1b – The Electric Field

Due: 11:59pm on Sunday, September 12, 2021

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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 is zero.

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 equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in

opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If

there is such a point, then select that region.

ANSWER:

Correct

Part B

A B C D E

nowhereFor the charge distribution provided, indicate the region (A to E) along the horizontal axis where a point exists at which the

net electric field is zero.

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 equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in

opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If

there is such a point, then select that region.

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:

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 an example, consider the point directly between the two charges. Which charge produces the largest

magnitude field directly between the two charges?

ANSWER:

ANSWER:

BCD

the charge on the right

the charge on the left

neither, because they have the same magnitudeCorrect

Part C

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 is zero.

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 equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in

opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If

there is such a point, then select that region.

ANSWER:

A B C D E

nowhereCorrect

Part D

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 is zero.

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 equal magnitudes. Therefore, first determine the region(s) where the two constituent electric fields point in

opposite directions. Then, in each region determine whether a point exists where the fields have equal magnitude. If

there is such a point, then select that region.

ANSWER:

A B C D E

nowhereCorrect

Magnitude and Direction of Electric Fields

A small object A, electrically charged, creates an electric field. At a point P located 0.250 directly north of A, the field has a

value of 40.0 directed to the south.

Part A

What is the charge of object A?

Hint 1. How to approach the problem

Recall that the electric field at a point P due to a point charge is proportional to the magnitude of the charge and

inversely proportional to the square of the distance of P from the charge. Furthermore, the direction of the field is

determined by the sign of the charge.

Hint 2. Find an expression for the charge

Which of the following expressions gives the correct magnitude of charge that produces an electric field of

magnitude at a distance from the charge? In the following expressions is a constant that has

units of .

Hint 1. Magnitude of the electric field of a point charge

Given a point charge , the magnitude of the electric field at a distance from the charge is given by

,

where the constant of proportionality is = 8.99×109 .

ANSWER:

A B C D E

Nowhere along the finite x axis

m

N/C

q

N/C r N ⋅ m2/C2

N ⋅ m2/C2

q E r

E = k |q|

r2

k N ⋅ m2/C2Hint 3. Find the sign of the charge

What is the sign of the charge that produces an electric field that points toward the charge?

ANSWER:

ANSWER:

Correct

Part B

If a second object B with the same charge as A is placed at 0.250 south of A (so that objects A and B and point P follow

a straight line), what is the magnitude of the total electric field produced by the two objects at P?

Hint 1. How to approach the problem

Since the electric field is a vector quantity, you need to apply the principle of superposition to find the total field at P.

The principle of superposition in terms of electric fields says that the total electric field at any point due to two or

more charges is the vector sum of the fields that would be produced at that point by the individual charges.

Hint 2. Find the vector sum of the electric fields

q = k E

d2

q = kEd2

q = k dE2

q = Ekd2

positive

negative

1.11×10−9 C

−1.11×10−9 C

2.78×10−10 C

−2.78×10−10 C

5.75×1012 C

−5.75×1012 C

mWhich of the following diagrams, where and are the electric fields produced by A and B, respectively,

correctly represents the situation described in this problem?

ANSWER:

Hint 3. Find the electric field produced by B at P

What is the magnitude of the electric field produced by the second object B at point P?

Express your answer in newtons per coulomb.

Hint 1. Magnitude of the electric field of a point charge

Given a point charge , the magnitude of the electric field at a distance from the charge is given by

,

where the constant of proportionality is = 8.99×109 .

Hint 2. Find the distance from P to the second object

How far ( ) is P from B? Recall that P is located 0.250 north of A and B is located 0.250 south of A.

Express your answer in meters.

ANSWER:

ANSWER:

E⃗PA E⃗PB

A B C D

EPB

q E r

E = k |q|

r2

k N ⋅ m2/C2

dPB m m

dPB = 0.500 mANSWER:

Correct

Mystery Charge

Consider the following configuration of fixed, uniformly charged spheres in :

a blue sphere fixed at the origin with positive charge ,

a red sphere fixed at the point with unknown charge

, and

a yellow sphere fixed at the point

with unknown charge .

The net electric force on the blue sphere has a magnitude and is

directed in the y direction.

Part A

What is the sign of the charge on the yellow sphere?

ANSWER:

Correct

EPB = 10.0 N/C

40.0 N/C

50.0 N/C

30.0 N/C

10.0 N/C

q

(d1, 0)

qred

(d2 cos(θ), −d2 sin(θ))

qyellow

F

−

positive

negativePart B

What is the sign of the charge on the red sphere?

