MP EM Ass 4: Electric Flux and Gauss's Law

Assignment 4: Flux and Gauss's Law

Due: 8:00am on Friday, January 20, 2012

Note: To understand how points are awarded, read your instructor's Grading Policy.

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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 is

the area of the surface element. The direction of is normal to the surface element, pointing out of the enclosed

volume.

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.

Calculating Electric Flux through a Disk

Suppose 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.

Part A

What is the electric flux through the surface of the disk that is facing right (the normal vector to this surface is

shown in the figure)? Assume that the presence of the disk does not interfere with the electric field.

Hint A.1 Definition of electric flux

Hint not displayed

Hint A.2 Simplifying the integrand

Hint not displayed

Hint A.3 Evaluate the scalar product

Hint not displayed

Express your answer in terms of , , and

ANSWER:

= Correct

± 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:

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 normal to the surface contributes to

the integral. That is, , where is the angle between and .

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 A

What is the electric flux through the annular ring, surface 3?

Hint A.1 Apply the definition of electric flux

Hint not displayed

Express your answer in terms of , , , and any constants.

ANSWER:

= 0

Correct

Part B

What is the electric flux through surface 1?

Hint B.1 Apply the definition of electric flux

Hint not displayed

Hint B.2 Find the area of surface 1

Hint not displayed

Express in terms of , , , and any needed constants.

ANSWER: =

Correct

Part C

What is the electric flux passing outward through surface 2?

Hint C.1 Apply the definition of electric flux

Hint not displayed

Hint C.2 Find the area of surface 2

Hint not displayed

Express in terms of , , , and any constants or other known quantities.

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 22.10

A point charge = 3.75 is located on the x-axis at = 1.85 , and a second point charge = -5.75 is on

the y-axis at = 1.25 .

Part A

What is the total electric flux due to these two point charges through a spherical surface centered at the origin and

with radius = 0.470 ?

ANSWER:

= 0

Correct

Part B



ANSWER:

= -650

Correct

Part C



ANSWER:

= -226

Correct

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 through each of the six faces of the cube?

Hint A.1 How to approach the problem

Hint not displayed

Hint A.2 Calculate the total electric flux

Hint not displayed

Hint A.3 Flux through a face

Hint not displayed

Use for the permittivity of free space.

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

Hint B.1 How to approach the problem

Hint not displayed

Use for the permittivity of free space.

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.

Note the similarity of the next problem with Discussion problem 22.34

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 A

Find the total electric flux through the surface of the cube.

Hint A.1 Definition of flux

Hint not displayed

Hint A.2 Flux through the face

Hint not displayed


Hint not displayed


Hint not displayed


Hint not displayed

Hint A.6 Putting it together

Hint not displayed

Express your answer in terms of , , , and .

ANSWER:

= Correct

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.

Correct

Part C

What is the net charge inside the cube?

Hint C.1 Gauss's law

Hint not displayed

Express your answer in terms of , , , , and .

ANSWER: =

Correct

Exercise 22.32

Two very large, nonconducting plastic sheets, each 10.0 thick, carry uniform charge densities and

on their surfaces, as shown in the following figure . These

surface charge densities have the values = -7.00 , , = 3.90 , and

. Use Gauss's law to find the magnitude and direction of the electric field at the following points,

far from the edges of these sheets.

Part A

What is the magnitude of the electric field at point , 5.00 from the left face of the left-hand sheet?

Express your answer to three significant figures and include the appropriate units.

ANSWER:

= 3.33×105

Correct

Part B

What is the direction of the electric field at point , 5.00 from the left face of the left-hand sheet?

ANSWER:

to the left.

to the right.

upwards.

downwards.

Correct

Part C

What is the magnitude of the electric field at point , 1.25 from the inner surface of the right-hand sheet?


ANSWER:

= 5.59×105

Correct

Part D

What is the direction of the electric field atpoint , 1.25 from the inner surface of the right-hand sheet?

ANSWER:

to the left.

to the right.

upwards.

downwards.

Correct

Part E

What is the magnitude of the electric field at point , in the middle of the right-hand sheet?


ANSWER:

= 1.19×105

Correct

Part F

What is the direction of the electric field at point , in the middle of the right-hand sheet?

ANSWER:

to the left.

to the right.

upwards.

downwards.

Correct

Exercise 22.28

A conductor with an inner cavity, like that shown in Fig.22.23c from the textbook, carries a total charge of

+5.40 . The charge within the cavity, insulated from the conductor, is -8.20 . How much charge is on

Part A

How much charge is on the inner surface of the conductor?

ANSWER:

=

8.20

Correct

Part B

How much charge is on the outer surface of the conductor?

ANSWER:

=

-2.80

Correct

Problem 22.46

A conducting spherical shell with inner radius and outer radius has a positive point charge located at its

center. The total charge on the shell is , and it is insulated from its surroundings

.

Part A

Derive the expression for the electric field magnitude in terms of the distance from the center for the region

.

Express your answer in terms of some or all of the variables , , , and appropriate constants.

ANSWER:

= Correct

Part B


.


ANSWER:

= 0

Correct

Part C


.


ANSWER:

= Correct

Part D

What is the surface charge density on the inner surface of the conducting shell?


ANSWER:

=

Correct

Part E

What is the surface charge density on the outer surface of the conducting shell?


ANSWER:

=

Correct

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MP EM Ass 4: Electric Flux and Gauss's Law