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Week 2 - Electric Fields

Due: 10:00pm on Tuesday, October 9, 2012

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

Electric Fields and Forces

Learning Goal:

To understand Coulomb's law, electric fields, and the connection between the electric field and the electric force.

Coulomb's law gives the electrostatic force acting between two charges. The magnitude of the force between two charges and

depends on the product of the charges and the square of the distance between the charges:

,

where . The direction of the force is along the line connecting the two charges. If the charges have the

same sign, the force will be repulsive. If the charges have opposite signs, the force will be attractive. In other words, opposite charges attract andlike charges repel.

Because the charges are not in contact with each other, there must be an intermediate mechanism to cause the force. This mechanism is theelectric field. The electric field at any location is equal to the force per unit charge experienced by a charge placed at that location. In other words,if a charge experiences a force , the electric field at that point is

.

The electric field vector has the same direction as the force vector on a positive charge and the opposite direction to that of the force vector on anegative charge.

An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point charge has magnitude

.

The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can befound by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by ,

will not significantly affect the electric field if it is small compared to .

Imagine an isolated positive point charge with a charge (many times larger than the charge on a single electron).

Part A

There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?

Check all that apply.

ANSWER:

Physics 210 Q1 2012 ( PHYSICS210BRIDGE )

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1 of 16 9/7/2012 1:10 PM

According to Coulomb's law, the force between two particles depends on the charge on each of them and the distance between them.

Part B

For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?

Check all that apply.

ANSWER:

The electrostatic force cannot exist unless two charges are present. The electric field, on the other hand, can be created by only onecharge. The value of the electric field depends only on the charge producing the electric field and the distance from that charge.

Part C

If the total positive charge is = 1.62×10−6 , what is the magnitude of the electric field caused by this charge at point P, a distance = 1.53

from the charge?

Enter your answer numerically in newtons per coulomb.

ANSWER:

the distance between the positive charge and the electron

the charge on the electron

the mass of the electron

the charge of the positive charge

the mass of the positive charge

the radius of the positive charge

the radius of the electron

the distance between the positive charge and the electron

the charge on the electron

the mass of the electron

the charge of the positive charge

the mass of the positive charge

the radius of the positive charge

the radius of the electron

= = 6220


2 of 16 9/7/2012 1:10 PM

Part D

What is the direction of the electric field at point P?

Enter the letter of the vector that represents the direction of .

ANSWER:

Part E

Now find the magnitude of the force on an electron placed at point P. Recall that the charge on an electron has magnitude

.

Enter your answer numerically in newtons.

Hint 1. Determine how to approach the problem

What strategy can you use to calculate the force between the positive charge and the electron?

ANSWER:

ANSWER:

Part F

What is the direction of the force on an electron placed at point P?

Enter the letter of the vector that represents the direction of .

Use Coulomb's law.

Multiply the electric field due to the positive charge by the charge on the electron.

Do either of the above.

Do neither of the above.

= = 9.95×10−16


3 of 16 9/7/2012 1:10 PM

ANSWER:

Electric Field Conceptual Question

Part A

For the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the net electric field iszero.

If no such region exists on the horizontal axis choose the last option(nowhere).

Hint 1. Zeros of the electric field

The net electric field can only be zero if the electric fields due to the two charges point in opposite directions and have equalmagnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in opposite directions. Then, in eachregion determine whether a point exists where the fields have equal magnitude. If there is such a point, then select that region.

ANSWER:

Part B



A

B

C

D

E

nowhere


4 of 16 9/7/2012 1:10 PM



Hint 2. Determine the regions where the electric fields could cancel

In which region(s) do the electric fields from the two source charges point in opposite directions?

List all the correct answers in alphabetical order.

ANSWER:

Since the two charges produce fields that point in opposite directions in these regions, if the magnitude of the fields are equal, thenet electric field will be zero.

Hint 3. Consider the magnitude of the electric field

For each of the three regions found in the previous hint, determine whether it is possible for the magnitudes to be equal. As anexample, consider the point directly between the two charges. Which charge produces the largest magnitude field directly between thetwo charges?

ANSWER:

Therefore, the point directly between the two charges is not the correct answer since the right charge dominates at this point.Check the other two possible regions.

ANSWER:

Part C



the charge on the right

the charge on the left

neither, because they have the same magnitude

A

B

C

D

E

nowhere


5 of 16 9/7/2012 1:10 PM



ANSWER:

Part D




ANSWER:

A

B

C

D

E

nowhere


6 of 16 9/7/2012 1:10 PM

Exercise 21.54

A straight, nonconducting plastic wire 7.00 long carries a charge density of 130 distributed uniformly along its length. It is lying on a

horizontal tabletop.

Part A

Find the magnitude and direction of the electric field this wire produces at a point 4.00 directly above its midpoint.

ANSWER:

Part B

ANSWER:

Part C

If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.00

directly above its center.

ANSWER:

Part D

ANSWER:

Problem 21.97

Negative charge is distributed uniformly around a quarter-circle of radius that lies in the first quadrant, with the center of curvature at the

origin.

A

B

C

D

E

Nowhere along the finite x axis

= = 3.85×104

electric field is directed upward

electric field is directed downward

= = 4.58×104

electric field is directed upward

electric field is directed downward


7 of 16 9/7/2012 1:10 PM

Part A

Find the x-component of the net electric field at the origin.

