Gauss\' law HW PDF

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Practice problems involving Gauss' Law...

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7/23/2021

Ch 22 HW

Ch 22 HW Due: 11:59pm on Sunday, June 6, 2021 To understand how points are awarded, read the Grading Policy for this assignment.

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: around the perimeter of a closed loop over the surface bounded by a closed loop over a closed surface

Correct In the integral for Gauss's law, the vector

represents an infinitesimal surface element. The magnitude of

the area of the surface element. The direction of volume.

is

is normal to the surface element, pointing out of the enclosed

Part B In Gauss's law, to what does

refer?

ANSWER:

the net charge inside the closed surface the charge residing on insulators inside the closed surface all the charge in the physical system any charge inside the closed surface that is arranged symmetrically

Correct The major use of Gauss's law is to determine an electric field when the charge distribution, both inside and outside the Gaussian surface, is symmetric. Of course, the electric field can always be found by adding up (or integrating) the contributions of all the charge in the problem. In highly symmetric situations, however, Gauss's law is much simpler computationally than dealing with all such contributions, and it provides better physical insight, too.

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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, th 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 t 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 charg 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

the magnitude of the electric field on the Gaussian surface, and

, where

is

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

.

You did not open hints for this part. ANSWER: =

Two dimensions

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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 cylindrica surface should cancel out of the expression for . Apply Gauss's law to this situation to find an expression for Express in terms of some or all of the variables , , and any needed constants.

You did not open hints for this part. ANSWER: =

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 infinit 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 . Express

for

in terms of some or all of the variables/constants , , and

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You did not open hints for this part. ANSWER: =

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 .

Express your answer in terms of ,

(Greek letter rho), and

.

You did not open hints for this part. ANSWER: =

Part B Complete previous part(s)

Flux out of a Cube A point charge of magnitude

is at the center of a cube with sides of length

.

Part A What is the electric flux Use

through each of the six faces of the cube?

for the permittivity of free space (not the EMF symbol

).

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Hint 1. How to approach the problem Since the magnitude and direction of changes over the entire surface, you cannot use the expression find the electric flux. Integration is possible, of course—but who likes to integrate if there is an easier way?

to

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. Hint 2. Calculate the total electric flux Calculate the total electric flux

coming out of the cube.

Hint 1. Calculate the flux for a sphere If the charge were in the center of a sphere of radius be the total flux out of the sphere?

instead of a cube with sides of length

, what would

ANSWER: =

Hint 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.

ANSWER: =

Correct

Hint 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.

ANSWER: =

Correct 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 Use

through a face of the cube if its sides were of length

?

for the permittivity of free space.

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Hint 1. 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?

ANSWER: =

Correct 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.

PSS 22.1: Gauss's Law Learning Goal: To practice Problem-Solving Strategy 22.1: Gauss's Law. An infinite cylindrical rod has a uniform volume charge density (where ). The cross section of the rod has radius the magnitude of the electric field at a distance from the axis of the rod. Assume that .

. Fin

Problem-Solving Strategy 22.1: Gauss's Law IDENTIFY the relevant concepts: Gauss’s law is most useful in situations where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly over a plane. In these situations we determine the direction of

from the symmetry of the charge distribution. If we

are given the charge distribution, we can use Gauss’s law to find the magnitude of . Alternatively, if we are given the field, we can use Gauss’s law to determine the details of the charge distribution. In either case, begin your analysis by asking the question: What is the symmetry? SET UP the problem using the following steps: 1. Select the surface that you will use with Gauss’s law. We often call it a Gaussian surface. If you are trying to find the field at a particular point, then that point must lie on your Gaussian surface. 2. The Gaussian surface does not have to be a real physical surface. Often the appropriate surface is an imaginary geometric surface. 3. If the charge distribution has cylindrical or spherical symmetry, choose the Gaussian surface to be a coaxial cylinder or a concentric sphere, respectively. EXECUTE the solution as follows: 1. Carry out the integral

The symmetry of the charge distribution and your careful choice of a Gaussian surface make the integration straightforward. 2. Often you can think of the closed Gaussian surface as being made up of several separate surfaces. The integral over the entire closed surface is always equal to the sum of the integrals over all the separate surfaces. 3. If is perpendicular (normal) at every point to a surface with area , if it points outward from the interior of the surface, and if it also has the same magnitude at every point on the surface, then , and over that surface is equal to

. If instead

is perpendicular and inward, then

and

. 4. If

is tangent to a surface at every point, then

and the integral over that surface is zero.

