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ELECTRIC CHARGES AND FORCES
25
Conceptual Questions
25.1. An insulator can be charged. Plastic is an insulator. A plastic rod can be charged by rubbing it with wool.
25.2. A conductor can be charged. A conductor can be charged by touching it with another charged object.
25.3. B and D are both neutral because they have no effect on each other and neutral is still attracted to either glass
or plastic. Since ball A has been touched by plastic it is also now plastic. Since ball C is attracted to plastic (A) and
neutral (B) then it must be glass.
25.4. (a) Like charges exert repulsive forces on each other, so the object must also have “plastic” charge. Therefore,
it will attract the glass rod, which has the opposite charge (i.e., “glass” charge).
(b) You cannot predict this because the object could be glass or neutral. Glass will repel the glass rod but neutral will
be attracted to the glass rod.
25.5. Upon touching the charged rod, the metal exchanges charge with the area of the rod touched by the sphere (we
are assuming the rod is an insulator). Some of the local excess charge on the rod will spread over the conducting
sphere, so that both the rod and the sphere will have an overall excess charge of the same type. Thus, the sphere and
the rod will repel each other.
25.6. Assume that the basic premise of “like charges repel, unlike charges attract” still holds. Suspend an object
with an excess of unknown charge from a string. First, approach a plastic-charged rod, then a glass-charged rod. An
object with charge X must be attracted by both of these. However, this is not sufficient because a neutral object
would also be attracted by both rods. To determine if the object is neutral or not, approach a neutral object. If the
object has charge X, it will be attracted, if the object is neutral, nothing will happen.
25.7. (a)
The negatively charged rod will repel the negative charges on the top of the electroscope, pushing more negative
charge down onto the leaves. The leaves will separate more.
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25-1
25-2
Chapter 25
(b)
The positively charged rod will attract more negative charges to the top of the electroscope. As they depart from the
leaves, the leaves will move closer together.
25.8.
The final state of each sphere and of the rod is neutral. The conducting rod allows the excess electrons in the
negatively charged sphere to move to the positively charged sphere and exactly neutralize the charge there, leaving
all three conductors neutral.
25.9. Each sphere ends up with one 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.
25.10.
The rod will polarize the charges in the combined conductor A + B, attracting negative charges to A and leaving B
with excess positive charge. The combined conductor A + B is still neutral, but A alone has net negative charge.
25.11.
Your finger becomes polarized. Positive charge is left on the tip of your finger when negative charges in the finger
are repelled by the ball. The excess positive charge in your finger is then nearer to the negatively charged ball than
the negative charge in your finger, resulting in a net attractive force that attracts the ball.
25.12.
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Electric Charges and Forces
25-3
25.13. (a) The magnitude of the force on A quadruples (increases by a factor of 4), since the force between the
charges is proportional to the product of the magnitudes of the charges. Therefore, FA′ = 4 F .
(b) By Newton’s third law, the force of A on B is equal in magnitude to the force of B on A; therefore the force on B
also quadruples; FB′ = FA′ = 4 F .
25.14. (a) We have E(r) = 1000 N/C. For a point charge, E ( r ) ∝
1
r2
. If the distance r to the charge is doubled,
⎛ 1 ⎞
E (2r ) ⎜⎝ 4r 2 ⎟⎠ 1
E (2r ) ∝
, so
=
=
= .
E (r )
⎛ 1 ⎞ 4
(2r ) 2 4r 2
⎜ 2⎟
⎝r ⎠
1
1
Therefore
1
E (2r ) = (1000 N/C) = 250 N/C.
4
1
⎛r⎞
(b) Similarly, E ⎜ ⎟ E ( r ) =
r 2
⎝2⎠
(2)
1
r2
= 4, so
⎛r⎞
E ⎜ ⎟ = 4(1000 N/C) = 4000 N/C.
⎝2⎠
25.15. Since the force on a charge in an electric field has magnitude F = qE, the new force is
3
⎛E⎞ 3
F ′ = (3q ) ⎜ ⎟ = qE = F .
2
⎝2⎠ 2
Exercises and Problems
Section 25.1 Developing a Charge Model
Section 25.2 Charge
25.1. Model: Use the charge model.
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 of atoms and
so are essentially not free to move. Thus, electrons have been added to the plastic rod to make it negatively charged.
(b) Because each electron has a charge of −1.60 × 10−19 C, the number of electrons added is
−12 × 10−9 C
− 1.60 × 10−19 C
= 7.5 × 1010
25.2. Model: Use the charge model.
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 the nuclei of atoms and so are
essentially not free to move. Thus, electrons have been removed from the glass rod to make it positively charged.
(b) Because each electron has a charge of −1.60 × 10−19 C, the number of electrons removed is
−8.0 × 10−9 C
= 5.0 × 1010
−1.60 × 10−19 C
where the numerator is negative because this is the charge that is removed, so the excess charge left behind is
8.0 × 10−9 C.
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25-4
Chapter 25
25.3. Model: Use the charge model and the model of a conductor as a material through which electrons move.
Solve: (a) The charge of the glass rod decreases from +12 nC to +8.0 nC. Because it is the electrons that are
transferred, −4.0 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 −1.60 × 10−19 C and a charge of −4.0 nC was transferred, number of
electrons transferred from the metal sphere to the glass rod is
−4.0 × 10−9 C
−1.60 × 10−19 C
= 2.5 × 1010
25.4. Model: Use the charge model and the model of a conductor as material through which electrons move.
Solve: (a) The charge of a plastic rod changes from −15 nC to −10 nC. That is, −5 nC charge has been removed from
the plastic. Because it is the negatively charged electrons that are transferred, −5 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
−1.60 × 10
−19
C
= 3.1 × 1010
25.5. Model: Use the charge model.
Solve: Each helium atom has 2 protons and there are 6.02 × 1023 helium molecules in 1.0 mole of helium. Because
each proton has a charge of +1.60 × 10−19 C, the amount of charge in 1.0 mole of oxygen is
(1.0 mol)(6.022 × 1023 atoms/mol)(2 protons/atom)(1.6 × 10−19 C/proton) = 1.9 × 105 C
25.6. Model: Use the charge model.
3
⎛ 1000 mL ⎞ ⎛ 1.0 cm ⎞ ⎛ 1.0 g ⎞
Solve: Since the density of water is 1.0 g/cm3 , the mass of 1.0 L of water is (1.0 L) ⎜
⎟⎜
⎟ ⎜⎜
⎟ = 1.0 kg.
L
⎝
⎠ ⎝ mL ⎟⎠ ⎝ cm3 ⎠
Each water molecule (H 2 0) has 10 protons (8 in the oxygen atom and one per hydrogen atom), and thus 10 electrons.
The number of moles is
1.0 × 103 g
= 100 moles. There are 6.02 × 1023 water molecules in 1.0 mole of water. Because
10 g/mole
one electron has a charge of −1.60 × 10−19 C, the amount of charge in 100 mole of water is
(100 mol)(6.022 × 1023 H 2O/mol)(10 electron/H 2O)( − 1.6 × 10−19 C/electron) = −9.6 × 107 C
Section 25.3 Insulators and Conductors
25.7. Model: Use the charge model and the model of a conductor as a material through which electrons move.
Visualize:
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Electric Charges and Forces
25-5
The charge carriers in a metal electroscope are the negative electrons. As the positive rod is brought near, electrons are
attracted toward it and move to the top of the electroscope. The electroscope leaves now have a net positive charge,
due to the missing electrons, and thus repel each other. At this point, the electroscope as a whole is still neutral (no net
charge) but has been polarized. On contact, some of the electrons move to the positive rod to neutralize some (but not
necessarily all) of the rod’s positive charge. After contact, the electroscope does have a net positive charge. When the
rod is removed, the net positive charge on the electroscope quickly covers the entire electroscope (note that no positive
charges move, but the electrons distribute themselves over the surface so that there is a net positive charge everywhere
on the surface). The net positive charge on the leaves causes them to continue to repel.
25.8. Model: Use the charge model.
Solve: (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 electric force and an upward frictional force balancing the gravitational force on the balloon.
(b)
25.9. Model: Use the charge model and the model of a conductor as a material through which electrons move.
Solve:
The first step shows two neutral metal spheres touching each other. In the second step, the negative rod repels the
negative charges which will retreat as far as possible from the top of the left sphere. Note that the two spheres are
touching and the net charge on these two spheres is still zero. While the rod is there on top of the left sphere, the right
sphere is moved away from the left sphere. Because the right sphere has an excess negative charge then, by charge
conservation, the left sphere has the same magnitude of positive charge. Upon separation, the negative charge is
trapped on the right sphere, as shown in the third step. As the two spheres are moved apart farther and the negatively
charged rod is moved away from the spheres, the charges on the two spheres redistribute uniformly over the entire
surface spheres. Thus, we are left with two oppositely charged spheres.