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

ANSWER:

Part B

Find the y-component of the net electric field at the origin.

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

ANSWER:

Problem 21.89 - Copy

Positive charge is distributed uniformly along the x-axis from to . A positive point charge is located on the positive x-axis at

, a distance to the right of the end of .

Part A

Calculate the x-component of the electric field produced by the charge distribution at points on the positive x-axis where .

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

ANSWER:

=

Also accepted: ,

=

Also accepted: ,

=

Also accepted: ,


8 of 16 9/7/2012 1:10 PM

Part B

Calculate the y-component of the electric field produced by the charge distribution at points on the positive x-axis where .

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

ANSWER:

Part C

Calculate the magnitude of the force that the charge distribution exerts on .

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

ANSWER:

Part D

Calculate the direction of the force that the charge distribution exerts on .

ANSWER:

Problem 21.91 - Copy

A charged line extends from y = 2.50 to y = -2.50 . The total charge distributed uniformly along the line is -7.50 .

= 0

=

Also accepted: ,

to the left

to the right


9 of 16 9/7/2012 1:10 PM

Part A

Find the magnitude of the electric field on the -axis at = 10.0 .

ANSWER:

Part B

Find the direction of the electric field on the -axis at = 10.0 .

ANSWER:

Part C

Is the magnitude of the electric field you calculated in part A larger or smaller than the electric field 10.0 from a point charge that has the

same total charge as this finite line of charge?

ANSWER:

Part D

At what distance does the result for the finite line of charge differ by 1.0% from that for the point charge?

Express your answer using two significant figures.

ANSWER:

Problem 21.100

Two very large parallel sheets are 5.00 apart. Sheet carries a uniform surface charge density of -8.00 , and sheet , which is to the

right of , carries a uniform charge of -12.2 . Assume the sheets are large enough to be treated as infinite.

Part A

Find the magnitude of the net electric field these sheets produce at a point 4.00 to the right of sheet .

ANSWER:

Part B

Find the direction of the net electric field these sheets produce at a point 4.00 to the right of sheet .

ANSWER:

= = 6550

= 180 counterclockwise from the -direction

larger

smaller

= 0.18

= = 2.37×105


10 of 16 9/7/2012 1:10 PM

Part C

Find the magnitude of the net electric field these sheets produce at a point 4.00 to the left of sheet .

ANSWER:

Part D

Find the direction of the net electric field these sheets produce at a point 4.00 to the left of sheet .

ANSWER:

Part E

Find the magnitude of the net electric field these sheets produce at a point 4.00 to the right of sheet .

ANSWER:

Part F

Find the direction of the net electric field these sheets produce at a point 4.00 to the right of sheet .

ANSWER:

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.

net electric field is directed to the right

net electric field is directed to the left

= = 1.14×106



= = 1.14×106




11 of 16 9/7/2012 1:10 PM

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.

Express your answer in terms of , , and

Hint 1. Definition of electric flux

The electric flux through a surface is given by , where is an infinitesimal element of area on the surface.

Hint 2. Simplifying the integrand

Note that the formula for electric flux can be expressed in the following way:

.

Since we are dealing with a constant electric field, which therefore does not vary across the surface of the disk, we can take

outside of the integral. This gives a much simpler expression:

.

Hint 3. Evaluate the scalar product

Find in terms of and .

ANSWER:

ANSWER:

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 aroundthe cube is given by .

Part A

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

=

=


12 of 16 9/7/2012 1:10 PM

Express your answer in terms of , , , and .

Hint 1. Definition of flux

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. Considerseparately the flux coming out of each of the six faces of the cube, and then add the results to obtain the net flux.

Hint 2. Flux through the face

Consider the face of the cube whose outward normal points in the positive x direction. What is the flux through this face?


Hint 1. Simplifying the integral

The field depends only on the spatial variable . On the face of the cube, , so is constant. Since is constant

over this entire surface, it can be pulled out of the integral:

.

Hint 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 3. Find the area of the face of the cube

This face of the cube is a square with sides of length . What is the area of this face?

ANSWER:

ANSWER:


Consider the face of the cube whose outward normal points in the positive y direction. What is the flux through this face?


ANSWER:

=

=

=


13 of 16 9/7/2012 1:10 PM


Consider the face of the cube whose outward normal points in the positive z direction. What is the flux through this face?

Hint 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:


Consider the face of the cube whose outward normal points in the negative x direction. What is the flux through this face?


Hint 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 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 positivewhen the field is coming out of the surface and negative when the field is going into the surface.

ANSWER:

Hint 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 toobtain the net flux.

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

=

= 0

=

=

=


14 of 16 9/7/2012 1:10 PM

,

then the flux through the cube would be zero. Why?

ANSWER:

Part C

What is the net charge inside the cube?

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

Hint 1. 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:

.

ANSWER:

Exercise 22.10

A point charge = 3.15 is located on the x-axis at = 2.25 , and a second point charge = -5.65 is on the y-axis at = 1.05 .

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 = 0.440 ?

ANSWER:

Part B


ANSWER:

Part C


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.

=

= 0

= = -638

= = -282


15 of 16 9/7/2012 1:10 PM

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