PH213 Chapter 28 – solutions – part 1 Gauss's Law in 3, 2, and 1 Dimension
Description: Review question on Gauss's law. Use Gauss's law to find the field due to a point, an
infinite line, and an infinte plane charge distribution. Compare the electric field's dependance on
distance in each case.
Gauss's law relates the electric flux
the surface:
through a closed surface to the total charge
enclosed by
.
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 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
is the magnitude of the electric field on the Gaussian surface, and
surface.
, where
is the area of the
Part A
Determine the magnitude
Hint
A.1
by applying Gauss's law.
Find the area of the surface
Find the area of a spherical surface of radius
Express your answer in terms of
surrounding the point charge.
and other constants.
ANSWER:
=
Express
in terms of some or all of the variables/constants
ANSWER:
=
Two dimensions
,
, and
.
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
SI system) of coulombs per meter.
, and it has units (in the
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
expression for
Hint
B.1
should cancel out of the
. Apply Gauss's law to this situation to find an expression for
.
Find the surface area of a Gaussian cylinder
Find
, the area of the Gaussian surface. Note that you do not need to include the ends of the
cylinder in the calculation of area. Since the electric field is radial (by symmetry), the ends of the
cylinder are parallel to the field, and there is no flux through them.
Express
constants.
in terms of the length
and radius
of the surface, as well as any needed
ANSWER:
=
Hint
B.2
Find the enclosed charge
What is the charge
contained within the Gaussian cylinder?
Express your answer in terms of
,
, and any needed constants.
Hint
B.2
Find the enclosed charge
What is the charge
contained within the Gaussian cylinder?
Express your answer in terms of
,
, and any needed constants.
ANSWER:
=
Express
in terms of some or all of the variables
,
, and any needed constants.
ANSWER:
=
One dimension
Now consider the case with one effective direction. In order to make a problem effectively onedimensional, 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
.
Hint
C.1
Find the total electric flux out of the cylinder
What is the total flux
out of the ends of the cylinder used as the Gaussian surface in this
problem? Note that since the electric field is directed along the z axis by symmetry, it is parallel to
the curved side walls of the cylinder, so there is no flux through these walls.
Express your answer in terms of the area
electric field, and any needed constants.
of the ends of the cylinder, the magnitude
of the
ANSWER:
=
Hint
C.2
Find the charge within the Gaussian surface
What is the charge
inside the Gaussian surface?
Express your answer in terms of
,
, and any needed constants.
ANSWER:
=
Express
for
in terms of some or all of the variables/constants
,
, and
.
ANSWER:
=
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, 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
Description: After two conceptual questions about the charge and field inside a conductor, find the
surface charge on a conductor, given the field just above the surface.
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.
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
of the conductor at that point?
Hint
C.1
on the surface
How to approach the problem
Which of the following is the best way to solve this problem?
ANSWER:
Use Coulomb's law for the electric field from each charge element.
Use Gauss's law with a short and flat cylindrical surface (picture a coin)
with one end just below (inside) and the other just above (outside) the
surface of the conductor.
Use Gauss's law with a long and thin cylindrical surface (picture a straw
capped at both ends) with one end far below (but inside the object) and
the other far above the surface of the conductor.
None of these: You need to know the surface charge distribution
everywhere on the surface.
None of these: You need to know the electric field at all points on the
surface.
A straightforward way to solve this problem is to choose a Gaussian surface one end of which is
just above and the other just below the surface of the conductor.
Hint
C.2
Calculate the flux through the top of the cylinder
Using a flat cylinder with large top and bottom each of area
surface of the conductor, find the flux
the electric field of magnitude
Express your answer in terms of
just above and just below the
generated through the top surface of the cylinder by
that points into the surface.
,
, and any needed constants.
ANSWER:
=
Hint
C.3
Calculate the flux through the bottom of the box
Using a flat cylinder with large top and bottom of area
just above and just below the surface of
the conductor, find the flux
generated through the bottom surface of the cylinder by the
electric field inside the conductor (keep in mind that positive flux is outward through the cylinder's
surface, which is downward into the conductor).
Answer in terms of
,
, and any needed constants.
ANSWER:
=
ANSWER:
=
Hint
C.4
What is the charge inside the Gaussian surface?
Find the net charge
inside this Gaussian surface.
Express your answer in terms of the charge density
and other given quantities.
ANSWER:
=
Notice that the surface charge density has dimensions of charge per area. Therefore, when
multiplied by an area, the result is charge, which is needed for Gauss's law.
Hint
C.5
is
Apply Gauss's law
Now apply Gauss's law, neglecting any contribution to the flux due to the very short sides of the
cylinder. Gauss's law states that
Express your answer in terms of
. The area
and
should cancel out of your result.
