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Physics 210 Q1 2012
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University Physics with Modern Physics, 13e
Young/Freedman
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Week 3 - Gauss's Law
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Week 3 - Gauss's Law
Due: 10:00pm on Monday, October 15, 2012
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy
Gauss's Law
Learning Goal:
To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable.
Gauss's law is usually written
where
is the permittivity of vacuum.
Part A
How should the integral in Gauss's law be evaluated?
ANSWER:
around the perimeter of a closed loop
over the surface bounded by a closed loop
over a closed surface
In the integral for Gauss's law, the vector
element. The direction of
represents an infinitesimal surface element. The magnitude of
is the area of the surface
is normal to the surface element, pointing out of the enclosed volume.
Part B
In Gauss's law, to what does
refer?
ANSWER:
the net charge inside the closed surface
the charge residing on insulators inside the closed surface
all the charge in the physical system
any charge inside the closed surface that is arranged symmetrically
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The major use of Gauss's law is to determine an electric field when the charge distribution, both inside and outside the Gaussian surface,
is symmetric. Of course, the electric field can always be found by adding up (or integrating) the contributions of all the charge in the
problem. In highly symmetric situations, however, Gauss's law is much simpler computationally than dealing with all such contributions,
and it provides better physical insight, too.
A Conducting Shell around a Conducting Rod
An infinitely long conducting cylindrical rod with a positive charge
per unit length is
surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge
per unit length of
and radius , as shown in the figure.
Part A
What is
, the radial component of the electric field between the rod and cylindrical shell as a function of the distance
from the axis of
the cylindrical rod?
Express your answer in terms of
, , and
, the permittivity of free space.
Hint 1. The implications of symmetry
Because the cylinder and rod are cylindrically symmetric, the magnitude of the electric field cannot vary as a function of angle around
the rod, nor as a function of longitudinal position along the rod (typically represented by the spatial variables and ). By symmetry,
the magnitude of the electric field can only depend on the distance from the axis of the rod (the spatial variable ).
Hint 2. Apply Gauss' law
Gauss's law states that
, where
is the electric flux through a Gaussian surface, and
surface. Construct a cylindrical Gaussian surface with radius
and length
is the total charge enclosed by the
coaxial with the rod, with
.
Hint 3. Find the charge inside the Gaussian surface
What is the total charge
enclosed by the surface?
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ANSWER:
=
Hint 4. Find the flux
What is
, the electric flux through the Gaussian surface?
Express your answer in terms of the magnitude of the electric field
and given variables.
ANSWER:
=
ANSWER:
=
Part B
What is
, the surface charge density (charge per unit area) on the inner surface of the conducting shell?
Hint 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:
= 0
Hint 2. Find the charge contribution from the surface
What is
, the total charge on the inner surface of the cylindrical shell that is contained within the Gaussian surface?
Express your answer in terms of
and
.
ANSWER:
=
To obtain the charge density per unit area, divide
within the Gaussian surface.
by the area of the inner surface of the conducting shell that is contained
ANSWER:
=
Part C
What is
, the surface charge density on the outside of the conducting shell? (Recall from the problem statement that the conducting shell
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has a total charge per unit length given by
.)
Hint 1. What is the charge on the cylindrical shell?
What is
, the total surface charge (the sum of charges on the inner and outer surfaces) of a portion of the shell of length
?
ANSWER:
=
Since the charge on the inner surface of the cylinder is
and the total charge on the cylinder is
, it is now easy to obtain
the charge on the outer surface of the cylinder. Then divide this result by the surface area of the portion of the cylinder that you
took to obtain your result.
ANSWER:
=
Part D
What is the radial component of the electric field,
, outside the shell?
Hint 1. How to approach the problem
Apply Gauss's law as you did to find the field between the rod and the shell. Again, choose the Gaussian surface to be a cylinder, with
length and radius , coaxial with the rod. This time, you need to take
.
Hint 2. Find the charge within the Gaussian surface
What is
, the total charge contained within the Gaussian surface?
ANSWER:
=
Now apply Gauss' law,
, using
for the enclosed charge.
Hint 3. Find the flux in terms of the electric field
What is
, the electric flux through the Gaussian surface?
Express your answer in terms of the magnitude of the electric field
and given variables.
ANSWER:
=
ANSWER:
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=
Charged Insulating Spheres
−2
Two small insulating spheres with radius 5.50×10
charged, with net charge -2.50
are separated by a large center-to-center distance of 0.480
, and the other sphere is positively charged, with net charge 3.95
. One sphere is negatively
. The charge is uniformly distributed within
the volume of each sphere.
Part A
What is the magnitude
of the electric field midway between the spheres?
Take the permittivity of free space to be
= 8.85×10
−12
.
Hint 1. 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 2. Using Gauss's law
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
. Also, be careful about the
radii of the spherical surfaces.
Hint 3. Calculate the field due to the negatively charged sphere
Calculate
, the magnitude of the electric field at the midway point due to the sphere with charge -2.50
Take the permittivity of free space to be
−12
= 8.85×10
only.
.
Hint 1. Using the flux to calculate the field
Since the flux through the surface will be
equation for the surface area of the (Gaussian) sphere,
(due to Gauss's law), you can solve for the electric field
using the
, and the value of the enclosed charge.
