Download Lesson 2 Flux and Gauss`s Law Charles Augustine de Coulomb

Lesson 2
Flux and Gauss’s Law
Charles Augustine de Coulomb (1736-1806) designed his famous experiment to
measure the force relationships between charged bodies: Coulomb’s law is the
resulting empirical statement. By contrast, Gauss’s law (Karl Friedrich Gauss, 17771855), which you will learn in this lesson, has a more abstract origin: it developed out of
mathematical theorem. Scientists in Gauss’s nineteenth century were much more
inclined than we are today to equate purely mathematical results with physical reality.
When it was realized that Gauss’s (mathematical) theorem could be applied to the
electric field concepts of Faraday to produce Gauss’s (physical) law, this extension was
eagerly accepted. The origins of the law, however, continued and still continue to lie in
the domain of pure logic; therefore they may be somewhat inaccessible to you within
the context of a beginning physics course. Your text and this lesson will use both
physical and mathematical arguments and examples to help you achieve a mastery of
these ideas and their applications.
2-1: Electric Flux
OBJECTIVES: Visualize the net flux through a surface in terms of the flux lines passing
through the surface.
Use the concepts of a vectorial surface area element and surface
integral to define the flux of a vector field through a given closed
surface.
Calculate electric flux through simple surfaces by evaluating integrals
over these surfaces.
PREREQUISITES: Using the definition of the dot product of two vectors
Reading Assignment
Study in your textbook Chapter 22, Sections 1 and 2.
Commentary
You have seen from your reading in Lesson 1 that you can picture electric lines of force
originating from each positive charge, following smooth curves, and continuing on until
they (possibly) end on some negative charge. You will recall that the tangent to a line of
force at any point gives the direction of the electric field at the point. We now carry
this pictorial representation of the electric field a little further, and specify the number of
lines to be drawn per unit of electric field strength.
This number is of course arbitrary since the lines of force are an abstraction.
Nonetheless, if we relate the number of lines through a unit surface area in space to the
average strength of an electric field over that surface, then we can treat “lines per unit
area” as a measure of electric field strength. In our SI system of units, we visualize
as representing an electric field of
.
By counting up the number of lines of force (often called lines of flux of flux lines in this
context) that pass through a surface, we can determine the total electric flux,
,
through the surface. This surface could be “open” or “closed”. A closed surface is one
that divides all space into two regions (inside and outside); any surface that does not
completely enclose a definite volume in this way is an open surface.
Note that the “counting up” of flux lines to determine total flux must be done
algebraically; if we add one for each flux line that passes through the surface in a given
direction, we must subtract one for each flux line that passes through the surface in
opposite direction. This gives the correct magnitude for
, but could leave some
ambiguity about whether the sum is positive or negative for an open surface. We can
easily avoid this ambiguity for closed surfaces, with which we are primarily concerned,
by specifying that we shall count as positive any flux line whose direction is outward
through the closed surface.
Lines of flux are quite useful for visualizing the flux through a surface (see Figures 22.2
and 22.13 in the text) but for quantitative calculations we define flux in terms of an
integral. In Figure 2-1(a) is shown a volume enclosed by a surface, much like the
surface of a distorted balloon. The surface can be divided into area elements
. You
may have done this in a calculus class for regular shapes such as cylinders, spheres,
and triangles. You probably have used only the magnitudes ( ) of these area
elements, however, rather than treating them as vectors. The direction of the vectorial
surface-area element
(or
) is perpendicular to the surfaces, and its sense is
toward the outside of the closed surface.
Now suppose you were told the value of the electric field
Each surface-area element
shown in Figure 2-1(b).
everywhere on the surface.
can then be associated with a particular value of
The product
, as
is found by a straightforward scalar
multiplication of vectors, and this product will be an infinitesimal scalar quantity;
.
Figure 2-1.
What if someone now asked you to find the total value of
volume? You would have to add up all of the
for the whole surface of the
’s associated with all of the
’s that
make up the surface. This sum is written as an integral:
.
all surface
of volume
all surface
of volume
This integral is called a surface integral, and we will use the notation
, with a
closed circle on the integral sign to indicate that it is taken over a closed surface.
Remember that is the net outward flux through such a closed surface. Study Sample
Problems 22.2 and 22.3 in the text before doing the practice exercise.
Practice Exercise 2-1
Write your solutions to the following problems in your notebook.
1. The electric field at the surface of a sphere (radius
) is constant in
magnitude and is everywhere inwardly perpendicular to the surface.
surface is
. Find the value of
for this
.
2. The diagram at right shows a closed
surface with six planar faces (the three
unseen faces are parallel to the
and
planes, respectively). The electric field
in this space is uniform and is described
by
. Evaluate
for each of
the six faces and then add these (scalar)
values to find
for the entire
closed surface.
