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ELECTRIC FIELDS AND FORCES
20
Q20.1. Reason: The statement of the question informs us that ball A is negative since charge is transferred to it from
a plastic rod that has been rubbed with wool. Recall that a plastic rod rubbed with wool has a negative charge.
The first statement informs us that balls B, C, and D are either neutral or positive. If they are neutral, their attraction
to ball A is due to charge polarization. If they are positive, their attraction to ball A is due to the fact that unlike
charges attract.
The second statement informs us that balls B and D are neutral. If both are charged, there will be attraction or
repulsion. If either one is charged, there will be attraction due to charge polarization of the other. Since there is
neither attraction nor repulsion, neither is charged.
The last statement informs us that ball C must be charged. It is attracted to ball B due to polarization of charges on
the neutral ball B.
Knowing that C is charged and attracted to A and that unlike charges attract, we can conclude that C has a positive
charge.
To recap, ball A is negative, balls B and D are neutral, and ball C is positive.
Assess: By careful application of our knowledge of electrostatics we have been able to account for all observations. In
some cases we had to combine two bits of information in order to make a conclusion regarding charge. All
conclusions are consistent with the knowledge that like charges repel, unlike charges attract, and charge polarization
can occur for a neutral object.
Q20.2. Reason: (a) Two like charges exert repulsive forces on each other. So the object must also have the
same charge as the plastic rod. Therefore it will attract the glass rod since the glass rod has the opposite charge from
the plastic rod.
(b) This time you cannot predict the outcome, because the object could either be charged like the glass rod (and
therefore repel the glass rod) or be neutral (and be attracted to the glass rod through polarization of the object). In
either case it is attracted to the plastic rod, so you can’t distinguish the two cases.
Assess: Don’t jump to conclusions that just because an object will attract one kind of rod that it will repel the other.
Neutral objects can be attracted to both kinds of rods through polarization.
Q20.3. Reason: (a) An insulator can be charged. We could rub it vigorously with either silk or wool. (b) A
conductor can be charged. We could touch it with an insulator that has been rubbed with either silk or wool and
hence charge it by transfer of charge.
Assess: The insulator obtained its charge by charge separation and the conductor obtained its charge by a transfer
of charge.
Q20.4. Reason: The clothes are charged by rubbing in the drier. Your body is neutral but the clothes can
polarize you a little and attract the opposite charge in your body.
Assess: This is the same way a balloon sticks to the wall after you rub it on your hair.
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20-1
20-2
Chapter 20
Q20.5. Reason: The negative charge is twice as far away from the dot as the positive charge is. But because
Coulomb’s law is an inverse square law the charge on the negative charge can’t just be double, it must be − 4Q.
Assess: At the dot the field due to the positive charge points toward the right and the field due to the negative
charge points to the left; those vectors must sum to zero.
Q20.6.
Reason: Each sphere ends up with 1 unit of negative charge. Once they touch, the two spheres become
essentially one conductor. The overall net charge is − 4 + 2 = −2. Charge is spread uniformly over the surface of a
conductor, so each sphere ends up with − 1.
Assess: The arithmetic is easy to compute the total charge. One must also realize that charge spreads out over
a conductor.
Q20.7. Reason: The following figure shows a representation of the charges on the metal sphere before and after
the positively charged rod is brought near the neutral metal sphere.
(a) The metal sphere is a conductor, so the charges are free to move. Negative charges are attracted to the positive
rod, but there are still equal numbers of positive and negative charges on the metal sphere since it is neutral.
In a real metal most of the inner electrons are still bound to their respective nucleus, but one or so of the outermost
electrons from each atom become free to move around the metal. The nuclei don’t move when the rod is broght near,
but the free electrons do, leaving the far side of the sphere short of electrons and therefore positively charged.
(b) Since the negative charges are concentrated on the region of the sphere closest to the rod, there will be a net
attractive force of the sphere to the rod.
Assess: This question uses our knowledge of charge polarization, the manner in which like and unlike charges
interact, and the fact that the electrons are the mobile charge carriers.
Q20.8.
Reason: (a) No, we cannot conclude that the wall is charged. Attractive electric forces occur between
(i) two opposite charges, or (ii) a charge and a neutral object that is polarized by the charge. Rubbing the balloon does
charge the balloon. Since the balloon is rubber, its charge is negative. As the balloon is brought near the wall, the wall
becomes polarized. The positive side of the wall is closer to the balloon than the negative side, so there is a net attractive
electric force between the wall and the balloon. This causes the balloon to stick to the wall, with a normal force
balancing the attractive force and an upward friction force balancing the very small weight of the balloon.
(b)
Assess: You have probably wondered at birthday parties how the balloons stick. Now you know. Tell everyone at
the next party, and explain why they eventually fall off, too.
Q20.9.
Reason: Put the two metal spheres in contact with each other. Then rub the glass rod on the silk to charge
the rod positively. Bring it near, but not touching, one of the spheres. This will make the near sphere negatively
charged and the far sphere equally positively charged. Separate the spheres and remove the rod.
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Electric Fields and Forces
20-3
Assess: We don’t want to touch the rod to the sphere and transfer charge, because it would be too easy to have both
spheres end up positive.
Q20.10. Reason: If the bees are charged and the grains of pollen are neutral, then the charge on the bees can
polarize the grains of pollen so that they are attracted to the charged bee with an attractive electric force. That is, the
charge on the pollen grain will be somewhat separated, with one side of the grain of pollen becoming slightly negative
(and therefore attracted to the positive bee) and the other side becoming slightly positive (repelled by the bee).
Assess: Bees also attract particles of dust and soil to their bodies the same way. In doing so they provide a chemical
survey of the area around their hive. This enables scientists to use bees to locate landmines and explosives.
Q20.11.
Reason: Charge is free to move throughout the rod-sphere combination. Negative charges move into
Rod A toward the positively charged rod, leaving the sphere positively charged.
Assess: Rod A ends up negatively charged.
Q20.12.
Reason:
(a) E1 < E2 , because the contributions from the two charges cancel at point 1 ( E1 = 0), but
the field contribution from each charge points to the right at point 2.
(b) E1 > E2 , because the contributions from each charge point in the same direction (to the right) at point 1 but point
in opposite directions and somewhat cancel at point 2 (the cancellation is not complete since point 2 is closer to the
negative charge than the positive one).
(c) E1 = E2 . The contributions from the different charges are in opposite directions at each of the two points, but in
each case the nearest charge to the point makes the stronger field to the left. The contribution from the other charge
reduces the field a bit, but there is still a net field to the left in each case. The magnitudes of the field strength are
equal.
(d) E1 < E2 , because the contributions from the two charges cancel at point 1 ( E1 = 0), but at point 2 both field
contributions have an upward component that doesn’t cancel.
(e) E1 < E2 . At point 1 the contributions are in opposite directions and partially cancel. While point 2 is farther from
the charge on the left, both contributions are in the same direction, so the field is stronger there.
(f) E1 > E2 . At point 1 the contributions are in the same direction (to the right), whereas at point 2 they partially
cancel because they are in opposite directions (not to mention that point 2 is farther away from the positive charge).
Assess: It is worth spending a few minutes to get comfortable with all these cases. There are various physics
software packages that allow you to map the fields around various charge distributions; they would be good to play
with also.
Q20.13.
Reason: Since like charges repel, negatively charged drug molecules should be placed near the
negative (black) electrode to be pushed through the skin.
Assess: The field lines point from the red electrode (positive) to the black electrode (negative), but negative
particles feel a force in the opposite direction from the local field lines.
Q20.14.
Reason: We should assume that both the 10 nC and 20 nC are small test charges that are not big enough
to affect the field around them. Under that assumption the field at the point in space remains 1200 N/C.
Assess: As long as the test charge is very small it doesn’t affect the field we are trying to measure.
Q20.15.
Reason: Let’s look at the charged straw from the end, divide it into a large number of small arc-length
segments, and represent the electric field of these segments at the center of the straw by a vector as shown in the
figure below. Next note that the two vectors add to zero. Around the straw, the electric field at the center from any
segment is cancelled by the electric field from a partner segment. Hence the electric field at the center is zero.
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20-4
Chapter 20
Assess: The answer seems reasonable in light of the symmetrical distribution of the charges.
Q20.16.
Reason: If we assume the size of the parallel plates is large compared to the separation, then the field
inside the capacitor is not changed by doubling the plate separation. So the force would stay the same.
Assess: This is not true of the plates are small compared to the separation, due to edge effects.
Q20.17.
Reason: Assume the distance between the particles is not changed. The force between two charged
particles depends directly on the charge of each particle. If the charge on each is doubled, then the force is doubled
twice. So the force increases by a factor of four.
Assess: This is true regardless of the sign of the two charges.
Q20.18.
Reason:
The operative equation is F = qE , but we’ll use the non-vector form since we only want the
magnitude.
The base case is F1 = qE1.
(a) The field is three times stronger: F3 = qE3 = q (3E1 ) = 3F1.
(b) Now the charge is three times bigger at point 1: F1′= q′E1 = (3q ) E1 = 3F1.
(c) Both the charge and the field are twice what they were in the base case: F2 = q2 E2 = (2q )(2 E1 ) = 4 F1.
(d) Both the charge and the field are twice what they were in the base case: F2 = q2 E2 = (− 2q )(2 E1 ) = − 4 F1. However,
the magnitude doesn’t have the negative sign and is still 4 F1.
Assess: Had we worried about the directional part of the vector forces, the force in part (d) would have pointed in
the opposite direction.
Q20.19.
Reason: When lightning strikes, there is a tremendous transfer of charge from the cloud to the
object struck.
When lightning strikes the metal plane, this charge is distributed over the surface of the plane.
There will initially be movement of charges over the plane (over a very short time interval) but a situation of static
equilibrium will quickly be established.
Since we have established that there is no electric field inside a conductor, the passengers will experience no change.
Assess: In like manner, if you are inside a car struck by lighting, you will be safe. The tires will most likely be
damaged as this tremendous amount of charge moves through them to ground.
Q20.20.
Reason: Opposite charges are attracted to each other.
(a) Since the microbes are positively charged, they will be attracted to the negatively charged plate.
(b) Once they are collected on the negative plate we could switch the charge on the capacitor plates so the negative
plate becomes positive, repelling and removing the microbes for analysis.
Assess: This is a clever application of simple electrical physics to a public health issue.
Q20.21.
Reason:
(a) Yes, the field would be zero at a point on the line between the two charges, closer to the 10 nC charge.
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Electric Fields and Forces
20-5
(b) In this case the contributions from the two charges are in the same direction on the line between the charges, so
there is no point between them at which the fields cancel each other.
Assess: In the first case the field contributions from the two charges are in opposite directions, so they can cancel
out when the magnitudes are the same.
Q20.22.
Reason: Because of Newton’s third law we shouldn’t choose choices D or E for any of the situations.