.
ANSWER:
=
A Conducting Shell around a Conducting Rod
Description: An infinite charged rod sits at the center of an infinite conducting cylindrical shell.
Determine the field between the rod and shell, the field outside the shell, and the surface charge on the
inner and outer surfaces of the shell.
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
shown in the figure.
and radius
, as
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
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
and
radius
).
, where
is the electric flux through a Gaussian surface,
is the total charge enclosed by the surface. Construct a cylindrical Gaussian surface with
and length
coaxial with the rod, with
.
Hint
A.3
Find the charge inside the Gaussian surface
What is the total charge
enclosed by the surface?
ANSWER:
=
Hint
A.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:
=
Express your answer in terms of
,
, and
, the permittivity of free space.
ANSWER:
=
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
surface, be?
, the total charge inside this Gaussian
ANSWER:
=
Hint
B.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
by the area of the inner surface of the
conducting shell that is contained within the Gaussian surface.
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 has a total charge per unit length given by
Hint C.1
What is
.)
What is the charge on the cylindrical shell?
, the total surface charge (the sum of charges on the inner and outer surfaces) of a
portion of the shell of length
?
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?
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,
Hint
D.1
, outside the shell?
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
you need to take
Hint
D.2
and radius
, coaxial with the rod. This time,
.
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 D.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
ANSWER:
=
ANSWER:
and given variables.
Express your answer in terms of the magnitude of the electric field
and given variables.
ANSWER:
=
ANSWER:
=
Charge Distribution on a Conductor with a Cavity
Description: Conceptual problem. Positive charge sits outside a conducting shell. What is the resulting
charge distribution on the inside and outside of the shell.
A positive charge is brought close to a fixed neutral conductor that has a cavity. The cavity is neutral;
that is, there is no net charge inside the cavity.
Part A
Which of the figures best represents the charge distribution on the inner and outer walls of the
conductor?
Hint
A.1
Conductors have no internal field
At steady state, conductors have no internal electric field (otherwise, charge would flow).
Therefore, the arrangement of charges on the surfaces of the conductor must exactly cancel out any
external electric field to ensure that the internal field is zero.
Hint
A.2
Charges on the cavity walls
Think about what the answer would be for a conductor without a cavity. Would there be net
charges on the surface of some imaginary sphere drawn inside of the conductor? Would this change
if you removed all of the material inside of that sphere?
Hint
A.2
Charges on the cavity walls
Think about what the answer would be for a conductor without a cavity. Would there be net
charges on the surface of some imaginary sphere drawn inside of the conductor? Would this change
if you removed all of the material inside of that sphere?
ANSWER:
1
2
3
Flux through a Cube
Description: Find the electric flux through a cube given an algebraic formula for the electric field in all
space. Then determine the charge enclosed by the cube using Gauss's law.
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
Hint
A.1
through the surface of the cube.
Definition of flux
The net electric flux
of a field
through a closed surface S is given by
,
.
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
A.2
Flux through the
face
Consider the face of the cube whose outward normal points in the positive x direction. What is the
flux
Hint
A.2.1
through this face?
Simplifying the integral
The field
depends only on the spatial variable
so
is constant. Since
integral:
. On the
face of the cube,
is constant over this entire surface, it can be pulled out of the
.
Hint
A.2.2
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
A.2.3
This
Find the area of the face of the cube
face of the cube is a square with sides of length
ANSWER:
=
Express your answer in terms of
ANSWER:
,
,
, and
.
. What is the area of this face?
,
ANSWER:
=
Express your answer in terms of
,
,
, and
.
ANSWER:
=
Hint
A.3
Flux through the
face
Consider the face of the cube whose outward normal points in the positive y direction. What is the
flux
through this face?
Express your answer in terms of
,
,
, and
.
ANSWER:
=
Hint
A.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 A.4.1
Consider the orientation of the field
The electric field has no z component (the field vector
lies entirely in the xy plane. What is
the dot product of a vector in the xy plane and a vector normal to the
face of the cube?
ANSWER:
=
Hint
A.5
Flux through the
face
Consider the face of the cube whose outward normal points in the negative x direction. What is the
flux
Hint
A.5.1
through this face?
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.,
). Find the x component
ANSWER:
=
of the electric field across this surface.
Hint
A.5.1
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.,
). Find the x component
of the electric field across this surface.
ANSWER:
=
Hint
A.5.2
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.
Express your answer in terms of
,
,
, and
.
ANSWER:
=
Hint
A.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.
Express your answer in terms of
,
,
, and
.
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
,
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
Hint
C.1
inside the cube?
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:
.