ANSWER:

Correct

Part C

Suppose that the magnitude of the charge on the yellow sphere is determined to be . Calculate the charge on the red

sphere.

Express your answer in terms of , , , and .

Hint 1. How to approach the problem

From the problem statement, you know that the x component of the net force acting on the blue sphere is zero. The

red sphere and the yellow sphere each exert a force on the blue sphere. You know the charge of the yellow sphere.

This allows you to calculate the x component of the force that the yellow sphere exerts on the blue sphere. You need

to find the appropriate charge for the red sphere such that the x components of the two forces sum to zero.

Hint 2. Find the x component of the force due to the yellow sphere

Find , the x component of the force that the yellow sphere exerts on the blue sphere.

Express your answer in terms of , , and . You may use for , where represents the permittivity

of free space.

Hint 1. How to approach this part

Use Coulomb’s law to find the force due to the yellow charge on the blue charge. Then find the x component

of the force.

ANSWER:

Hint 3. Find the force due to the red sphere

Find , the x component of the force that the red sphere exerts on the blue sphere.

Express your answer in terms of , , and the unknown charge . You may use for , where

represents the permittivity of free space.

ANSWER:

positive

negative

2q qred

q d1 d2 θ

qred

Fx,yellow

q d2 θ k 1/4πϵ0 ϵ0

Fx,yellow = k⋅2q2cos(θ)

(d2)2

Fx,red

q d1 qred k 1/4πϵ0 ϵ0ANSWER:

Correct

Placing Charges Conceptual Question

Below are free-body diagrams for three electric charges that lie in the same plane. Their relative positions are unknown. There

are two forces shown on each charge. These two forces represent the force exerted on the charge by each of the other two

charges.

Part A

Along which of the lines (A to H) in should charge 2 be placed so

that the free-body diagrams of charge 1 and charge 2 are

consistent? Note that only one of the forces on each charge will

be consistent. The other force on each charge will be addressed

in Part B with the introduction of charge 3.

Hint 1. How to approach the problem

Newton’s 3rd law states that the forces exerted by a pair of objects on each other are always equal in magnitude and

opposite in direction. Identifying the forces that correspond to 3rd-law pairs in the free-body diagrams will enable you

to place the particles in their proper relative position.

Hint 2. Placing charge 2

The two forces acting on charge 2 correspond to the forces exerted on it by charge 1 and charge 3. This means that

one of these forces must pair with a force on charge 1 of equal magnitude and opposite direction and the other must

pair with a force on charge 3 of equal magnitude and opposite direction. Also note that charge 2 should be repelled

by charge 1, since both are negative. Therefore, the vector that represents the force of charge 1 on charge 2 must

point away from charge 1. This information is all you need to place charge 2 in its correct position.

ANSWER:

Fx,red = −kqredq

d1 2

qred = 2qcosθ( d d1 2 )2Correct

Part B

Along which of the lines (A to H) in should charge 3 be placed so

that the free-body diagrams of charge 1, charge 2, and charge 3

are consistent?

ANSWER:

Correct

Part C

Along which of the lines (A to H) in should charge 2 be placed so

that the free-body diagrams of charge 1 and charge 2 are

consistent? Note that only one of the forces on each charge will

be consistent. The other force on each charge will be addressed

in Part D with the introduction of charge 3.

ANSWER:

C DCorrect

Part D

Along which lines (A to H) in should charge 3 be placed so that

the free-body diagrams of charge 1, charge 2, and charge 3 are

consistent?

ANSWER:

Correct

Problem 20.23

Part A

What is the strength of the electric field 4.0 from a small glass bead that has been charged to 8.0 ?

Express your answer in newtons per coulomb.

ANSWER:

Correct

Part B

What is the direction of the electric field 4.0 from a small glass bead that has been charged to 8.0 ?

H F

cm + nC

E = 4.5×104 N/C

cm nCANSWER:

Correct

± Suspending Charged Particles Using Electric Fields

Part A

What must the charge (sign and magnitude) of a particle of mass 1.44 be for it to remain stationary when placed in a

downward-directed electric field of magnitude 680 ?

Use 9.80 for the magnitude of the free-fall acceleration.

Hint 1. How to approach the problem

Find the charge needed to create a force that will be equal but opposite to the force of gravity. Keep a close eye on

the signs as you work, taking upward to be the positive direction for reference.