25.10. Model: Use the charge model and the model of a conductor as a material through which electrons move.
Solve:
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25-6
Chapter 25
Charging two neutral spheres with like charges of exactly equal magnitude can be achieved through the following six
steps. (i) Bring a charged rod (say, negative) near a neutral metal sphere. (ii) Touch the neutral sphere with the
negatively charged rod, so that the rod-sphere system has a net negative charge. (iii) Move the rod away from the
sphere. The sphere is now negatively charged. (iv) Bring this negatively charged sphere close to the second neutral
sphere. (v) Touch these two spheres. The excess negative charge is distributed evenly over the two spheres. (vi)
Separate the spheres. The excess charge will have the same sign as the charge on the charging rod and will be evenly
distributed between the two spheres.
25.11. Model: Use the charge model and the model of a conductor as a material through which electrons move.
Solve:
Charging two neutral spheres with opposite charges of equal magnitude can be done through the following four steps.
(i) Touch the two neutral metal spheres together. (ii) Bring a charged rod (say, positive) close (but not touching) to
one of the spheres (say, the left sphere). Note that the two spheres are still touching and the net charge on the pair is
zero. The right sphere has an excess positive charge of exactly the same magnitude as the left sphere’s negative
charge. (iii) Separate the spheres while the charged rod remains close to the left sphere, so the separated charge
remains on the spheres. (iv) Take the charged rod away from the two spheres. The separated charges redistribute
uniformly over the metal sphere surfaces.
Section 25.4 Coulomb’s Law
25.12. Model: Model the charged masses as point charges.
Visualize:
Solve: (a) 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. Using Coulomb’s law,
F1 on 2 = F2 on 1 =
K q1 q1
r122
=
(9.0 × 109 N m 2 /C2 )(10 × 10−6 C)(10 × 10−6 C)
(1.0 m) 2
= 0.90 N
(b) Applying Newton’s second law on either q1 or q2 gives
F1 on 2 = m1a1 ⇒ a1 =
0.90 N
= 0.90 m/s 2
1.0 kg
Assess: Even a micro-Couomb is a lot of charge. That is why F1 on 2 (or F2 on 1) is a measurable force.
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Electric Charges and Forces
25-7
25.13. Model: Model the plastic spheres as point charges.
Visualize:
Solve: (a) The charge q1 = −50.0 nC exerts a force F1 on 2 on q2 = −50.0 nC to the right, and the charge q2 exerts a
force F2 on 1 on q1 to the left. Using Coulomb’s law,
F1 on 2 = F2 on 1 =
K q1 q2
r122
=
(9.0 × 109 N m 2 /C2 )(50.0 × 10−9 C)(50.0 × 10−9 C)
(2.0 × 10−2 m) 2
= 0.056 N
(b) The ratio of the electric force to the weight is
F1 on 2
0.056 N
=
= 2.9
mg
(2.0 × 10−3 kg)(9.8 m/s 2 )
25.14. Model: Model the glass bead and the ball bearing as point charges.
Visualize:
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,
F1 on 2 = K
q1 q2
r122
⇒ 0.018 N =
(9.0 × 109 N m 2 /C2 )(20 × 10−9 C) q2
(1.0 × 10−2 m) 2
⇒
q2 = 1.0 × 10−8 C
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.
25.15. Model: The protons are point charges.
Solve: (a) The electric force between the protons is
FE = K
q1 q2
r2
=
(9.0 × 109 N m 2 /C2 )(1.60 × 10−19 C)(1.60 × 10−19 C)
(2.0 × 10−15 m) 2
= 58 N
(b) The gravitational force between the protons is
G m1m2 (6.67 × 10211 N m 2 /kg 2 )(1.67 × 10−27 kg)(1.67 × 10−27 kg)
=
= 4.7 × 10−35 N
FG =
(2.0 × 10−15 m)2
r2
(c) The ratio of the electric force to the gravitational force is
FE
58 N
=
= 1.2 × 1036
FG 4.7 × 10−35 N
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25-8
Chapter 25
25.16. Model: Charges A, B, and C are point charges.
Visualize: Please refer to Figure EX25.16. 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 = K
FC on A = K
qA qB
r2
qC qA
r
2
=
=
(9.0 × 109 N m 2 /C2 )(1.0 × 10−9 C)(1.0 × 10−9 C)
(1.0 × 10−2 m)2
(9.0 × 109 N m 2 /C2 )(1.0 × 10−9 C)(4.0 × 10−9 C)
(2.0 × 10
−2
m)
2
= 9.0 × 10−5 N
= 9.0 × 10−5 N
The net force on A is
Fon A = FB on A + FC on A = (9.0 × 10−5 N)iˆ + (9.0 × 10−5 N)(−iˆ) = 0.0 N
25.17. Model: Charges A, B, and C are point charges.
Visualize: Please refer to Figure EX25.17.
Solve: The force on B from charge A is directed downward since two negative charges repel. Coulomb’s law gives
the magnitude of the force as
FA on B = K
qA qB
r2
=
(9.0 × 109 Nm 2 /C2 )(1.0 × 10−9 C)(2.0 × 10−9 C)
(2.0 × 10−2 m ) 2
= 4.5 × 10−5 N
So FA on B = (−4.5 × 10−5 ˆj ) N
The force on B from charge C is directed downwards since the two opposite charges attract. Coulomb’s law gives the
magnitude of the force as
q q
(9.0 × 109 Nm 2 /C2 )(2.0 × 10−9 C)(2.0 × 10−9 C)
= 3.6 × 10−4 N
FC on B = K C 2 B =
(1.0 × 10−2 m) 2
r
So FC on B = −(3.6 × 10−4 ˆj ) N
The net electric force on charge A is
FA = FB on A + FC on A = −(4.5 × 10−5 ˆj ) N − (3.6 × 104 ˆj ) N
= −(4.1 × 10−4 ˆj ) N
25.18. Model: Objects A and B are point charges.
Visualize:
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.
Solve: Coulomb’s law gives the magnitude of the forces between the charge:
FA on B = FB on A =
(9.0 × 109 N m 2 /C2 )(8.0 × 10−9 C)(4.0 × 10−9 C)
(2.0 × 10−2 m) 2
= 7.2 × 10−4 N
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Electric Charges and Forces
25-9
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 = + (7.2 × 10−4 N) ˆj
and
FA on B = −(7.2 × 10−4 N) ˆj
Assess: By Newton’s third law, the two forces have equal magnitudes but opposite directions because they form an
action-reaction pair, just as we found.
25.19. Model: Assume the plastic bead, the proton, and the electron are point charges.
Visualize:
Solve: Coulomb’s law gives
(9.0 × 109 N m 2 /C2 )(15 × 10−9 C)(1.60 × 10−19 C)
= 2.16 × 10−13 N
(1.0 × 10−2 m)2
(a) Becaue the bead is much more massive than both the electron and the proton, we can ignore any acceleration of
the bead. Newton’s second law is F = ma, so
Fbead on proton 2.16 × 10−13 N
aproton =
=
= 1.3 × 1014 m/s 2
mproton
1.67 × 10−27 kg
Fbead on electron = Fbead
on proton
=
Because opposite charges attract,
aproton = (1.3 × 1014 m/s 2 , toward bead)
(b) Similarly,
aelectron =
Fbead on electron 2.16 × 10−13 N
=
= 2.4 × 1017 m/s 2
−31
melectron
9.11 × 10 kg
Thus aelectron = (2.4 × 1017 m/s 2 , away from bead).
Assess: Although the force on the proton has the same magnitude as the force on the electron, the electron has a
much greater acceleration because it has a much smaller mass.
Section 25.5 The Field Model
25.20. Model: Protons and electrons produce electric fields.
Solve: (a) The electric field of the proton is
−19 ⎤
C
9
2 2 ⎡ +1.60 × 10
ˆ
r
=
(9.0
×
10
N
m
/C
)
rˆ = (1.4 × 10−3 N/C, away from proton)
⎢
−3
2
2 ⎥
r
⎣⎢ (1.0 × 10 m) ⎦⎥
(b) The electron carries the opposite charge, so the electric field of the electron is
E = (1.4 × 10−3 N/C)(− rˆ) = (1.4 × 10−3 N/C, toward electron)
E=K
q
25.21. Model: Model the proton and the electron as point charges.
Solve: (a) The force that an electric field E exerts on a charge q is F = qE. A proton has q = e. Thus,
Fproton = e(400iˆ + 100 ˆj ) N/C = (6.4iˆ + 1.6 ˆj ) × 10−17 N
where we used e = 1.602 × 10−19 C.
(b) The charge on an electron is q = −e. Thus, Felectron = − Fproton = ( −6.4iˆ − 1.6 ˆj ) × 10−17 N
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25-10
Chapter 25
(c) From Newton’s second law,
aproton =
Fproton
mproton
=
Fx2 + Fy2
mproton
=
6.605 × 10−17 N
1.67 × 10
−27
kg
= 4.0 × 1010 m/s 2
(d) The electron experiences a force of the same magnitude but it has a different mass. Thus,
6.605 × 10−17 N
= 7.3 × 1013 m/s 2
aelectron =
9.11 × 10−31 kg
The forces may be the same, but the electron has a much larger acceleration due to its much smaller mass.
Assess: The two forces in parts (a) and (b) are equal in magnitude but opposite in direction.
