MasteringPhysics: Print View with Answers
1 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Help
Logged in as Julie Alexander, Instructor
Physics 210 Q1 2012
My Courses
Assignments
Week 2 - Electric Fields
Overview
( PHYSICS210BRIDGE )
Course Settings
Course Home
Summary View
Log Out
University Physics with Modern Physics, 13e
Young/Freedman
Roster
Gradebook
Instructor Resources
Item Library
eText
Study Area
[ Edit ]
Diagnostics View
Print View with Answers
Week 2 - Electric Fields
Due: 10:00pm on Tuesday, October 9, 2012
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy
Electric Fields and Forces
Learning Goal:
To understand Coulomb's law, electric fields, and the connection between the electric field and the electric force.
Coulomb's law gives the electrostatic force
acting between two charges. The magnitude
depends on the product of the charges and the square of the distance
of the force between two charges
and
between the charges:
,
where
. The direction of the force is along the line connecting the two charges. If the charges have the
same sign, the force will be repulsive. If the charges have opposite signs, the force will be attractive. In other words, opposite charges attract and
like charges repel.
Because the charges are not in contact with each other, there must be an intermediate mechanism to cause the force. This mechanism is the
electric field. The electric field at any location is equal to the force per unit charge experienced by a charge placed at that location. In other words,
if a charge experiences a force , the electric field
at that point is
.
The electric field vector has the same direction as the force vector on a positive charge and the opposite direction to that of the force vector on a
negative charge.
An electric field can be created by a single charge or a distribution of charges. The electric field a distance
from a point charge
has magnitude
.
The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be
found by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by ,
will not significantly affect the electric field if it is small compared to
Imagine an isolated positive point charge with a charge
.
(many times larger than the charge on a single electron).
Part A
There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?
Check all that apply.
ANSWER:
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
2 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
the distance between the positive charge and the electron
the charge on the electron
the mass of the electron
the charge of the positive charge
the mass of the positive charge
the radius of the positive charge
the radius of the electron
According to Coulomb's law, the force between two particles depends on the charge on each of them and the distance between them.
Part B
For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?
Check all that apply.
ANSWER:
the distance between the positive charge and the electron
the charge on the electron
the mass of the electron
the charge of the positive charge
the mass of the positive charge
the radius of the positive charge
the radius of the electron
The electrostatic force cannot exist unless two charges are present. The electric field, on the other hand, can be created by only one
charge. The value of the electric field depends only on the charge producing the electric field and the distance from that charge.
Part C
If the total positive charge is
= 1.62×10
−6
, what is the magnitude of the electric field caused by this charge at point P, a distance
= 1.53
from the charge?
Enter your answer numerically in newtons per coulomb.
ANSWER:
=
= 6220
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
3 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Part D
What is the direction of the electric field at point P?
Enter the letter of the vector that represents the direction of
.
ANSWER:
Part E
Now find the magnitude of the force on an electron placed at point P. Recall that the charge on an electron has magnitude
.
Enter your answer numerically in newtons.
Hint 1. Determine how to approach the problem
What strategy can you use to calculate the force between the positive charge and the electron?
ANSWER:
Use Coulomb's law.
Multiply the electric field due to the positive charge by the charge on the electron.
Do either of the above.
Do neither of the above.
ANSWER:
=
= 9.95×10−16
Part F
What is the direction of the force on an electron placed at point P?
Enter the letter of the vector that represents the direction of
.
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
4 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
ANSWER:
Electric Field Conceptual Question
Part A
For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is
zero.
If no such region exists on the horizontal axis choose the last option
(nowhere).
Hint 1. Zeros of the electric field
The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal
magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each
region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region.
ANSWER:
A
B
C
D
E
nowhere
Part B
For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is
zero.
If no such region exists on the horizontal axis choose the last option
(nowhere).
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
5 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Hint 1. Zeros of the electric field
The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal
magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each
region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region.
Hint 2. Determine the regions where the electric fields could cancel
In which region(s) do the electric fields from the two source charges point in opposite directions?
List all the correct answers in alphabetical order.
ANSWER:
Since the two charges produce fields that point in opposite directions in these regions, if the magnitude of the fields are equal, the
net electric field will be zero.
Hint 3. Consider the magnitude of the electric field
For each of the three regions found in the previous hint, determine whether it is possible for the magnitudes to be equal. As an
example, consider the point directly between the two charges. Which charge produces the largest magnitude field directly between the
two charges?
ANSWER:
the charge on the right
the charge on the left
neither, because they have the same magnitude
Therefore, the point directly between the two charges is not the correct answer since the right charge dominates at this point.
Check the other two possible regions.