Hint 2. Calculate the gravitational force

Calculate the force of gravity on the particle, including the sign, assuming that the positive direction is upward.

Use 9.80 for the magnitude of the free-fall acceleration.

ANSWER:

Hint 3. Determine the sign of the charge needed

What should the sign of the charge be to counteract the effects of gravity in this system?

ANSWER:

ANSWER:

toward the bead

away from the bead

g

N/C

m/s2

Fg

m/s2

Fg = −1.41×10−2 N

positive

negative

−2.08×10−5 CCorrect

Part B

What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Use 1.67×10−27 for the mass of a proton, 1.60×10−19 for the magnitude of the charge on an electron, and 9.80

for the magnitude of the free-fall acceleration.

Hint 1. How to approach the problem

Find the electric field needed to create a force that will be equal but opposite to the force of gravity. For this section,

the sign is not needed in your answer.

Hint 2. Calculate the gravitational force

Calculate , the magnitude of the force of gravity on the proton.

ANSWER:

ANSWER:

Correct

This is the magnitude of the electric field needed to counteract the proton’s weight, but in what direction should it

point? Since gravity points downward, and a positively charged particle (such as a proton) will experience an

electric force in the same direction as the electric field to which it is exposed, the field should point upward.

Electric Field due to Two Point Charges

Two point charges are placed on the x axis as shown in . The first

charge, = 8.00 , is placed a distance 16.0 from the origin

along the positive x axis; the second charge, = 6.00 , is placed

a distance 9.00 from the origin along the negative x axis.

kg C

m/s2

Fg

Fg = 1.64×10−26 N

1.02×10−7 N/C

q1 nC m

q2 nC

mPart A

Find the x-component of the electric field at the origin, point O.

Express your answer in newtons per coulomb to three significant figures, keeping in mind that an x component

that points to the right is positive.

Hint 1. How to approach the problem

Find the contributions to the electric field at the origin separately for and , then add them together (using

vector addition) to find the total electric field at the origin.

Hint 2. Determine the directions of the electric fields

Which of the following describes the directions for the electric fields and created by charges and ,

respectively?

ANSWER:

Hint 3. Calculate the electric field due to

What is the magnitude of the electric field at the origin due to charge only?

Express your answer in newtons per coulomb to three significant figures.

ANSWER:

Hint 4. Calculate the electric field due to

What is the magnitude of the electric field at the origin due to charge only?

Express your answer in newtons per coulomb to three significant figures.

ANSWER:

ANSWER:

nC nC

E⃗O1 E⃗O2 nC nC

Both E⃗O1 and E⃗O2 point to the right.

Both E⃗O1 and E⃗O2 point to the left.

E⃗O1 points to the right and E⃗O2 points to the left.

E⃗O1 points to the left and E⃗O2 points to the right.

nC

nC

EO1 = 0.281 N/C

nC

nC

EO2 = 0.666 N/C

EOx = 0.385 N/CCorrect

Part B

Now, assume that charge is negative; , as shown in . What is the x-component of the net electric field at the

origin, point O?

Express your answer in newtons per coulomb to three

significant figures, keeping in mind that an x component that

points to the right is positive.

Hint 1. How to approach the problem

Find the contributions to the electric field at the origin separately for and , then add them together (using

vector addition) to find the total electric field at the origin.

Hint 2. Determine the directions of the electric fields

Which of the following describes the directions for the electric fields and created by charges and ,

respectively?

ANSWER:

ANSWER:

Correct

q2 q2 = −6 nC

nC nC

E⃗O1 E⃗O2 nC nC

Both E⃗O1 and E⃗O2 point to the right.

Both E⃗O1 and E⃗O2 point to the left.

E⃗O1 points to the right whereas E⃗O2 points to the left.

E⃗O1 points to the left whereas E⃗O2 points to the right.

EOx = -0.947 N/CElectric 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 and like 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 the electric 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 a negative 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 be found 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:

F⃗ F

q1 q2 r

F = k |q1 q2|

r2

k = 1/(4πϵ0) = 8.99 × 109 N ⋅ m2/C2

q F⃗ E⃗

E⃗ = F ⃗

q

r

q′

E = k |q′|

r2

q

q′ q q′

Q

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 electronCorrect

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:

Correct

The electrostatic force cannot exist unless two charges are present. The electric field, on the other hand, can be

created by only one charge. 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

Q C

d mCorrect

Part D

What is the direction of the electric field at this point (previously marked P)?