Check your answers with the key and review Section 2-1 as necessary before going on
to the next section.
2-2: Gauss’s Law
OBJECTIVES: State Gauss’s law in integral form and explain the meaning of all
symbols used in this statement.
Show the equivalence of Gauss’s law and Coulomb’s law.
Recognize cases where Gauss’s law does not yield a simple analytic
solution for the electric field due to a static charge distribution, and
explain why it is not useful in these cases.
Reading Assignment
Study in your textbook, Chapter 22, Section 3.
Commentary
A Gaussian surface is simply a hypothetical closed surface enclosing a region of space.
Gauss’s law states that the net charge enclosed by any such surface is related to the
flux of the electric field through the surface:
.
Although Gauss’s law is always true, it is not always useful. The integral
can be solved simply and analytically for
only if
can be factored out of the
integral. If we can somehow determine (from the symmetry of the charge arrangement)
a surface for which
is constant, we may write Gauss’s law as
and solve for
as
You can see that in order for
to factor out,
must have a constant component in
the direction perpendicular to the surface, that is, parallel to every
vector on the
surface of integration. Or, as you saw in problem 2 of the last practice exercise, it may
be possible to divide a surface into smaller surfaces over which
has a fixed
relationship to
. We can then evaluate the flux integrals for these surfaces and take
their algebraic sum to find the total flux through the Gaussian surface, and hence the
net enclosed charge from
Let us return to lines of flux for a more visual and geometric interpretation of Gauss’s
law. Recall that we draw
in the appropriate direction to represent an electric
field strength of
. On a spherical surface of
radius centered on a
positive
point charge, the (constant) electric field strength is
This is equivalent to (
) radial lines per square meter through the
area of the sphere; thus by convention we must draw
surface
flux lines originating on the
positive charge and we draw an identical number of lines terminating on a negative
charge.
The equivalence of Gauss’s law and Coulomb’s law is perhaps most easily understood
in terms of flux lines. Coulomb’s law postulates that the electrostatic force between two
point charges is inversely proportional to the square of the distance between them and
the electric field surrounding an isolated charge
therefore has, by definition, an
identical inverse-square-law dependence.
In our Gauss’s-law picture, the radial flux lines originating or terminating on an isolated
charge extend indefinitely, so that the number of lines passing through any spherical
surface centered on this charge remains constant as long as no other charge is
enclosed. The number of lines through a unit of surface area (and hence the field
strength) is thus inversely proportional to the area of the spherical surface; that is, to .
This simple geometric argument should help you to see how the same basic statement
is embodied in both Coulomb’s law and Gauss’s law: the electric field strength due to a
point charge depends only on distance from the charge and the amount of charge. Any
violation of the inverse-square dependence of Coulomb’s law would be equivalent to a
flux line originating or terminating in the absence of any charge; that is, to a violation of
Gauss’s law.
Consider Figure 2-2(a). Assuming that the flux lines entering and leaving the cube are
properly drawn relative to the field strength, we can count the net number of lines
through the cubical surface, multiply by , and equate this with (net) coulombs of
charge inside the cube. We can do so even if we cannot actually “see” inside the cube
[Figure 2-2(b)].
Figure 2-2.
In practice, we can “count lines” (i.e., evaluate the integral
) only if we can infer,
from the symmetry of the charge arrangement, a reasonably simple surface over which
is constant; we will consider this procedure in more detail in the next section.
They symmetry constraint should suggest to you why Gauss’s law is of limited practical
use for determining the field due to a given charge distribution: we must, in effect, be
able to assume all characteristics of (except magnitude) before we can use Gauss’s
law to find
. You can then check analytically your assumptions about the field due to
a given arrangement of charges by writing the equation(s) for your proposed Gaussian
surface and using Coulomb’s law (in vector form) to verify that the net component of
the force on a test charge is the same at any point on your surface (i.e., that
indeed constant).
Practice Exercise 2-2
Write your solutions to the following problems in your
notebook.
1. State Gauss’s law in integral form and explain all of
the symbols.
2. The three figures in the figure at right are fixed at the
vertices of an equilateral triangle. Explain why
is
Gauss’s law is not useful for finding
at any nearby point.
Check your answers with the key and review Section 2-2 as necessary before going on
to the next section.
2-3: Applications of Gauss’s Law
OBJECTIVES:
Use Gauss’s law to (a) determine the electric field due to certain
symmetric charge distributions; or (b) determine the net charge inside
volumes where the electric field is known everywhere on the surface of
the volume.
Reading Assignment
Study in your textbook Chapter 22, Section 4.
Commentary
Even though, as you saw in problem 2 of Practice Exercise 2-2, Gauss’s law generally
does not yield simple solutions for the local field due to arrangement of point charges, it
is a useful tool for analyzing the fields that arise from certain simple continuous charge
distributions. Such distributions can be characterized by a charge density, which may
be either three-dimensional (symbolized by , where
), two-dimensional
(surface charge density
;
) or one-dimensional (linear charge density
;
).