Express your answer in terms of
,
,
,
, and
.
ANSWER:
=
The Charge Inside a Conductor
Description: A spherical conductor has a cavity containing a fixed charge. Conceptual questions about
how charge is arranged on the surfaces of a conductor to cancel the field due to the fixed charge and
how the charge arrangement would change if an additional external charge were brought near the
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
cavity)?
Hint A.1
on the interior surface of the conductor (i.e., on the wall of the
Gauss's law and properties of conductors
The net electric field in the interior of the conducting material must be zero. (The electric field in
the cavity, however, need not be zero.) Knowing this, you can use Gauss's law to find the net
charge on the interior surface of the cavity. Use the following Gaussian surface: an imaginary
sphere, centered at the cavity, that has an infinitesimally larger radius than that of the cavity, so that
it encompasses the inner surface of the cavity. This Gaussian surface lies within the conductor, so
the field on the Gaussian surface must be zero. Thus, by Gauss's law, the net charge inside the
Gaussian surface must be zero as well. But you know that there is a point charge within the
Gaussian surface. If the net charge within the Gaussian surface must be zero, how much charge
must be present on the surface of the cavity?
ANSWER:
=
Part B
What is the total surface charge
Hint B.1
on the exterior surface of the conductor?
Properties of the conductor
In the problem introduction you are told that the conducting sphere is neutral. Furthermore, recall
that the free charges within a conductor always accumulate on the conductor's surface (or surfaces,
in this case). You found the net charge on the conductor's interior surface in Part A. If the
conductor is to have zero net charge (as it must, since it is neutral), how much charge must be
present on its exterior surface?
ANSWER:
=
In the problem introduction you are told that the conducting sphere is neutral. Furthermore, recall
that the free charges within a conductor always accumulate on the conductor's surface (or surfaces,
in this case). You found the net charge on the conductor's interior surface in Part A. If the
conductor is to have zero net charge (as it must, since it is neutral), how much charge must be
present on its exterior surface?
ANSWER:
=
Part C
What is the magnitude
the point charge? Let
Hint
C.1
of the electric field inside the cavity as a function of the distance
, as usual, denote
from
.
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
and
be
distributed for this to happen?
Here's a clue: the first two contributions above cancel each other out, outside the cavity. Then the
electric field produced by
must
inside the spherical conductor must separately be zero also. How
be distributed for this to happen?
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.
Hint
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
cavity?
ANSWER:
ANSWER:
produces inside the
zero
the same as the field produced by a point charge
of the sphere
0
located at the center
the same as the field produced by a point charge located at the position
of the charge in the cavity
ANSWER:
0
Part D
What is the electric field
Hint
D.1
outside the conductor?
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
and
be
distributed for this to happen?
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
the sphere
located at the center of
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
conductor.
ANSWER:
do not penetrate into the
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
± The Charge on a Thundercloud
Description: ± Includes Math Remediation. Estimate the charge on a thundercloud just before a
lightning strike, given the breakdown voltage of air.
In a thunderstorm, charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can
be considered to be distributed uniformly throughout the cloud. For the purposes of this problem, take
the cloud to be a sphere of diameter 1.00 kilometer. The point of this problem is to estimate the
maximum amount of charge that this cloud can contain, assuming that the charge builds up until the
electric field at the surface of the cloud reaches the value at which the surrounding air breaks down.
This breakdown means that the air becomes highly ionized, enabling it to conduct the charge from the
cloud to the ground or another nearby cloud. The ionized air will then emit light due to the
recombination of the electrons and atoms to form excited molecules that radiate light. In addition, the
large current will heat up the air, resulting in its rapid expansion. These two phenomena account for the
appearance of lightning and the sound of thunder. Take the breakdown electric field of air to be
.
Part A
Estimate the total charge
Hint
A.1
on the cloud when the breakdown of the surrounding air is reached.
Use Gauss's law
This problem calls for the use of Gauss' law:
. Choose a Gaussian surface
that lies just on the surface of the cloud and apply Gauss's law to find the charge enclosed.
Hint
A.2
Evaluate Gauss's law
Solve Gauss's law for the enclosed charge
. Keep in mind that the electric field produced by
a uniform spherical charge distribution is in the same direction as the electric field of a point
charge. Therefore, the integral should be very easy to evaluate.
Express your answer in terms of the magnitude of the electric field,
Gaussian surface,
; and the permittivity constant,
; the surface area of the
.
ANSWER:
=
Express your answer numerically, to three significant figures, using
.
ANSWER:
=
Coulombs
Flux out of a Cube
Description: If a point charge is in the center of a cube, calculate the flux through one of the faces, and
then do the same for a smaller cube.
