is positively charged, and there is a strongly attractive force between ball A and the test charge, what must be the nature of the net charge of ball A?
ANSWER:
positive
negative
zero
Correct
Hint 2. What is the net charge on ball B?
Since the test charge is positively charged, and there is a strongly repulsive force between ball B and the test charge, what must be the nature of the net charge of ball B?
ANSWER:
positive
negative
zero
Correct
ANSWER:
strongly attractive
strongly repulsive
weakly attractive
neither attractive nor repulsive
Correct
Part B
What is the nature of the force between balls A and C?
Hint 1. What is the charge on ball C?
Recall that ball C is composed of insulating material, which means that it can be polarized, but the charges inside are otherwise not free to move around inside the ball. Since the test charge experiences only a weak force due to ball C, if we compare to ball A we conclude that the charge on ball C must be
ANSWER:
+

zero
Correct
ANSWER:
strongly attractive
strongly repulsive
weakly attractive
neither attractive nor repulsive
Correct
Recall that ball C is composed of insulating material, which can be polarized in the presence of an external charged object such as ball A. Once polarized, there will be a weak attraction between balls A and C, because the positive and negative charges in ball C are at slightly different average distances from ball A. If ball C had a very small negative charge the test charge would have the same response (weakly attractive) but it would have a weak repulsive interaction with ball A. However, a smaller negative charge is not one of the options.
Part C
What is the nature of the force between balls A and D?
Hint 1. What are the surface charges on ball D?
Recall that copper is a conductor, in which charges can freely flow. When ball D is brought close to ball A, what will be the nature of the surface charge density on the side of ball D that is closest to ball A?
ANSWER:
positive
negative
zero
Correct
The negatively charged ball A (see Part A) will exert an attractive force on the positive charges in ball D and a repulsive force on the negative charges (namely, the electrons). Since ball D is made of copper, which is a conductor, the electrons will be repelled from negatively charged ball A and will migrate to the side of ball D farthest from ball A. The deficit of electrons on the side of ball D that is closest to ball A results in a positive net surface charge density on that side of ball D. Because the positive charge on ball D is much closer to ball A than the negative charge, the attractive force that ball A experiences due to the positive charges on ball D is stronger than the repulsive force ball A experiences due to the negative charges on ball D.
ANSWER:
attractive
repulsive
neither attractive nor repulsive
Correct
Part D
What is the nature of the force between balls D and C?
ANSWER:
attractive
repulsive
neither attractive nor repulsive
Correct
Because the test charge T is neither strongly attracted to nor repelled from ball C, ball C must have zero net charge. Since ball D also has zero net charge, there will not be any force between the two balls.
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 yaxis 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 yaxis at position . What is the new net force on particle 0, from particle 1 and particle 2?
Express your answer (a vector) using any or all of , , , , , , , , and .
ANSWER:
=
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, with positive charge , fixed in the yzplane at . What is the net force on particle 0 due solely to this charge?
Express your answer (a vector) using , , , , , , and . Include only the force caused by particle 3.
Hint 1. Find the magnitude of force from particle 3
What is the magnitude of the force on particle 0 from particle 3, fixed at ?
Express your answer using , , , .
Hint 1. Distance to particle 3
Use the Pythagorean theorem to find the straight line distance between the origin and .
ANSWER:
=
Hint 2. Vector components
The force vector points from to . Because is symmetrically located between the yaxis and the zaxis, the angle between , the unit vector pointing from particle 3 to particle 0, and the yaxis is radians. You have already calculated the magnitude of the vector above. Now break up the force vector into its y and z components.
ANSWER:
=
Correct
The Trajectory of a Charge in an Electric Field
An charge with mass and charge is emitted from the origin, . A large, flat screen is located at . There is a target on the screen at y position , where . In this problem, you will examine two different ways that the charge might hit the target. Ignore gravity in this problem.
Part A
Assume that the charge is emitted with velocity in the positive x direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive y direction. What should the magnitude of the electric field be if the charge is to hit the target on the screen?
Express your answer in terms of , , , , and .
