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Chapter 28 Homework
Due: 8:00am on Wednesday, March 3, 2010
Note: To understand how points are awarded, read your instructor's Grading Policy.
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Calculating Electric Flux through a Disk
Suppose a disk with area
is placed in a uniform electric field of magnitude
. The disk is oriented so that the vector normal to its surface, , makes an angle
with the electric field, as shown in the figure.
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.
Hint A.1
Definition of electric flux
Hint not displayed
Hint A.2
Simplifying the integrand
Hint not displayed
Hint A.3
Evaluate the scalar product
Hint not displayed
Express your answer in terms of
ANSWER:
,
, and
=
Correct
Problem 28.9
Part A
What is the electric flux through the surface shown in the figure?
ANSWER:
1.00
Correct
Problem 28.13
A
rectangle lies in the xz-plane. What is the electric flux through the rectangle if
Part A
ANSWER:
0
Correct
Part B
ANSWER:
6.00×10−2
Correct
Flux out of a Cube
A point charge of magnitude
is at the center of a cube with sides of length
.
Part A
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What is the electric flux
Hint A.1
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through each of the six faces of the cube?
How to approach the problem
Hint not displayed
Hint A.2
Calculate the total electric flux
Hint not displayed
Hint A.3
Flux through a face
Hint not displayed
Use
for the permittivity of free space.
ANSWER:
=
Correct
The shape of the surface enclosing a charge, in this case a cube, does not affect the total electric flux through the surface. The flux depends only on the total enclosed charge.
Part B
What would be the flux
Hint B.1
through a face of the cube if its sides were of length
?
How to approach the problem
Gauss's law states that the flux through a closed surface depends only on the charge enclosed by that surface. Does changing the size of the cube affect the total charge enclosed by the cube?
Use
for the permittivity of free space.
ANSWER:
=
Correct
Just as the shape of the surface does not affect the total electric flux coming out of that surface, its size does not make any difference in the total electric flux either. The only relevant quantity is the total
enclosed charge.
Problem 28.4
The electric field is constant over each face of the cube shown in the figure .
Part A
Does the box contain positive charge, negative charge, or no charge?
ANSWER:
positive charge
negative charge
no charge
Correct
Problem 28.22
The net electric flux through a closed surface is
1600
.
Part A
How much charge is enclosed within the surface?
ANSWER:
-14.2
nC
Correct
Charged Insulating Spheres
Two small insulating spheres with radius 5.50×10−2
positively charged, with net charge 4.50
are separated by a large center-to-center distance of 0.535
. One sphere is negatively charged, with net charge -1.00
, and the other sphere is
. The charge is uniformly distributed within the volume of each sphere.
Part A
What is the magnitude
Hint A.1
of the electric field midway between the spheres?
How to approach the problem
Draw a diagram of the system to keep track of the directions of the fields. Calculate the electric field at the point midway between the charged spheres separately for each sphere, using Gauss's law, and use
vector addition to determine the net electric field.
Hint A.2
Using Gauss's law
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You need to construct two separate Gaussian surfaces, each of which will surround one of the charged spheres. These surfaces also need to contain the point midway between the spheres so that the electric
field at that point can be found. Therefore, for convenience, you should use a spherical Gaussian surface with one of the charged spheres in the center, since that way the electric field will be uniform and
perpendicular to the surface at all points, so that the flux will be given by
Hint A.3
Calculate
. Also, be careful about the radii of the spherical surfaces.
Calculate the field due to the negatively charged sphere
, the magnitude of the electric field at the midway point due to the sphere with charge -1.00
Hint A.3.1
only.
Using the flux to calculate the field
Since the flux through the surface will be
(due to Gauss's law), you can solve for the electric field
using the equation for the surface area of the (Gaussian) sphere,
, and the
value of the enclosed charge.
Take the permittivity of free space to be
ANSWER:
Hint A.4
= 8.85×10−12
.
5
= 1.26×10
Correct
Determine the direction of the electric field from the first sphere
Hint not displayed
Hint A.5
Calculate the field due to the positively charged sphere
Hint not displayed
Hint A.6
Determine the direction of the electric field from the positively charged sphere
Hint not displayed
Hint A.7
Vector addition
Hint not displayed
= 8.85×10−12
Take the permittivity of free space to be
ANSWER:
.