25.22. Model: 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
q
q
E = K 2 ⇒ 1.0 N/C = (9.0 × 109 N m 2 /C2 )
⇒ q = 1.1 × 10−10 C = 0.11 nC
(1.0 m) 2
r
25.23. Model: The electric field is that of a negative charge on the plastic bead. Model the small bead as a point
charge.
Solve: The electric field is
q
−8.0 × 10−9 C
E = K 2 rˆ = (9.0 × 109 N m 2 /C2 )
rˆ = −4.5 × 104 rˆ N/C
2
−2
(4.0 × 10 m)
r
where r̂ is the unit vector from the charge to the point at which we calculate the field.
Assess: The direction of the electric field is toward the bead, which it should be since the bead is negative.
25.24. Model: Treat the charge on the object is a point charge.
Solve: The electric field at a distance r from a point charge q is E = K
q
r2
rˆ. Because the electric field points away
from the object, E = (270,000 N/C)rˆ. Thus,
K
q =
q
r2
rˆ = 270,000 N/C rˆ
(270,000 N/C)(2.0 × 10−2 m) 2
9
2
2
= 1.2 × 10−8 C = 12 nC
(9.0 × 10 N m /C )
Assess: Since the field points away from it, we know that the charge must be positive.
25.25. Model: A field is the agent that exerts an electric force on a charge.
Visualize:
Solve: Newton’s second law on the plastic ball is Σ( Fnet ) y = Fon q − FG . To balance the gravitational force with the
electric force,
Fon q = FG
⇒
q E = mg
⇒
E=
mg (1.0 × 10−3 kg)(9.8 N/kg)
=
= 3.3 × 106 N/C
q
3.0 × 10−9 C
Because Fon q must be upward and the charge is negative, the electric field at the location of the plastic ball must be
pointing downward. Thus E = (3.3 × 106 N/C, downward).
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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 Charges and Forces
25-11
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.
25.26. Model: The electric field is that of a positive point charge located at the origin.
Visualize:
The positions (5.0 cm, 0.0 cm), (−5.0 cm, 5.0 cm), and (−5.0 cm, −5.0 cm) are denoted by A, B, and C, respectively.
Solve: (a) The electric field for a positive charge is
⎛ q
⎞
E = ⎜ K 2 , away from q ⎟
⎝ r
⎠
Using K = 9.0 × 109 N m 2 /C2 and q = 12 × 10−9 C,
⎛ 108 N m 2 /C
⎞
E =⎜
, away from q ⎟
2
⎜
⎟
r
⎝
⎠
The electric fields at points A, B, and C are
EA =
EB =
EC =
108 N m 2 /C ˆ
i = 4.3 × 104 iˆ N/C
2
−2
(5.0 × 10 m)
108 N m 2 /C
(−5.0 × 10
−2
⎡ 1
⎤
(−iˆ + ˆj ) ⎥ = (−1.5 × 104 iˆ + 1.5 × 104 ˆj ) N/C
⎢
m) 2 + (5.0 × 10−2 m) 2 ⎣ 2
⎦
108 N m 2 /C
(−5.0 × 10
−2
2
m) + (−5.0 × 10
−2
⎡ 1
⎤
(−iˆ − ˆj ) ⎥ = ( −1.5 × 104 iˆ − 1.5 × 104 ˆj ) N/C
⎢
m) ⎣ 2
⎦
2
(b) The three vectors are shown in the diagram.
Assess: The vectors EA , EB , and EC are pointing away from the positive charge.
25.27. Model: The electric field is that of a negative point charge located at (x, y) = (1.0 cm, 0 cm).
Visualize:
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25-12
Chapter 25
Solve: The electric field for a positive charge is
E=K
q
r2
rˆ
Using K = 9.0 × 109 N m 2 /C2 and q = 12 × 10−9 C,
E=
108 N m 2 /C
r2
The electric fields at points A, B, and C (see figure above) are
rˆ
EA =
108 N m 2 /C ˆ
i = −6.8 × 104 iˆ N/C
(4.0 × 10−2 m)2
EC =
108 N m 2 /C ˆ
i = 3.0 × 104 iˆ N/C
(6.0 × 10−2 m) 2
We need to take components to find EB:
EB =
108 N m 2 /C
(5.0 × 10−2 m) 2 + (1.0 × 10−2
⎡⎛
⎞ ⎛
⎞ ⎤
1.0 × 10−2 m
5.0 × 10−2 m
⎢⎜
⎟ iˆ − ⎜
⎟ ˆj ⎥
m) 2 ⎢⎜ (5.0 × 10−2 m) 2 + (1.0 × 10−2 m) 2 ⎟ ⎜ (5.0 × 10−2 m) 2 + (1.0 × 10−2 m) 2 ⎟ ⎥
⎠ ⎝
⎠ ⎦
⎣⎝
= (8.1 × 103 iˆ − 4.1 × 104 ˆj ) N/C
Assess: Since the field points toward the negative charge, it should have a positive x-component and a negative
y-component, as we have found.
25.28. Model: Use the charge model.
Solve: The number of moles in the penny is
n=
M
3.1 g
=
= 0.04882 mol
A 63.5 g/mol
The number of copper atoms in the penny is
N = nN A = (0.04882 mol)(6.02 × 1023 mol−1 ) = 2.939 × 1022
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.
25.29. Model: The beads are point charges.
Visualize:
Solve: The beads are oppositely charged so are attracted to one another. The force on each is the same by Newton’s
third law. Coulomb’s law give the force as
F=K
q1 q2
r2
= (9.0 × 10−9 Nm 2 /C2 )
(4.0 × 10−9 C)(8.0 × 10−9 C)
(2.0 × 10−2 m) 2
= 7.2 × 10−4 N
The beads accelerate at different rates because their masses are different. By Newton’s second law, the acceleration
of the plastic bead is
F (7.2 × 10−4 N)
aplastic = =
= 0.36 m/s 2 toward glass bead.
m (2.0 × 10−3 kg)
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Electric Charges and Forces
25-13
For the glass bead, the acceleration is
aglass =
25.30. Model: The
(7.2 × 10−4 N)
(4.0 × 10
−3
kg)
= 0.18 m/s 2 toward plastic bead.
125
Xe nucleus and the proton will be treated as point charges. That is, all the charge on the Xe
nucleus is assumed to be at its center.
Visualize:
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 /C2 )(54 × 1.60 × 10−19 C)(1.60 × 10−19 C)
Fnucleus on proton =
=
= 5.0 × 102 N
r2
(5.0 × 10−15 m) 2
(b) Applying Newton’s second law to the proton,
5.0 × 102 N
Fon proton = mproton aproton ⇒ aproton =
= 3.0 × 1029 m/s 2
1.67 × 10−27 kg
25.31. Model: Treat the two charged spheres as point charges.
Solve: The electric force on one charged sphere due to the other charged sphere is equal to the sphere’s mass times
its acceleration. Because the spheres are identical and equally charged, m1 = m2 = m and q1 = q2 = q. We have
F2 on 1 = F1 on 2 =
q2 =
Kq1q2
r
2
=
Kq 2
r2
= ma
mar 2 (1.0 × 10−3 kg)(150 m/s 2 )(2.0 × 10−2 m) 2
=
= 6.7 × 10−15 C2
K
9.0 × 109 N m 2 /C2
q = 8.2 × 10−8 C = 82 nC
25.32. Model: Objects A and B are point charges.
Visualize:
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.
Coulomb’s law gives
FB on A = FA on B = 0.45 N =
qA =
KqA qB
r2
=
K (qA )( 12 qA )
r2
2(0.45 N)r 2
2(0.45 N)(10 × 10−2 m)2
=
= 1.0 × 10−6 C
9
2 2
K
(9.0 × 10 N m /C )
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25-14
Chapter 25
(b) Newton’s second law is FB on A = mA aA . Hence,
aA =
FB on A FA on B
0.45 N
=
=
= 4.5 m/s 2
mA
mB
0.100 kg
25.33. Model: The charges are point charges.
Visualize: Please refer to Figure P25.33.
Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1, where q1 is the 1.0 nC
charge, q2 is the 2.0 nC charge, and q3 is the other 2.0 nC charge. We have
⎛ K q1 q2
⎞
, away from q2 ⎟
F2 on 1 = ⎜
2
r
⎝
⎠
⎛ (9.0 × 109 N m 2 /C2 )(1.0 × 10−9 C)(2.0 × 10−9 C)
⎞
=⎜
, away from q2 ⎟
2
2
−
⎜
⎟
(1.0 × 10 m)
⎝
⎠
−4
−4
ˆ
= (1.8 × 10 N, away from q2 ) = (1.8 × 10 N)[cos(60°)i + sin(60°) ˆj ]
⎛ K q1 q3
⎞
, away from q3 ⎟ = (1.8 × 10−4 N, away from q3 ) = (1.8 × 10−4 N)[− cos(60°)iˆ + sin(60°) ˆj ]
F3 on 1 = ⎜
r2
⎝
⎠
Fon 1 = F2 on 1 + F3 on 1 = 2(1.8 × 10−4 N)sin(60°) ˆj = (3.1 × 10−4 ˆj ) N
The force on the 1.0 nC charge is 3.1 × 10−4 N directed upward.
25.34. Model: The charges are point charges.
Visualize: Please refer to Figure P25.34.
Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1, where q1 is the 1.0 nC
charge, q2 is the 2.0 nC charge, and q3 is the −2.0 nC charges. We have
⎛ K q1 q2
⎞
, away from q2 ⎟
F2 on 1 = ⎜
2
⎝ r
⎠
9
2
2
⎛ (9.0 × 10 N m /C )(1.0 × 10−19 C)(2.0 × 10−19 C)
⎞
, away from q2 ⎟
=⎜
−2
2
⎜
⎟
(1.0 × 10 m)
⎝
⎠
= (1.80 × 10−4 N, away from q2 ) = (1.80 × 10−4 N)[cos(60°)iˆ + sin(60°) ˆj ]
⎛ K q1 q2
⎞
F3 on 1 = ⎜
, toward q3 ⎟ = (1.80 × 10−4 N, toward q3 ) = (1.80 × 10−4 N)[cos(60°)iˆ − sin(60°) ˆj ]
2
⎝ r
⎠
Fon 1 = F2 on 1 + F3 on 1 = 2(1.80 × 10−4 N)cos(60°)iˆ = 1.8 × 10−4 iˆ N
So, the force on the 1.0 nC charge is 1.8 × 10−4 N and it is directed to the right.
25.35. Model: The charges are point charges.
Visualize:
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Electric Charges and Forces
25-15
Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have
⎛ K q1 q2
⎞
, away from q2 ⎟
F2 on 1 = ⎜
2
r
⎝
⎠
⎛ (9.0 × 109 N m 2 /C2 )(10 × 10−9 C)(5.0 × 10−9 C)
⎞
, away from q2 ⎟
=⎜
−
2
2
⎜
⎟
(1.0 × 10 m)
⎝
⎠
−3
−3 ˆ
= (4.5 × 10 N, away from q2 ) = −4.5 × 10 j N
⎛ K q1 q2
⎞
, toward q3 ⎟
F3 on 1 = ⎜
2
⎝ r
⎠
9
2
2
⎛ (9.0 × 10 N m /C )(10 × 10−9 C)(15 × 10−9 C)
⎞
, toward q3 ⎟
=⎜
−2
−2
2
2
⎜
⎟
(3.0 × 10 m) + (1.0 × 10 m)
⎝
⎠
−3
−3
= (1.35 × 10 N, toward q3 ) = (1.35 × 10 N)(− cosθ iˆ + sin θ ˆj )
From the geometry of the figure,
⎛ 1.0 cm ⎞
⎟ = 18.4°
⎝ 3.0 cm ⎠
θ = tan −1 ⎜
This means cos θ = 0.949 and sin θ = 0.316. Therefore,
F3 on 1 = (−1.28 × 10−3 iˆ + 0.43 × 10−3 ˆj ) N
Fon 1 = F2 on 1 + F3 on 1 = ( −1.28 × 10−3 iˆ − 4.07 × 10−3 ˆj ) N
The magnitude and direction of the resultant force vector are
Fon 1 = (−1.28 × 10−3 N )2 + (−4.07 × 10−3 N )2 = 4.3 × 10−3 N
tanφ =
4.07 × 10−3 N
1.28 × 10−3 N
= 3.180 ⇒ φ = tan −1 (3.180) = 73° below the −x axis,
or Fon 1 points 253° counterclockwise from the +x-axis.
25.36. Model: The charges are point charges.
Visualize:
Solve: The point charges are q1 = −10 nC, q2 = +8.0 nC, and q3 = 10 nC. The electric force on charge q1 is the
vector sum of the forces F2 on 1 and F3 on 1. We have
⎛ K q1 q2
⎞
F2 on 1 = ⎜
, toward q2 ⎟
2
⎝ r
⎠
9
2
2
⎛ (9.0 × 10 N m /C )(10 × 10−9 C)(8.0 × 10−9 C)
⎞
, toward q2 ⎟
=⎜
2
−2
⎜
⎟
(1.0 × 10 m)
⎝
⎠
= (7.2 × 10−3 N, toward q2 ) = (7.2 × 10−3 N) ˆj
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25-16
Chapter 25
⎛ K q1 q3
⎞
F3 on 1 = ⎜
, toward q3 ⎟
2
r
⎝
⎠
⎛ (9.0 × 109 N m 2 /C 2 )(10 × 10−9 C)(10 × 10−9 C)
⎞
, toward q3 ⎟
=⎜
2
−2
⎜
⎟
(3.0 × 10 m)
⎝
⎠
−3
= (1.0 × 10 N, toward q3 ) = −(1.0 × 10−3 N ) iˆ
Fon 1 = F2 on 1 + F3 on 1 = ( −1.0 × 10−3 iˆ + 7.2 × 10−3 ˆj ) N
The magnitude and direction of the resultant force vector are
Fon 1 = (−1.0 × 10−3 N) 2 + (7.2 × 10−3 N) 2 = 7.3 × 10−3 N
tan φ =
7.2 × 10−3 N
1.0 × 10
−3
N
⇒ φ = tan −1(7.2) = 82°above the −x-axis,
or Fon 1 points 98° counterclockwise from the +x-axis.
25.37. Model: The charges are point charges.
Visualize:
Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have
⎛ K q1 q2
⎞
F2 on 1 = ⎜
, away from q2 ⎟
2
r
⎝
⎠
⎛ (9.0 × 109 N m 2 /C2 )(5 × 10−9 C)(10 × 10−9 C)
⎞
, away from q2 ⎟
=⎜
−
−
2
2
2
2
⎜
⎟
(4.0 × 10 m) + (3.0 × 10 m)
⎝
⎠
−4
−4
ˆ
ˆ
= (1.8 × 10 N, away from q2 ) = (1.8 × 10 N)(− cosθ i − sin θ j )
From the geometry of the figure,
4.0 cm
tan θ =
3.0 cm
⇒ θ = 53.13° ⇒
F2 on 1 = (1.8 × 10−4 N)(−0.60iˆ − 0.80 ˆj )
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Electric Charges and Forces
25-17
⎛ (9.0 × 109 N m 2 /C 2 )(5 × 1029 C)(5 × 1029 C)
⎞
F3 on 1 = ⎜
, toward q3 ⎟ = (2.5 × 10−4 N, toward q3 ) = 2.5 × 10−4 iˆ N
2
−2
⎜
⎟
(3.0 × 10 m)
⎝
⎠
Fon 1 = F2 on 1 + F3 on 1 = (1.42 × 10−4 iˆ − 1.44 × 10−4 ˆj ) N
The magnitude and direction of the resultant force vector are
Fon 1 = (1.42 × 10−4 N) 2 + ( −1.44 × 10−4 N) 2 = 2.0 × 10−4 N
⎛ 1.44 × 10−4 N ⎞
= 45° clockwise from the +x-axis.
⎜ 1.42 × 10−4 N ⎟⎟
⎝
⎠
φ = tan −1 ⎜
25.38. Model: The charges are point charges.
Visualize:
Solve: The charges are q1 = +5.0 nC, q2 = +10 nC, and q3 = −10 nC. The electric force on q1 is the vector sum of
the forces F2 on 1 and F3 on 1. We have
⎛ K q1 q2
⎞
F2 on 1 = ⎜
, away from q2 ⎟
2
⎝ r
⎠
9
2
2
⎛ (9.0 × 10 N m /C )(5 × 10−9 C)(10 × 10−9 C)
⎞
, away from q2 ⎟
=⎜
2
−2
⎜
⎟
(4.0 × 10 m)
⎝
⎠
−4
−4
ˆ
(2.81
10
N,
away
from
q
)
(2.81
10
N)
j
=
×
×
2 =−
⎛ K q1 q3
⎞
, toward q3 ⎟
F3 on 1 = ⎜
2
⎝ r
⎠
9
2
2
⎛ (9.0 × 10 N m /C )(5.0 × 10−9 C)(10 × 10−9 C)
⎞
=⎜
, toward q3 ⎟
2
2
−2
−2
⎜
⎟
(4.0 × 10 m) + (3.0 × 10 m)
⎝
⎠
= (1.80 × 10−4 N)(cosθ iˆ + sin θ ˆj )
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25-18
Chapter 25
From the geometry of the figure,
⎛ 4⎞
⇒ θ = tan −1 ⎜ ⎟ = 53.13°
⎝3⎠
ˆ
N)[cos(53.13°) i + sin (53.13°) ˆj ] = (1.80 iˆ + 1.44 ˆj ) × 10−4 N
tan θ =
F3 on 1 = (1.80 × 10−4
4.0 cm
3.0 cm
Therefore,
Fon 1 = F2 on 1 + F3 on 1 = (1.80 × 10−4 iˆ − 1.37 × 10−4 ˆj ) N
The magnitude and direction of the resultant force vector are
Fon 1 = (1.80 × 10−4 N)2 + (−1.37 × 10−4 N) 2 = 1.7 × 10−4 N
tan φ =
1.37 × 10−4 N
1.80 × 1024 N
⇒ φ = 52° clockwise from the +x-axis.
25.39. Model: The charges are point charges.
Visualize: Please refer to Figure P25.39.