The fields we can analyze easily with Gauss’s law are those resulting from known
charge distributions that have sufficient symmetry. This includes distributions having
radial symmetry as well as uniform distributions that extend “infinitely” in one or two
dimensions. (We can approximate as infinite a distribution whose extent is much
greater than the area in which we want to analyze the field.) The following steps outline
the general procedure for using Gauss’s law to evaluate the field due to a given charge
distribution.
Step 1. Deduce the direction of
from the symmetry of the charge distribution and
Coulomb’s law. For example, for a spherically symmetric distribution,
must be radial,
i.e., must point away from (or toward) the symmetry center of the distribution.
Step 2. Use the symmetry of the charge distribution to determine the locus of points for
which must be constant in magnitude; e.g., for spherical symmetry the magnitude of
is necessarily the same as all points on the surface of any sphere with symmetry center.
Step 3. With Steps 1 and 2 as guides, determine a closed (Gaussian) surface such that
at each point on the surface, either (a) is perpendicular to the surface and of constant
magnitude
, or (b) is in the plane of the surface; i.e., has no component normal to the
surface.
Step 4. Let
be the area of that portion of the Gaussian surface for which
is normal
and of constant magnitude . The electric flux for this part of the surface is
; the
flux for the remaining portion of the surface (if there is any such part) is then zero since
has no normal component. Thus the surface integral over the Gaussian surface is
.
Step 5. Now set the electric flux found in Step 4 equal to the net enclosed charge
multiplied by
:
or
This gives the magnitude of
.
at any point on that part of the surface where
is normal
to the surface.
Study the following example before doing the practice exercise.
EXAMPLE 2-1: Charge is distributed with a uniform density
(infinite) cylindrical rod of radius
throughout a long
as in the figure below.
Let
measure the distance from the symmetry axis of the cylinder to a point.
Determine
.
Solution: The Gaussian surface is depicted on the next page, followed by the steps
used in determining this surface and
.
Step 1. From the symmetry of the charge distribution,
must be radial, i.e.,
perpendicular to the symmetry axis.
Step 2. The charge symmetry ensures that
points a distance
will have the same magnitude at all
from the axis.
Step 3. The Gaussian surface is a cylinder of length
with the symmetry axis. On the curved surface
constant in magnitude (Step 2). On the flat ends,
and radius
concentric
is perpendicular (Step 1) and
has no normal component
(Step 1).
Step 4. The area of the curved part of the Gaussian surface is
this part that
is parallel to
. It is on
and of constant magnitude. Therefore the electric
flux for the curved part is
.
Since
is perpendicular to
on the two flat ends of the cylinder, the flux
through these two “caps” is zero. Therefore
Gaussian surface.
is the total flux through the
Step 5. Cleary the charge enclosed by the Gaussian surface depends upon
whether
or
(see diagram). For
, the surface encloses all of the
charge in a length
of the rod, namely,
;
but for
at radius
, the surface encloses only that charge inside the Gaussian surface
and length , namely,
.
Now using Gauss’s law and equating the electric flux to
times the net
enclosed charge we obtain
for
and
for
, so that
and
.
A graph of the magnitude of
versus
is shown at right. At each point
the indicated magnitude and points radially outward if
.
Practice Exercise 2-3
Write your solutions to the following problems in
your notebook.
1. Two long, thin, statically charged coaxial
cylinders are shown in the figure below.
The surface charge densities on these
cylinders (in units of coulombs per square
has
and radially inward if
. Use Gauss’s law to find :
meter) have the relationship
a. between the cylinders (show your choice of Gaussian surface);
b. outside the larger cylinder (show your choice of Gaussian surface).
2. The diagram below shows an infinite slab of charge (thickness
) with a uniform
charge density . There are no other charges in this region of space, so that the
field must be symmetric about the plane
Gauss’s law to find
for
, that is,
. Be sure to indicate clearly the (closed
Gaussian surface you are using. (b) Which way does
3. The diagram at right represents an infinitely
long cylindrical shell of charge. The uniform
charge density in this region is , in units of
. A wire carrying a uniform linear charge
density , in units of
, is coaxial with the
cylinder as shown. If
is the radial distance
from the wire, use Gauss’s law to find:
. (a) Use
point if
?
a.
in the region
b.
in the region
c.
in the region
(show your choice of Gaussian surface);
(show your choice of Gaussian surface);
(show your choice of Gaussian surface).
4. A spherical region of
radius contains a uniform charge density
. The charge density outside this region is zero. Use
Gauss’s law to find :
a.
from the center of the sphere (indicate your choice of Gaussian
surface);
b.
from the center of the sphere (indicate your choice of Gaussian
surface).