ANSWER:
=
= 3.90×105
Hint 4. Determine the direction of the electric field from the first sphere
What will be the direction of the electric field due to the negatively charged sphere only?
ANSWER:
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toward the center of the negatively charged sphere
away from the center of 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
Note that the electric field will point toward a negatively charged sphere, just as it would point toward a negative point charge.
Hint 5. Calculate the field due to the positively charged sphere
Calculate
, the magnitude of the electric field at the midway point due to the sphere of charge 3.95
Take the permittivity of free space to be
−12
= 8.85×10
only.
.
Hint 1. Using the flux to calculate the field
Since the flux through the surface will be
equation for the surface area of the (Gaussian) sphere,
(due to Gauss's law), you can solve for the electric field
using the
, and the value of the enclosed charge.
ANSWER:
=
= 6.17×105
Hint 6. Determine the direction of the electric field from the positively charged sphere
What will be the direction of the electric field due to the positively charged sphere only?
ANSWER:
toward the positively charged sphere
away from the positively charged sphere
upward perpendicular to the line connecting the centers of the spheres
downward perpendicular to the line connecting the centers of the spheres
Note that the electric field will point away from a positively charged sphere, just as it would point away from a positive point
charge.
Hint 7. Vector addition
Keep in mind that you need to use vector addition in adding the electric fields from the two spheres. Also, keep in mind that the two
fields point along the same line.
ANSWER:
=
= 1.01×106
Part B
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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
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.
Exercise 22.21
A hollow, conducting sphere with an outer radius of 0.246
. A charge of -0.830
and an inner radius of 0.207
−6
has a uniform surface charge density of +6.24×10
is now introduced into the cavity inside the sphere.
Part A
What is the new charge density on the outside of the sphere?
ANSWER:
= 5.15×10−6
=
Part B
Calculate the strength of the electric field just outside the sphere.
ANSWER:
= 5.82×105
=
Part C
What is the electric flux through a spherical surface just inside the inner surface of the sphere?
ANSWER:
=
= −9.38×104
Exercise 22.22
A point charge of -2.00
is located in the center of a spherical cavity of radius 6.54
−4
density in the solid is 7.35×10
inside an insulating spherical charged solid. The charge
.
Part A
Calculate the magnitude of the electric field inside the solid at a distance of 9.49
from the center of the cavity.
Express your answer using two significant figures.
ANSWER:
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= 2.3×105
=
Part B
Find the direction of the electric field inside the solid at a distance of 9.49
from the center of the cavity.
ANSWER:
The electric field points radially outward.
The electric field points radially inward.
The electric field is zero.
Problem 22.34
A cube has sides of length
given by
= 0.300
. It is placed with one corner at the origin as shown in the figure . The electric field is not uniform but is
-4.93
2.49
.
Part A
Find the electric flux through each of the six cube faces
, and
.
Enter your answers in ascending order separated by commas.
ANSWER:
=
,
, , ,
,
= 0, 6.72×10−2, 0, 0, -0.133, 0
Part B
Find the total electric charge inside the cube.
ANSWER:
=
= −5.83×10−13
Problem 22.46
A conducting spherical shell with inner radius
is
and outer radius
has a positive point charge
located at its center. The total charge on the shell
, and it is insulated from its surroundings .
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Part A
Derive the expression for the electric field magnitude in terms of the distance
Express your answer in terms of some or all of the variables
from the center for the region
.
, , , and appropriate constants.
ANSWER:
=
Also accepted:
Part B
Derive the expression for the electric field magnitude in terms of the distance
Express your answer in terms of some or all of the variables
from the center for the region
.
, , , and appropriate constants.
ANSWER:
= 0
Part C
Derive the expression for the electric field magnitude in terms of the distance
Express your answer in terms of some or all of the variables
from the center for the region
.
, , , and appropriate constants.
ANSWER:
=
Also accepted:
Part D
What is the surface charge density on the inner surface of the conducting shell?
Express your answer in terms of some or all of the variables
, , , and appropriate constants.
ANSWER:
=
Part E
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What is the surface charge density on the outer surface of the conducting shell?
Express your answer in terms of some or all of the variables
, , , and appropriate constants.
ANSWER:
=
Problem 22.58
A nonuniform, but spherically symmetric, distribution of charge has a charge density
given as follows:
for
for
where
is a positive constant.
Part A
Find the total charge contained in the charge distribution.
Express your answer in terms of the variables ,
,
, and appropriate constants.
ANSWER:
= 0
Part B
Obtain an expression for the electric field in the region
Express your answer in terms of the variables ,
.
,
, and appropriate constants.
ANSWER:
= 0
Part C
Obtain an expression for the electric field in the region
Express your answer in terms of the variables ,
.
,
, and appropriate constants.
ANSWER:
=
Also accepted:
Part D
Find the value of
at which the electric field is maximum.
Express your answer in terms of the variables ,
,
, and appropriate constants.
ANSWER:
=
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Part E
Find the value of that maximum field.
Express your answer in terms of the variables ,
,
, and appropriate constants.
ANSWER:
=
Also accepted:
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