Solve: Placing the 1.0 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 )(2.0 × 10−9 C)(1.0 × 10−9 C)
(0.50 × 10−2 m) 2 + (0.50 × 10−2 m) 2
= 3.6 × 10−4 N
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 ). In component form,
{
Fon 1 = Fon 1 [− cos(45°) iˆ − sin (45°) ˆj ] + −[cos(45°) iˆ + sin (45°) ˆj ]
}
+[− cos(45°) iˆ − sin (45°) ˆj ] + [ − cos(45°) iˆ + sin (45°) ˆj ]
= (3.6 × 10−4 N)[−4cos(45°) iˆ] = −1.0 × 10−3 iˆ N
Assess: By symmetry, we see that the vertical forces must cancel, and that the horizontal force must be in the
negative direction, which agrees with the calculation.
25.40. Model: The charges are point charges.
Visualize: Please refer to Figure P25.40.
Solve: Placing the 1.0 nC charge at the origin and calling it q1, the −6.0 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 three charges q2 , q3 , and q4 . We have
⎛ K q1 q2
⎞
F2 on 1 = ⎜
, away from q2 ⎟
2
⎝ r
⎠
9
2
2
⎛ (9.0 × 10 N m /C )(1.0 × 10−9 C)(2.0 × 10−9 C)
⎞
, away from q2 ⎟
=⎜
−2
2
⎜
⎟
(5.0 × 10 m)
⎝
⎠
= (0.72 × 10−5 N, away from q2 ) = (0.720 × 10−5 N)[− cos(45°) iˆ − sin (45°) ˆj ]
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Electric Charges and Forces
25-19
⎛ K q1 q3
⎞
F3 on 1 = ⎜
, toward q3 ⎟
2
⎝ r
⎠
9
2
2
⎛ (9.0 × 10 N m /C )(1.0 × 10−9 C)(6.0 × 10−9 C)
⎞
=⎜
, toward q3 ⎟
2
−2
⎜
⎟
(5.0 × 10 m)
⎝
⎠
= (2.16 × 10−5 N, away from q3 ) = 2.16 × 10−5 ˆj N
⎛ K q1 q4
⎞
F4 on 1 = ⎜
, away from q4 ⎟ = (0.720 × 10−5 N)[cos(45°) iˆ − sin (45°) ˆj ]
2
⎝ r
⎠
Summing these forces vectorially gives
Fon 1 = F2 on 1 + F3 on 1 + F4 on 1 = [(2.16 × 10−5 N) − 2(0.720 × 10−5 N)sin (45°)] ˆj = 1.1 × 10−5 ˆj N
Assess: By symmetry, we see that the horizontal forces must cancel, and that the vertical force must be upward,
which agrees with the calculation.
25.41. Model: The charges are point charges.
Visualize: Please refer to Figure P25.41.
Solve: Place the 1.0 nC charge at the origin and call it q1; the −6.0 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 . We have
⎛ K q1 q2
⎞
F2 on 1 = ⎜
, toward q2 ⎟
2
⎝ r
⎠
9
2
2
⎛ (9.0 × 10 N m /C )(1.0 × 10−9 C)(2.0 × 10−9 C)
⎞
, toward q2 ⎟
=⎜
2
−2
⎜
⎟
(5.0 × 10 m)
⎝
⎠
= (0.72 × 10−5 N, toward q2 ) = (0.72 × 10−5 N)[cos(45°) iˆ + sin (45°) ˆj ]
⎛K
F3 on 1 = ⎜
⎝
⎛K
F4 on 1 = ⎜
⎝
⎞
, toward q3 ⎟ = (2.16 × 10−5 N, toward q3 ) = 2.16 × 10−5 ˆj N
r
⎠
⎞
q1 q4
, away from q4 ⎟ = (0.72 × 10−5 N)[cos(45°) iˆ − sin (45°) ˆj ]
2
r
⎠
q1 q3
2
Adding these components together vector-wise gives
Fon 1 = F2 on 1 + F3 on 1 + F4 on 1 = 2(0.72 × 10−5 N)cos(45°) iˆ + (2.16 × 10−5 N) ˆj
= (1.02 × 10−5 iˆ + 2.2 × 10−5 ˆj ) N
25.42. Model: The charged particles are point charges.
Visualize:
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25-20
Chapter 25
Solve: (a) The mathematical problem is to find the position for which the forces F1 on p and F2 on p are equal in
magnitude and opposite in direction. If the proton is at position x, it is a distance x from q1 and d − x from q2 ,
where d = 1.0 cm. The magnitudes of the forces are
F1 on p =
K q1 qp
r1p2
=
K q1 qp
x
F2 on p =
2
K q2 qp
( d − x) 2
Equating the two forces and using d = 1.0 cm,
K q1 qp
x
2
=
K q2 qp
(d − x)
2
⇒
2.0 nC
x
2
=
4.0 nC
( d − x) 2
⇒ 1 + x2 − 2x = 2x2
⇒
x2 + 2 x − 1 = 0
The solutions to the equation are x = +0.414 cm and −2.41 cm. Both are points where the magnitudes of the two
forces are equal, but +0.414 cm is a point where the magnitudes are equal and the directions are the same. The
solution we want is that the proton should be placed at x = −2.4 cm.
(b) Yes, the net force on the electron located at x = −2.4 cm will also be zero. This is because the solution in part (a)
does not depend specifically on the type of the charge that experiences zero force from the other two charges.
Furthermore, if Fon p is zero for a proton, the electric field at that point must be zero. Thus, there will be no force on
any charged particle at that point.
25.43. Model: The charged particles are point charges.
Visualize:
Solve: The two 2.0 nC charges exert an upward force on the 1.0 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.
The force of charge 2 on charge 1 is
⎛ K q1 q2
⎞
F2 on 1 = ⎜
, away from q2 ⎟
⎜ r2
⎟
12
⎝
⎠
=
(9.0 × 109 N m 2 /C2 )(1.0 × 10−9 C)(2.0 × 10−9 C)
(0.020 m) 2 + (0.030 m)2
(cosθ iˆ + sin θ ˆj )
From the figure, θ = tan −1 (2/3) = 33.69°. Thus
F2 on 1 = (1.152 × 10−5 iˆ + 0.768 × 10−5 ˆj ) N
From symmetry, F3 on 1 is the same except the x-component is reversed. 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 ˆj N
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Electric Charges and Forces
25-21
Fq on 1 must have the same magnitude, pointing in the − ĵ direction, so
Fq on 1 = 1.536 × 10−5 N =
K q q1
r2
⇒ q=
(1.536 × 10−5 N)(0.020 m) 2
(9.0 × 109 N m 2 /C2 )(1.0 × 10−9 C)
= 0.68 nC
A positive charge q = 0.68 nC will cause the net force on the 1.0 nC charge to be zero.
25.44. Model: The charged particles are point charges.
Visualize: Please refer to Figure P25.44.
Solve: The charge q2 is in static equilibrium, so the net electric field at the location of q2 is zero. We have
Enet = Eq1 + E5 nC = K
−9
(5.0 × 10 C) ˆ
( ±iˆ) + K
(i ) = 0 N/C
2
(0.20 m)
(0.10 m) 2
q1
We have used the ± sign to indicate that a positive charge on q1 leads to an electric field along +iˆ and a negative
charge on q1 leads to an electric field along −iˆ. Because the above equation can only be satisfied if we use −iˆ, we
infer that the charge q1 is a negative charge. Thus,
−9
5.0 × 10 C ˆ
( −iˆ) +
(i ) = 0 N/C ⇒
2
(0.20 m)
(0.10 m) 2
q1
q1 = 20 nC ⇒ q1 = −20 nC
25.45. Model: The charged particles are point charges.
Visualize:
Solve: The force on q is the vector sum of the force from −Q and +Q. We have
⎛ −Q + q
⎞
KQq
F2Q on + q = ⎜⎜ K 2
, toward −Q ⎟⎟ = 2
( − cosθ iˆ − sin θ ˆj )
2
2
a
y
a
y
+
+
⎝
⎠
⎛ K +Q + q
⎞
KQq
, away from + Q ⎟⎟ = 2
(− cosθ iˆ + sin θ ˆj )
F+ Q on + q = ⎜⎜ 2
2
2
+
+
a
y
a
y
⎝
⎠
KQq
Fnet = 2
(−2cosθ ) iˆ + 0 ˆj N
a + y2
From the figure we see that cosθ = a
a 2 + y 2 . Thus
( Fnet ) x =
−2 KQqa
(a 2 + y 2 )3/2
Assess: Note that (Fnet ) x = Fnet because the y-components of the two forces cancel each other out.
25.46. Model: The charged particles are point charges.
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25-22
Chapter 25
Visualize:
Solve: (a) The force on q is the vector sum of the force from −Q and Q
 . We have
⎛ K +Q + q
⎞
KQq
, away from + Q ⎟⎟ =
( −iˆ)
F+ Q on + q = ⎜⎜
2
2
⎝ (a − x)
⎠ ( a − x)
⎛ K −Q + q
⎞
KQq
F−Q on + q = ⎜⎜
, toward −Q ⎟⎟ =
(−iˆ)
2
2
⎝ (a + x)
⎠ (a + x)
⎡ 1
1 ⎤
2KQq (a 2 + x 2 )
( Fnet ) x = − KQq ⎢
+
=
−
⎥
2
(a + x)2 ⎦
(a 2 − x 2 )2
⎣ (a − x)
To arrive at the final expression we used (a − x) 2 (a + x) 2 = [(a − x)(a + x)]2 = (a 2 − x 2 ) 2 .