Check your answer with the key and review Section 2-3 as necessary before going on
to the next section.
2-4: An Insulated Conductor
OBJECTIVES: Given a conductor with a static charge distribution, use the properties of
a conductor and/or Gauss’s law to
(a) explain why the electric field is necessarily perpendicular to the surface of the
conductor
(b) explain why the electric field is necessarily zero inside the conductor,
(c) explain why the excess charge is on the surface of the conductor, and
(d) solve problems involving charges applied to symmetrical conductors.
PREREQUISITES:
Distinguishing between conductors and insulators from their
behavior in an electric field
Reading Assignment
Study in your textbook Chapter 22, Section 5.
Commentary
The defining characteristic of an electrical conductor is the presence of charges that are
free to move in response to the forces they feel due to non-zero electric fields.
Electrostatic equilibrium can exist inside a piece of conducting material only if the
electric field is zero at all points within the piece. Gauss’s law tells us that there can be
a zero field only if no net charge resides within the conductor. We know from
experiment that a net electric charge can be transferred to a conductor, and conclude
that such a charge must necessarily reside on the surface of the conductor.
In Section 2-2 you also saw that the equivalence of Gauss’s law and Coulomb’s law is
established by simple geometry (along with the rule that flux lines can only begin and
end on charges). Make sure that you understand how and why the relationships among
conductors, Gauss’s law, and spatial geometry allow us to verify the exponent ( ) in
Coulomb’s law by testing the Gauss’s law prediction concerning the distribution of
charge in a conductor. The following example shows how fields due to charges on
symmetrical conductors can be analyzed quantitatively with Gauss’s law.
EXAMPLE 2-2: A point charge
carrying charges
is surrounded by two spherical, thin metal shells
and
, as shown in the diagram below (only one
quadrant of the spheres is shown). Point
distance
from the origin. Point
is between the shells a
distance
from the origin. Point
is outside the shells a
distance
Use Gauss’s law to find
,
is inside the inner shell a
, and
from the origin. The charges are
at
. Indicate your
Gaussian surfaces.
Solution: Point : A spherical shell of radius
Point
will be a good Gaussian surface.
is located on this surface; it is the locus of points with constant
is parallel to
everywhere on this surface.
is outward since
, and
is
positive. From Gauss’s law,
,
.
Thus,
.
Point B: Since the charged shell will not alter the spherical symmetry of this
problem, another spherical shell of radius
is picked for the Gaussian
surface.
Now
the
total
charge
inside
. As before,
this
surface
is
.
The direction of
is again radially outward since the net charge inside the
Gaussian surface is positive.
Point C: Now the net charge inside a spherical Gaussian surface of radius
is zero.
cannot be perpendicular to
; this would violate the
spherical symmetry. Thus
.
Practice Exercise 2-4
Write your solutions to the following problems in your notebook.
1. a. Do you need to use Gauss’s law to show that
is perpendicular to the
surface of a conductor with a static charge distribution?
b. Do you need to use Gauss’s law to show that
is zero inside a conductor
with a static charge distribution?
c. Do you need to use Gauss’s law to show that the excess charge is on the
surface of a conductor with a static charge distribution?
2. A large, statically charged,
flat conducting plate is
shown in the figure at right.
The charge density is (in
units of coulombs
square meter).
per
a. Why is the charge density specified in units of
b. Use Gauss’s law to find
instead of
?
outside the plate. Show your choice of Gaussian
surface.
Check your answers with the key and review Section 2-4 as necessary before doing the
self-check test. To see if you have achieved the objectives in this lesson, try to solve
the problems in the Self-Check Test 2 without using any reference materials.
Self-Check Test 2
Write your solutions to the following problems in you notebook.
1. State Gauss’s law and briefly explain all its symbols.
2. Gauss’s law is always true but not always useful. What mathematical condition
must be met in order for Gauss’s law to be a useful tool for determining the
electric field caused by a static charge distribution?
3. Use Gauss’s law to answer these questions:
a. If a Gaussian surface encloses zero net charge, is it necessarily true that
at all points on the surface?
b. If
everywhere on a Gaussian surface, is the net charge inside
necessarily zero?
4. The top and bottom of the
cylindrical can at right each have
an area of
. In this region
of space there is a layer of charge,
some of which is inside the can, so
that
is larger than
; but an
additional field is superimposed so
that the resultant field points up
everywhere. If
and
,
how
much charge is contained in the can?
5. You are shown a conductor with a static charge distribution. Use the properties
of a conductor and/or Gauss’s law to:
a. Explain why
is parallel or anti-parallel to
at the surface of the conductor.
b. Explain why the excess charge lies on the surface of the conductor.
Check your answers with the key and review Lesson 2 as necessary.
Assignment 2
When you have demonstrated mastery of the content of this lesson, log into Mastering
Physics and work Assignment 2.