Hint 1. How to approach the problem
Once you determine the force on the charge due to the electric field, this becomes a standard twodimensional kinematics problem. To solve the problem, first determine the equations of motion in both the x and y directions. Then use the fact that at some final time you know that the position of the charge is to obtain two equations in terms of the two unknowns and . Eliminate and solve for .
Hint 2. Find the equation of motion in the x direction
Find an expression for , the charge's x position as a function of time.
Express your answer in terms of as well as any of the given variables and constants.
Hint 1. Find the force in the x direction
What net force does the charge experience in the x direction?
Hint 1. Formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
ANSWER:
=
0
Hint 2. A helpful kinematic equation
Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is
.
ANSWER:
=
Hint 3. Find the equation of motion in the y direction
Find an expression for , the charge's y position as a function of time.
Express your answer in terms of as well as any of the given variables and constants.
Hint 1. Find the force in the y direction
What is the net force acting on the charge in the y direction?
Express your answer in terms of the given variables and constants.
Hint 1. Formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
ANSWER:
=
Hint 2. A helpful kinematic equation
Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is
.
ANSWER:
=
Hint 4. Combine Your Results
At some final time , you have and . Starting with these two equations, eliminate and solve for .
Hint 5. Find
Use the equation for the motion of the charge in the x direction to find .
Express your answer in terms of the variables and .
ANSWER:
=
ANSWER:
=
Correct
Part B
Now assume that the charge is emitted with velocity in the positive y direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive x direction. What should the magnitude of the electric field be if the charge is to hit the target on the screen?
Express your answer in terms of , , , , and .
Hint 1. How to approach the problem
Just as in the previous part, once you determine the force on the charge due to the electric field, this becomes a standard twodimensional kinematics problem. To solve the problem, first determine the equations of motion in both the x and y directions. Then use the fact that at some final time you know that the position of the charge is to obtain two equations in terms of the two unknowns and . Eliminate and solve for .
Hint 2. Find the equation of motion in the y direction
Find an expression for , the charge's y position as a function of time.
Express your answer in terms of as well as any of the given variables and constants.
Hint 1. Find the force in the y direction
What net force does the charge experience in the y direction?
Hint 1. Formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
ANSWER:
=
0
Hint 2. A helpful kinematic equation
Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is
.
ANSWER:
=
Hint 3. Find the equation of motion in the x direction
Find an expression for , the charge's x position as a function of time.
Express your answer in terms of as well as any of the given variables and constants.
Hint 1. Find the force in the x direction
What is the net force acting on the charge in the x direction?
Express your answer in terms of the given variables and constants.
Hint 1. Formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
ANSWER:
=
Hint 2. A helpful kinematic equation
Recall that kinematic equation that gives the distance travelled in terms of the initial velocity , the acceleration and elapsed time is
.
ANSWER:
=
Hint 4. Combine your results
At some final time , you have and . Starting with these two equations, eliminate and solve for .
Hint 5. Find
Use the equation for the motion of the charge in the y direction to find .
Express your answer in terms of the variables and .
ANSWER:
=
ANSWER:
=
Correct
The equations of motion for this part are identical to the equations of motion for the previous part, with and interchanged. Thus it is no surprise that the answers to the two parts are also identical, with and interchanged.
Charged Ring
Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius and positive charge distributed evenly along its circumference.
Part A
What is the direction of the electric field at any point on the z axis?
Hint 1. How to approach the problem
Approach 1
In what direction is the field due to a point on the ring? Add to this the field from a point on the opposite side of the ring. In what direction is the net field? What if you did this for every pair of points on opposite sides of the ring?
Approach 2
Consider a general electric field at a point on the z axis, i.e., one that has a z component as well as a component in the xy plane. Now imagine that you make a copy of the ring and rotate this copy about its axis. As a result of the rotation, the component of the electric field in the xy plane will rotate also. Now you ask a friend to look at both rings. Your friend wouldn't be able to tell them apart, because the ring that is rotated looks just like the one that isn't. However, they have the component of the electric field in the xy plane pointing in different directions! This apparent contradiction can be resolved if this component of the field has a particular value. What is this value?
Does a similar argument hold for the z component of the field?
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?
Use in your answer, where .
