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Hwork 01 Electric Force and Field
Hwork 01 Electric Force and Field
Due: 11:59pm on Thursday, June 11, 2020
To understand how points are awarded, read the Grading Policy for this assignment.
Problem 1
Learning Goal:
To practice Problem-Solving Strategy 21.1 Coulomb's Law.
Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.7 cm . Two of the particles have a
negative charge: q1 = -8.3 nC and q2 = -16.6 nC . The remaining particle has a positive charge, q3 = 8.0 nC . What is the net electric force acting on
particle 3 due to particle 1 and particle 2?
Problem-Solving Strategy: Coulomb's law IDENTIFY the relevant concepts:
Coulomb’s law comes into play whenever you need to know the electric force acting between charged particles.
SET UP the problem using the following steps:
1. Make a drawing showing the locations of the charged particles, and label each particle with its charge.
2. If three or more particles are present and they do not all lie on the same line, set up an xy coordinate system.
3. Often you will need to find the electric force on just one particle. If so, identify that particle.
EXECUTE the solution as follows:
1. For each particle that exerts a force on the particle of interest, calculate the magnitude of that force using F
=
1
4πϵ0
|q1 q2 |
r
2
.
2. Sketch a free-body diagram showing the electric force vectors acting on the particle(s) of interest due to each of the other particles. Recall
that the force exerted by particle 1 on particle 2 points from particle 2 toward particle 1 if the two charges have opposite signs, but points
from particle 2 directly away from particle 1 if the charges have the same sign.
3. Calculate the total electric force on the particle(s) of interest. Recall that the electric force, like any force, is a vector.
4. As always, using consistent units is essential. If you are given non-SI units, don’t forget to convert!
5. If there is a continuous distribution of charge along a line or over a surface, divide the total charge distribution into infinitesimal pieces, use
Coulomb’s law for each piece, and then integrate to find the vector sum.
6. In many situations, the charge distribution will be symmetrical. Whenever possible, exploit any symmetries to simplify the problem-solving
process.
EVALUATE your answer:
Check whether your numerical results are reasonable, and confirm that the direction of the net electric force agrees with the principle that like charges
repel and opposite charges attract.
IDENTIFY the relevant concepts
To determine the angle of the force vector on a single charged particle, you will need to calculate the vector sum of all the forces on that particle due to the
presence of other charged particles. To do this, you will need to use Coulomb's law.
SET UP the problem using the following steps
Part A
Identify the most appropriate xy coordinate system.
ANSWER:
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Hwork 01 Electric Force and Field
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You are asked to find the net force acting on particle 3. Centering the xy coordinate system on particle 3 will make this easier.
EXECUTE the solution as follows
Part B
Find the net force ΣF 3⃗ acting on particle 3 due to the presence of the other two particles. Report you answer as a magnitude ΣF3 and a direction θ
measured from the positive x axis.
Express the magnitude in newtons and the direction in degrees to three significant figures.
Hint 1. How to approach the problem
To calculate the electric force acting on particle 3, you should begin by drawing a free-body diagram indicating the forces acting on particle 3
due to particle 1 and particle 2. You know that
⃗
⃗
⃗
ΣF 3 = F 1 on 3 + F 2 on 3 .
Use Coulomb's law to calculate the magnitude of each of these forces. Apply vector algebra to find the component forces in the x and the y
directions. Then, sum the component forces for each direction:
(ΣF3 )x
=
(F1
on 3
)x + (F2
on 3
)x
(ΣF3 )y
=
(F1
on 3
)y + (F2
on 3
)y
.
From (ΣF3 )x and (ΣF3 )y you can find the magnitude and direction of the resulting electric force vector.
Hint 2. Draw a free-body diagram
Identify the forces on the positively charged particle 3.
Draw your vectors starting at the origin. The orientation of your vectors will be graded but their precise length will not.
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Hwork 01 Electric Force and Field
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Hint 3. Calculate the force on particle 3 due to particle 1
Using the equation for Coulomb's law, calculate the magnitude of the force on particle 3 due to particle 1. Keep in mind that
1
4πϵ0
= 8.988 × 10
9
N⋅m
C
2
2
.
Express your answer in newtons using three significant figures.
ANSWER:
F1
on 3
= 4.36×10−4
N
Hint 4. Calculate the component forces on particle 3 due to particle 1
Calculate the x component and the y component forces acting on particle 3 due to particle 1, using simple trigonometry. The angle between
particle 1 and particle 3 is 60∘ :
Enter the components of the force in newtons separated by a comma.
ANSWER:
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Hwork 01 Electric Force and Field
(F1
on 3
)x
, (F1
on 3
)y
= 2.18×10−4,3.78×10−4
N
,N
Hint 5. How to calculate the component forces on particle 3 due to particle 2
Because particles 2 and 3 both lie on the x axis, there will be no y component to calculate. The x component of force will therefore be equal to
the value calculated from Coulomb's law, and the y component will be zero.
Hint 6. How to determine the magnitude and direction of a vector from its components
If a vector F ⃗ has components Fx and Fy , the magnitude F and direction θ are given by
−
−−
−
−
−
−
−
−−
−
2
F
=
√ (Fx )
θ
=
arctan(Fy /Fx )
2
+ (Fy )
,
where
ANSWER:
ΣF3
, θ = 1.15×10−3,19.1
N
,∘
Correct
EVALUATE your answer
Part C
Assume that particle 3 is no longer fixed to a corner of the triangle and is now allowed to move. In what direction would particle 3 move the instant
after being released?