(b) There are two cases when x > a. For x > a,
⎛ K +Q + q
⎞
KQq
F+ Q on + q = ⎜⎜
, away from + Q ⎟⎟ =
( +iˆ)
2
2
⎝ ( x − a)
⎠ ( x − a)
⎛ K −Q + q
⎞
KQq
, toward −Q ⎟⎟ =
( −iˆ)
F−Q on + q = ⎜⎜
2
2
⎝ ( x + a)
⎠ ( x + a)
⎡ 1
1 ⎤ −4 KQqax
( Fnet ) x = KQq ⎢
−
⎥=
2
( x + a)2 ⎦ ( x 2 − a 2 )2
⎣ ( x − a)
For x < − a (that is, for negative values of x),
KQq
(−iˆ)
F+ Q on + q =
( x − a)2
F−Q on + q =
( Fnet ) x = −
KQq
(a + x)2
( +iˆ)
4 KQqax
( x2 − a 2 )2
That is, the net force is to the right when x > a and to the right when x < − a. We can combine these two cases into a
single equation for x > a:
( Fnet ) x =
4 KQqa x
( x 2 − a 2 )2
Here, the force is always to the right when x > a.
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Electric Charges and Forces
25-23
25.47. Model: The charges are point charges.
Solve:
We will denote the charges −Q, 4Q and −Q by 1, 2, and 3, respectively.
⎛ K −Q q
⎞ KQq
F1 on q = ⎜
, toward −Q ⎟ = 2 ( −iˆ)
2
L
L
⎝
⎠
⎛ K 4Q q
⎞ 4 KQq
2 KQq
= 2 F1 on q
, away from 4Q ⎟⎟ =
[+ cos (45°) iˆ + sin (45°) ˆj ] ⇒ F2 on q =
F2 on q = ⎜⎜
2
2
L2
2L
⎝ ( 2 L)
⎠
⎛ K −Q q
⎞ KQq
F3 on q = ⎜
, toward −Q ⎟ = 2 (− ˆj )
2
L
L
⎝
⎠
The net electric force on the charge +q is the vector sum of the electric forces from the other three charges. The net
force is
ˆj ⎞ KQq
KQq
KQq
KQq
2 KQq ⎛ iˆ
+
+
Fnet = 2 ( −iˆ) +
⎟ + 2 (− ˆj ) = − 2 iˆ(1 − 2) − 2 ˆj (1 − 2)
2 ⎜
2
2⎠
L
L ⎝
L
L
L
2
2
KQq
⎡ KQq
⎤ ⎡ KQq
⎤
Fnet = ⎢ 2 (1 − 2) ⎥ + ⎢ 2 (1 − 2) ⎥ = (2 − 2) 2
L
⎣ L
⎦ ⎣ L
⎦
25.48. Model: The charges are point charges.
Solve: Label the charges q, −q, and Q with numbers 1, 2, and 3, respectively. The positive charge q exerts an
attractive force on the charge −q at the orgin that is given by
q q
q2
F1 on 2 = K 1 2 2 ˆj = K 2 ˆj
L
L
The charge Q exerts an attractive force on −q at the origin that is given by
Q q2 ˆ
α q2 ˆ
=
F3 on 2 = K
i
K
i
(2 L) 2
4 L2
Because the net force is at 45° to the x-axis (or y-axis), the the iˆ and ĵ components must be of equal magnitude.
Setting them equal an solving for α gives
K
q2
L2
=K
α q2
4 L2
⇒ α =4
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25-24
Chapter 25
25.49. Model: The charges are point charges.
Visualize:
We must first identify the region of space where the third charge q3 is located. You can see from the figure that the
forces can’t possibly add to zero if q3 is above or below the axis or outside the charges. However, at some point on
the x-axis between the two charges the forces from the two charges will be oppositely directed.
Solve: The mathematical problem is to find the position for which the forces F1 on 3 and F2 on 3 are equal in
magnitude. If q3 is the distance x from q1, it is the distance L − x from q2 . The magnitudes of the forces are
F1 on 3 =
Equating the two forces gives
Kq q3
x
2
K q1 q3
r132
=
Kq q3
=
x
K (4q ) q3
( L − x)
F2 on 3 =
2
⇒ ( L − x)2 = 4 x 2
2
K q2 q3
=
2
r23
⇒
x=
K (4q) q3
( L − x)2
L
and − L
3
The solution x = −L is not allowed as you can see from the figure. To find the magnitude of the charge q3 , we apply
the equilibrium condition to charge q1:
K q2 q1
F2 on 1 = F3 on 1 ⇒
2
L
=
K q3 q1
( )
1L
3
2
⇒ 4q = 9 q3
4
q3 = q
9
⇒
We are now able to check the static equilibrium condition for the charge 4q (or q2 ):
F1 on 2 = F3 on 2
⇒
K
q1 q2
2
L
=
K q3 q2
( L − x)
2
⇒
q
2
L
=
4q
9
( 23 L )
2
=
q
L2
The sign of the third charge q3 must be negative. A positive sign on q3 will not have a net force of zero either on
the charge q or the charge 4q. In summary, a charge of − 94 q placed x = 13 L from the charge q will cause the
3-charge system to be in static equilibrium.
25.50. Model: Use the charge model and assume the copper spheres are point objects with point charges.
Solve: (a) The mass of the copper sphere is
⎛ 4π 3 ⎞
⎡ 4π
⎤
M = ρV = ⎜
r ⎟ = (8920 kg/m3 ) ⎢ (1.0 × 10−3 m)3 ⎥ = 3.736 × 10−5 kg = 0.03736 g
⎝ 3 ⎠
⎣ 3
⎦
The number of moles in the sphere is
M 0.03742 g
n=
=
= 5.884 × 10−4 mol
A 63.5 g/mol
The number of copper atoms in the sphere is
N = nN A = (5.884 × 10−4 mol)(6.02 × 1023 mol−1 ) = 5.542 × 1020
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Electric Charges and Forces
25-25
The number of electrons in the copper sphere is thus 29 × 3.542 × 1020 = 1.027 × 1022. The total positive or negative
charge in the sphere is (1.027 × 1022 )(1.60 × 10−19 C) = 1643 C. Hence, the spheres will have a net charge of
10−9 × 1643 C = 1.643 × 10−6 C. The force between two such spheres is
Kq1q2
(9.0 × 109 N m 2 /C2 )(1.643 × 10−6 C) 2
= 2.4 × 102 N
(1.0 × 10−2 m) 2
(1.0 × 10−2 m) 2
(b) This is a force that is easily detectable. Since we don’t observe such forces, any difference between the proton
charge and the electron charge must be smaller than 1 part in 109 .
F=
=
25.51. Model: The electron and the proton are point charges.
Solve: The electric Coulomb force between the electron and the proton provides the centripetal acceleration for the
electron’s circular motion. Thus,
K (e)(e)
r
f =
Ke
2
mr
3
=
2
=
mv 2
= mrω 2
r
(9.0 × 10 N m /C )(1.60 × 10−19 C) 2
9
(9.11 × 10
2
−31
2
kg)(5.3 × 10
−11 3
)
⎛ 1 rev ⎞
15
= (4.12 × 1016 rad/s) ⎜
⎟ = 6.6 × 10 rev/s
⎝ 2π rad ⎠
25.52. Model: Model the charged balls as point charges and ignore friction.
Solve: Combining Coulomb’s law and Newton’s second law for the balls at an arbitrary distance d apart gives their
acceleration:
q2
q2
F = K 2 = ma ⇒ a = K
d
md 2
Thus, if we plot acceleration as a function of inverse distance squared, the magnitude of the charge can be calculated
from the slope s:
Kq 2
sm
s=
⇒ q=
m
K
The fit gives a slope of s = 2.97 × 10−4 N m 2 /kg, so the magnitude of the charge is
q=
(2.97 × 10−4 N m 2 /kg)(2.0 × 10−3 kg)
(9.0 × 109 N m 2 /C2 )
= 8.1 nC
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25-26
Chapter 25
25.53. Model: Model the bee as a point charge.
Solve: (a) The force on the bee due to gravity is FG = mg , and the electric force on the bee is Fe = Eq. The ratio of
the electric force to the bee’s weight is
Fe Eq
(100 N/C)(23 × 10−12 C)
=
=
= 2.3 × 10−6
FG mg (0.10 × 10−3 kg)(9.8 m/s 2 )
(b) For the bee to be suspended by the electric force, this force must have the same magnitude as the force due to
gravity and be directed upward. Equating the magnitudes of the forces and solving for E give
Eq = mg
⇒
E=
mg (0.10 × 10−3 kg)(9.8 m/s 2 )
=
= 4.3 × 107 N/C
q
(23 × 10−12 C)
The electric field must be directed upward so that the force on a positive charge is upward.
