1b – The Electric Field
Due: 11:59pm on Sunday, September 12, 2021
You will receive no credit for items you complete after the assignment is due. Grading Policy
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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