We ignore the gravitational attraction between the balls as it is many orders of magnitude smaller than usual
electrical forces.
(a) Like charges repel, so the correct choice is B.
(b) Opposite charges attract so the correct choice is C.
(c) Like charges repel, so the correct choice is B.
(d) The changed rod will polarize the neutral rod and they will attract each other, so the correct choice is C.
Assess: It is important to remember that F12 = − F21.
Q20.23.
Reason: In the following figure, let’s label the charges 1 and 2 so we can keep track of the electric field
due to each of them. Next, in order to make a quantitative comparison, define a unit of electric field strength as
E = kq / r 2 . For example, at a distance or 2r from a charge q, the magnitude of the field is E = kq /(2r ) 2 =
(1/4)(kq / r 2 ) = (1/4) unit.
Note that the electric field strengths at the site of interest due to the charges are either one unit or (1/4) unit. The
direction of the electric field is determined by the sign of the charge. As the figure points out, the electric field at the
dot is greatest for case A.
Assess: In this case the calculation is simplified by defining a unit of electric field and then determining the electric field
at the dot due to each charge in terms of this definition. This is a useful technique when you want to compare values.
Q20.24.
Reason: It might be helpful to take two colored pencils and draw arrows representing the contributions
to the field from the two charges. Of course, the total field is the vector sum of the contributions.
In choice A the two contributions are in the same direction, but one of the charges is farther away and doesn’t
contribute as much.
In choice B both charges contribute field in the same direction (to the right) and both charges are close. The field at
the dot will be strongest in this situation.
In choice C the contributions are in opposite directions and partially cancel out.
In choice D the contributions are equal in magnitude and opposite in direction and the total field at the dot is zero.
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20-6
Chapter 20
The correct choice is B.
Assess: The total field is the vector superposition (sum) of the contributions from the various charges in the
situation.
Q20.25.
Reason: In the following figure, let’s label the charges 1, 2, and 3 so we can keep track of the electric
field due to each of them. Let’s also represent the separation by r as shown in the figure. Next, in order to make a
quantitative comparison, define a unit of electric field strength as E = kq / r 2 . For example, at a distance of 2r from a
charge q, the magnitude of the field is E = kq /(2r ) 2 = (1/4)(kq / r 2 ) = (1/4) unit.
Note that in case A the electric field due to 2 and 3 add to zero and in case C the electric field due to 1 and 2 add to
zero. As the figure points out, the electric field at the dot is greatest for case D.
Assess: In this case the calculation is simplified by defining a unit of electric field and then determining the electric
field at the dot due to each charge in terms of this definition. This is a useful technique when you want to compare
values.
Q20.26.
Reason:
Call the charge on the glass bead q1 = 3 . 5 nC and the charge on the plastic bead q2 . We need
to solve Coulomb’s law (Equation 20.1) for q2:
F=K
| q1|| q2 |
r2
Where F = 8 . 0 × 10−4 N and r = 2 . 9 cm = 0 . 029 m, and K = 9 .0 × 109 N ⋅ m 2 /C2.
| q2 | =
Fr 2
(8 . 0 × 10−4 N)(0 . 029 m) 2
=
= 21 nC
K | q1| (9 . 0 × 109 N ⋅ m 2 /C 2 )(3 . 5 × 10−9 C)
We know q2 must be a positive charge since it is repelled by another positive charge.
The correct choice is C.
Assess: The order of magnitude of q2 is in the same ballpark as q1 , so the number is probably reasonable. Checking
your number-crunching in your calculator twice never hurts either. As for units, the N and m 2 cancel out, leaving C.
Q20.27. Reason: First, let’s make sure we have a good mental picture of the physical situation. This is best done
with a simple sketch as shown in the following figure.
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Electric Fields and Forces
20-7
Next determine the magnitude of the electric field due to each charge: E1 = kq1/ r12 = 30 × 104 N/C to the right and
E2 = kq2 / r22 = 8 × 104 N/C to the right.
Since the electric field due to both charges is in the same direction, the magnitude of the resultant electric field is
E = 3.8 × 105 N/C and the direction of this field is to the right.
The correct answer is D.
Assess: The magnitude of this field is not out of line with other electric fields determined in the chapter.
Q20.28.
Reason:
The contributions from the +3 charge and the −12 charge will cancel (!) because while the −12
charge is twice as far away (and we square r to give a reduction by a factor of 4), the charge is also 4 times larger (in
magnitude); and the directions of the contributions of those two charges are opposite. So at the dot there is no field
contribution from the combination of the +3 charge and the − 12 charge.
That leaves the − 1 charge’s contribution, which will be a field vector toward the charge.
The correct choice is A.
Assess: The electric field is the sum of the contributions to the field by each of the three charges.
Q20.29. Reason: The positive end of the dipole is repelled by the lone positive charge, while the negative end
of the dipole is attracted to the lone positive charge, so the dipole will initially rotate in a clockwise direction. The
correct answer is A.
Assess: If the lone positive charge is fixed then the dipole would rotate clockwise 180 degrees and then rotate back
180 degrees counterclockwise.
Problems
P20.1. Prepare: We will use the charge model. An electron has a negative charge of magnitude 1.6 × 10−19 C.
Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the
other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So, electrons
have been removed from the glass rod to make it positively charged.
(b) The number of electrons removed is
5 × 10−9 C
10
= 3.1 × 10
1.6 × 10−19 C
Assess: A large number of electrons are needed to create a modest charge.
P20.2. Prepare: Use the charge model. An electron has a negative charge of magnitude 1.6 × 10−19 C.
Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred into the
other because they are relatively free to move. Protons, on the other hand, are tightly bound in the nuclei. So,
electrons have been added to the plastic rod to make it negatively charged.
(b) The number of electrons added is
20 × 10−9 C
= 1.25 × 1011
1.6 × 10−19 C
which to two significant figures is 1.3 × 1011.
Assess: To create a modest charge of 20 nC, a large number of electrons need to be transferred from one material to
the other.
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20-8
Chapter 20
P20.3. Prepare: Use the charge model. Each oxygen molecule has 16 protons (8 per atom), and there are
6.02 × 1023 oxygen molecules in 1.0 mole of oxygen. The proton has a positive charge of magnitude 1.6 × 10−19 C.
Solve: The amount of positive charge in 1.0 mole of oxygen is
6
6.02 × 1023 × (16 × 1.6 × 10−19 C) = 1.5 × 10 C
Assess: Coulomb is a “big” unit of charge, so 1 mole of oxygen has a lot of positive charge.
P20.4. Prepare: We will use the charge model and the model of a conductor as material through which electrons
move. An electron has a negative charge of magnitude 1.60 × 10−19 C.
Solve: (a) The charge of a plastic rod decreases from − 15.0 nC to − 10.0 nC. That is, − 5.0 nC charge has been
removed from the plastic. Because it is the negatively charged electrons that are transferred, − 5.0 nC has been added
to the metal sphere.
(b) Because each electron has a charge of 1.60 × 10−19 C and a charge of 5.0 nC was transferred, the number of
electrons transferred from the plastic rod to the metal sphere is
5.0 × 10−9 C
= 3.1 × 1010
1.60 × 10−19 C
Assess: A modest charge of 5 nC contains over 30 billion electrons!
P20.5. Prepare: Use the charge model and the model of a conductor as a material through which electrons move.
An electron has a negative charge of magnitude 1.6 × 10−19 C.
Solve: (a) The charge of the glass rod decreases from +12 nC to +8 nC. Because it is the electrons that are
transferred, − 4 nC of electrons has been added to the glass rod. Thus, electrons are removed from the metal sphere
and added to the glass rod.
(b) Because each electron has a charge of 1.6 × 10−19 C and a charge of 4 nC was transferred, the number of electrons
transferred from the metal sphere to the glass rod is
4 × 10−9 C
= 2.5 × 1010
1.6 × 10−19 C
Assess: 25 billion electrons constitute a charge of 4 nC.
P20.6. Prepare: We will use the charge model and the model of a conductor as a material through which
electrons move. An electron has a negative charge of magnitude 1.6 × 10−19 C.
Solve: Because the metal spheres are identical, the total charge is split equally between the two spheres. That is,
11
qA = qB = 5.0 × 1011 electrons. Thus, the charge on metal sphere A and B is (5.0 × 10 )(− 1.6 × 10−19 C) = − 80 nC.
Assess:
Flow of charge from one charged conductor to another occurs when they come into contact.
P20.7. Prepare: Use the charge model and the model of a conductor as a material through which electrons
move. An electron has a negative charge of magnitude 1.6 × 10−19 C.
Solve: Plastic is an insulator and does not transfer charge from one sphere to the other. The charge of metal sphere A
12
is (1.0 × 10 )(− 1.6 × 10−19 C) = −160 nC and the charge of metal sphere B is 0 C.
Assess: Flow of charge does not occur between a charged conductor and an insulator when they are brought into
contact.
P20.8. Prepare: When two identical conducting spheres are in contact the charge is evenly distributed between them.
Solve: (a) The following grid shows the initial charge on each of the spheres and the charge after each event.
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Electric Fields and Forces
A
B
C
Initial charge
q
−q / 2
0
C is touched to B
q
−q / 4
−q / 4
C is touched to A
3q /8
−q / 4
3q /8
20-9
(b) The following grid shows the initial charge on each of the spheres and the charge after each event.
q
−q / 2
0
C is touched to A
q/2
−q / 2
q/2
C is touched to B
q/2
0
0
Initial charge
Assess: The end result depends on the order in which the various events occur.
P20.9. Prepare: We will model the charged masses as point charges. A visual overview of the forces and the
coordinate system is shown. The charge q1 exerts a force F1 on 2 on q2 to the right, and the charge q2 exerts a force
F2 on 1 on q1 to the left.
Solve: (a) Using Coulomb’s law,
9
2
K | q1|| q2| (9.0 × 10 N ⋅ m2 /C )(1.0 × 10−6 C) (1.0 × 10−6 C)
−3
=
= 9.0 × 10 N
F1 on 2 = F2 on 1 =
(1.0 m) 2
r122
(b) Newton’s second law on either q1 or q2 is
9.0 × 10−3 N
−3
= 9.0 × 10 m/s 2
1.0 kg
Assess: A relatively small force on a relatively large mass causes small acceleration.
F1 on 2 = m1a1 ⇒ a1 =
P20.10. Prepare: We will model the charged masses as point charges. A visual overview of the forces and the
coordinate system is shown. The charge q1 exerts a force F1 on 2 on q2 upward, and the charge q2 exerts a force
F2 on 1 on q1 downward.