25.54. Model: Model the metal plate and yourself as point charges.
Solve: At the beginning, both you and the metal plates are neutral. As the electrons are pumped from the metal plate
into you, the plate gains as much positive charge as you gain negative charge. When enough charge difference builds
up between you and the metal plate, the gravitational force on you will be counterbalanced by the upward electrical
force. You will begin to hang suspended in the air when
K q −q
Fplate on you = FG ⇒
= mg
(2.0 m) 2
q =
mg (2.0 m) 2
(60 kg)(9.8 N/kg)(2.0 m) 2
=
= 5.11 × 10−4 C
K
9.0 × 109 N m 2 /C2
Dividing this charge by the charge on an electron yields 3.2 × 1015 electrons.
25.55. Model: The charged plastic beads are point charges and the spring is an ideal spring that obeys Hooke’s law.
Solve: Let q be the charge on each plastic bead. The repulsive force between the beads pushes the beads apart. The
spring is stretched until the restoring spring force on either bead is equal to the repulsive Coulomb force:
Kq 2
r2
= k Δx ⇒ q =
k Δx r 2
K
The spring constant k is obtained by noting that the weight of a 1.0 g mass stretches the spring 1.0 cm. Thus
mg = k (1.0 × 10−2 m) ⇒ k =
q=
(1.0 × 10−3 kg)(9.8 N/kg)
1.0 × 10−2 m
= 0.98 N/m
(0.98 N/m)(4.5 × 10−2 m − 4.0 × 10−2 m)(4.5 × 10−2 m)2
9.0 × 109 N m 2 /C2
= 33 nC
25.56. Model: Take the center of mass of the dipole to be midway between the two charges.
Solve: The torque about the center of mass due to each charge is τ = Fe ( 2s ) and each is in the same direction, so the
total torque is τ = Fe s. The electric force Fe = Eq, so the torque may be written as τ = Eqs = pE , where we have
used p ≡ qs in the last step.
25.57. Solve: (a) Kinetic energy is K = 12 mv 2 , so the velocity squared is v 2 = 2 K/m. From kinematics, a particle
moving through distance Δx with acceleration a, starting from rest, finishes with v 2 = 2aΔx. To gain K = 2 × 10−18 J
of kinetic energy in Δx = 2.0 μm requires an acceleration of
v2
2 K/ m
K
2.0 × 10−18 J
a=
=
=
=
= 1.10 × 1018 m/s 2 ≈ 1.1 × 1018 m/s 2
2Δx 2Δx mΔx (9.11 × 10−31 kg)(2.0 × 10−6 m)
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Electric Charges and Forces
25-27
(b) The force that produces this acceleration is
F = ma = (9.11 × 10−31 kg)(1.10 × 1018 m/s 2 ) = 1.00 × 10−12 N ≈ 1.0 × 10−12 N
(c) The electric field required is
E=
F 1.00 × 10−12 N
=
= 6.3 × 106 N/C
e 1.602 × 10−19 C
(d) The force on an electron due to a charge q is F = K q e/r 2 . To have breakdown, the force on the electron must be
at least 1.0 × 10−12 N. The minimum charge that could cause a breakdown will be the charge that causes exactly a
force of 1.0 × 10−12 N:
F =
K qe
r
2
= 1.0 × 10−12 N ⇒
(0.010 m) 2 (1.0 × 10−12 N)
r 2F
=
= 6.9 × 10−8 C = 69 nC
Ke (9.0 × 109 N m 2 /C2 )(1.6 × 10−19 C)
q =
25.58. Model: The charged spheres behave as point charges.
Visualize:
Each sphere is in static equilibrium and the string makes an angle θ with the vertical. The three forces acting on each
sphere are the electric force, the gravitational force on the sphere, and the tension force.
Solve: In static equilibrium, Newton’s first law gives Fnet = T + FG + Fe = 0. In component form,
( Fnet ) x = Tx + ( FG ) x + ( Fe ) x = 0 N
−T sinθ + 0 N +
T sinθ =
Kq
2
d2
Kq 2
d
2
=
=0N
( Fnet ) y = Ty + ( FG ) + ( Fe ) y = 0 N
T cosθ − mg + 0 N = 0 N
Kq 2
(2 L sinθ ) 2
Tcosθ = + mg
Dividing the two equations gives
sin 2 θ tanθ =
Kq 2
4 L2mg
=
(9.0 × 109 N m 2 /C2 )(100 × 10−9 C) 2
4(1.0 m)2 (5.0 × 10−3 kg)(9.8 N/kg)
= 4.59 × 10−4
For small-angles, tanθ ≈ sinθ . With this approximation we obtain sinθ = 0.07714 rad, , so θ = 4.4°.
25.59. Model: The charged spheres behave as point charges.
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25-28
Chapter 25
Visualize:
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 gravitational force on the sphere, and the tension force.
Solve: In the static equilibrium, Newton’s first law gives Fnet = T + ( FG ) + Fe = 0. In component form, we have
( Fnet ) x = Tx + ( FG ) x + ( Fe ) x = 0 N
−T sinθ + 0 N +
T sinθ =
Kq 2
d2
Kq 2
d
2
=
( Fnet ) y = Ty + ( FG ) + ( Fe ) y = 0 N
=0N
+ Kq 2
(2 L sinθ ) 2
T cosθ − mg + 0 N = 0 N
T cosθ = + mg
Dividing the two equations and solving for q gives
q=
4sin 2 θ tanθ L2mg
4(sin 2 20° tan 20°)(1.0 m) 2 (3.0 × 10−3 kg)(9.8 N/kg)
=
= 0.75 μ C
K
9.0 × 109 N m 2 /C2
25.60. Model: The electric field is that of a positive point charge located at the origin.
Visualize: Please refer to Figure P25.60. Place the +10 nC charge at the origin.
Solve: The electric field is
9
2 2
−9
⎞ ⎛ 90 N m 2 /C
⎞
⎛ q
⎞ ⎛ (9.0 × 10 N m /C )(10 × 10 C)
,
away
from
, away from q ⎟
E = ⎜ K 2 , away from q ⎟ = ⎜
q
⎟⎟ = ⎜⎜
2
2
⎜
⎟
r
r
⎝ r
⎠ ⎝
⎠ ⎝
⎠
The electric field at each of the three points is
⎛ 90.0 N m 2 /C
⎞
E1 = ⎜
, away from q ⎟ = 1.0 × 105 ˆj N/C
⎜ (3.0 × 10−2 m) 2
⎟
⎝
⎠
⎛ 90.0 N m 2 /C
⎞
, away from q ⎟ = (3.6 × 104 N/C)(cosθ iˆ + sinθ ˆj )
E2 = ⎜
⎜ (5.0 × 10−2 m) 2
⎟
⎝
⎠
4
3 ˆ
4
ˆ
= (3.6 × 10 N/C) i + j = (2.9 × 104 iˆ + 2.2 × 104 ˆj ) N/C
(5
5
)
⎛ 90.0 N m 2 /C
⎞
E3 = ⎜
, away from q ⎟ = 5.6 × 104 iˆ N/C
⎜ (4.0 × 10−2 m) 2
⎟
⎝
⎠
25.61. Model: The electric field is that of a negative point charge.
Visualize: Please refer to Figure P25.61. Place point 1 at the origin.
Solve: The electric field is
⎞
⎛ q
⎞ ⎛ (9.0 × 109 N m 2 /C2 )(2.0 × 10−9 C)
, toward q ⎟
E = ⎜ K 2 , toward q ⎟ = ⎜
2
⎜
⎟
r
⎝ r
⎠ ⎝
⎠
⎛ 18.0 N m 2 /C
⎞
=⎜
, toward q ⎟
2
⎜
⎟
r
⎝
⎠
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Electric Charges and Forces
25-29
The electric fields at the two points are
⎛ 18.0 N m 2 /C
⎞
,
toward
E1 = ⎜
q
⎟⎟
⎜ (1.0 × 10−2 m) 2
⎝
⎠
= (1.8 × 105 N/C, 60° counter clockwise from the +x -axis or 60° north of east)
⎛ 18.0 N m 2 /C
⎞
, toward q ⎟
E2 = ⎜
⎜ (1.0 × 10−2 m) 2
⎟
⎝
⎠
= (1.8 × 105 N/C, 60° clockwise from the −x -axis or 60° north of west)
25.62. Model: The electric field is that of a positive point charge located at the origin.
Visualize: Please refer to Figure P25.62. Place the 5.0 nC charge at the origin.