Draw the velocity vector for particle 3 below. The orientation of your vector will be graded, but not its length.
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Hwork 01 Electric Force and Field
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Specifically, from Newton's 2nd law, F ⃗
= ma⃗
, you know that a mass accelerates in the same direction as the net force acting upon it. Therefore,
at the instant after being released, particle 3 accelerates in the same direction as ΣF 3⃗ . Moreover, since particle 3 starts from rest, its velocity at
that instant will be v ⃗ = at⃗ . In other words, the initial direction of particle 3 is the same direction as its acceleration, and therefore the same
direction as the applied net force.Let us interpret this result in terms of electric forces. In general, like charges repel and unlike charges attract. If
particle 3 were free to move, it would move toward the negative charges q1 and q2 . If q1 and q2 were the same size, particle 3 would start to
move toward them along a direction equidistant from each charge, that is, at an angle of 30 ∘ from the positive x axis. Instead, |q2 | > |q1 |, so
particle 3 will be more strongly attracted toward particle 2 and will move off in a direction less than 30 ∘ .
Problem 2
Coulomb's law for the magnitude of the force F between two particles with charges
Q
and
′
Q
separated by a distance d is
′
|F | = K
|QQ |
d
where K
=
1
4πϵ0
, and ϵ0
= 8.854 × 10
−12
2
2
C /(N ⋅ m )
2
,
is the permittivity of free space.
Consider two point charges located on the x axis: one charge, q1 = -20.0 nC , is located at x1 = -1.740 m ; the second charge, q2 = 37.5 nC , is at the
origin (x = 0).
Part A
What is (Fnet 3 )x , the x-component of the net force exerted by these two charges on a third charge q3 = 46.0 nC placed between q1 and q2 at
x3 = -1.170 m ?
Your answer may be positive or negative, depending on the direction of the force.
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Hwork 01 Electric Force and Field
Express your answer numerically in newtons to three significant figures.
Hint 1. How to approach the problem
First, draw a diagram of the system. Next, find the magnitudes of the forces exerted on the third charge by each of the other charges. Then
determine the direction of each of these forces. Finally, use vector addition to find the net force on the third charge.
Hint 2. Calculate the force on the third charge by the first charge
Calculate the magnitude |F1 | of the force that the first charge exerts on the third charge.
Express your answer numerically in newtons to three significant figures.
Hint 1. Coulomb's law
Coulomb's law for the magnitude of the force F between two particles with charges q1 and q2 separated by a distance d is
|F | =
where ϵ0
= 8.854 × 10
−12
2
2
C /(N ⋅ m )
1
|q1 q2 |
4πϵ0
d
2
,
is the permittivity of free space.
ANSWER:
|F1 |
= 2.55×10−5
N
Hint 3. Calculate the force on the third charge by the second charge
Calculate the magnitude |F2 | of the force that the second charge exerts on the third charge.
Express your answer numerically in newtons to three significant figures.
Hint 1. Coulomb's law
Coulomb's law for the magnitude of the force F between two particles with charges q1 and q2 separated by a distance d is
|F | =
where ϵ0
= 8.854 × 10
−12
2
2
C /(N ⋅ m )
1
|q1 q2 |
4πϵ0
d
2
,
is the permittivity of free space.
ANSWER:
|F2 |
= 1.13×10−5
N
Hint 4. What are the directions of the forces?
In what directions do the forces on the third charge point in our system? Note that since all the forces are in the x direction, you can drop the
vector notation. Let F1 be the force on q3 due to q1 , and let F2 be the force on q3 due to q2 .
ANSWER:
F1
and F2 both point along the +x direction.
F1
and F2 both point along the -x direction.
F1
points along the +x direction and F2 points along the -x direction.
F1
points along the -x direction and F2 points along the +x direction.
Hint 5. Relating the net force and the forces between pairs of charges
Since all the forces are in the x direction, you can drop the vector notation. Let Fnet be the net force on q3 . Similarly, let F1 be the force on q3
due to q1 , and F2 be the force on q3 due to q2 . Then Fnet = F1 + F2 .
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ANSWER:
(Fnet
3 )x
= −3.68×10−5
N
Correct
Problem 3
Each of the four parts of this problem depicts a motion diagram showing the position and velocity of a charged particle at equal time intervals as it moves
through a region of uniform electric field. For each part, draw a vector representing the direction of the electric field.
Part A
Draw a vector representing the direction of the electric field. The orientation of the vector will be graded. The location and length of the
vector will not be graded.
Hint 1. Relationship between electric field and electric force
The relationship between the electric force that acts on a particle and the electric field at the location of the particle isF ⃗ = qE⃗ .This formula
indicates that the force and the electric field point in the same direction for a positively charged particle, and in opposite directions for a
negatively charged particle.
Hint 2. Determining the direction of the electric field
The acceleration of the particle can be determined from the change in its velocity. By Newton’s 2nd law, the force acting on the particle is
parallel to its acceleration. Finally, since this is a positively charged particle, the electric field is parallel to the force. Putting this all together
results in an electric field that is parallel to the particle’s acceleration.