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20-10
Chapter 20
Solve: (a) Using Coulomb’s law, the electrical force between the spheres is
9
2
K | q1|| q2| (9.0 × 10 N ⋅ m2 /C )( − 23 × 10−9 C) ( − 23 × 10−9 C)
−4
−4
F1 on 2 = F2 on 1 =
=
= 4.76 × 10 N ≈ 4.8 × 10 N
r122
(0.100 m) 2
(b) For Newton’s second law on q2 we must also consider the downward gravitational force
Fnet = F1 on 2 − m2 g = m2 a2 ⇒ a2 =
F1 on 2 − m2 g 4.76 × 10−4 N − (0.15 g)(9.80 m/s 2 )
=
= − 6.6 m/s 2
m2
0.15 g
The negative sign indicates the acceleration is downward; the magnitude of the initial acceleration is 6.6 m/s 2 .
Assess: The upward electrical force is less than the downward gravitational force, so the particle accelerates
downward, but at less than g. The answer obtained is only the initial acceleration. As the particle falls the upward
electrical force becomes stronger, so the acceleration will decrease.
P20.11. Prepare: We need to solve Coulomb’s law (Equation 20.1) for r :
F=K
| q1|| q2 |
r2
where F = 8 . 2 × 10−4 N and q1 = − 5 .0 nC, q2 = − 12 nC, and K = 9 .0 × 10 N ⋅ m 2 /C2.
9
Solve:
r2 = K
| q1|| q2 |
F
r= K
| q1|| q2 |
| − 0.5 × 10−9 C|| −12 × 10−9 C|
9
= (9 . 0 × 10 N ⋅ m 2 /C2 )
= 0 . 026 m = 2 . 6 cm
8.2 × 10−4 N
F
Assess: Notice the N and C cancel out leaving units of m.
P20.12. Prepare: We are given that qA + qB = 25 nC. Solve Coulomb’s law for qA qB , dropping the absolute
values because everything is positive.
Solve:
qA qB =
Fr 2 (5 . 4 × 10−4 N)(0 . 050 m) 2
=
= 1 . 5 × 10−16 C2
k
9 .0 × 109 N ⋅ m 2 /C2
Use qB = 25 nC − qA .
qA(25 nC − qA ) = 1. 5 × 10−16 C2
This is a quadratic equation that is solved to yield qA = 10 nC or qA = 15 nC. Then qB is the other one. We don’t
know which is which, but the charges are 10 nC and 15 nC.
Assess: These are typical charges and they do add up to the required 25 nC.
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Electric Fields and Forces
20-11
P20.13. Prepare: We will model the glass bead and the ball bearing as point charges. A visual overview of the
forces and the coordinate system is shown. The ball bearing experiences a downward electric force F1 on 2 . By
Newton’s third law, F2 on 1 = F1 on 2.
Solve: Using Coulomb’s law,
| q || q |
(9.0 × 109 N ⋅ m2 /C 2 )(20 × 10−9 C) | q2|
F1 on 2 = K 1 2 2 ⇒ 0.018 N =
⇒ | q2| = 1.0 × 10−8 C
r12
(1.0 × 10−2 m) 2
Because the force F1 on 2 is attractive and q1 is a positive charge, the charge q2 is a negative charge. Thus,
q2 = −1.0 × 10−8 C = −10 nC.
P20.14. Prepare: Charges A, B, and C are point charges. Charge A experiences an electric force FB on A due to
charge B and an electric force FC on A due to charge C. The force FB on A is directed to the right, and the force FC on A
is directed to the left.
Solve: Coulomb’s law yields:
FB on A =
(9 × 109 N ⋅ m2 /C2 )(1.0 × 10−9 C)(1.0 × 10−9 C)
= 9.0 × 10−5 N
(1.0 × 10−2 m) 2
FC on A =
(9 × 10 N ⋅ m2 /C2 )(1.0 × 10−9 C)(4.0 × 10−9 C)
= 9.0 × 10−5 N
(2.0 × 10−2 m)2
9
The net force on A is
Fon A = FB on A + FC on A = (9.0 × 10−5 N, + x-direction) + (9.0 × 10−5 N, − x-direction) = 0 N
Assess: The force on A by C is the same (but in the opposite direction) as that of B on C because C has four times
the charge and is twice the distance away compared to B. Check this statement against Coulomb’s law!
P20.15. Prepare:
The charged particles are point charges. The charge q2 is in static equilibrium, so the net
force on q2 is zero. If q2 is positive, q1 will have to be positive to make the net force zero on q2 . And, if q2 . is
negative, q1 will still have to be positive for q2 to be in equilibrium. We will assume that the charge q2 is positive.
For this situation, the force on q2 by the − 2 nC charge is to the left and by q1 is to the right.
Solve: We have
⎛ 1
⎞ ⎛ 1 (2 × 10−9 C)| q2|
⎞
| q1|| q2|
Fnet on q2 = Fq1 on q2 + F−2 nC on q2 = ⎜
,
x
-direction
, − x-direction ⎟ = 0 N/C
+
+⎜
⎟
2
2
⎝ 4πε 0 (0.2 m)
⎠ ⎝ 4πε 0 (0.10 m)
⎠
Thus,
q1
2 × 10−9 C
−
= 0 N/C ⇒ q1 = 8.0 nC
2
(0.2 m)
(0.10 m) 2
Assess: If the charge q2 is assumed negative, the force on q2 by the − 2 nC charge is to the right and by q1 is to
the left. The magnitude of q1 remains unchanged.
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20-12
Chapter 20
P20.16. Prepare: Objects A and B are point charges. Because there are only two charges A and B, the force on
charge A is due to charge B only, and the force on B is due to charge A only. A visual overview of the forces and the
coordinate system is shown.
Solve: Coulomb’s law gives the magnitude of the forces between the charge. Thus,
9
(9 × 10 N ⋅ m 2/C 2)(20.0 × 10−9 C)(10.0 × 10−9 C)
FA on B = FB on A
= 4.5 × 10−3 N
(2.0 × 10−2 m) 2
Because the charge on object A is positive and on object B is negative, FB on A is upward and FA on B is downward.
Thus,
FB on A = (4.5 × 10−3 N, + y -direction)
FA on B = (4.5 × 10−3 N, − y -direction)
Assess: You would expect this above result because of Newton’s third law.
P20.17. Prepare: Assume the glass bead, the proton, and the electron are point charges. A visual overview of
the forces and the coordinate system is shown.
Solve: Coulomb’s law gives
9
Fbead on electron = Fbead on proton =
(9 × 10 N ⋅ m2/C2 )(20 × 10−9 C)(1.60 × 10−19 C)
= 2.88 × 10−13 N
(1.0 × 10−2 m) 2
(a) Newton’s second law is F = ma, so
aproton =
Fbead on proton
mproton
=
2.88 × 10−13 N
= 1.7 × 1014 m/s2
1.67 × 10−27 kg
In vector form
aproton = (1.7 × 1014 m/s2 , away from bead)
(b) Similarly,
aelectron =
Fbead on electron 2.88 × 10−13 N
=
= 3.2 × 1017 m/s2
melectron
9.11 × 10−31 kg
Thus aelectron = (3.2 × 1017 m/s 2, toward bead ).
Assess:
Because of their small mass, electrons and protons are accelerated tremendously by electric fields.
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Electric Fields and Forces
20-13
P20.18. Prepare: The electric field is due to a charge and extends to all points in space.
Solve: The magnitude of the electric field at a distance r from a charge q is
1 |q|
|q|
E=
⇒ 1.0 N/C = (9.0 × 109 N ⋅ m2/C2 )
⇒ | q | = 1.11 × 10−10 C = 0.11 nC
2
4πε 0 r
(1.0 m)2
P20.19. Prepare: The electric field is that of a positive charge on the glass bead. The charge is assumed to be a
point charge.
Solve: The electric field is
⎛
⎞
6.0 × 10−9 C
E = ⎜ (9.0 × 109 N ⋅ m2 /C 2 )
, away from bead ⎟ = (1.4 × 105 N/C, away from bead)
2
−2
×
(2.0
10
m)
⎝
⎠
Assess: This is a typical electric field near objects that are charged by rubbing.
P20.20. Prepare: Use the equation F = qE . We are not given directions, so we’ll drop the vectors.
Solve:
E=
F 0.035 N
=
= 1.2 × 106 N/C
q
30 nC
Assess: This is a reasonable field strength.
P20.21. Prepare: Protons and electrons are point charges and produce electric fields, according to Equation
20.6. The charge on a proton is positive and on an electron is negative.
Solve: (a) The electric field of the proton is
−19
⎞
⎛ 1 |q|
⎞ ⎛
C⎤
9
2 ⎡ + 1.6 × 10
2
E =⎜
q
,
away
from
(9.0
10
N
m
/C
)
, away from q ⎟
=
×
⋅
⎜
⎟
⎢
2 ⎥
−3
2
⎜
⎟
⎣ (1.0 × 10 m) ⎦
⎝ 4πε 0 r
⎠ ⎝
⎠
−3
= (1.4 × 10 N/C, away from proton)
(b) The electric field of the electron is
⎞
⎛ 1 |q|
⎞ ⎛
⎡ + 1.6 × 10−19 C ⎤
9
E =⎜
, toward q ⎟ = ⎜ (9.0 × 10 N ⋅ m2 /C 2) ⎢
, toward q ⎟
2 ⎥
−3
2
⎜
⎟
⎣ (1.0 × 10 m) ⎦
⎝ 4πε 0 r
⎠ ⎝
⎠
−3
= (1.4 × 10 N/C, toward electron)
Assess: An electron charge is very small, so its electric field at a point 1 mm away was expected to be small.
P20.22. Prepare: The electric field is that of a positive point charge located at the origin, as shown in the
following figure.
The positions (5 cm, 0 cm), (− 5 cm, 5 cm), and (− 5 cm, − 5 cm) are denoted by A, B, and C, respectively.
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20-14
Chapter 20
Solve: (a) The electric field strength for a positive charge is
E=K
q
r2
Using K = 9.0 × 10 9 N ⋅ m2/C2 and q = 10.0 × 10−9 C,
E=
90.0 N ⋅ m2/C
r2
The electric fields at points A, B, and C are
90.0 N ⋅ m2 /C
4
EA =
= 3.6 × 10 N/C
(5.0 × 10−2 m) 2
EB =
90.0 N ⋅ m2/C
4
= 1.8 × 10 N/C
( −5.0 × 10−2 m)2 + (5.0 × 10−2 m) 2
90.0 N ⋅ m2/C
4
= 1.8 × 10 N/C
(− 5.0 × 10−2 m) 2 + (− 5.0 × 10−2 m) 2
(b) The three vectors are shown in the diagram.
Assess: The vectors EA, EB , and EC are pointing away from the positive charge.
EC =
P20.23. Model: The electric field is that of a negative charge located at the origin as shown below. We will use
Equation 20.6.
The positions (0 cm, 5 cm), (− 5 cm, −5 cm), and (− 5 cm, 5 cm) are denoted by A, B, and C, respectively.