5
= 6.91×10
Correct
Part B
What is the direction of the electric field midway between the spheres?
ANSWER:
toward the positively charged sphere
toward the negatively charged sphere
upward perpendicular to the line connecting the centers of the spheres
downward perpendicular to the line connecting the centers of the spheres
Correct
Since the electric field will point toward a negative charge and away from a positive charge, the electric field from each sphere separately will point toward the negatively charged sphere, and so the total
field will also point in that direction.
The Electric Field of a Ball of Uniform Charge Density
A solid ball of radius
has a uniform charge density .
Part A
What is the magnitude of the electric field
Hint A.1
at a distance
from the center of the ball?
Gauss's law
Hint not displayed
Hint A.2
Find
Hint not displayed
Express your answer in terms of ,
, , and
.
ANSWER:
=
Correct
Notice that this result is identical to that reached by applying Coulomb's law to a point charge centered at the origin with
. The field outside of a uniformly charged sphere does not depend on the
size of the sphere, only on its charge. A uniformly charged sphere generates an electric field as if all the charge were concentrated at its center.
Part B
What is the magnitude of the electric field
Hint B.1
at a distance
from the center of the ball?
How does this situation compare to that of the field outside the ball?
Hint not displayed
Express your answer in terms of , ,
, and
.
ANSWER:
=
Correct
Part C
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Let
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represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?
Hint C.1
Plot the electric field
The figure shows a plot of the electric field as a function of .
Check all that apply.
ANSWER:
.
.
.
The maximum electric field occurs when
.
The maximum electric field occurs when
.
The maximum electric field occurs as
Answer Requested
A Conducting Shell around a Conducting Rod
An infinitely long conducting cylindrical rod with a positive charge
infinitely long) with a charge per unit length of
and radius
per unit length is surrounded by a conducting cylindrical shell (which is also
, as shown in the figure.
Part A
What is
, the radial component of the electric field between the rod and cylindrical shell as a function of the distance
Hint A.1
from the axis of the cylindrical rod?
The implications of symmetry
Hint not displayed
Hint A.2
Apply Gauss' law
Hint not displayed
Hint A.3
Find the charge inside the Gaussian surface
Hint not displayed
Hint A.4
Find the flux
Hint not displayed
Express your answer in terms of
, , and
, the permittivity of free space.
ANSWER:
=
Correct
Part B
What is
, the surface charge density (charge per unit area) on the inner surface of the conducting shell?
Hint B.1
Apply Gauss's law
The magnitude of the net force on charges within a conductor is always zero. This implies that the magnitude of the electric field within the conductor is zero. Think about a cylindrical Gaussian surface of
length whose radius lies at the middle of the outer cylindrical shell. Since the electric field inside a conductor is zero and the Gaussian surface lies within the conductor, the electric flux across the Gaussian
surface must be zero. What, then, must , the total charge inside this Gaussian surface, be?
ANSWER:
Hint B.2
What is
=0
Correct
Find the charge contribution from the surface
, the total charge on the inner surface of the cylindrical shell that is contained within the Gaussian surface?
Express your answer in terms of
ANSWER:
and .
= Answer not displayed
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ANSWER:
=
Correct
Part C
What is
, the surface charge density on the outside of the conducting shell? (Recall from the problem statement that the conducting shell has a total charge per unit length given by
Hint C.1
.)
What is the charge on the cylindrical shell?
Hint not displayed
ANSWER:
=
Correct
Part D
What is the radial component of the electric field,
Hint D.1
, outside the shell?
How to approach the problem
Hint not displayed
Hint D.2
Find the charge within the Gaussian surface
Hint not displayed
Hint D.3
Find the flux in terms of the electric field
Hint not displayed
ANSWER:
=
Correct
Score Summary:
Your score on this assignment is 103.6%.
You received 62.69 out of a possible total of 70 points, plus 9.8 points of extra credit.
3/2/2010 12:23 PM