ANSWER:
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The motion diagram shows that the particle's acceleration points to the right. Because the particle has positive charge, the electric field
should point to the right.
Part B
Draw a vector representing the direction of the electric field. The orientation of the vector will be graded. The location and length of the
vector will not be graded.
Hint 1. Relationship between electric field and electric force
The relationship between the electric force that acts on a particle and the electric field at the location of the particle isF ⃗ = qE⃗ .This formula
indicates that the force and the electric field point in the same direction for a positively charged particle, and in opposite directions for a
negatively charged particle.
Hint 2. Determining the direction of the electric field
The acceleration of the particle can be determined from the change in its velocity. By Newton’s 2nd law, the force acting on the particle is
parallel to its acceleration. Finally, since this is a negatively charged particle, the electric field is directed opposite to the force. Putting this all
together results in an electric field that is directed opposite to the particle’s acceleration.
ANSWER:
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Hwork 01 Electric Force and Field
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The motion diagram shows that the particle's acceleration points to the right. Because the particle has negative charge, the electric field
should point to the left.
Part C
Draw a vector representing the direction of the electric field. The orientation of the vector will be graded. The location and length of the
vector will not be graded.
Hint 1. Relationship between electric field and electric force
The relationship between the electric force that acts on a particle and the electric field at the location of the particle isF ⃗ = qE⃗ .This formula
indicates that the force and the electric field point in the same direction for a positively charged particle, and in opposite directions for a
negatively charged particle.
Hint 2. Determining the direction of the electric field
The acceleration of the particle can be determined from the change in its velocity. By Newton’s 2nd law, the force acting on the particle is
parallel to its acceleration. Finally, since this is a positively charged particle, the electric field is parallel to the force. Putting this all together
results in an electric field that is parallel to the particle’s acceleration.Because the electric field is uniform, you can find the direction of the
particle's acceleration by subtracting any two consecutive velocity vectors graphically. If v i⃗ and v f⃗ are any two consecutive velocities, you
can subtract v i⃗ from v f⃗ by placing −v i⃗ at the tip of v f⃗ . v f⃗ − v i⃗ is the vector that starts at the tail of v f⃗ and ends at the tip of v i⃗ .To find the
direction of the particle's acceleration graphically, use two unlabeled vectors to represent −v i⃗ and v f⃗ − v i⃗ . Pick any two vectors v i⃗ and v f⃗
that would make your subtraction easier; you can verify your result by subtracting any other pair of consecutive vectors.
ANSWER:
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Hwork 01 Electric Force and Field
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Part D
Draw a vector representing the direction of the electric field. The orientation of the vector will be graded. The location and length of the
vector will not be graded.
Hint 1. Relationship between electric field and electric force
The relationship between the electric force that acts on a particle and the electric field at the location of the particle is F ⃗ = qE⃗ .This formula
indicates that the force and the electric field point in the same direction for a positively charged particle, and in opposite directions for a
negatively charged particle.
Hint 2. Determining the direction of the electric field
The acceleration of the particle can be determined from the change in the illustrated velocity vectors. By Newton’s 2nd law, the force acting on
the particle is parallel to its acceleration. Finally, since this is a negatively charged particle, the electric field is directed opposite to the electric
force. Putting this all together results in an electric field that is directed opposite to the particle’s acceleration.Because the electric field is
uniform, you can find the direction of the particle's acceleration by subtracting any two consecutive velocity vectors graphically. If v i⃗ and v f⃗
are any two consecutive velocities, you can subtract v i⃗ from v f⃗ by placing −v i⃗ at the tip of v f⃗ . v f⃗ − v i⃗ is the vector that starts at the tail of v f⃗
and ends at the tip of v i⃗ .To find the direction of the particle's acceleration graphically, use two unlabeled vectors to represent −v i⃗ and v f⃗ − v i⃗ .
Pick any two vectors v i⃗ and v f⃗ that would make your subtraction easier; you can verify your result by subtracting any other pair of consecutive
vectors.
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Hwork 01 Electric Force and Field
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Problem 4
A uniform electric field exists in the region between two oppositely charged parallel plates 1.61 cm apart. A proton is released from rest at the surface of
the positively charged plate and strikes the surface of the opposite plate in a time interval 1.58×10−6 s .
Part A
Find the magnitude of the electric field.
Use 1.60×10−19 C for the magnitude of the charge on an electron and 1.67×10−27 kg for the mass of a proton.
Hint 1. How to approach the problem
Use the equation for the motion over time of a particle under constant acceleration, and use the acceleration you calculate to find the electric
field needed to create that acceleration for the proton.
Hint 2. A relationship between electric force and electric field
Recall that a charged particle in an electric field will have an electric force on it equal to F ⃗
⃗
= qE
. This means that the force will point in the
same direction as E⃗ if the particle is positively charged, and in the opposite direction as E⃗ if the particle is negatively charged.
Hint 3. Calculate the acceleration of the proton
Calculate the acceleration a of the proton, given the time taken to get from one plate to the other and the separation of the plates.
ANSWER:
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Hwork 01 Electric Force and Field
= 1.29×1010
a
Hint 4. Calculate the force on the proton
Calculate the force F on the proton needed to create the acceleration calculated in Part A.3.
