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
therefore reduced to
, where
field on the Gaussian surface, and
. The flux integral is
is the magnitude of the electric
is the area of the surface.
Part A
Determine the magnitude
Express
by applying Gauss's law.
in terms of some or all of the variables/constants , , and
.
=1/(4*pi*epsilon_0)*q/r^2Correct
Two dimensions
Part B
Now consider the case that the point charge has been extended along the z axis. This is
generally called a _____ charge.
Answer with one word.
lineCorrect
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 C
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
this situation to find an expression for
Express
. Apply Gauss's law to
.
in terms of some or all of the variables , , and any needed constants.
=1/(2*pi*epsilon_0) * lambda/rCorrect
One dimension
Now consider the case with one effective direction.
Part D
In order to make a problem effectively one-dimensional, it is necessary to extend a point
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 usually
called a _______ charge.
Answer in one word.
planeCorrect
The usual variable for a sheet or surface charge is , and the charge density has units (in
the SI system) of coulombs per meter squared.
Part E
In solving for the magnitude of the electric field
produced by a sheet charge with
charge density , one may use a Gaussian surface in the shape of a rectangular box two of
whose faces lie a distance above and below the sheet of charge. The area
of these
faces must then be calculated; they 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
both
and
Express
for
for
.
in terms of some or all of the variables/constants , , and
.
=1/(2*epsilon_0) * etaCorrect
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
is, the strength of the field is independent of the distance from the sheet!
; that
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.
Part A
Which of the following describes the electric field inside this conductor?
It is always zero.Correct
The electric field inside a conductor is always zero. If the 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:
0
Correct
You already know that there is a zero 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?
Express your answer in terms of
=-epsilon_0*ECorrect
and
.
The Charge Inside a Conductor
Part A
What is the total surface charge
wall of the cavity)?
=-qCorrect
Part B
What is the total surface charge
=qCorrect
on the interior surface of the conductor (i.e., on the
on the exterior surface of the conductor?
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
.
Correct
Part D
What is the electric field
outside the conductor?
the same as the field produced by a point charge located at the center of the sphere
Correct
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,
:
would not changeCorrect
Part F
The total surface charge on the exterior of the conductor,
would not changeCorrect
:
Part G
The electric field within the cavity,
would not changeCorrect
:
Part H
The electric field outside the conductor,
would changeCorrect
:
A Conducting Shell around a Conducting Rod
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.
=lambda/(2*pi*r*epsilon_0)Correct
Part B
What is
, the surface charge density (charge per unit area) on the inner surface of the
conducting shell?
=-lambda/(2*pi*r_1)Correct
Part C
What is
, the surface charge density on the outside of the conducting shell? (Recall
from the problem statement that the conducting shell has a total charge per unit length
given by
.)
=-lambda/(2*PI*r_1)Correct
Part D
What is the radial component of the electric field,
=-lambda/(2*pi*r*epsilon_0)Correct
, outside the shell?