Hint 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.
Use in your answer, where . You may also use some or all of the variables , , and .
ANSWER:
=
Hint 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 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.
ANSWER:
=
Hint 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
.
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
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 .
Express your answer in terms of given charges, dimensions, and constants.
Hint 1. Simple harmonic motion
Recall the nature of simple harmonic motion of an object attached to a spring. Newton's second law for the system states that
, leading to oscillation at a frequency of
(here, the prime on the symbol representing the spring constant is to distinguish it from ). The solution to this differential equation is a sinusoidal function of time with angular frequency . Write an analogous equation for the ball near the charged ring in order to find the term.
Hint 2. Find the force on the charge
What is , the z component of the force on the ball on the ball at the point ? Use the approximation .
Express your answer in terms of , , , , and .
Hint 1. A formula for the force on a charge in an electric field
The formula for the force on a charge in an electric field is
.
Therfore, in particular,
.
You have already found in Part B. Use that expression in the equation above to find an expression for the z component of the force on the ball at the point . Don't forget to use the approximation given.
ANSWER:
=
ANSWER:
=
Correct
The Electric Field Produced by a Finite Charged Wire
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 1. Consider opposite ends of the wire
Consider the sum of the electric fields produced by the two equivalent charge elements at opposite ends of the wire. What is the direction of the resulting electric field?
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 .
Express your answer in terms of , , , and .
Hint 1. How to approach the problem
To compute the field at point , divide the rod into infinitesimal segments and find the electric field produced by each. Then integrate to find the total electric field. The electric field due to an arbitrary infinitesimal segment of wire has both x and y components. You could integrate these two components separately, but because of symmetry, you already know that the total electric field points only in the y direction. So you need to work only with the y component of the field.
Hint 2. Find the field due to an infinitesimal segment
Find , the y component of the electric field produced at point by the infinitesimal segment of the wire between and .
Express your answer in terms of , , , , and .
Hint 1. Find the total electric field
Find the magnitude of the electric field at point produced by the infinitesimal wire segment between and .
Express your answer in terms of , , , , and the constant .
Hint 1. Formula for electric field due to a point charge
The electric field at position away from an infinitesimal charge element is given by .
Hint 2. Find
What is the distance from an infinitesimal segment of wire at position to the point ?
Express your answer in terms of and .
ANSWER:
=
Hint 3. Find an expression for
The wire has charge per unit length , and the length of an infinitesimal charge element is . Write an expression for , the total charge on the infinitesimal element .
Express your answer in terms of and .
ANSWER:
=
ANSWER:
=
Hint 2. Find the y component of
What is , the y component of the electric field at point due to an infinitesimal segment of wire located at position ? Assume that the magnitude of the electric field at point due to the infinitesimal wire segment is .
Express your answer in terms of , , and .
ANSWER:
=
ANSWER:
=
Hint 3. A necessary integral
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 apart. The dipole has moment of inertia about its center of mass. The dipole is released from angle , and it is allowed to rotate freely.
Part A
What is , the magnitude of the dipole's angular velocity when it is pointing along the y axis?
Express your answer in terms of quantities given in the problem introduction.
Hint 1. How to approach the problem
Because there is no dissipation (friction, air resistance, etc.), you can solve this problem using conservation of energy. When the dipole is released from rest, it has potential energy but no kinetic energy. When the dipole is aligned with the y axis, it is rotating, and therefore has both kinetic and potential energy. The sum of potential and kinetic energy will remain constant.
Hint 2. Find the potential energy
Find the dipole's potential energy due to its interaction with the electric field as a function of the angle that the dipole's positive end makes with the positive y axis. Define the potential energy to be zero when the dipole is oriented perpendicular to the field: .
Express your answer in terms of , , , and .
Hint 1. The formula for the potential energy of a dipole
The general formula for the potential energy of an electric dipole with dipole moment in the presence of a uniform electric field is .
Hint 2. The dipole moment
The dipole moment of the electric dipole , when it makes an angle with the positive y axis can be written as
.