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5. If at every point on a surface, the integral is zero. 6. In , is always the perpendicular component of the total electric field at each point on the closed Gaussian surface. In general, this field may be caused partly by charges within the surface and partly by charges outside it. Even when there is no charge within the surface, the field at points on the Gaussian surface is not necessarily zero. In that case, however, the integral over the Gaussian surface—that is, the total electric flux through the Gaussian surface—is always zero. 7. Once you have evaluated the integral, use the equation variable.

to solve for your target

EVALUATE your answer: Often your result will be a function that describes how the magnitude of the electric field varies with position. Examine this function with a critical eye to see whether it makes sense. IDENTIFY the relevant concepts The charge distribution described in this problem has cylindrical symmetry. The electric field must point away from the positive charges distributed on the rod. Since you are given information on the charge distribution, use Gauss’s law to determine the magnitude of

at the required position.

SET UP the problem using the following steps

Part A Considering the symmetry of the charge distribution, choose one of the following options as the most appropriate choice of Gaussian surface to use in this problem. ANSWER: a finite closed cylinder whose axis coincides with the axis of the rod and whose cross section has radius a finite closed cylinder whose axis coincides with the axis of the rod and whose cross section has radius a sphere of radius

whose center lies on the axis of the rod

a finite closed cylinder whose axis coincides with the axis of the rod and whose cross section has radius an infinite cylinder whose axis coincides with the axis of the rod and whose cross section has radius

Part B Complete previous part(s) EXECUTE the solution as follows

Part C Complete previous part(s) EVALUATE your answer

Part D Complete previous part(s)

± Calculating Flux for Hemispheres of Different Radii Learning Goal: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9221581

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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 the electric field in which the surface lies. There are several important points to consider in this expression:

, and

i

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

normal to the surface contributes to the

is the angle between

and

When you compute flux, try to pick a surface that is either parallel or perpendicular to compute.

.

, so that the dot product is easy to

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, is the unit vector in the radial direction.

, and

Part A What is the electric flux

through the annular ring, surface 3?

Express your answer in terms of

,

,

, and any constants.

Hint 1. Apply the definition of electric flux The integrand in the equation defining flux (in the problem introduction) can be calculated by noting the following: . Since the electric field is everywhere parallel to the surface of the annular ring, the element electric field, and thus . Therefore, .

is normal to the

ANSWER: = 0

Correct

Part B https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=9221581

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What is the electric flux

through surface 1?

Express

,

in terms of

,

, and any needed constants.

Hint 1. Apply the definition of electric flux The integrand in the equation defining flux (in the problem introduction) can be calculated by noting the following:

Since the electric field is everywhere perpendicular to surface 2, . Therefore, the integral simply becomes the magnitude of the electric field multiplied by the surface area of surface 2. Hint 2. Find the area of surface 1 Find the area Express

of the hemisphere that is surface 1.

in terms of

and other given or known quantities.

ANSWER: =

ANSWER: =

Correct

Part C What is the electric flux

passing outward through surface 2?

Express

,

in terms of

,

, and any constants or other known quantities.

Hint 1. Apply the definition of electric flux The integrand in the equation defining flux (in the problem introduction) can be calculated by noting the following:

Since the electric field is everywhere perpendicular to surface 2, . Therefore, the integral simply becomes the magnitude of the electric field multiplied by the surface area of surface 2. Hint 2. Find the area of surface 2 Find the area

of the hemisphere that is surface 2.

Express your answer in terms of

, and any needed constants.

ANSWER: =

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ANSWER: =

Correct Observe that the electric flux through surface 1 is the same as that through surface 2, despite the fact that surface 2 has a larger area. If you think in terms of field lines, this means that there is the same number of field lines passing through both surfaces. This is because of the inverse square,

, behavior of the electric field

surrounding a point particle. A good rule of thumb is that the flux through a surface is proportional to the number of field lines that pass through the surface.

Exercise 22.9 - Enhanced - with Feedback A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 of -59.0 .

, giving it a charg

Part A Find the electric field just inside the paint layer. Express your answer with the appropriate units. Enter positive value if the field is directed away from the center of the sphere and negative value if the field is directed toward the center of the sphere. ANSWER: = 0

Correct

Part B Find the electric field just outside the paint layer. Express your answer with the appropriate units. Enter positive value if the field is directed away from the center of the sphere and negative value if the field is directed towar...