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Eðlisfræði 2, vor 2007
22. Gauss' Law
Assignment is due at 2:00am on Wednesday, January 31, 2007
Credit for problems submitted late will decrease to 0% after the deadline has passed.
The wrong answer penalty is 2% per part. Multiple choice questions are penalized as described in the online help.
The unopened hint bonus is 2% per part.
You are allowed 4 attempts per answer.
Gauss' Law
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: Answer not displayed
Part B
Part not displayed
Gauss's Law in 3, 2, and 1 Dimension
Gauss's law relates the electric flux
through a closed surface to the total charge
enclosed by the surface:
.
You can use Gauss's law to determine the charge enclosed inside a closed surface on which the electric field is
known. However, Gauss's law is most frequently used to determine the electric field from a symmetric charge
distribution.
The simplest case in which Gauss's law can be used to determine the electric field is that in which the charge is
localized at a point, a line, or a plane. When the charge is localized at a point, so that the electric field radiates in
three-dimensional space, the Gaussian surface is a sphere, and computations can be done in spherical coordinates.
Now consider extending all elements of the problem (charge, Gaussian surface, boundary conditions) infinitely
along some direction, say along the z axis. In this case, the point has been extended to a line, namely, the z axis,
and the resulting electric field has cylindrical symmetry. Consequently, the problem reduces to two dimensions,
since the field varies only with x and y, or with and in cylindrical coordinates. A one-dimensional problem may
be achieved by extending the problem uniformly in two directions. In this case, the point is extended to a plane,
and consequently, it has planar symmetry.
Three dimensions
Consider a point charge in three-dimensional space. Symmetry requires the electric field to point directly away
from the charge in all directions. To find
, the magnitude of the field at distance from the charge, the logical
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Gaussian surface is a sphere centered at the charge. The electric field is normal to this surface, so the dot product of
the electric field and an infinitesimal surface element involves
. The flux integral is therefore reduced to
, where
is the magnitude of the electric field on the Gaussian surface, and
is the area of the
surface.
Part A
Determine the magnitude
by applying Gauss's law.
Part A.1 Find the area of the surface
Part not displayed
Express
in terms of some or all of the variables/constants , , and .
ANSWER:
=
Two dimensions
Now consider the case that the charge has been extended along the z axis. This is generally called a line charge.
The usual variable for a line charge density (charge per unit length) is , and it has units (in the SI system) of
coulombs per meter.
Part B
By symmetry, the electric field must point radially outward from the wire at each point; that is, the field lines lie
in planes perpendicular to the wire. In solving for the magnitude of the radial electric field
produced by a line
charge with charge density , one should use a cylindrical Gaussian surface whose axis is the line charge. The
length of the cylindrical surface should cancel out of the expression for
. Apply Gauss's law to this situation
to find an expression for
.
Part B.1 Find the surface area of a Gaussian cylinder
Part not displayed
Part B.2 Find the enclosed charge
Part not displayed
Express
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in terms of some or all of the variables , , and any needed constants.
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=
One dimension
Now consider the case with one effective direction. In order to make a problem effectively one-dimensional, it is
necessary to extend a charge to infinity along two orthogonal axes, conventionally taken to be x and y. When the
charge is extended to infinity in the xy plane (so that by symmetry, the electric field will be directed in the z
direction and depend only on z), the charge distribution is sometimes called a sheet charge. The symbol usually
used for two-dimensional charge density is either , or . In this problem we will use . has units of coulombs
per meter squared.
Part C
In solving for the magnitude of the electric field
produced by a sheet charge with charge density , use the
planar symmetry since the charge distribution doesn't change if you slide it in any direction of xy plane parallel to
the sheet. Therefore at each point, the electric field is perpendicular to the sheet and must have the same magnitude
at any given distance on either side of the sheet. To take advantage of these symmetry properties, use a Gaussian
surface in the shape of a cylinder with its axis perpendicular to the sheet of charge, with ends of area which will
cancel out of the expression for
in the end. The result of applying Gauss's law to this situation then gives an
expression for
for both
and
.