Solve: (a) The electric field strength for a charge q is
E=K
q
r2
−9
Using K = 9.0 × 10 N ⋅ m2/C and | q | = 10.0 × 10 C,
9
2
E=
90.0 N ⋅ m2 /C
r2
The electric field strengths at points A, B, and C are
90.0 N ⋅ m2 /C
EA =
= 3.6 × 10 4 N/C
2
−2
(5.0 × 10 m)
EB =
90.0 N ⋅ m2 /C
= 1.8 × 10 4 N/C
( −5.0 × 10−2 m) 2 + (5.0 × 10−2 m) 2
90.0 N ⋅ m2 /C
= 1.8 × 10 4 N/C
( −5.0 × 10 m) 2 + (− 5.0 × 10−2 m) 2
(b) The three vectors are shown in the diagram.
Assess: Note that the vectors EA , EB , and EC are pointing toward the negative charge.
EC =
−2
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Electric Fields and Forces
20-15
P20.24. Prepare: The electric field is that of the two charges placed on the y-axis. We denote the upper charge by
q1 and the lower charge by q2 . Because both the charges are positive, their electric fields at P are directed away
from the charges.
Solve: The electric field strength of q1 is
E1 = K
| q1| (9.0 × 10 9 N ⋅ m2 /C2 )(1 × 10−9 C)
=
= 1800 N/C
(0.050 m) 2 + (0.050 m) 2
r 12
Similarly, the electric field strength of q2 is
E2 = K
| q2| (9.0 × 10 9 N ⋅ m2 /C2 )(1 × 10−9 C)
=
= 1800 N/C
(0.050 m) 2 + (0.050 m) 2
r 22
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 in the
fourth quadrant and that due to q2 is away from q2 in the first quadrant. Their components are
E1x = E1 cos 45 °
E1 y = − E1 sin 45 °
E2 x = E2 cos 45 °
E2 y = E2 sin 45 °
The x and y components of the net electric field are:
( Enet ) x = E1x + E2 x = E1 cos 45° + E2 cos 45° = 2500 N/C
( Enet ) y = E1 y + E2 y = −E1 sin 45 ° + E2 sin 45 ° = 0 N/C
⇒ Enet at dot = (2500 N/C, along + x axis)
Thus, the strength of the electric field is 2500 N/C and its direction is horizontal.
Assess: Because the charges are located symmetrically on either side of the y-axis and are of equal value, the
y-components of their fields will cancel when added.
P20.25. Prepare: The electric field is that of the two charges placed on the y-axis. We denote the upper charge by
q1 and the lower charge by q2 . The electric field at the dot due to the positive charge is directed away from the
charge and making an angle of 45° below the + x axis, but the electric field due to the negative charge is directed
toward it making an angle of 45° below the − x axis.
Solve: The electric field strength of q1 is
E1 = K
| q1| (9.0 × 109 N ⋅ m2/C2 )(1 × 10−9 C)
=
= 1800 N/C
r 12
(0.050 m) 2 + (0.050 m) 2
Similarly, the electric field strength of q2 is
E2 = K
| q2| (9.0 × 109 N ⋅ m 2/C2 )(1 × 10−9 C)
=
= 1800 N/C
r 22
(0.050 m) 2 + (0.050 m) 2
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 in the
fourth quadrant and that due to q2 is toward q2 in the third quadrant. Their components are
E1x = E1 cos 45°
E1 y = −E1 sin 45°
E2 x = −E2 cos 45°
E2 y = −E2 sin 45°
The x and y components of the net electric field are:
( Enet ) x = E1x + E2 x = E1 cos 45 ° − E2 cos 45 ° = 0 N/C
( Enet ) y = E1 y + E2 y = − E1 sin 45 ° − E2 sin 45 ° = −2500 N/C
⇒ Enet at dot = (2500 N/C, along − y axis)
Thus, the strength of the electric field is 2500 N/C and its direction is vertically downward.
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20-16
Chapter 20
Assess: A quick visualization of the components of the two electric fields shows that the horizontal components
cancel.
P20.26. Prepare: A field is the agent that exerts an electric force on a charge. Because the weight of the proton
and the electron act downward, the electric force must act upward.
Solve: (a) To balance the weight of a proton Σ( Fnet ) y = Fon p − w = 0 N. This means
Fon p = w ⇒ | q | E = mg ⇒ E =
mg (1.67 × 10−27 kg)(9.8 N/kg)
=
= 1.0 × 10−7 N/C
|q|
1.60 × 10−19 C
Because Fon p must be upward and the proton charge is positive, the electric field at the location of the proton must
also be pointing upward. Thus E = (1.0 × 10−7 N/C, upward).
(b) In the case of the electron,
E=
mg (9.11 × 10−31 kg)(9.8 N/kg)
=
= 5.6 × 10−11 N/C
|q|
1.60 × 10−19 C
Because Fon e must be upward and the electron has a negative charge, the electric field at the location of the electron
must be pointing downward. Thus E = (5.6 × 10−11 N/C, downward).
Assess:
F = qE means the sign of the charge q determines the direction of F or E . For positive q, E and F are
pointing in the same direction. But E and F point in opposite directions when q is negative.
P20.27. Prepare: The charged plastic bead is a point charge. The bead hangs suspended in the air when the net
force acting on the bead is zero. The two forces that act on the bead are the electric force and the weight. Because the
bead is negatively charged, the electric field must be pointed downward to cause an upward force, which will balance
the weight.
Solve: For the bead to be in static equilibrium,
( Fnet ) y = qE − mg = 0 N ⇒ E =
mg (0.10 × 10−3 kg)(9.8 N/kg)
=
= 6.1 × 105 N/C
q
(1.0 × 1010 )(1.60 × 10−19 C)
Thus the required field is E = (6.1 × 105 N/C, down).
Assess:
F = qE means the sign of the charge q determines the direction of F or E . For positive q, E and F are
pointing in the same direction. But E and F point in opposite directions when q is negative.
P20.28. Prepare:
Equation 20.7.
The electric field magnitude between the two parallel plates of the capacitor is given by
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Electric Fields and Forces
20-17
Solve: The electric field is E = Q/(ε 0 A). So,
Ef Qf / Af ⎛ Qf ⎞⎛ Ai ⎞
=
= ⎜ ⎟⎜ ⎟
Ei Qi / Ai ⎝ Qi ⎠⎝ Af ⎠
(a) If Q is doubled (with area remaining the same), the ratio of the final and initial electric field strengths will be
doubled.
(b) If the length L of the plates is doubled, the area increases by a factor of 4. Thus, with the charge on the plates
remaining the same, a doubling of length will decrease the ratio by a factor of 4.
(c) The electric field does not depend on the separation between the plates, so the ratio will be 1.
P20.29. Prepare: The electric field is uniform in a region of space between closely spaced capacitor plates and
is given by Equation 20.7.
Solve: The electric field inside a capacitor is E = Q/(ε 0 A). Thus, the charge needed to produce a field of strength E is
Q = ε 0 AE = (8.85 × 10−12 C 2 /N ⋅ m 2 )(0.04 m × 0.04 m)(1.0 × 106 N/C) = 14 nC
Thus, one plate has a charge of 14 nC and the other has a charge of − 14 nC.
Assess:
Note that the capacitor as a whole has no net charge.
P20.30. Prepare: The electric field in a region of space between two charged circular disks is uniform.
Solve: The electric field strength inside the capacitor is E = Q/(ε 0 A). Thus, the area is
Q
(1.5 × 10 9 )(1.6 × 10−19 C)
D2
4A
=
= 2.71 × 10−4 m2 = π
⇒D=
= 1.86 cm ≈ 1.9 cm
2
5
−12
2
ε 0 E (8.85 × 10 C /N ⋅ m )(1.0 × 10 N/C)
π
4
Assess: As long as the spacing is much less than the plate dimensions, the electric field is independent of the
spacing and depends only on the diameter of the plates.
A=
P20.31. Prepare: For the negatively charged bead to be suspended at rest the net force on it must be zero. There
is a downward gravitational force, so there must be an upward electric force.
Solve: (a) For the negatively charged bead to be suspended it must be repelled by a negative plate below and
attracted to a positive plate above. So the upper plate is positively charged.
(b) The electric force has the same magnitude as the gravitational force: qE = mg , where q is the 6.0 nC charge on
the bead and E = Q/(ε 0 A). We seek Q, the charge on the upper plate.
qE = mg ⇒ E =
Q
mg
=
⇒
ε0 A
q
(ε 0 A)(mg ) (8.85 × 10−12 C2/N ⋅ m2 ) π (6.0 cm)2 (1.0 g)(9.8 m/s 2 )
=
= 1.6 × 10−7 C
q
6.0 nC
Assess: This is a reasonable charge for a parallel plate capacitor.
Q=
P20.32. Prepare: The high part of the bottom conductor is closest to the upper conductor, so the positive charges
like to congregate there.
Solve:
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20-18
Chapter 20
(c) We see from the figures that the field is strongest at the high point of the bottom conductor (the earth); this is
where the air is mostly likely to break down and become conducting and allow the opposite charges to rush toward
each other (we call this lightning).
Assess: This is why we are told to stay away from high and tall things in a lightning storm. Stay away from tall
things and crouch down low and hug your knees if you must remain outside. Inside metal cars is usually a safe place
to be.
P20.33. Prepare: The positively charged bottom plate will attract negative charges in the neutral sphere, as the
negative charges in the top plate will attract positive charges in the neutral sphere. Hence the sphere, while neutral,
becomes polarized. The sphere has a large influence on the field lines in the space between the plates.
Solve:
Assess:
Remember that the field lines must be perpendicular to the surface of a conductor.
P20.34. Prepare: The maximum torque occurs when the dipole is perpendicular to the field. In such a situation
the moment arm for the torque is 1 2 the distance L between the charges. Also use F = qE .
Solve: There is a force on each end of the dipole, so the total torque is
τ = 2 × ( 12 L) F = L(qE ) = (100 × 10−6 m)(3.5 × 10−15 C)(4.0 × 105 N/C) = 1.4 × 10−13 N ⋅ m
Assess: This is a tiny torque, but all that is needed to turn the tiny spheres.
P20.35. Prepare: Equation 20.8 tells us the force on a charged object in an electric field: Fon q = qE.
We are given q = 30e and E = 1500 N/C.
Solve:
Fon q = qE = (30)(1 . 6 × 10−19 C)(1500 N/C) = 7 . 2 × 10−15 N
Assess: Notice the C’s cancel out leaving units of N. The answer is very small, but that is what we expect for such a
small charge.
P20.36. Prepare: Equation 20.8 tells us the force on a charged object in an electric field: Fon q = qE.
We are given q = e and E = 1 .0 × 107 N/C.
Solve:
Fon q = qE = (1 .6 × 10−19 C)(1 . 0 × 107 N/C) = 1 . 6 × 10−12 N
Assess: Notice the C’s cancel out leaving units of N. The answer is very small, but that is what we expect for such
a small charge.