ANSWER:
F
= 2.16×10−17
N
ANSWER:
135
N/C
Correct
Remember that the electric field will point from the positively charged plate to the negatively charged plate, and that the positively charged
proton moves in the same direction as the electric field.
Part B
Find the speed of the proton at the moment it strikes the negatively charged plate.
Hint 1. How to approach the problem
Use the acceleration calculated in Part A and the equations of motion to find the final velocity.
ANSWER:
2.04×104
m/s
Correct
Problem 5
Part A
What must the charge (sign and magnitude) of a particle of mass 1.50 g be for it to remain stationary when placed in a downward-directed electric
field of magnitude 610 N/C ?
Use 9.80 m/s2 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 Fg on the particle, including the sign, assuming that the positive direction is upward.
Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
ANSWER:
Fg
= −1.47×10−2
N
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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:
positive
negative
ANSWER:
−2.41×10−5
C
Correct
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 kg for the mass of a proton, 1.60×10−19 C for the magnitude of the charge on an electron, and 9.80 m/s2 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 Fg , the magnitude of the force of gravity on the proton.
ANSWER:
Fg
= 1.64×10−26
N
ANSWER:
1.02×10−7
N/C
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.
Problem 6
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
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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.
Hint 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:
A
B
C
D
Correct
Part B
In , what is wrong with panel B? (Pick only those statements that apply to panel B.)
Check all that apply.
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ANSWER:
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.
Correct
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:
A
B
C
D
Correct
Part D
In , what is wrong with panel D? (Pick only those statements that apply to panel D.)
Check all that apply.
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ANSWER:
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.
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, QA and QB . Which
of the following could represent the magnitudes and signs of QA and QB ?
In the following, take q to be a positive quantity.
ANSWER:
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Hwork 01 Electric Force and Field
,
QA = +q QB = −q
,
QA = +7q QB = −3q
,
QA = +3q QB = −7q
,
QA = −3q QB = +7q
,
QA = −7q QB = +3q
Correct
Very far from the two charges, the system looks like a single charge with value QA
will be indistinguishable from the field lines due to a single point charge +4q .
+ QB = +4q
. At large enough distances, the field lines
Problem 7
Learning Goal:
To understand the spatial distribution of the electric field for a variety of simple charge configurations.
For this problem, use the PhET simulation Charges and Fields. This simulation allows you to place multiple positive and negative point-charges in any
configuration and look at the resulting electric field.
Start the simulation. You can click and drag positive charges (red) or negative charges (blue) into the main screen. If you select Electric Field in the
menu, arrows will appear, showing the direction of the electric field. Faint arrows indicate that the electric field is weaker than at locations where the
arrows are brighter (this simulation does not use arrow length as a measure of field magnitude).
Feel free to play around with the simulation. When you are done, click the Reset button before beginning Part A.
Part A
Select Electric Field and Grid in the green menu. Drag one positive charge and place it near the middle of the screen, right on top of two intersecting
bold grid lines. You should see something similar to the figure below.
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Hwork 01 Electric Force and Field
Which of the following describes the electric field produced by the positive charge?
ANSWER:
The electric field is directed radially toward the charge at all locations near the charge.
The electric field is directed radially away from the charge at all locations near the charge.
The electric field wraps circularly around the positive charge.
Correct
This means that another positive charge, if placed near the original charge, would experience a force directed radially away from the
original charge.
Part B
Now, let's look at how the distance from the charge affects the magnitude of the electric field. Select Values on the menu, and then click and drag
one of the yellow E-Field Sensors. You will see the magnitude of the electric field given in units of V/m (volts per meter, which is the same as
newtons per coulomb). Place the E-Field Sensor 1 m away from the positive charge (1 m is two bold grid lines away if going in a horizontal or vertical
direction), and look at the resulting field strength.
Consider the locations to the right, left, above, and below the positive charge, all 1 m away. For these four locations, the magnitude of the electric
field is________________.
ANSWER:
greatest to the left of the charge.
greatest above the charge.
the same.
greatest below the charge.
greatest to the right of the charge.
Correct
This result implies that the strength of the electric field due to one point charge depends solely on the distance away from the charge.
Mathematically, we say the electric field is spherically symmetric.
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Hwork 01 Electric Force and Field
Part C
The magnitude of the electric field 1 m away from the positive charge is ________________ the magnitude of the electric field 2 m away.
Hint 1. How to approach the problem
Use an E-Field Sensor to determine the field strength both at 1 m away and at 2 m away from the charge. Then, take the ratio of the two field
strengths.
ANSWER:
one-half
two times
four times
equal to
one-quarter
Correct
The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is
proportional to the inverse of the distance squared (E ∝ 1/r2 , where r is the distance from the charge). You should verify this by looking
at the field strength 3 or 4 meters away. This is consistent with Coulomb's law, which states that the magnitude of the force between two
charged particles is F = kQ1 Q2 /r2 .
Part D
If the field strength is E = 9 V/m a distance of 1 m from the charge, what is the field strength E a distance of 3 m from the charge?
Hint 1. How to approach the problem
The magnitude of the electric field is inversely proportional to distance squared (E ∝ 1/r2 ). So if the distance is increased by a factor of
three, the field strength must decrease by a factor of three squared. You could use the simulation to make a measurement (you might have to
drag the charge away from the center so you have enough room to get 3 m away).