ANSWER:
=
Hint 3. Find the total energy at the moment of release
Find , the total energy (kinetic plus potential) at the moment the dipole is released from rest at angle with respect to the y axis. Use the convention that the potential energy is zero when the dipole is oriented perpendicular to the field: .
Express your answer in terms of some or all of the variables , , , and .
ANSWER:
=
Hint 4. Find the total energy when
Find an expression for , the total energy (kinetic plus potential) at the moment when the dipole is aligned with the y axis. Use the convention that the potential energy is zero when the dipole is oriented perpendicular to the field: .
Express your answer in terms of quantities given in the problem introduction and .
Hint 1. What is kinetic energy as a function of angular velocity?
What is the kinetic energy of a body rotating with angular velocity around an axis about which the moment of inertia is ?
ANSWER:
=
ANSWER:
=
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?
Express your answer in terms of and quantities given in the problem introduction.
Hint 1. How to approach the problem
The equation of motion for a simple harmonic oscillator can always be written in the standard form . To solve this problem, you need to write the equation of motion for the dipole in the standard form with replaced by the angular variable . This will allow you to read off the expression for , which has a simple relationship to the period of oscillation. (Note: Here, the variable does not represent the angular velocity of the dipole; rather, it denotes the frequency of the dipole's oscillation.) Start with the angular analogue of Newton's second law: . Recall that , the angular acceleration, is equal to the second derivative of , just as linear acceleration is equal to the second derivative of position.
Hint 2. Compute the torque
What is the magnitude of the torque that the electric field exerts about the center of mass of the dipole when the dipole is oriented at an angle with respect to the electric field?
Express the magnitude of the torque in terms of quantities given in the problem introduction and .
Hint 1. Formula for torque on a dipole
The torque on a dipole with dipole moment in an electric field is given by . Alternatively, the torque can be related to the potential energy by .
Hint 2. The dipole moment
When it makes an angle with the positive y axis, the dipole moment of the electric dipole can be written as
.
ANSWER:
=
Hint 3. The smallangle approximation
Because is small, you can apply the smallangle approximation to the expression for torque, and take the torque to be .
Up to this point we have been interested only in the magnitude of the torque. Now let's think about the direction. After all, torque is a vector quantity. For a system to oscillate, the torque must be a restoring torque; that is, the torque and the (small) angular displacement must be in opposite directions. (Recall that small angular displacements can be treated as vectors, since they obey vector addition, while large angles do not.) If you did the vector algebra carefully, you would find that the correct vector equation is
.
For future purposes we will write this as , keeping in mind that now represents the component of in the direction, rather than the magnitude of .
Hint 4. Find the oscillation frequency
Putting together what you have so far yields
.
Compare this to the standard form for a simple harmonic oscillator to obtain the oscillation frequency for the motion of the dipole.
Express your answer in terms of quantities given in the problem introduction.
ANSWER:
=
Hint 5. The relationship between (angular) oscillation frequency and period
The relationship between , the angular oscillation frequency of the dipole, and the period of oscillation is given by
.
ANSWER:
=
Correct
Charging an Insulator
This problem explores the behavior of charge on realistic (i.e. nonideal) insulators. We take as an example a long insulating rod suspended by insulating wires. Assume that the rod is initially electrically neutral. For convenience, we will refer to the left end of the rod as end A, and the right end of the rod as end B . In the answer options for this problem, "weakly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two balls, one of which is charged, and the other acquires a small induced charge". An attractive/repulsive force greater than this should be classified as "strongly attracted/repelled".
Part A
A small metal ball is given a negative charge, then brought near (i.e., within a few millimeters) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Select the expected behavior.
Hint 1. What is an insulator?
An insulator is a material which does not allow charge/current to flow easily through it.
Hint 2. Charge at end A
Keeping in mind that like charges repel each other, and opposite charges attract each other, what sort of charge is induced at end A of the (nonideal) insulating rod?
ANSWER:
A small positive charge
A small negative charge
ANSWER:
strongly repelled
strongly attracted
weakly attracted
weakly repelled
neither attracted nor repelled
Correct
Currently, you can think of this in the following way: When the sphere is brought near the rod, a positive charge is induced at end A (and correspondingly, end B acquires a negative induced charge). This means that some charge must have flowed from A to B. Since charge flow is inhibited in an insulator, the induced charges are typically small. Later you will learn how to model insulators more accurately and formulate a slightly more accurate argument.