P20.37. Prepare: Assume the external electric field is uniform. The CO molecule is an electric dipole
with q = 3. 4 × 10−21 C and L = 0 . 11 × 10−9 m.
We are given E = 15 000 N/C and the field is vertical.
Solve: (a) The force on one end of the dipole is equal in magnitude and opposite in direction to the force on the
other end of the dipole; hence the net force on the dipole in a uniform field is zero.
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Electric Fields and Forces
20-19
(b) The maximum torque on a dipole in a uniform electric field is τ = qEL.
τ = qEL = (3 . 4 × 10−21 C)(15 000 N/C)(0 .11 × 10−9 m) = 5. 6 × 10−27 N ⋅ m
Assess: There is no net force on a dipole in an electric field, but there can be a torque (although the torque depends
on the orientation of the molecule).
P20.38. Prepare: Use the charge model. The number of electrons per atom is the atomic number, and both the
atomic number (29) and the average atomic mass (63.5 g) are taken from the periodic table in the textbook.
Solve: The mass of copper in a 2.0-mm-diameter copper ball is
⎛ 4π 3 ⎞
⎡ 4π
⎤
M = ρV = ρ ⎜
r ⎟ = (8900 kg/m3 ) ⎢ (1.0 × 10−3 m)3 ⎥ = 3.728 × 10−5 kg = 0.03728 g
3
3
⎝
⎠
⎣
⎦
The number of moles in the ball is
M
0.03728 g
=
= 5.871 × 10−4 mol
n=
A 63.5 g/mol
The number of copper atoms in the ball is
N = nNA = (5.871 × 10−4 mol)(6.02 × 1023 mol−1 ) = 3.534 × 1020
The number of electrons in the copper ball is thus 29 × 3.534 × 10 20 = 1.025 × 1022. The number of electrons
removed from the copper ball is
50 × 10 −9 C
= 3.125 × 1011
1.60 × 10−19 C
So, the fraction of electrons removed from the copper ball is
3.125 × 1011
= 3.0 × 10−11
1.025 × 1022
Assess: This is indeed a very small fraction of the available number of electrons in the copper ball.
P20.39. Prepare: Use the charge model. The number of electrons per atom is the atomic number, and both the
atomic number (29) and the average atomic mass (63.5 g) are taken from the periodic table in the textbook.
Solve: The number of moles in the penny is
M
3.1 g
n=
=
= 0.04882 mol
A 63.5 g/mol
The number of copper atoms in the penny is
N = nNA = (0.04882 mol)(6.02 × 1023 mol−1 ) = 2.939 × 10 22
Since each copper atom has 29 electrons and 29 protons, the total positive charge in the copper penny is
(29 × 2.939 × 1022 )(1.60 × 10 −19 C) = 1.4 × 105 C
Similarly, the total negative charge is −1.4 × 105 C.
Assess: Total positive and negative charges are equal in magnitude.
P20.40. Prepare: The protons are point charges and carry positive charge. The protons have mass, so they also
have gravitational force between them.
Solve: (a) The electric force between the protons is
|q | |q | (9.0 ×109 N ⋅ m2 /C2 )(1.60 ×10−19 C)(1.60 × 10−19 C)
FE = K 1 2 2 =
= 57.6 N ≈ 58 N
r
(2.0 ×10−15 m)2
(b) The gravitational force between the protons is
Gm1m2
(6.67 × 10−11 N·m 2 /kg 2 )(1.67 × 10−27 kg)(1.67 × 10−27 kg)
FG =
=
= 4.7 × 10−35 N
2
r
(2.0 × 10−15 m) 2
(c) The ratio of the electric force to the gravitational force is
FE
57.6 N
=
= 1.2 × 1036
FG 4.7 × 10−35 N
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20-20
Chapter 20
Assess: The smallness of this ratio implies that one can normally neglect the force of gravitation between small
masses as encountered in atoms and nuclei.
P20.41. Prepare: The
125
Xe nucleus and the proton are point charges. That is, all the charge on the Xe nucleus
is assumed to be at its center. Electron charge is 1.60 × 10 −16 C.
Solve: (a) The magnitude of the force between the nucleus and the proton is given by Coulomb’s law:
K|qnucleus || qproton | (9.0 × 109 N·m 2 /C 2 )(54 × 1.60 × 10−19 C)(1.60 × 10−19 C)
=
= 498 N
Fnucleus on proton =
r2
(5.0 × 10−15 m) 2
which we will report as 500 N.
(b) Applying Newton’s second law to the proton,
498 N
Fon proton = mproton aproton ⇒ aproton =
= 2.98 × 10 29 m/s2
1.67 × 10 −27 kg
which we will report as 3.0 × 1029 m/s 2 .
Assess: A relatively large force on such a small object as a proton will cause tremendous acceleration.
P20.42. Prepare: The two charged spheres are point charges. The electric force on one charged sphere due to the
other charged sphere is equal to the sphere’s mass multiplied by its acceleration. Because the spheres are identical
and equally charged, m1 = m2 = m and q1 = q2 = q.
Solve: We have
Kq1q2
Kq 2
= ma
r2
r2
mar 2 (1.0 ×10−3 kg)(225 m/s2 )(2.0 ×10−2 m)2
=
= 1.0 × 10−14 C 2
⇒ q2 =
K
9.0 ×109 N ⋅ m2/C 2
F2 on 1 = F1 on 2 =
=
⇒ q = 1.0 ×10−7 C = 100 nC
P20.43. Prepare: Positively charged objects A and B are point charges. The repulsive force by A on B and by B
on A are equal and given by Coulomb’s law.
Solve: (a) It is given that FA on B = 0.45 N. By Newton’s third law, FB on A = FA on B = 0.45 N.
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Electric Fields and Forces
20-21
(b) Coulomb’s law is
FB on A = FA on B = 0.45 N =
⇒ qB =
KqA qB
r2
(0.45 N)r 2
=
2K
=
K (2qB )(qB )
r2
(0.45 N)(10 × 10−2 m) 2
= 5.0 ×10−7 C ⇒ qA = 2qB = 1.0 ×10−6 C
2(9.0 ×109 N ⋅ m2 /C 2 )
Assess: A relatively large force of 0.45 N between charges separated by 10 cm must mean significant charge on the
objects.
P20.44. Prepare: The distances to the observation points are large compared to the size of the dipole, so we will
model the field as that of a dipole moment. The dipole consists of charges ± q along the y-axis. The electric field is
that of the two charges placed on the y-axis. We denote the upper charge by q1 and the lower charge by q2 . The
electric field at the point shown in (a) due to the positive charge is directed away from the charge and making an
angle of θ below the + x-axis, but the electric field due to the negative charge is directed toward it making an angle
of θ below the − x-axis. The angle θ is obtained from tan θ = 0.001 m/(0.010 m), which turns out to be equal to
0.10 rad. On the other hand, the electric field at the point shown in (b) points vertically up for the positive charge and
vertically down for the negative charge.
Solve: (a) The electric field strength of q1 = + q = 1.0 nC is
E1 = K
|q1|
r 12
=
(9.0 × 109 N ⋅ m 2 /C 2 )(1.0 × 10−9 C)
= 89 109 N/C
(0.0010 m) 2 + (0.010 m) 2
Similarly, the electric field strength of q2 = −q = 1.0 nC is
E2 = K
|q2|
r 22
=
(9.0 × 109 N ⋅ m 2 /C2 )(1.0 × 10−9 C)
= 89 109 N/C
(0.0010m) 2 + (0.010m) 2
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 in the
fourth quadrant and that due to q2 is toward q2 in the third quadrant. Their components are
E1x = E1 cosθ
E1 y = −E1 sinθ
E2x = −E2 cosθ
E2 y = −E2 sinθ
The x and y components of the net electric field are:
(Enet ) x = E1x + E2x = 0 N/C
(Enet ) y = E1y + E2y = − E1 sin θ − E2 sin θ = −178 218 sin θ N/C = − 18 000 N/C
The electric field strength at (10 mm, 0 mm) is thus 18 000 N/C.
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20-22
Chapter 20
(b) The electric field strength of q1 = + q = 1.0 nC is
E1 = K
| q1| (9.0 × 109 N ⋅ m 2 /C 2 )(1.0 × 10−9 C)
=
= 111111N/C
(0.0090m) 2
r12
Similarly, the electric field strength of q2 = −q = 1.0 nC is
| q2 | (9.0 × 109 N ⋅ m 2 /C2 )(1 × 10−9 C)
=
= 74 380 N/C
(0.0110m) 2
r22
The electric field strength at (0 mm, 10 mm) is thus 111 111 N/C − 74 380 N/C = 36 730 N/C ≈ 37 000 N/C.
E2 = K
P20.45. Prepare: The electric field is that of the two 1 nC charges located on the y-axis. We denote the
top 1 nC charge by q1 and the bottom 1 nC charge by q2 . The electric fields ( E1 and E2 ) of both the positive
charges are directed away from their respective charges. With vector addition, they yield the net electric field Enet at
the point P indicated by the dot.
Solve: The electric fields from q1 and q2 are
⎛ |q |
⎞ ⎛ (9.0 × 109 N.m 2 /C 2 )(1 × 10−9 C)
⎞
E1 = ⎜ K 21 , along + x-axis⎟ = ⎜
, along + x-axis⎟
(0.05 m) 2
⎠
⎝ r1
⎠ ⎝
= (3600 N/C, along + x-axis)
⎛ 1 | q2 |
⎞
E2 = ⎜
, θ above + x-axis⎟ = (720 N/C, θ above + x-axis)
2
⎝ 4πε 0 r2
⎠
Because tan θ = 10 cm/5 cm, θ = tan −1 (2) = 63.43°.
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 along + x
and that due to q2 is away from q2 in the first quadrant. Their components are
E1x = E1
E1y = 0
E2x = E2 cos63.45°
E2y = E2 sin63.45°
The x and y components of the net electric field are:
(Enet ) x = E1x + E2x = E1 + E2 cos63.45° = 3922 N/C
(Enet ) y = E1y + E2 y = 0 + E2 sin 63.45° = 644 N/C
Thus, the strength of the electric field at P is
Enet = (3922 N/C)2 + (644 N/C)2 = 3975 N/C
which will be reported as 4000 N/C.
To find the angle this net vector makes with the x-axis, we calculate
644 N/C
⇒ φ = 9.3°
tan φ =
3922 N/C
Assess:
Because of the inverse square dependence on distance, E2 < E1. Additionally, because the point P has no
special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.
P20.46. Prepare: The electric field is that of the two 1 nC charges located on the y-axis. We denote the top
1 nC charge by q1 and the bottom 1 nC charge by q2 . The electric field E1 of the positive charge is directed away
from the charge and the electric field E2 due to the negative charge is directed toward it. With vector addition, they
yield the net electric field Enet at the point P indicated by the dot.