ANSWER:
A
B
C
D
E
nowhere
Part C
For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is
zero.
If no such region exists on the horizontal axis choose the last option
(nowhere).
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
6 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Hint 1. Zeros of the electric field
The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal
magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each
region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region.
ANSWER:
A
B
C
D
E
nowhere
Part D
For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field is
zero.
Hint 1. Zeros of the electric field
The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equal
magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in each
region determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region.
ANSWER:
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
7 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
A
B
C
D
E
Nowhere along the finite x axis
Exercise 21.54
A straight, nonconducting plastic wire 7.00
long carries a charge density of 130
distributed uniformly along its length. It is lying on a
horizontal tabletop.
Part A
Find the magnitude and direction of the electric field this wire produces at a point 4.00
directly above its midpoint.
ANSWER:
= 3.85×104
=
Part B
ANSWER:
electric field is directed upward
electric field is directed downward
Part C
If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.00
directly above its center.
ANSWER:
= 4.58×104
=
Part D
ANSWER:
electric field is directed upward
electric field is directed downward
Problem 21.97
Negative charge
is distributed uniformly around a quarter-circle of radius
that lies in the first quadrant, with the center of curvature at the
origin.
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
8 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Part A
Find the x-component of the net electric field at the origin.
Express your answer in terms of the variables
,
and appropriate constants.
ANSWER:
=
Also accepted:
,
Part B
Find the y-component of the net electric field at the origin.
Express your answer in terms of the variables
,
and appropriate constants.
ANSWER:
=
Also accepted:
,
Problem 21.89 - Copy
Positive charge
is distributed uniformly along the x-axis from
, a distance
to the right of the end of
to
. A positive point charge
is located on the positive x-axis at
.
Part A
Calculate the x-component of the electric field produced by the charge distribution
Express your answer in terms of the variables
at points on the positive x-axis where
.
, , , and and appropriate constants.
ANSWER:
=
Also accepted:
,
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
9 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Part B
Calculate the y-component of the electric field produced by the charge distribution
Express your answer in terms of the variables
, ,
at points on the positive x-axis where
.
, and appropriate constants.
ANSWER:
= 0
Part C
Calculate the magnitude of the force that the charge distribution
Express your answer in terms of the variables
exerts on .
, , , , and appropriate constants.
ANSWER:
=
Also accepted:
,
Part D
Calculate the direction of the force that the charge distribution
exerts on .
ANSWER:
to the left
to the right
Problem 21.91 - Copy
A charged line extends from y = 2.50
to y = -2.50
. The total charge distributed uniformly along the line is -7.50
.
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
10 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Part A
Find the magnitude of the electric field on the
-axis at
= 10.0
.
ANSWER:
=
= 6550
Part B
Find the direction of the electric field on the
-axis at
= 10.0
.
ANSWER:
= 180
counterclockwise from the
-direction
Part C
Is the magnitude of the electric field you calculated in part A larger or smaller than the electric field 10.0
from a point charge that has the
same total charge as this finite line of charge?
ANSWER:
larger
smaller
Part D
At what distance
does the result for the finite line of charge differ by 1.0% from that for the point charge?
Express your answer using two significant figures.
ANSWER:
= 0.18
Problem 21.100
Two very large parallel sheets are 5.00
right of
apart. Sheet
, carries a uniform charge of -12.2
carries a uniform surface charge density of -8.00
, and sheet
, which is to the
. Assume the sheets are large enough to be treated as infinite.
Part A
Find the magnitude of the net electric field these sheets produce at a point 4.00
to the right of sheet
.
ANSWER:
=
= 2.37×105
Part B
Find the direction of the net electric field these sheets produce at a point 4.00
to the right of sheet
.
ANSWER:
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
11 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
net electric field is directed to the right
net electric field is directed to the left
Part C
Find the magnitude of the net electric field these sheets produce at a point 4.00
to the left of sheet
.
ANSWER:
= 1.14×106
=
Part D
Find the direction of the net electric field these sheets produce at a point 4.00
to the left of sheet
.
ANSWER:
net electric field is directed to the right
net electric field is directed to the left
Part E
Find the magnitude of the net electric field these sheets produce at a point 4.00
to the right of sheet
.
ANSWER:
= 1.14×106
=
Part F
Find the direction of the net electric field these sheets produce at a point 4.00
to the right of sheet
.
ANSWER:
net electric field is directed to the right
net electric field is directed to the left
Calculating Electric Flux through a Disk
Suppose a disk with area
makes an angle
is placed in a uniform electric field of magnitude
. The disk is oriented so that the vector normal to its surface,
,
with the electric field, as shown in the figure.