Now consider what happens when the small metal ball is repeatedly given a negative charge and then brought into contact with end A of the rod
Part B
After several contacts with the charged ball, how is the charge on the rod arranged?
Select the best description.
Hint 1. What is an insulator?
An insulator is a material which does not allow charge/current to flow easily through it.
ANSWER:
positive charge on end B and negative charge on end A
negative charge spread evenly on both ends
negative charge on end A with end B remaining almost neutral
positive charge on end A with end B remaining almost neutral
none of the above
Correct
When the sphere is touched to end A, some of its negative charge will be deposited there. However, since charge cannot flow easily through an insulator, most of this charge will just sit at end A and will not distribute itself over the rod, as it would if the rod was a conductor.
Part C
How does end A of the rod react when the ball approaches it after it has already made several contacts with the rod, such that a fairly large charge has been deposited at end A?
Select the expected behavior.
ANSWER:
strongly repelled
strongly attracted
weakly attracted
weakly repelled
neither attracted nor repelled
Correct
More on insulators
You may have learnt that any material is made of atoms, which in turn consist of a nucleus and electrons. In the atoms of some materials, some of the electrons are "bound" to the nucleus very weakly, which leaves them free to move around the volume of the material. Such electrons are called "free" electrons, and such materials are called conductors, because the charge (i.e. electrons) can move around easily. In insulators, all the electrons in the atom are bound quite tightly to the nucleus, i.e. there are no free electrons available to move through the insulator.
Exercise 21.19
Three point charges are arranged along the xaxis. Charge = +3.00 is at the origin, and charge = 5.00 is at = 0.200 . Charge = 8.00 .
Part A
Where is located if the net force on is 7.00 in the direction?
ANSWER:
=
0.144
Correct
Exercise 21.38
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.56distant from the first, in a time interval of 1.43×10−6 .
Part A
Find the magnitude of the electric field.
ANSWER:
=
159
Correct
Part B
Find the speed of the proton when it strikes the negatively charged plate.
ANSWER:
=
2.18×104
Correct
Exercise 21.53
Positive electric charge is distributed along the yaxis with charge per unit length .
Part A
Consider the case where charge is distributed only between the points and . For points on the axis, find the component of the electric field as a function of .
Express your answer in terms of the variables , , and appropriate constants.
ANSWER:
=
Part B
Consider instead the case where charge is distributed along the entire yaxis with the same charge per unit length . Find the component of the electric field as a function of for values.
ANSWER:
=
Exercise 21.57
Point charges 4.00 and 4.00 are separated by a distance of 3.80 , forming an electric dipole.
Part A
Find the magnitude of the electric dipole moment.
ANSWER:
=
1.52×10−11
Correct
Part B
Find the direction of the electric dipole moment?
ANSWER:
from to
from to
Correct
Part C
The charges are in a uniform electric field whose direction makes an angle of 36.6 with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.50×10−9 ?
ANSWER:
=
828
Correct
Problem 21.82
Two tiny spheres of mass = 9.00 carry charges of equal magnitude, 72.0 , but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 . When a horizontal uniform electric field that is directed to the left is turned on, the spheres hang at rest with the angle between the strings equal to in the following figure.
Part A
Which ball (the one on the right or the one on the left) has positive charge?
ANSWER:
Which ball (the one on the right or the one on the left) has positive charge?
The one on the right
The one on the left
Part B
What is the magnitude of the field?
Express your answer with the appropriate units.
ANSWER:
=
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Your score on this assignment is 85.4%.
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why most of the answers are not written?
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Wendy Alcius
commented on a video on YouTube.I bought a EPX2800 but the power switch is orange instead of green. I dont know why
Wendy Alcius
commented on a video on YouTube.lol she said somebody is getting a ticket lmao
Wendy Alcius
commented on a video on YouTube.would you tell me how you came up with the 15 in the triangle? is that a formula you used?
Wendy Alcius
commented on a video on YouTube.how come somebody dislikes that vedeo
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