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Electric Fields and Forces
20-23
Solve: The electric fields from q1 and q2 are
⎛ |q |
⎞ ⎛ (9.0 × 109 N.m 2 /C 2 )(1 × 10−9 C)
⎞
E1 = ⎜ K 21 , along + x -axis ⎟ = ⎜
, along + x-axis ⎟
2
r
(0.05
m)
⎠
⎝ 1
⎠ ⎝
= (3600 N/C, along + x-axis)
⎛ 1 | q2 |
⎞
E2 = ⎜
, θ below − x -axis ⎟ = (720 N/C, θ below − x -axis)
2
⎝ 4πε 0 r2
⎠
−1
Because tan θ = 10 cm/5 cm, θ = tan (2) = 63.43°.
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 along + x
and that due to q2 is toward q2 in the third quadrant. Their components are
E1x = E1
E1y = 0
E2x = − E2 cos 63.45°
E2y = − E2 sin 63.45°
The x and y components of the net electric field are
(Enet ) x = E1x + E2x = E1 − E2 cos 63.45° = 3278 N/C
(Enet ) y = E1y + E2y = 0 − E2 sin 63.45° = −644 N/C
Thus, the strength of the electric field at P is
Enet = (3278 N/C)2 + (−644 N/C)2 = 3300 N/C
To find the angle this net vector makes with the horizontal, we calculate
|(E ) | 644 N/C
⇒ φ = 11°
tan φ = net y =
|(Enet x )| 3278 N/C
Thus, the strength of the net electric field at P is 3300 N/C and Enet makes an angle of 11° below the +x-axis.
Assess: Because of the inverse square dependence on distance, E2 < E1. Additionally, because the point P has no
special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.
P20.47. Prepare: The charges are point charges. The electric force on charge q1 is the vector sum of the forces
F2 on 1 and F3 on 1 , where q1 is the 1 nC charge, q2 is the left 2 nC charge, and q3 is the right 2 nC charge.
Solve: We have
⎛ K | q1|| q2 |
⎞
F 2 on 1 = ⎜
, away from q2 ⎟
⎝ r2
⎠
⎛ (9.0 × 109 N ⋅ m 2 / C2 )(1 × 10−9 C)(2 × 10−9 C)
⎞
, away from q2 ⎟
=⎜
2
−2
(1 × 10 m)
⎝
⎠
= (1.8 × 10−4 N, away from q2 )
( F 2 on 1 ) x = (1.8 × 10−4 N)(cos 60 °) = (0.9 × 10−4 N)
( F 2 on 1 ) y = (1.8 × 10−4 N)(sin 60 °) = (1.56 × 10−4 N)
⎛ K | q1|| q3 |
⎞
F 3 on 1 = ⎜
, away from q3 ⎟ = (1.8 × 10−4 N, away from q3 )
⎝ r2
⎠
( F 3 on 1 ) x = − (1.8 × 10−4 N)(cos 60°) = − (0.9 × 10−4 N)
( F 3 on 1 ) y = (1.8 × 10−4 N)(sin 60 °) = (1.56 × 10−4 N)
( F on 1 ) x = ( F 2 on 1 ) x + ( F 3 on 1 ) x = 0
( F on 1 ) y = ( F 2 on 1 ) y + ( F 3 on 1 ) y = 3.12 × 10−4 N
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20-24
Chapter 20
So the force on the 1 nC charge is 3.1 × 10−4 N directed upward.
Assess:
The magnitude and symmetry of q2 and q3 ensure that their x-component of the net force is zero.
P20.48. Prepare: The charges are point charges. The electric force on charge q1 is the vector sum of the forces
F2 on 1 and F3 on 1, where q1 is the 1 nC charge, q2 is the left 2 nC charge, and q3 is the right 2 nC charge.
Solve: We have
⎛ K | q1|| q2 |
⎞
F 2 on 1 = ⎜
, away from q2 ⎟
⎝ r2
⎠
⎛ (9.00 × 109 N ⋅ m 2 /C 2 )(1 × 10 −9 C)(2 × 10−9 C)
⎞
, away from q2 ⎟
=⎜
2
−2
(1 × 10 m)
⎝
⎠
= (1.8 × 10−4 N, away from q2 )
( F 2 on 1 ) x = (1.8 × 10−4 N)(cos 60 °) = (0.9 × 10−4 N)
( F 2 on 1 ) y = − (1.8 × 10−4 N)(sin 60 °) = − (1.56 × 10−4 N)
⎛ K | q1|| q3 |
⎞
, towards q3 ⎟ = (1.8 × 10−4 N, towards q3 )
F 3 on 1 = ⎜
⎝ r2
⎠
( F 3 on 1 ) x = (1.8 × 10−4 N)(cos 60 °) = (0.9 × 10−4 N)
( F 3 on 1 ) y = (1.8 × 10−4 N)(sin 60 °) = (1.56 × 10−4 N)
( F on 1 ) x = ( F 2 on 1 ) x + ( F 3 on 1 ) x = 1.8 × 10−4 N
( F on 1 ) y = ( F 2 on 1 ) y + ( F 3 on 1 ) y = 0 N
So the force on the 1 nC charge is 1.8 × 10−4 N directed horizontally to the right.
Assess:
The magnitudes and symmetry of q2 and q3 ensure that their y-component of the net force is zero.
P20.49. Prepare: The charges are point charges. Placing the 1 nC charge at the origin and calling it q1 , the q2
charge is in the first quadrant, the q3 charge is in the fourth quadrant, the q4 charge is in the third quadrant, and the
q5 charge is in the second quadrant. The electric force on q1 is the vector sum of the forces F2 on 1, F3 on 1, F4 on 1,
and F5 on 1.
Solve: The magnitude of the four forces is the same because all four charges are equal and equidistant from q1. So,
F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 =
(9.0 × 109 N ⋅ m 2 /C2 )(2 × 10−9 C)(1 × 10−9 C)
= 3.6 × 10−4 N
(5.0 × 10−2 m)2 + (0.5 × 10−2 m) 2
Thus, Fon 1 = (3.6 × 10−4 N, toward q2 ) + (3.6 × 10−4 N, toward q3 ) + (3.6 × 10−4 N, toward q4 ) + (3.6 × 10−4 N,
toward q5 ). It is now easy to see that the net force on q1 is zero.
Assess:
Look at the symmetry of the charges. It is no surprise that the net force on charge q1 is zero.
P20.50. Prepare: The charges are point charges.
Solve: Placing the 1 nC charge at the origin and calling it q1 , the q2 charge is in the first quadrant, the q3 charge is in
the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. The electric
force on q1 , is the vector sum of the electric forces from the other four charges q2 , q3 , q4 , and q5 . The magnitude of
these four forces is the same because all four charges are equal in magnitude and are equidistant from q1. So,
F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 =
(9.0 × 109 N ⋅ m 2 /C2 )(2 × 10−9 C)(1 × 10−9 C)
= 3.6 × 10−4 N
(5.0 × 10−2 m)2 + (0.5 × 10−2 m) 2
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Electric Fields and Forces
20-25
Thus, Fon 1 = (3.6 × 10−4 N, away from q2 ) + (3.6 × 10−4 N, away from q3 ) + (3.6 × 10−4 N, toward q4 ) + (3.6 × 10−4 N,
toward q5 ).
( F 2 on 1 ) x = − (3.6 × 10−4 N)(cos 45 °) = −(2.55 × 10−4 N)
( F 2 on 1 ) y = − (3.6 × 10−4 N)(sin 45 °) = − (2.55 × 10−4 N)
( F 3 on 1 ) x = − (3.6 × 10−4 N)(cos 45 °) = −(2.55 × 10−4 N)
( F 3 on 1 ) y = (3.6 × 10−4 N)(sin 45°) = (2.55 × 10−4 N)
( F 4 on 1 ) x = − (3.6 × 10−4 N)(cos 45 °) = −(2.55 × 10−4 N)
( F 4 on 1 ) y = − (3.6 × 10−4 N)(sin 45 °) = − (2.55 × 10−4 N)
( F 5 on 1 ) x = − (3.6 × 10−4 N)(cos 45 °) = −(2.55 × 10−4 N)
( F 5 on 1 ) y = (3.6 × 10−4 N)(sin 45°) = (2.55 × 10−4 N)
( F on 1 ) x = ( F 2 on 1 ) x + ( F 3 on 1 ) x + ( F 4 on 1 ) x + ( F 5 on 1 ) x = − 1.0 × 10−3 N
( F on 1 ) y = ( F 2 on 1 ) y + ( F 3 on 1 ) y + ( F 4 on 1 ) y + ( F 5 on 1 ) y = 0 N
So the force on the 1 nC charge is 1.0 × 10−3 N directed to the left.
P20.51. Prepare: The charges are point charges. Placing the 1 nC charge at the origin and calling it q1 , the
−6 nC is q3 , the q2 charge is in the first quadrant, and the q4 charge is in the second quadrant. The net electric force
on q1 is the vector sum of the electric forces from the other three charges q2 , q3 , and q4 .
Solve: We have
⎛ K | q1|| q2 |
⎞
F 2 on 1 = ⎜
, away from q2 ⎟
⎝ r2
⎠
⎛ (9.0 × 109 N ⋅ m 2 /C 2 )(1 × 10−9 C)(2 × 10−9 C)
⎞
, away from q2 ⎟
=⎜
(5.0 × 10−2 m) 2
⎝
⎠
= (0.72 × 10−5 N, away from q2 )
⎛ K | q1|| q3 |
⎞
F 3 on 1 = ⎜
, toward q3 ⎟
⎝ r2
⎠
⎛ (9.0 × 109 N ⋅ m 2 /C2 )(1 × 10−9 C)(6 × 10−9 C)
⎞
=⎜
, toward q3 ⎟
(5.0 × 10−2 m) 2
⎝
⎠
= (2.16 × 10−5 N, away from q3 )
⎛ K | q1|| q4 |
⎞
, away from q4 ⎟ = (0.72 × 10−5 N, away from q4 )
F 4 on 1 = ⎜
⎝ r2
⎠
( F 2 on 1 ) x = − (0.72 × 10−5 N)(cos 45 °) = − (0.509 × 10−5 N)
( F 2 on 1 ) y = − (0.72 × 10−5 N)(sin 45 °) = − (0.509 × 10−5 N)
( F 3 on 1 ) x = 0 N
( F 3 on 1 ) y = (2.16 × 10−5 N)
( F 4 on 1 ) x = (0.72 × 10−5 N)(cos 45 °) = (0.509 × 10−5 N)
( F 4 on 1 ) y = − (0.72 × 10−5 N)(sin 45 °) = − (0.509 × 10−5 N)
( F on 1 ) x = ( F 2 on 1 ) x + ( F 3 on 1 ) x + ( F 4 on 1 ) x = 0 N
( F on 1 ) y = ( F 2 on 1 ) y + ( F 3 on 1 ) y + ( F 4 on 1 ) y = 1.14 × 10−5 N
So the force on the 1 nC charge is 1.1 × 10−5 N directed vertically up.
