MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assig... 1 of 13 17/4/07 14:59 [ Assignment View ] 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 [
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 LawLearning 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 AHow should the integral in Gauss's law be evaluated?
ANSWER: Answer not displayed
Part BPart 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
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 ADetermine the magnitude by applying Gauss's law.
Part A.1 Find the area of the surfacePart 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 BBy 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 cylinderPart not displayed
Part B.2 Find the enclosed chargePart not displayed
Express in terms of some or all of the variables , , and any needed constants.
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 zdirection 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 CIn 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 tothe 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 cylinderPart not displayed
Part C.2 Find the charge within the Gaussian surfacePart not displayed
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
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 AWhich 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 BThe charge density inside the conductor is:
ANSWER: 0non-zero; but uniformnon-zero; non-uniforminfinite
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 CAssume 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 problemWhich 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 cylinderPart not displayed
Part C.3 Calculate the flux through the bottom of the boxPart not displayed
Part C.4 What is the charge inside the Gaussian surface?
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 AWhich of the following statements is true of the electric field at the surface of one side of the slab?
ANSWER: Answer not displayed
Part BPart not displayed
Part CPart not displayed
Part DPart not displayed
Concept and Exercises on Electric Flux
Calculating Electric Flux through a DiskSuppose 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.
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:
It is an integral over a surface, involving the electric field at the surface.1. 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.2.
The scalar (dot) product implies that only the component of normal to the surface contributes to the integral. That is, , where is the angle between and .
3.
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, , and is the unit vector in the radial direction.
Part AWhat is the electric flux through the annular ring, surface 3?
Hint A.1 Apply the definition of electric fluxHint not displayed
Express your answer in terms of , , , and any constants.
ANSWER: = Answer not displayed
Part BWhat is the electric flux through surface 1?
Hint B.1 Apply the definition of electric fluxHint not displayed
Part B.2 Find the area of surface 1Part not displayed
Express in terms of , , , and any needed constants.
What is the electric flux passing outward through surface 2?
Hint C.1 Apply the definition of electric fluxHint not displayed
Part C.2 Find the area of surface 2Part not displayed
Express in terms of , , , and any constants or other known quantities.
ANSWER: = 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 AFind the total electric flux through the surface of the cube.
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 AWhat 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 conductorsHint not displayed
ANSWER: =
Part BWhat is the total surface charge on the exterior surface of the conductor?
Hint B.1 Properties of the conductorHint not displayed
ANSWER: =
Part CWhat 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 problemThe net electric field inside the conductor has three contributions:
from the charge ;1.from the charge on the cavity's walls ;2.from the charge on the outer surface of the spherical conductor .3.
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 inside the spherical conductor must separately be zero also. How must 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.
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: zerothe same as the field produced by a point charge located at the center of the spherethe same as the field produced by a point charge located at the position of the charge in the
cavity
ANSWER: 0
Part DWhat is the electric field outside the conductor?
Hint D.1 How to approach the problemThe net electric field inside the conductor has three contributions:
from the charge ;1.from the charge on the cavity's walls ;2.from the charge on the outer surface of the spherical conductor .3.
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: zerothe same as the field produced by a point charge located at the center of the spherethe 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 EThe 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.
Part FThe 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 GThe electric field within the cavity, :
ANSWER: would change would not change
Part HThe 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 DensityA solid ball of radius has a uniform charge density .
Part AWhat is the magnitude of the electric field at a distance from the center of the ball?
Hint A.1 Gauss's lawHint not displayed
Part A.2 Find Part not displayed
Express your answer in terms of , , , and .
ANSWER: = Answer not displayed
Part BWhat 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
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.The maximum electric field occurs when .4.The maximum electric field occurs when .5.The maximum electric field occurs as 6.
Hint C.1 Plot the electric fieldHint 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 AWhat 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 symmetryBecause 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' lawGauss'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 coaxial with the rod, with .
Part BWhat is , the surface charge density (charge per unit area) on the inner surface of the conducting shell?
Part B.1 Apply Gauss's lawThe 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 surfaceWhat 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 CWhat 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 DWhat is the radial component of the electric field, , outside the shell?
Hint D.1 How to approach the problemHint not displayed
Part D.2 Find the charge within the Gaussian surfacePart not displayed
Part D.3 Find the flux in terms of the electric fieldPart not displayed