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
12 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Part A
What is the electric flux
through the surface of the disk that is facing right (the normal vector to this surface is shown in the figure)?
Assume that the presence of the disk does not interfere with the electric field.
Express your answer in terms of
,
, and
Hint 1. Definition of electric flux
The electric flux
through a surface is given by
, where
is an infinitesimal element of area on the surface.
Hint 2. Simplifying the integrand
Note that the formula for electric flux can be expressed in the following way:
.
Since we are dealing with a constant electric field, which therefore does not vary across the surface of the disk, we can take
outside of the integral. This gives a much simpler expression:
.
Hint 3. Evaluate the scalar product
Find
in terms of
and .
ANSWER:
=
ANSWER:
=
Flux through a Cube
A cube has one corner at the origin and the opposite corner at the point
. The
sides of the cube are parallel to the coordinate planes. The electric field in and around
the cube is given by
.
Part A
Find the total electric flux
through the surface of the cube.
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
13 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Express your answer in terms of , , , and
.
Hint 1. Definition of flux
The net electric flux
of a field
through a closed surface S is given by
,
where the differential vector
has magnitude proportional to the differential area and is oriented outward and normal (perpendicular)
to the surface. In some cases with simple geometry (like this one), you can break up the integral into manageable pieces. Consider
separately the flux coming out of each of the six faces of the cube, and then add the results to obtain the net flux.
Hint 2. Flux through the
face
Consider the face of the cube whose outward normal points in the positive x direction. What is the flux
Express your answer in terms of , , , and
through this face?
.
Hint 1. Simplifying the integral
The field
depends only on the spatial variable
. On the
face of the cube,
, so
is constant. Since
is constant
over this entire surface, it can be pulled out of the integral:
.
Hint 2. Evaluate the scalar product
The scalar (dot) product yields the component of the field that is in the direction of the normal (i.e., perpendicular to the surface).
Evaluate the dot product
.
Express your answer in terms of , , ,
, and
.
ANSWER:
=
Hint 3. Find the area of the face of the cube
This
face of the cube is a square with sides of length
. What is the area of this face?
ANSWER:
=
ANSWER:
=
Hint 3. Flux through the
face
Consider the face of the cube whose outward normal points in the positive y direction. What is the flux
Express your answer in terms of , , , and
through this face?
.
ANSWER:
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
14 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
=
Hint 4. Flux through the
face
Consider the face of the cube whose outward normal points in the positive z direction. What is the flux
through this face?
Hint 1. Consider the orientation of the field
The electric field has no z component (the field vector
xy plane and a vector normal to the
lies entirely in the xy plane. What is the dot product of a vector in the
face of the cube?
ANSWER:
= 0
Hint 5. Flux through the
face
Consider the face of the cube whose outward normal points in the negative x direction. What is the flux
Express your answer in terms of , , , and
through this face?
.
Hint 1. What is the electric field?
The face of the cube whose outward normal points in the negative x direction lies in the yz plane (i.e.,
component
). Find the x
of the electric field across this surface.
ANSWER:
=
Hint 2. Direction of flux
Remember to take note of whether the electric field is going into the surface or out of the surface. Flux is defined as positive
when the field is coming out of the surface and negative when the field is going into the surface.
ANSWER:
=
Hint 6. Putting it together
Using similar calculations to those above, you should be able to find the flux through each of the six faces. Add the six quantities to
obtain the net flux.
ANSWER:
=
Part B
Notice that the flux through the cube does not depend on
or . Equivalently, if we were to set
, so that the electric field becomes
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
15 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
,
then the flux through the cube would be zero. Why?
ANSWER:
does not generate any flux across any of the surfaces.
The flux into one side of the cube is exactly canceled by the flux out of the opposite side.
Both of the above statements are true.
Part C
What is the net charge
inside the cube?
Express your answer in terms of , , ,
, and
.
Hint 1. Gauss's law
Gauss's law states that the net flux of an electric field through a surface is proportional to the net charge inside that surface:
.
ANSWER:
=
Exercise 22.10
A point charge
= 3.15
is located on the x-axis at
= 2.25
, and a second point charge
= -5.65
is on the y-axis at
= 1.05
.
Part A
What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius
= 0.440
?
ANSWER:
= 0
Part B
What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius
= 1.70
?
= 2.95
?
ANSWER:
=
= -638
Part C
What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius
ANSWER:
=
= -282
9/7/2012 1:10 PM
MasteringPhysics: Print View with Answers
16 of 16
http://session.masteringphysics.com/myct/assignmentPrintView?assignm...
Copyright © 2012 Pearson. All rights reserved.
Legal Notice
Privacy Policy
Permissions
Support
9/7/2012 1:10 PM