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20-26
Chapter 20
P20.52. Prepare: For the particles to experience no net force at a point, the field must be zero there. The field due
to the charge at the origin must cancel the 4500 N/C uniform field, so we need to know where the field strength due
to the 5.0 nC charge is 4500 N/C, button to the left. Both the proton and the electron will experience no net force at
this same point where the field is zero.
Solve: Solve the field equation for r.
r=
kq
=
E
(9.0 × 109 N ⋅ m 2 /C 2 )(5.0 × 10−9 C)
= 0.10m
4500 N/C
For the field from this charge at the origin to cancel the uniform field the point needs to be left of the origin at
x = − 10 cm.
(a) − 10 cm.
(b) − 10 cm.
Assess: The fields would add, not cancel, at x = +10 cm, so the particles would feel a net force there, although in
different directions.
P20.53. Prepare:
The charged particles are point charges. The two 2 nC charges exert an upward force on the
1 nC charge. Since the net force on the 1 nC charge is zero, the unknown charge must exert a downward force of
equal magnitude. This implies that q is a positive charge.
Solve: The force of charges 2 on charge 1 is
⎛ K | q1|| q2 |
⎞
, away from q2 ⎟
F 2 on 1 = ⎜
2
⎝ r
⎠
12
⎛ (9.0 × 109 N ⋅ m 2 /C2 )(1.0 × 10−9 C)(2.0 × 10−9 C)
⎞
=⎜
, away from q2 ⎟
2
2
(0.020 m) + (0.030 m)
⎝
⎠
= (1.385 × 10−5 N, away from q2 )
From the figure, θ = tan −1 (2 / 3) = 33.69 ° . So,
( F 2 on 1 ) x = (1.385 × 10−5 N)(cos 33.69 °) = (1.152 × 10−5 N)
( F 2 on 1 ) y = (1.385 × 10−5 N)(sin 33.69°) = (0.768 × 10−5 N)
From symmetry, F3 on 1 is the same except the x-component is reversed. So, when we add F2 on 1 and F3 on 1, the
x-components cancel and the y-components add to give
F 2 on 1 + F 3 on 1 = 1.536 × 10−5 N
Fq on 1 must have the same magnitude, pointing in the vertically downward direction, so
(1.536 × 10−5 N)(0.020 m)2
= 0.68 nC
r
(9.0 × 109 N ⋅ m 2/C2 )(1.0 × 10−9 C)
A positive charge q = 0.68 nC will cause the net force on the 1 nC charge to be zero.
Fq on 1 = 1.536 × 10−5 N =
K |q||q1|
2
⇒q=
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Electric Fields and Forces
20-27
P20.54. Prepare: The force from q on Q must be cancelled by the force from the third particle on Q.
Solve: The third particle is half as far from Q as q is, so because of the inverse square law it must be 1/4 the charge
of q but opposite in sign. So the charge on the third particle must be −q / 4.
Assess: The field from q and the third particle would cancel at the position of Q, so no charged particle would feel
a net electrical force at that position.
P20.55. Prepare: The charges are point charges. We will denote the charges counterclockwise starting at the
upper left by 1, 2, and 3, respectively.
There is enough symmetry in this problem that we can simply the calculation. We need to show that in the x-direction
F1 on q = − F2 on q cos 45° and solve for Q. We do not need to re-do the calculation in the y-direction as the result would
be the same because of the symmetry.
Solve: First write the force of charge 1 on q and then find the x-component of it
⎛ K | − 10 nC|| q |
⎞
F1 on q = ⎜
, toward − Q ⎟
L2
⎝
⎠
KQq
( F1 on q ) x = − 2
L
Now do the same for the force of charge 2 (i.e., Q) on q. The distance is now 2 L.
⎛ K | Q|| q|
⎞
F 2 on q = ⎜
, away from 4Q⎟
2
⎝ ( 2 L)
⎠
( F 2 on q ) x =
KQq
cos 45°
2 L2
Set the sum of the x-components to zero.
( F 1 on q ) x = − ( F 2 on q ) x
−
K (− 10 nC)q
KQq
=−
cos 45
L2
2 L2
Cancel the common factors and solve for Q.
(10 nC) =
Q
Q⎛ 2⎞
4
(cos 45 °) = ⎜
(10 nC) = 28 nC
⇒Q=
⎟
2
2⎝ 2 ⎠
2
This same Q will also do the job in the y-direction to cancel the force from charge 3.
Assess: Because the x- and y-components of the net force are equal, the net force is away from +Q connecting
it to + q.
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20-28
Chapter 20
P20.56. Prepare: Assume all the atoms are C-12. The strategy will be to find the mass of one carbon grain, then
determine how many protons it has and the charge in coulombs. Then calculate the force between two such grains
with Coulomb’s law.
Solve: First find the mass in grams of a grain of carbon.
⎛4
⎞ ⎛ 1000 g ⎞
= 1.204 × 10−6 g
m = ρV = (2300 kg/m3 ) ⎜ π (0.050 mm)3 ⎟ ⎜
⎝3
⎠ ⎝ 1 kg ⎟⎠
Now find the charge in coulombs on the grain of carbon.
⎛ 6.02 × 1023 atoms ⎞ ⎛ 6 protons ⎞ ⎛ 1.60 × 10−9 C ⎞
q = (1.204 × 10−6 g) ⎜
⎟⎠ ⎜⎝ atom ⎟⎠ ⎜⎝
⎟⎠ = 0.0580 C
12.0 g
e
⎝
Apply Coulomb’s law for two such carbon grains 1.0 m apart.
| q || q |
(0.0580 C) 2
F=K
= (8.99 × 109 N ⋅ m 2 / C2 )
= 3.0 × 107 N
2
(1.0 m) 2
r
This is about 1/30 the weight of an aircraft carrier.
Assess: This is an impressively large force; it is good to remember that a charged object just has a small excess of
charge, and most of the protons are countered by electrons.
P20.57. Prepare: The electron and the proton are point charges. The electric Coulomb force between the
electron and the proton provides the centripetal acceleration for the electron’s circular motion. ω = 2 π f .
Solve:
K (e)(e) mv2
=
= mrω 2 = mr(2π f )2
r
r2
⇒ f =
Assess:
1
2π
Ke2
=
mr 3
(9.0 × 109 N ⋅ m2/C 2 )(1.60 × 10 −19 C) 2
1
=
(4.12 × 1016 rev/s) = 6.6 × 1015 rev/s
2π
(9.11 × 10 −31 kg)(5.3 × 10 −11 )3
This is an extremely fast motion, nearly 7 × 1015 rpm!
P20.58. Prepare: We have sufficient information to determine the electric force on the charged bee and the
weight of the bee. Knowing these two forces we can determine their ratio. When these two forces are equal in
magnitude and opposite in direction, the bee will hang suspended in air.
Solve: (a) The ratio of the electric force to the weight is determined by:
FE qE
=
= 2.3 × 10 −6
w mg
(b) The bee will hang suspended when the electric force is equal to the weight: FE = w or qE = mg.
This gives an electric field of E = mg/q = 4.3 × 107 N/C.
Assess: Table 20.2 informs us that an electric field of 106 N/C will create a spark in air. Note that the required
electric field is greater than the air breakdown electric field. As a result we don’t expect the bees to just hang in the
air due to the charge acquired while flying; they will have provide some of the lift.
P20.59. Prepare: The field due to the third charge (−10 nC) must be equal in magnitude and opposite in
direction to the vector sum of the contributions to the field at the origin by the other two charges.
We’ll use Equation 20.6 for the field from a point charge:
K|q|
E= 2
r
with the direction away from q if q > 0 and toward q if q < 0.
This problem is most elegantly done if we use letters to represent the values; this will save a lot of writing, and will
actually make the algebra easier. Say that q = 5.0 nC so that the charge on the x-axis is −q and the charge on the
y-axis is 2q . Also let a stand for the distance to the closest charge, 5.0 cm; this means the distance to the + 10 nC
charge is 2 a .
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Electric Fields and Forces
20-29
Solve: The field due to the first two charges is:
| q|
Ex = K 2
(due to the −5.0 nC charge)
a
in the positive x-direction; and
|2 q|
| q|
Ey = K
=K 2
(due to the +10 nC charge)
(2a ) 2
2a
in the negative y-direction. You can see that Ex = 2E y . The magnitude of the total field due to the first two charges is
then
2
E=
Ex2 + E y2 = K
|q| 2 ⎛ 1 ⎞
|q|
1 +⎜ ⎟ = K 2
a2
a
⎝ 2⎠
5
|q| 5
=K 2
4
a 2
Because Ex = 2E y , the direction of the total field is α = tan −1 ( − E y / Ex ) = tan −1 (− 1/ 2) = −26.57 ° .
We now have the magnitude and direction of the total field from the first two charges.
The field due to the third charge must have the same magnitude and opposite direction. The third charge has charge
−2q, and is r3 away from the origin. We need to solve for r3:
E3 = K
| q| 5
| −2q|
=K 2
a2 2
r3
This implies that
1 5
2
= 2
a2 2
r3
r 23 =
r3 =
4a 2
5
2a
4
5
The direction of E3 is θ = α + 180° = 153.43°.
We now have the distance from the origin and direction of the third charge; we can use these to get the x- and
y-coordinates.
2q
2(5.0 cm)
x = r3 cos θ = 4 cos θ =
cos 153.43° = − 5.98 cm ≈ − 6.0 cm
4
5
5
2a
2(5.0 cm)
sin 153.43 ° = 2.00 cm ≈ 3.0 cm
y = r3 sin θ = 4 sin θ =
4
5
5
The final result is that (x, y) for the − 10 nC charge is (− 6.0 cm, 3.0 cm).
Assess: A quick sketch of the coordinate plane with the three charges and their field contributions at the origin
should convince you that we have the right answer.
The fact that we never need to plug in the actual value of q anywhere shows that result is independent of q as long as
the ratios of the q’s are the same. Furthermore, since K also canceled, then the result would hold for any inversesquare field, such as a gravitational field if you could find both positive and negative masses.
P20.60. Prepare: The disks form a parallel-plate capacitor. The electric field inside a parallel-plate capacitor is a
uniform field, so the proton will have a constant acceleration.