ANSWER:
E
= 1
V/m
Correct
Correct. Since E
∝ 1/r
2
, if the distance is increased by a factor of three, the electric field is decreased by a factor of nine.
Part E
Remove the positive charge by dragging it back to the box at the bottom, and drag a negative charge (blue) toward the middle of the screen.
Determine how the electric field is different from that of the positive charge. Which statement best describes the differences in the electric field due to
a negative charge as compared to a positive charge?
ANSWER:
The electric field changes direction (now points radially inward), but the electric field strength does not change.
Nothing changes; the electric field remains directed radially outward, and the electric field strength doesn't change.
The electric field changes direction (now points radially inward), and the magnitude of the electric field decreases at all locations.
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Correct
The electric field is now directed toward the negative charge, but the field strength doesn't change. The electric field of a point charge is
given by E⃗ = (kQ/r2 )r^ . Because of the sign of the charge, the field produced by a negative charge is directed opposite to that of a
positive charge but the magnitude of the field is the same.
Part F
Now, remove the negative charge, and drag two positive charges, placing them 1 m apart, as shown below.
Let's look at the resulting electric field due to both charges. Recall that the electric field is a vector, so the net electric field is the vector sum of the
electric fields due to each of the two charges.
Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)?
ANSWER:
The electric field is nonzero everywhere on the screen.
The electric field is roughly zero near the midpoint of the two charges.
The electric field is zero at any location along a vertical line going through the point directly between the two charges.
Correct
Directly between the two charges, the electric fields produced by each charge are equal in magnitude and point in opposite directions, so
the two vectors add up to zero.
Part G
Consider a point 0.5 m above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to
only one of the positive charges is about 18 V/m . What is the magnitude of the total electric field due to both charges at this location?
ANSWER:
36 V/m
16 V/m
25 V/m
zero
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Correct
Notice that this number is less than twice the magnitude of the field due to each charge. This occurs because the horizontal components of
the electric field due to each charge exactly cancel out (add to zero). Only the vertical components of the fields add together.
Part H
Make an electric dipole by replacing one of the positive charges with a negative charge, so the final configuration looks like the figure shown below.
The electric field at the midpoint is ________________.
ANSWER:
zero.
directed to the left.
directed to the right.
Correct
The electric field due to the positive charge is directed to the right, as is the electric field due to the negative charge. So the net electric field,
which is the sum of these two fields, is also to the right.
Part I
Make a small dipole by bringing the two charges very close to each other, where they are barely touching. The midpoint of the two charges should still
be on one of the grid point intersections (see figure below).
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Measure the strength of the electric field 0.5 m directly above the midpoint as well as 1 m directly above. Does the strength of the electric field
decrease as 1 over distance squared (1/r2 )?
Hint 1. How to approach the problem
If the strength of the field is decreasing as 1/r2 , then the ratio of the magnitudes of the electric field measured at two distances, say 0.5 m
away and 1 m away, would beE r=0.5/E r=1 = (1/0.5)2 /(1/1)2 = 4 .Compare this value to the value you measure with an E-Field Sensor.
ANSWER:
No, it decreases more quickly with distance.
No, it decreases less quickly with distance.
Yes, it does.
Correct
In fact, it turns out that the strength of the electric field decreases roughly as 1/r3 ! So the field 1 m above the midpoint is roughly eight
times weaker than at 0.5 m above the midpoint. The important lesson here is that, in general, a distribution of charges produces an electric
field that is very different from that of a single charge.
PhET Interactive Simulations
University of Colorado
http://phet.colorado.edu
Problem 8
A very long straight wire has charge per unit length 1.60×10−10 C/m .
Part A
At what distance from the wire is the magnitude of the electric field equal to 2.41 N/C ?
Use 8.85×10−12 C2 /(N ⋅ m2 ) for the permittivity of free space, and use π
.
= 3.14159
Hint 1. Equation for the electric field from a long wire
A very long wire can usually be approximated as an infinitely long wire. More specifically, you can approximate the wire as infinitely long if you
are computing the field at a point whose distance from the axis of the wire is much smaller than the length of the wire, and you are computing
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the field at a point whose distance from the axis of the wire is much smaller than its distance from the end of the wire. The magnitude of the
electric field due to an infinite wire is given by E
=
λ
2πϵ0 r
.
ANSWER:
1.19
m
Correct
Problem 9
A particle has a charge of -6.05
nC
.
Part A
Find the magnitude of the electric field due to this particle at a point 0.350
m
directly above it.
Express your answer with the appropriate units.
ANSWER:
E
= 444
N
C
Correct
Part B
Find the direction of this electric field.
ANSWER:
up, away from the particle
down, toward the particle
Correct
Part C
At what distance from this particle does its electric field have a magnitude of 11.0
N/C
?
Express your answer with the appropriate units.
ANSWER:
L
= 2.22 m
Correct
Problem 10
Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius a and positive charge q distributed evenly along its
circumference.
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Part A
What is the direction of the electric field at any point on the z axis?
Hint 1. How to approach the problem
Approach 1
In what direction is the field due to a point on the ring? Add to this the field from a point on the opposite side of the ring. In what direction is the
net field? What if you did this for every pair of points on opposite sides of the ring?