A point charge of magnitude
Part A
is at the center of a cube with sides of length
.
What is the electric flux
Hint
A.1
through each of the six faces of the cube?
How to approach the problem
Since the magnitude and direction of
changes over the entire surface, you cannot use the
expression
to find the electric flux. Integration is possible, of course—but who likes
to integrate if there is an easier way?
In this case, there is a way indeed. Just use Gauss's law to find the total electric flux through the
cube and then exploit the symmetry of the problem.
Calculate the total electric flux
Hint
A.2
Calculate the total electric flux
Hint
A.2.1
coming out of the cube.
Calculate the flux for a sphere
If the charge were in the center of a sphere of radius
what would be the total flux
Use
instead of a cube with sides of length
out of the sphere?
for the permittivity of free space.
ANSWER:
=
Hint
A.2.2
Flux and surface shape
The total flux through any closed surface depends solely on the total amount of charge enclosed
by that surface, not on its shape.
Use
for the permittivity of free space.
ANSWER:
=
Hint
A.3
Flux through a face
Since the charge is at the center of the cube, by symmetry the flux through each face is the same.
Use
for the permittivity of free space.
ANSWER:
=
,
Since the charge is at the center of the cube, by symmetry the flux through each face is the same.
Use
for the permittivity of free space.
ANSWER:
=
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:
=
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 27.26
Description: A point charge Q_1 is distance r from the center of a dipole consisting of charges pm q_2
separated by distance s. The charge is located in the plane that bisects the dipole. At this instant,
assuming r gg s, (a) what is the magnitude of the force on ...
A point charge
distance
is distance
from the center of a dipole consisting of charges
separated by
. The charge is located in the plane that bisects the dipole. At this instant, assuming
,
Part A
what is the magnitude of the force on the dipole?
In the space provided, enter the factor that multiplies
of
,
,
ANSWER:
,
, and
.
in your answer. Express this factor in terms
ANSWER:
Part B
what is the magnitude of the torque on the dipole?
In the space provided, enter the factor that multiplies
of
,
,
,
, and
in your answer. Express this factor in terms
.
ANSWER:
Problem 27.35
Description: Three charges are on the y-axis. Charges - q are at y= pm d and charge + 2q is at y =0. (a)
Determine the electric field E_vec along the x-axis. (b) Verify that your answer to part A has the
expected behavior as x becomes very small. (c) Verify ...
Three charges are on the y-axis. Charges
are at
Part A
Determine the electric field
ANSWER:
Part B
along the x-axis.
and charge
is at
.
Verify that your answer to part A has the expected behavior as
Express your answer in terms of
,
and constant
becomes very small.
.
ANSWER:
=
Part C
Verify that your answer to part A has the expected behavior as
Express your answer in terms of
,
,
and constant
becomes very large.
.
ANSWER:
=
Problem 27.37
Description: The figure is a cross section of two infinite lines of charge that extend out of the page. The
linear charge densities are pm lambda. (a) Find an expression for the electric field strength E at height y
above the midpoint between the lines. (b) Draw ...
The figure
is a cross
section of two infinite lines of charge that extend out of the page. The linear charge densities are
Part A
.
Find an expression for the electric field strength
lines.
Express your answer in terms of the variables
at height
,
,
above the midpoint between the
and appropriate constants.
ANSWER:
=
Part B
Draw a graph of
versus
.
ANSWER:
View
Problem 27.51
Description: Two parallel plates d apart are equally and oppositely charged. An electron is released
from rest at the surface of the negative plate and simultaneously a proton is released from rest at the
surface of the positive plate. (a) How far from the...
Two parallel plates 0.600
apart are equally and oppositely charged. An electron is released from
rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface
of the positive plate.
Part A
How far from the negative plate is the point at which the electron and proton pass each other?
ANSWER:
Problem 27.60
Description: An electric field can induce an electric dipole in a neutral atom or molecule by pushing the
positive and negative charge in opposite directions. The dipole moment of an induced dipole is directly
proportional to the electric field. That is, p_vec =...
An electric field can induce an electric dipole in a neutral atom or molecule by pushing the positive and
negative charge in opposite directions. The dipole moment of an induced dipole is directly proportional
to the electric field. That is,
, where
is called the polarizability of the molecule. A bigger
field stretches the molecule farther and causes a larger dipole moment.
Part A
What are the units of
?
ANSWER:
Part B
An ion with charge
is distance
the magnitude force
from a molecule with polarizability
.
Express your answer in terms of the variables
ANSWER:
=
Part C
What is the direction of this force?
ANSWER:
Toward ion
Away from ion
. Find an expression for
,
,
and appropriate constants.
ANSWER:
Toward ion
Away from ion