Enter the letter of the vector that represents the direction of

.

ANSWER:

Correct

Part E

Now find the magnitude of the force on an electron placed at this same point. 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:

EP = 6220 N/C

⃗EP G

e = 1.60 × 10−19 C

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.ANSWER:

Correct

Part F

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

Enter the letter of the vector that represents the direction of

.

ANSWER:

Correct

Problem 20.41 – Enhanced – with Video Tutor Solution

A 0.15 plastic bead is charged by the addition of excess electrons.

Part A

What electric field (strength) will cause the bead to hang suspended in the air?

Express your answer in newtons per coulomb.

ANSWER:

Correct

F = 9.95×10−16 N

⃗F C g

1.0 × 1010

E⃗

E = 9.2×105 N/CPart B

What electric field (direction) will cause the bead to hang suspended in the air?

ANSWER:

Correct

For the steps and strategies involved in solving this problem, you may view a Video Tutor Solution.

Visualizing Electric Fields

Learning Goal:

To understand the nature of electric fields and how to draw field lines.

Electric field lines are a tool used to visualize electric fields. A field line is drawn beginning at a positive charge and ending at a

negative charge. Field lines may also appear from the edge of a picture or disappear at the edge of the picture. Such lines are

said to begin or end at infinity. The field lines are directed so that the electric field at any point is tangent to the field line at that

point.

shows two different ways to visualize an electric field. On the left,

vectors are drawn at various points to show the direction and

magnitude of the electric field. On the right, electric field lines depict

the same situation. Notice that, as stated above, the electric field lines

are drawn such that their tangents point in the same direction as the

electric field vectors on the left. Because of the nature of electric

fields, field lines never cross. Also, the vectors shrink as you move

away from the charge, and the electric field lines spread out as you

move away from the charge. The spacing between electric field lines

indicates the strength of the electric field, just as the length of vectors

indicates the strength of the electric field. The greater the spacing

between field lines, the weaker the electric field. Although the

advantage of field lines over field vectors may not be apparent in the

case of a single charge, electric field lines present a much less

cluttered and more intuitive picture of more complicated charge

arrangements.

Part A

Which of the following panels (labelled A, B, C, and D) in correctly depicts the field lines from an infinite uniformly negatively

charged sheet? Note that the sheet is being viewed edge-on in all pictures.

E⃗

up

downHint 1. Description of the field

Recall that the field around an infinite charged sheet is always perpendicular to the sheet and that the field strength

does not change, regardless of distance from the sheet.

ANSWER:

Correct

Part B

In , what is wrong with panel B? (Pick only those statements that

apply to panel B.)

Check all that apply.

ANSWER:

A B C DCorrect

Part C

Which of the following panels (labelled A, B, C, and D) in shows

the correct electric field lines for an electric dipole?

ANSWER:

Correct

Part D

In , what is wrong with panel D? (Pick only those statements that apply to panel D.)

Check all that apply.

Field lines cannot cross each other.

The field lines should be parallel because of the sheet’s symmetry.

The field lines should spread apart as they leave the sheet to indicate the weakening of the field with distance.

The field lines should always end on negative charges or at infinity.

A B C DANSWER:

Correct

Even in relatively simple setups as in the figure shown,

electric field lines are quite helpful for understanding the field qualitatively (understanding the general direction in

which a certain charge will move from a specific position, identifying locations where the field is roughly zero or

where the field points a specific direction, etc.). A good figure with electric field lines can help you to organize your

thoughts as well as check your calculations to see whether they make sense.

Part E

In , the electric field lines are shown for a system of two point charges, and . Which of the following could represent

the magnitudes and signs of and ?

In the following, take to be a positive quantity.

Field lines cannot cross each other.

The field lines should turn sharply as you move from one charge to the other.

The field lines should be smooth curves.

The field lines should always end on negative charges or at infinity.

QA QB

QA QB

qANSWER:

Correct

Very far from the two charges, the system looks like a single charge with value . At large

enough distances, the field lines will be indistinguishable from the field lines due to a single point charge .

Score Summary:

Your score on this assignment is 99.0%.

You received 43.58 out of a possible total of 44 points.

QA = +q, QB = −q

QA = +7q, QB = −3q

QA = +3q, QB = −7q

QA = −3q, QB = +7q

QA = −7q, QB = +3q

QA + QB = +4q

+4q

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