Approach 2
Consider a general electric field at a point on the z axis, i.e., one that has a z component as well as a component in the xy plane. Now imagine
that you make a copy of the ring and rotate this copy about its axis. As a result of the rotation, the component of the electric field in the xy
plane will rotate also. Now you ask a friend to look at both rings. Your friend wouldn't be able to tell them apart, because the ring that is rotated
looks just like the one that isn't. However, they have the component of the electric field in the xy plane pointing in different directions! This
apparent contradiction can be resolved if this component of the field has a particular value. What is this value?
Does a similar argument hold for the z component of the field?
ANSWER:
parallel to the x axis
parallel to the y axis
parallel to the z axis
in a circle parallel to the xy plane
Correct
Part B
What is the magnitude of the electric field along the positive z axis?
Use k in your answer, where k
=
1
4πϵ0
.
Hint 1. Formula for the electric field
You can always use Coulomb's law, F
= k
q1 q2
r
2
, to find the electric field (the Coulomb force per unit charge) due to a point charge. Given
the force, the electric field say at q2 due to q1 is E
=
F
q2
= k
q1
r
2
.
In the situation below, you should use Coulomb's law to find the contribution dE to the electric field at the point (0, 0, z) from a piece of
charge dq on the ring at a distance r away. Then, you can integrate over the ring to find the value of E . Consider an infinitesimal piece of the
ring with charge dq. Use Coulomb's law to write the magnitude of the infinitesimal dE at a point on the positive z axis due to the charge dq
shown in the figure.
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Use k in your answer, where k
=
1
4πϵ0
. You may also use some or all of the variables dq, z, and a.
ANSWER:
dE
kdq
=
z
2
+a
2
Hint 2. Simplifying with symmetry
By symmetry, the net field must point along the z axis, away from the ring, because the horizontal component of each contribution of
magnitude dE is exactly canceled by the horizontal component of a similar contribution of magnitude dE from the other side of the ring.
Therefore, all we care about is the z component of each such contribution. What is the component dE z of the electric field caused by the
charge on an infinitesimally small portion of the ring dq in the z direction?
Express your answer in terms of dE, the infinitesimally small contribution to the electric field; z, the coordinate of the point on the z
axis; and a, the radius of the ring.
ANSWER:
dE z
=
dEz
−
−−
−
−
√z 2 +a2
Hint 3. Integrating around the ring
If you combine your results from the first two hints, you will have an expression for dE z , the vertical component of the field due to the
infinitesimal charge dq. The total field is
⃗
^
^
E = E z k = k∮
dE z
.
ring
If you are not comfortable integrating dq over the ring, change to a spatial variable. Since the total charge q is distributed evenly about the
ring, convince yourself that
2π
∮
ring
q
dq = ∫
0
dθ .
2π
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ANSWER:
kzq
E (z)
=
3
(z 2 +a2 )
2
Correct
Notice that this expression is valid for both positive and negative charges as well as for points located on the positive and negative z axis. If
the charge is positive, the electric field should point outward. For points on the positive z axis, the field points in the positive z direction,
which is outward from the origin. For points on the negative z axis, the field points in the negative z direction, which is also outward from the
origin. If the charge is negative, the electric field should point toward the origin. For points on the positive z axis, the negative sign from the
charge causes the electric field to point in the negative z direction, which points toward the origin. For points on the negative z axis, the
negative sign from the z coordinate and the negative sign from the charge cancel, and the field points in the positive z direction, which also
points toward the origin. Therefore, even though we obtained the above result for postive q and z, the algebraic expression is valid for any
signs of the parameters. As a check, it is good to see that if |z| is much greater than a the magnitude of E (z) is approximately k
independent of the size of the ring: The field due to the ring is almost the same as that due to a point charge q at the origin.
q
z
2
,
Part C
Imagine a small metal ball of mass m and negative charge −q0 . The ball is released from rest at the point (0, 0, d) and constrained to move along
the z axis, with no damping. If 0 < d ≪ a , what will be the ball's subsequent trajectory?
ANSWER:
repelled from the origin
attracted toward the origin and coming to rest
oscillating along the z axis between z
circling around the z axis at z
= d
and z
= −d
= d
Correct
Part D
The ball will oscillate along the z axis between z = d and z = −d in simple harmonic motion. What will be the angular frequency ω of these
oscillations? Use the approximation d ≪ a to simplify your calculation; that is, assume that d 2 + a2 ≈ a2 .
Express your answer in terms of given charges, dimensions, and constants.
Hint 1. Simple harmonic motion
Recall the nature of simple harmonic motion of an object attached to a spring. Newton's second law for the system states that
−
−
−
Fx = m
d
2
dt
x
2
′
= −k x
, leading to oscillation at a frequency of
(here, the prime on the symbol representing the spring constant is to distinguish it from k
ω = √
k
′
m
=
1
4πϵ0
). The solution to this differential equation is
a sinusoidal function of time with angular frequency ω. Write an analogous equation for the ball near the charged ring in order to find the ω
term.
Hint 2. Find the force on the charge
What is Fz , the z component of the force on the ball on the ball at the point (0, 0, d) ? Use the approximation d 2
2
+a
2
≈ a
.
Express your answer in terms of q0 , k, q , d, and a.
Hint 1. A formula for the force on a charge in an electric field
The formula for the force F ⃗ on a charge q in an electric field E⃗ is
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⃗
⃗
F = qE
.
Therfore, in particular,
Fz = q E z .