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20-30
Chapter 20
Solve: (a) The field strength inside a capacitor is
Q
10 × 10−9 C
E=
=
= 3.60 × 106 N/C
−12
ε 0 A (8.85 × 10 C2 /N ⋅ m 2 )π (0.01 m) 2
(b) The electric field points toward the negative plate, so in the coordinate system of the figure
E = (3.60 × 106 N/C, vertically down). The field exerts a force F = qproton E = eE on the proton, causing an
acceleration with a y-component, that is
F eE y (1.6 × 10−19 C)( −3.60 × 106 N/C)
=
= − 3.45 × 1014 m/s 2
ay = =
1.67 × 10−27 kg
m
m
After the proton is launched, this acceleration will cause it to lose speed. To just barely reach the positive plate, it
should reach vf = 0 m/s at yf = 1 mm. The kinematic equation of motion is
v 2f = 0 m2/s 2 = v 2i + 2a y Δy ⇒ vi = −2a y Δy = −2(−3.45 × 1014 m/s2 )(0.001 m) = 8.3 × 105 m/s
Assess: The acceleration of the proton in the electric field is enormous in comparison to the gravitation acceleration
g. That is why we did not explicitly consider g in our calculations.
P20.61. Prepare: The electric field is uniform, so the electrons will have a constant acceleration. The net force
on the electron in the electric field is F = qE = ma. Using kinematics and the given information, we can find the
field strength E.
Solve: A constant-acceleration kinematic equation of motion is
v 2 − v 2 (5.0 × 107 m/s)2 − (0 m/s) 2
= 1.042 × 1017 m/s 2
v 2f = v 2i + 2a(Δx) ⇒ a = f i =
2Δx
2(1.2 × 10 −2 m)
Thus,
ma (9.11 × 10−31 kg)(1.042 × 1017 m/s 2 )
E=
=
= 5.9 × 105 N/C
1.60 × 10−19 C
q
Hence, the field strength is 5.9 × 105 N/C .
Assess: The acceleration of the electron in the electric field is so high compared with the gravitational acceleration
(g), we did not consider g explicitly in our calculations.
P20.62. Prepare: The suspended bead is not accelerating so the net force on it must be zero. Use a free-body
diagram and Newton’s second law in both directions. The electric force is purely horizontal.
Solve:
mg
Σ Fy = T cos 45 ° − mg = 0 ⇒ T =
cos 45 °
Σ Fx = k
q=
q2
r2
− T sin 45 ° = 0 ⇒ q 2 = T sin 45 °
2
r
k
r2
T sin 45 ° =
k
= (0.050 m)
r 2 ⎛ mg ⎞
mg
sin 45 ° = r
tan 45°
⎜
⎟
k ⎝ cos 45 ° ⎠
k
(0.020 × 10−3 kg)(9.8 m/s 2 )
(1) = 7.4 nC
9.0 × 109 N ⋅ m 2 /C2
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Fields and Forces
20-31
Assess: This is a typical charge for a plastic bead.
P20.63. Prepare: Just as the bead is lifted off the table the normal force from the table will be zero. The freebody diagram will have only the downward gravitational force and the upward (because the charges have opposite
sign) electric force, and they will be equal in magnitude.
Solve: Set the magnitude of the gravitational force equal to the magnitude of the electric force and solve for r.
k
| q1|| q2 |
= mg ⇒ r =
r2
k | q1|| q2 |
=
mg
(8.99 × 109 N ⋅ m 2 / C2 )(2.5 nC)(5.6 nC)
= 5.7 cm
(4.0 mg)(9.8 m/s 2 )
Assess: This is a reasonable distance between beads.
P20.64. Model: The charged spheres are point charges. Each sphere is in static equilibrium when the string
makes an angle of 20° with the vertical. The three forces acting on each sphere are the electric force, the weight of
the sphere, and the tension force. A visual overview of the problem is shown.
Solve: In the static equilibrium, Newton’s first law is Fnet = T + w + Fe = 0. In component form,
( Fnet ) x = Tx + wx + ( Fe ) x = 0 N (Fnet ) y = Ty + wy + ( Fe ) y = 0 N
Kq 2
= 0 N T cos θ − mg + 0 N = 0 N
d2
Kq 2
+Kq 2
T cos θ = +mg
⇒ T sin θ = 2 =
d
(2 L sin θ ) 2
Dividing the two equations and solving for q,
⇒ −T sin θ + 0 N +
q=
4 sin 2 θ tan θ L2 mg
=
K
4(sin 2 20° tan 20°)(1.0 m)2 (3.0 × 10−3 kg)(9.8 N/kg)
= 750 nC
9.0 × 109 N ⋅ m 2 /C2
P20.65. Prepare: The charged ball attached to the string is a point charge. The ball is in static equilibrium in the
external electric field when the string makes an angle θ = 20 ° with the vertical. The three forces acting on the
charged ball are the electric force due to the field, the weight of the ball, and the tension force.
Solve: In static equilibrium, Newton’s second law for the ball is Fnet = T + w + Fe = 0. In component form,
(Fnet ) x = Tx + 0 N + qE = 0 N
( Fnet ) y = Ty − mg + 0 N = 0 N
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20-32
Chapter 20
The two previous equations simplify to
T sin θ = qE T cos θ = mg
Divide the first equation by the second to get
qE
mg tan θ (5.0 × 10−3 kg)(9.8 N/kg) tan 20 °
tan θ =
⇒q=
=
= 1.78 × 10−7 C = 180 nC
mg
E
100 000 n/C
to two significant figures.
P20.66. Prepare: The charged ball attached to the string is the point charge. The charged ball is in static
equilibrium in the external electric field when the string makes an angle θ with the vertical. The three forces acting
on the charge are the electric force due to the electric field, the weight of the ball, and the tension force.
Solve: In static equilibrium, Newton’s second law for the charged ball is Fnet = T + w + Fe = 0. In component form,
(Fnet ) x = Tx + 0 N + qE = 0 N
( Fnet ) y = Ty − mg + 0 N = 0 N
These two equations become T sin θ = qE and T cosθ = mg . Dividing the equations gives
tan θ =
qE (25 × 10−9 C)(200 000 N/C)
=
= 0.255 ⇒ θ = 14°
mg
(2.0 × 10−3 kg)(9.8 N/kg)
P20.67. Prepare: The acceleration due to gravity is of comparable size to the acceleration given in the problem, so
we will not neglect gravity. The forces on the bead are mg down and qE up. Use Newton’s second law.
Solve:
m(a + g ) (0.0010 kg)(20 m/s 2 + 9.8 m/s 2 )
=
= 150 nC
E
200 000 N/C
Assess: The answer is in the ballpark of charges for plastic beads.
∑ F = qE − mg = ma ⇒ q =
P20.68. Prepare: Since the bead is in equilibrium the sum of the forces on it is zero. Assume the rod is frictionless
so the only forces are gravity and the electric force.
Solve: Solve for r in the application of Newton’s second law.
qq
qq
Σ F = K 1 2 2 − mg = 0 ⇒ r 2 = K 1 2
r
mg
r= K
q1q2
(15 nC)(10 nC)
= (9.0 × 109 N ⋅ m 2 /C 2 )
= 5.2 cm
mg
(5.0 × 10−5 kg)(9.8 m/s 2 )
Assess: 5.2 cm seems to be a reasonable distance.
P20.69. Prepare: When the bead is at rest the sum of the forces on it is zero, so the magnitude of the force from
the 4q charge at the right must be the same as the magnitude of the force from the q charge at the left. This can be
solved with Coulomb’s law and some algebra, but it is easier to just reason about it.
Solve: Since Coulomb’s law is an inverse square law, the 4q charge needs to be twice as far away from the free-tomove charge as the q on the left is. The place that is twice as far away from x = 4.0 cm as the origin is one-third the
way from the origin, or at x = 1.3 cm.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Fields and Forces
20-33
Assess: The charge that is twice as far away has four times the charge to produce the same magnitude force.
P20.70. Prepare: The gravitational force is so much smaller than the electrical force that it can be neglected.
Therefore, the net force is the electrical force. The acceleration will be constant, so the kinematic equations will
Q
.
apply; we’ll solve Δx = 12 a(Δt ) 2 for Δ t. The electric field in a parallel plate capacitor is E =
ε0 A
Solve: Apply Newton’s second law and the kinematic equations.
Fnet = ma = qE ⇒
a=
q
q Q
(1.60 × 10−19 C)
(8.3 nC)
E=
=
= 3.21 × 1013 m/s 2
m
m ε 0 A (1.67 × 10−27 kg) (8.85 × 10−12 C2 /N ⋅ m 2 )(28 cm 2 )
Plug this into the kinematic equation.
Δt =
2Δ x
2(3.0 mm)
=
= 1.4 × 10−8 s
a
(3.21 × 1013 m/s 2 )
Assess: The acceleration is huge because the mass is so small. The time required to traverse the capacitor is only 14 ns.
P20.71. Prepare: The laser beam probes the solution and establishes the different types of cells. Based on the
type of cell, the laser can direct the electronics to charge the color (positive or negative) or leave the color uncharged.
As the drop (containing one cell) is formed and leaves the liquid stream, if the laser determines for example that it is
type “A,” it might cause the color to become negative; if it is type “B,” it might cause the color to become positive;
and if it is type “C,” the color would remain uncharged. If the color is positive, as the drop is formed it will be
polarized with the bottom of the drop being negative and the top positive. When the drop breaks off the liquid stream,
since the bottom is more negative, the drop will have a net negative charge. This negative charge will be deflected in
a predictable manner by the deflection plates and into the collection well.
Solve: When the color is positive, the net charge on the drops will be negative. The correct answer is B.
Assess: The underlying principle for the operation of this device is that like charges repel and unlike charges
attract. It is impressive the mileage we can get out of such a simple basic phenomenon.
P20.72. Prepare: Like charges attract and opposite charges repel.
Solve: Positively charged drops will be repelled by the positive plate and attracted to the negative plate, so they will
be directed to the right, with the greater charges feeling a stronger force.
Negative charges, on the other hand, are attracted by the positive plate and repelled by the negative plate, so they will
be directed toward the left; drops with more charge will feel a stronger force and be deflected more ( F = qE ).
The correct choice is D.
Assess: One might think that it is somehow harder to deflect drops with a greater charge, but that is not the case. It
would be harder to deflect them if they had a greater mass (because of F = ma ), but the greater charge actually
makes them experience a greater force in the same field. And the mass of a doubly charged drop is not appreciably
different from the mass of a singly charged drop.
P20.73. Prepare: The dipole moment points from the negative to the positive charge. The negative charge will
be attracted to the positive plate and the positive charge will be attracted to the negative plate.
Solve: Since the negative charge is attracted to the positive plate and the positive charge is attracted to the negative
plate, and the electric dipole moment goes from the negative to the positive charge, the dipole moment must point in
the same direction as the electric field, or, in this case, to the right. The correct choice is D.
Assess: This result is consistent with our knowledge of electrostatics, electric fields, and electric dipole moments.
P20.74. Prepare: As drawn in the figure, there are several drops between the plates at the same time.
Solve: If all the drops have the same charge, then they will all feel the same force between the plates, so they would
not be sorted.
The correct choice is A.
Assess: It is not likely that the forces between drops would dominate the deflecting force.
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