Part C.1 Find the total electric flux out of the cylinder
Part not displayed
Part C.2 Find the charge within the Gaussian surface
Part not displayed
Express
ANSWER:
for
in terms of some or all of the variables/constants , , and
.
=
In this problem, the electric field from a distribution of charge in 3, 2, and 1 dimension has been found using
Gauss's law. The most noteworthy feature of the three solutions is that in each case, there is a different relation
of the field strength to the distance from the source of charge. In each case, the field strength varies inversely as
an integral power of the distance from the charge. In the case of a point charge (spherical symmetry, field in
three dimensions), the field strength varies as . In the case of a line charge (cylindrical symmetry, field in
two dimensions), the field strength varies as . Finally, in the case of a sheet charge (planar symmetry, field
in one dimension), the field varies as
; that is, the strength of the field is independent of the distance from
the sheet!
If you visualize the electric field using field lines, this result shows that as the number of directions in which
the electric field can point is reduced, the field lines have one dimension fewer in which to to spread out, and
the field therefore falls off less rapidly with distance. In a one-dimensional problem (sheet charge), the
extension of the charge in the xy plane means that all field lines are parallel to the z axis, and so the field
strength does not change with distance. Such a situation, of course, is impossible in the real world: In reality,
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the planar charge is not infinite, so the field will in fact fall off over long distances.
The Electric Field and Surface Charge at a Conductor
Learning Goal: To understand the behavior of the electric field at the surface of a conductor, and its relationship
to surface charge on the conductor.
A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed
within it. The conductor is completely isolated from any source of current or charge.
Part A
Which of the following describes the electric field inside this conductor?
ANSWER:
It is in the same direction as the original external field.
It is in the opposite direction from that of the original external field.
It has a direction determined entirely by the charge on its surface.
It is always zero.
The net electric field inside a conductor is always zero. If the net electric field were not zero, a current would
flow inside the conductor. This would build up charge on the exterior of the conductor. This charge would
oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
Part B
The charge density inside the conductor is:
ANSWER:
0
non-zero; but uniform
non-zero; non-uniform
infinite
You already know that there is a zero net electric field inside a conductor; therefore, if you surround any internal
point with a Gaussian surface, there will be no flux at any point on this surface, and hence the surface will
enclose zero net charge. This surface can be imagined around any point inside the conductor with the same
result, so the charge density must be zero everywhere inside the conductor. This argument breaks down at the
surface of the conductor, because in that case, part of the Gaussian surface must lie outside the conducting
object, where there is an electric field.
Part C
Assume that at some point just outside the surface of the conductor, the electric field has magnitude and is
directed toward the surface of the conductor. What is the charge density on the surface of the conductor at that
point?
Part C.1 How to approach the problem
Which of the following is the best way to solve this problem?
ANSWER: Answer not displayed
Part C.2 Calculate the flux through the top of the cylinder
Part not displayed
Part C.3 Calculate the flux through the bottom of the box
Part not displayed
Part C.4 What is the charge inside the Gaussian surface?
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Part not displayed
Hint C.5 Apply Gauss's law
Hint not displayed
Express your answer in terms of
ANSWER:
and .
=
The Electric Field inside and outside a Charged Insulator
A slab of insulating material of uniform thickness , lying between
to
along the x axis, extends infinitely in the y and z directions, as shown in
the figure. The slab has a uniform charge density . The electric field is zero in
the middle of the slab, at
.
Part A
Which of the following statements is true of the electric field
at the surface of one side of the slab?
ANSWER: Answer not displayed
Part B
Part not displayed
Part C
Part not displayed
Part D
Part not displayed
Concept and Exercises on Electric Flux
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
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Calculating Flux for Hemispheres of Different Radii
Learning Goal: To understand the definition of electric flux, and how to calculate it.
Flux is the amount of a vector field that "flows" through a surface. We now discuss the electric flux through a
surface (a quantity needed in Gauss's law):
, where is the flux through a surface with differential area
element
, and
is the electric field in which the surface lies. There are several important points to consider in
this expression:
1. It is an integral over a surface, involving the electric field at the surface.
2.
is a vector with magnitude equal to the area of an infinitesmal surface element and pointing in a direction
normal (and usually outward) to the infinitesmal surface element.