You have already found E z (z) in Part B. Use that expression in the equation above to find an expression for the z component of the
force Fz on the ball at the point (0, 0, d) . Don't forget to use the approximation given.
ANSWER:
Fz
=
−kqq0 d
a
3
ANSWER:
−−
−−
−
ω
=
√[
kqq0
3
a m
]
Correct
Problem 11
The two charges q1 and q2 shown in the figure have equal magnitudes. Sketch the direction of
the net electric field due to these two charges at points A (midway between the charges), B ,
and C .
Part A
If both charges are negative.
Draw the net electric field vectors at points A , B , and C . The orientation of your vectors will be graded. The exact lengths of your vectors
will not be graded.
ANSWER:
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No elements selected
Select the elements from the list and add them to the canvas setting the appropriate attributes.
Correct
Part B
If both charges are positive.
Draw the net electric field vectors at points A , B , and C . The orientation of your vectors will be graded. The exact lengths of your vectors
will not be graded.
ANSWER:
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No elements selected
Select the elements from the list and add them to the canvas setting the appropriate attributes.
Correct
Part C
If q1 is positive and q2 is negative.
Draw the net electric field vectors at points A , B , and C . The orientation of your vectors will be graded. The exact lengths of your vectors
will not be graded.
ANSWER:
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No elements selected
Select the elements from the list and add them to the canvas setting the appropriate attributes.
Correct
Problem 12
A very long, straight wire has charge per unit length 3.80×10−10
C/m
.
Part A
At what distance from the wire is the electricfield magnitude equal to 2.90
N/C
?
Express your answer with the appropriate units.
ANSWER:
d
= 2.36 m
Correct
Problem 13
Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00nC is on the x -axis at x = 3.00cm. Point P is on the y-axis at y = 4.00cm.
Part A
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Calculate the electric fields E⃗ 1 and E⃗ 2 at point P due to the charges q1 and q2 . Express your results in terms of unit vectors (see example 21.6 in
the textbook).
^
Express your answer in terms of the unit vectors ^
i , j . Enter your answers separated by a comma.
ANSWER:
⃗
⃗
E1 , E2
=
^
^
^
−28125 j , −6480 i + 8640 j
N/C
Correct
Part B
Use the results of part (a) to obtain the resultant field at P , expressed in unit vector form.
^
Express your answer in terms of the unit vectors ^
i , j.
ANSWER:
⃗
E
=
^
^
−6480 i − 19485 j
N/C
Correct
Problem 14
A charge of -6.50 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.10 cm .
You may want to review (Pages 699 - 704) .
For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Field of a uniformly charged disk.
Part A
Find the magnitude of the electric field this disk produces at a point P on the axis of the disk a distance of 2.00 cm from its center.
ANSWER:
E
= 1.20×105
N/C
Correct
IDENTIFY: We must use the appropriate electric field formula: a uniform disk in (a), a ring in (b) because all the charge is along the rim of
the disk, and a point-charge in (c).
SET UP: First find the surface charge density (Q/A ), then use the formula for the field due to a disk of charge,
Ex =
σ
2ϵ0
1
[1−
−
−−
−−−
−
−
]
.
2
√(R/x) +1
EXECUTE: The surface charge density is σ
Ex =
σ
2ϵ0
−
−−
−−−
−
−
2
√(R/x) +1
E x = 1.20 × 10
5
A
−5
1
[1−
Q
=
N/C
]=
1.710×10
−12
2(8.85×10
−9
Q
=
πr
C/m
2
=
6.50×10
2
2
2
2
C /N⋅m )
C
= 1.710 × 10
−5
C/m
2
.
π(0.0110 m)
[1−
1
−
−−
−−−
−−−−−
−
2
√(
1.10 cm
2.00 cm
)
]
+1
, toward the center of the disk.
Part B
Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude of the
electric field at point P .
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ANSWER:
E
= 9.84×104
N/C
Correct
SET UP: For a ring of charge, the field is E
=
Qx
1
4πϵ0
2
2
3/2
.
(x +a )
EXECUTE: Substituting into the electric field formula gives
E =
4πϵ0
9
(9.00×10
Qx
1
2
2
3/2
4
N/C
−9
C)(0.0200 m)
2
(x +a )
E = 9.84 × 10
2
2
N⋅m /C )(6.50×10
=
2
[(0.0200 m) +(0.0110 m) ]
3/2
, toward the center of the disk.
Part C
If the charge is all brought to the center of the disk, find the magnitude and direction of the electric field at point P .
ANSWER:
E
= 1.46×105
N/C
Correct
SET UP: For a point charge, E
EXECUTE: E
= (9.00 × 10
9
= (1/4πϵ0 )q/r
2
2
.
2
N ⋅ m /C )(6.50 × 10
−9
2
C)/(0.0200 m)
= 1.46 × 10
5
N/C
.
Part D
Why is the field in part (a) stronger than the field in part (b)?
ANSWER:
3702 Character(s) remaining
The electric field of the disk in part A is stronger than the
electric field of the ring in part b
Submitted, grade pending
With the ring, more of the charge is farther from P than with the disk. Also with the ring the component of the electric field parallel to the plane of
the ring is greater than with the disk, and this component cancels.
Part E
Why is the field in part (c) the strongest of the three fields?