3. The scalar (dot) product
implies that only the component of
integral. That is,
, where is the angle between
normal to the surface contributes to the
and
.
When you compute flux, try to pick a surface that is either parallel or perpendicular to , so that the dot product is
easy to compute.
Two hemispherical surfaces, 1 and 2, of respective radii and , are centered at
a point charge and are facing each other so that their edges define an annular
ring (surface 3), as shown. The field at position due to the point charge is:
where is a constant proportional to the charge,
vector in the radial direction.
, and
is the unit
Part A
What is the electric flux
through the annular ring, surface 3?
Hint A.1 Apply the definition of electric flux
Hint not displayed
Express your answer in terms of
ANSWER:
,
,
, and any constants.
= Answer not displayed
Part B
What is the electric flux
through surface 1?
Hint B.1 Apply the definition of electric flux
Hint not displayed
Part B.2 Find the area of surface 1
Part not displayed
Express
ANSWER:
in terms of
,
,
, and any needed constants.
= Answer not displayed
Part C
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What is the electric flux
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passing outward through surface 2?
Hint C.1 Apply the definition of electric flux
Hint not displayed
Part C.2 Find the area of surface 2
Part not displayed
Express
ANSWER:
in terms of
,
,
, and any constants or other known quantities.
= Answer not displayed
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.
Hint A.1 Definition of flux
Hint not displayed
Part A.2 Flux through the
face
Part not displayed
Part A.3 Flux through the
face
Part not displayed
Part A.4 Flux through the
face
Part not displayed
Part A.5 Flux through the
face
Part not displayed
Hint A.6 Putting it together
Hint not displayed
Express your answer in terms of , , , and .
ANSWER:
= Answer not displayed
Part B
Part not displayed
Part C
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What is the net charge inside the cube?
Hint C.1 Gauss's law
Hint not displayed
Express your answer in terms of , , ,
ANSWER:
, and
.
= Answer not displayed
The Charge Inside a Conductor
A spherical cavity is hollowed out of the interior of a neutral conducting
sphere. At the center of the cavity is a point charge, of positive charge .
Part A
What is the total surface charge
on the interior surface of the conductor (i.e., on the wall of the cavity)?
Hint A.1 Gauss's law and properties of conductors
Hint not displayed
ANSWER:
=
Part B
What is the total surface charge
on the exterior surface of the conductor?
Hint B.1 Properties of the conductor
Hint not displayed
ANSWER:
=
Part C
What is the magnitude
of the electric field inside the cavity as a function of the distance from the point
charge? Let , as usual, denote
.
Hint C.1 How to approach the problem
The net electric field inside the conductor has three contributions:
1. from the charge ;
2. from the charge on the cavity's walls ;
3. from the charge on the outer surface of the spherical conductor
.
However, the net electric field inside the conductor must be zero. How must
happen?
and
be distributed for this to
Here's a clue: the first two contributions above cancel each other out, outside the cavity. Then the electric field
produced by inside the spherical conductor must separately be zero also. How must be distributed for this to
happen?
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After you have figured out how and are distributed, it will be easy to find the field in the cavity, either by
adding field contributions from all charges, or using Gauss's Law.
Part C.2 Charge distributions and finding the electric field
and are both uniformly distributed. Unfortunately there is no easy way to determine this, that is why a clue
was given in the last hint. You might hit upon it by assuming the simplest possible distribution (i.e., uniform)
or by trial and error, and check that it works (gives no net electric field inside the conductor).
If is distributed uniformly over the surface of the conducting sphere, it will not produce a net electric field
inside the sphere. What are the characteristics of the field produces inside the cavity?
ANSWER:
ANSWER:
zero
the same as the field produced by a point charge located at the center of the sphere
the same as the field produced by a point charge located at the position of the charge in the
cavity
0
Part D
What is the electric field
outside the conductor?