ANSWER:
3603 Character(s) remaining
The electric field due to a point charge in part C is greater
than the other two cases because the total field vectors
area adding with no cancellation, ant the total charge is
closer to a point P.
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Submitted, grade pending
With the point charge in (c), all the field vectors add with no cancellation, and all the charge is closer to point P than in the other two cases.
Problem 15
Two small, copper spheres each have radius 3.00 mm .
Part A
How many atoms does each sphere contain?
ANSWER:
N
= 9.54×1021 atoms
Correct
Part B
Assume that each copper atom contains 29 protons and 29 electrons. We know that electrons and protons have charges of exactly the same
magnitude, but let's explore the effect of small differences (see also Problem 21.83). Suppose charge of a proton is +e and the magnitude of the
charge of an electron is 0.100% smaller. What is the net charge of each sphere?
ANSWER:
qnet
= 44.3
C
Correct
Part C
What force would one sphere exert on the other if they were separated by 3.00 m ?
ANSWER:
F
= 1.96×1012
N
Correct
Problem 16
Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.62×10−11 m around a stationary proton.
Part A
What is the speed of the electron in its orbit?
Express your answer with the appropriate units.
ANSWER:
v
= 2.12×106
m
s
Correct
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Problem 17
Positive charge Q is distributed uniformly along the positive y-axis between y
distance x from the origin (the figure ).
= 0
and y
= a
. A negative point charge −q lies on the positive x-axis, a
Part A
Calculate the x-component of the electric field produced by the charge distribution Q at points on the positive x-axis.
Express your answer in terms of the variables Q, x, y , a and appropriate constants.
ANSWER:
Ex
=
Q
1
−
−−
−−
4πϵ0 x √ 2
2
x +a
Correct
Part B
Calculate the y-component of the electric field produced by the charge distribution Q at points on the positive x-axis.
Express your answer in terms of the variables Q, x, y , a and appropriate constants.
ANSWER:
Ey
=
−
Q
4πϵ0 a
[
1
x
−
1
−
−−
−−
√x 2 +a2
]
Correct
Part C
Calculate the x-component of the force that the charge distribution Q exerts on q .
Express your answer in terms of the variables Q, q , x, y , a and appropriate constants.
ANSWER:
Fx
=
−qQ
1
−
−−
−−
4πϵ0 x √ 2
2
x +a
Correct
Part D
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Calculate the y-component of the force that the charge distribution Q exerts on q .
Express your answer in terms of the variables Q, q , x, y , a and appropriate constants.
ANSWER:
Fy
qQ
=
4πϵ0 a
(
1
x
−
1
−
−−
−−
√x 2 +a2
)
Correct
Problem 18
A small sphere with mass m carries a positive charge q and is attached to one end of a silk fiber of length L . The other end of the fiber is attached to a
large vertical insulating sheet that has a positive surface charge density σ.
Part A
Assume that the sphere is in equilibrium and find the angle that fiber makes with the vertical sheet.
Express your answer in terms of the variables q , σ, m and appropriate constants.
ANSWER:
α
=
arctan (
σq
2mgϵ0
)
Correct
Problem 19
A semicircle of radius a is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around
the left half of the semicircle, and negative charge −Q is distributed uniformly around the right half of the semicircle in the following figure.
Part A
What is the magnitude of the net electric field at the origin produced by this distribution of charge?
Express your answer in terms of the variables Q, a, constant π, and electric constant ϵ0 .
ANSWER:
|E|
=
Q
2
π ϵ0 a
2
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Correct
Part B
What is the direction of the net electric field at the origin produced by this distribution of charge?
ANSWER:
+x
direction
−x
direction
+y
direction
−y
direction
another direction
Correct
Problem 20
A thin disk with a circular hole at its center, called an annulus, has inner radius R1 and outer radius R2 . The disk has a uniform positive surface charge
density σ on its surface.
Part A
Determine the total electric charge on the annulus.
Express your answer in terms of the variables R1 , R2 , σ and appropriate constants.
ANSWER:
Q
=
σπ (R2
2
− R1
2
)
Correct
Part B
The annulus lies in the yz-plane, with its center at the origin. For an arbitrary point on the x-axis (the axis of the annulus), find the magnitude of the
electric field E⃗ . Consider points above the annulus in the figure.
Express your answer in terms of the variables R1 , R2 , σ, x and appropriate constants.
ANSWER:
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6/22/2020
Hwork 01 Electric Force and Field
E(x)
σx
=
2ϵ0
[
1
−
−
−
−−
−
−
√x 2 +R1 2
−
1
−
−
−
−−
−
−
√x 2 +R2 2
]
Correct
Part C
Find the direction of the electric field E⃗ . Consider points above the annulus in the figure.
ANSWER:
+
x-direction
−
x-direction
Correct
Part D
A point particle with mass m and negative charge −q is free to move along the x-axis (but cannot move off the axis). The particle is originally placed
at rest at x = 0.01R1 and released. Find the frequency of oscillation of the particle.
Express your answer in terms of the variables R1 , R2 , σ, m , q , and appropriate constants.
ANSWER:
−−−−−−−−−−−−−
−
1
2π
√[
qσ
2ϵ0 m
(
1
R1
−
1
R2
)]
Correct
Score Summary:
Your score on this assignment is 95.6%.
You received 95.57 out of a possible total of 100 points.
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=8407459
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