Hint D.1 How to approach the problem
The net electric field inside the conductor has three contributions:
1. from the charge ;
2. from the charge on the cavity's walls ;
3. from the charge on the outer surface of the spherical conductor
.
However, the net electric field inside the conductor must be zero. How must
happen?
and
be distributed for this to
Here's a helpful clue: the first two contributions above cancel each other out, outside the cavity. Then the electric
field produced by inside the spherical conductor must be separately be zero also. How must be distributed
for this to happen? What sort of field would such a distribution produce outside the conductor?
Hint D.2 The distribution of
If is distributed uniformly over the surface of the conducting sphere, it will not produce a net electric field
inside the sphere. What are the characteristics of the field it produces outside the sphere?
ANSWER:
zero
the same as the field produced by a point charge located at the center of the sphere
the same as the field produced by a point charge located at the position of the charge in the
cavity
Now a second charge, , is brought near the outside of the conductor. Which of the following quantities would
change?
Part E
The total surface charge on the wall of the cavity,
:
Hint E.1 Canceling the field due to the charge
The net electric field inside a conductor is always zero. The charges on the inner conductor cavity will always
arrange themselves so that the field lines due to charge do not penetrate into the conductor.
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would change would not change
Part F
The total surface charge on the exterior of the conductor,
:
Hint F.1 Canceling the field due to the charge
The net electric field inside a conductor is always zero. The charges on the outer surface of the conductor will
rearrange themselves to shield the external field completely. Does this require the net charge on the outer surface
to change?
ANSWER:
would change would not change
Part G
The electric field within the cavity,
ANSWER:
:
would change would not change
Part H
The electric field outside the conductor,
ANSWER:
:
would change would not change
Finding E-Fields Using Gauss' law
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
at a distance
from the center of the ball?
Hint A.1 Gauss's law
Hint not displayed
Part A.2 Find
Part not displayed
Express your answer in terms of ,
ANSWER:
, , and
.
= Answer not displayed
Part B
What is the magnitude of the electric field
at a distance
from the center of the ball?
Part B.1 How does this situation compare to that of the field outside the ball?
Part not displayed
Express your answer in terms of , ,
ANSWER:
, and
.
= Answer not displayed
Part C
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Let
represent the electric field due to the charged ball throughout all of space. Which of the following
statements about the electric field are true?
1.
.
2.
.
3.
.
4. The maximum electric field occurs when
5. The maximum electric field occurs when
6. The maximum electric field occurs as
.
.
Hint C.1 Plot the electric field
Hint not displayed
Enter t (true) or f (false) for each statement. Separate your answers with commas.
ANSWER: Answer not displayed
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?
Hint A.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 A.2 Apply Gauss' law
Gauss's law states that
, where
is the electric flux through a Gaussian surface, and is the total charge
enclosed by the surface. Construct a cylindrical Gaussian surface with radius and length
with
.
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Part B
What is
, the surface charge density (charge per unit area) on the inner surface of the conducting shell?
Part 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:
= 0
Part B.2 Find the charge contribution from the surface
What is
surface?
, the total charge on the inner surface of the cylindrical shell that is contained within the Gaussian
Express your answer in terms of
ANSWER:
and .
=
To obtain the charge density per unit area, divide
that is contained within the Gaussian surface.
ANSWER:
by the area of the inner surface of the conducting shell
=
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
.)
Part C.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 D.1 How to approach the problem
Hint not displayed
Part D.2 Find the charge within the Gaussian surface
Part not displayed
Part D.3 Find the flux in terms of the electric field
Part not displayed
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=
A Charged Sphere with a Cavity
An insulating sphere of radius , centered at the origin, has a uniform volume charge density .
Part A
Find the electric field
inside the sphere (for < ) in terms of the position vector .
Hint A.1 How to approach the problem
Hint not displayed
Part A.2 Determine the enclosed charge
Part not displayed
Part A.3 Calculate the integral over the Gaussian surface
Part not displayed
Express your answer in terms of
ANSWER:
, , and
.
= Answer not displayed
Part B
Part not displayed
Summary
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4 of 11 problems complete (35.13% avg. score)
19.32 of 20 points
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