Solve: The electric field is
9
2 2
−9
⎞
⎛ q
⎞ ⎛ (9 × 10 N m /C )(5.0 × 10 C)
E = ⎜ K 2 , away from q ⎟ = ⎜
, away from q ⎟
2
⎜
⎟
r
⎝ r
⎠ ⎝
⎠
⎛ 45 N m 2 /C
⎞
=⎜
, away from q ⎟
2
⎜
⎟
r
⎝
⎠
The electric field at the three points is
⎛
⎞
45 N m 2 /C
E1 = ⎜
, away from q ⎟ = (9.0 × 104 N/C)(cosθ iˆ + sinθ ˆj )
⎜ (2.0 × 10−2 m) 2 + (1.0 × 10−2 m) 2
⎟
⎝
⎠
⎛ 1 ˆ 2 ˆ⎞
= (9.0 × 104 N/C) ⎜
i+
j ⎟ = (4.0 × 104 iˆ + 8.0 × 104 ˆj ) N/C
5 ⎠
⎝ 5
⎛ 45 N m 2 /C
⎞
E2 = ⎜
, away from q ⎟ = 4.5 × 105 iˆ N/C
⎜ (1.0 × 10−2 m)2
⎟
⎝
⎠
2
⎛
⎞
45 N m /C
E3 = ⎜
, away from q ⎟ = (4.0 × 104 iˆ − 8.0 × 104 ˆj ) N/C
−
−
2
2
2
2
⎜ (2.0 × 10 m) + (1.0 × 10 m)
⎟
⎝
⎠
25.63. Model: The electric field is that of a negative charge at (x, y) = (2.0 cm, 1.0 cm).
Visualize:
Solve: (a) The electric field of a negative charge points toward the charge, so we can roughly locate where the
field has a particular value by inspecting the signs of Ex and E y . At point a, the electric field has no y-component
and the x-component points to the left, so its location must be to the right of the charge along a horizontal line.
Using the equation for the field of a point charge,
Ex = E =
Kq
ra2
⇒ ra =
Kq
Ex
=
(9.0 × 109 N m 2 /C2 )(10.0 × 10−9 C)
= 0.020 m = 2.0 cm
225,000 N/C
Thus, point a is at the position ( xa , ya ) = (4.0 cm, 1.0 cm).
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25-30
Chapter 25
(b) Point b is above and to the left of the charge. The magnitude of the field at this point is
E = E x2 + E y2 = (161,000 N/C) 2 + (80,500 N/C) 2 = 180,000 N/C
Using the equation for the field of a point charge,
E=
Kq
Kq
(9.0 × 109 N m 2 /C2 )(10 × 10−9 C)
=
= 2.236 cm
E
180,000 N/C
⇒ rb =
rb2
This gives the total distance but not the horizontal and vertical components. However, we can determine the angle θ
because Eb points straight toward the negative charge. Thus,
⎛ Ey ⎞
80,500 ⎞
−1 ⎛ 1 ⎞
⎟ = tan −1 ⎛⎜
⎟ = tan ⎜ ⎟ = 26.57°
⎜ Ex ⎟
161,000 ⎠
⎝2⎠
⎝
⎝
⎠
The horizontal and vertical distances are then d x = rb cosθ = 2.00 cm and d y = rb sinθ = 1.00 cm. Thus, point b is at
θ = tan −1 ⎜
the position ( xb , yb ) = (0.0 cm, 2.0 cm).
(c) Point c, which is below and to the left of the charge, is calculated by following a similar procedure. We first find
that E = 36,000 N/C. From this we find that the total distance rc = 5.00 cm. The angle φ is
⎛ Ey ⎞
21,600 ⎞
⎟ = tan −1 ⎛⎜
⎟ = 36.87°
⎜ Ex ⎟
⎝ 28,000 ⎠
⎝
⎠
which gives the distances d x = rc cosφ = 4.00 cm and d y = rc sinφ = 3.00 cm. Thus point c is at position (xc , yc ) =
φ = tan −1 ⎜
( − 2.0 cm, − 2.0 cm).
25.64. Model: The electric field is that of a positive charge at (x, y) = (1.0 cm, 2.0 cm).
Visualize:
Solve: (a) The electric field of a positive charge points straight away from the charge, so we can roughly locate the
points of interest based simply on whether the signs of Ex and E y are positive or negative. For point a, the electric
field has no y-component and the x-component points to the left, so point a must be to the left of the charge along a
horizontal line. Using the field of a point charge,
Ex = E =
Kq
ra2
⇒ ra =
Kq
Ex
=
(9.0 × 109 N m 2 /C2 )(10.0 × 10−9 C)
= 0.020 m = 2.0 cm
225,000 N/C
Thus, (xa , ya ) = ( − 1.0 cm, 2.0 cm).
(b) Point b is above and to the right of the charge. The magnitude of the field at this point is
E = E x2 + E y2 = (161,000 N/C) 2 + (80,500 N/C)2 = 180,000 N/C
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Electric Charges and Forces
25-31
Using the field of a point charge,
E=
Kq
rb2
⇒ rb =
Kq
E
=
(9.0 × 109 N m 2 /C2 )(10.0 × 10−9 C)
= 2.236 cm
180,000 N/C
This gives the total distance but not the horizontal and vertical components. However, we can determine the angle θ
because Eb points straight away from the positive charge. Thus,
⎛ Ey ⎞
⎛ 80,500 N/C ⎞
−1 ⎛ 1 ⎞
⎟ = tan −1 ⎜
⎟ = tan ⎜ ⎟ = 26.57°
⎜ Ex ⎟
161,000
N/C
⎝2⎠
⎝
⎠
⎝
⎠
θ = tan −1 ⎜
The horizontal and vertical distances are then d x = rb cosθ = (2.236 cm)cos 26.57° = 2.00 cm and d y = rb sinθ = 1.00 cm.
Thus, point b is at position (xb , yb ) = (3.0 cm, 3.0 cm).
(c) To calculate point c, which is below and to the right of the charge, a similar procedure is followed. We first find
E = 36,000 N/C from which we find the total distance rc = 5.00 cm. The angle φ is
⎛ Ey ⎞
28,800 ⎞
⎟ = tan −1 ⎛⎜
⎟ = 53.13°
⎜ Ex ⎟
⎝ 21,600 ⎠
⎝
⎠
which gives distances d x = rc cos φ = 3.00 cm and d y = rc sin φ = 4.00 cm. Thus, point c is at position ( xc , yc ) =
φ = tan −1 ⎜
(4.0 cm, −2.0 cm).
25.65. Model: The electric field is that of three point charges.
Visualize:
Solve: (a) In the figure, the distances are r1 = r3 = (1.0 cm) 2 + (3.0 cm)2 = 3.162 cm and the angle is θ = tan −1 (1.0/3.0) =
18.43°. Using the equation for the field of a point charge,
E1 = E3 =
K q1
r12
=
(9.0 × 109 N m 2 /C2 )(1.0 × 10−9 C)
(0.03162 m) 2
= 9.0 kN/C
We now use the angle θ to find the components of the field vectors:
E1 = E1 cosθ iˆ − E1 sinθ ˆj = (8540 iˆ − 2840 ˆj ) N/C = (8.5 iˆ − 2.8 ˆj ) kN/C
E3 = E3 cosθ iˆ + E3 sinθ ˆj = (8540 iˆ + 2840 ˆj ) N/C = (8.5 iˆ + 2.8 ˆj ) kN/C
E2 is easier since it has only an x-component. Its magnitude is
E2 =
K q2
r22
=
(9.0 × 109 N m 2 /C2 )(1.0 × 10−9 C)
(0.0300 m) 2
= 10,000 N/C ⇒
E2 = E2 iˆ = 10 iˆ kN/C
(b) The electric field is defined in terms of an electric force acting on charge q: E = F/q. Since forces obey a
principle of superposition ( Fnet = F1 + F2 +
) it follows that the electric field due to several charges also obeys a
principle of superposition.
(c) The net electric field at a point 3.0 cm to the right of q2 is Enet = E1 + E2 + E3 = 27 iˆ kN/C. The y-components of
E1 and E2 cancel, giving a net field pointing along the x-axis.
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25-32
Chapter 25
25.66. Model: The charged ball attached to the string is a point charge.
Visualize:
The ball is in static equilibrium in the 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 gravitational force on the ball, and
the tension force.
Solve: In static equilibrium, Newton’s second law for the ball gives Fnet = T + FG + Fe = 0. In component form,
( Fnet ) x = Tx + 0 N + qE = 0 N
( Fnet ) y = Ty − mg + 0 N = 0 N
The above two equations simplify to
T sinθ = qE
T cosθ = mg
Dividing the equations, we get
qE
mg tanθ (5.0 × 10−3 kg)(9.8 N/kg) tan 20°
tanθ =
⇒ q=
=
= 1.78 × 10−7 C = 0.18 μ C
100,000 N/C
mg
E
25.67. Model: The charged ball attached to the string is the point charge.
Visualize:
The charged ball is in static equilibrium in the 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 gravitational force on the ball, and
the tension force.
Solve: In static equilibrium, Newton’s second law for the charged ball gives Fnet = T + FG + F3 = 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 sinθ = qE and T cosθ = mg . Dividing the equations gives
tanθ =
qE (25 × 1029 C)(200,000 N/C)
=
= 0.255 ⇒ θ = 14°
mg
(2.0 × 10−3 kg)(9.8 N/kg)
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Electric Charges and Forces
25-33
25.68. Solve: (a) How many excess electrons on a dust particle produce an electric field of magnitude 1.0 × 106 N/C
a distance of 1.0 μ m from the dust particle?
(b) The number of electrons is
N=
(1.5 × 106 N/C)(1.0 × 10−6 m) 2
(9.0 × 109 N m 2 /C2 )(1.60 × 10−19 C)
= 1.0 × 103
25.69. Solve: (a) Two equal charges separated by 1.50 cm exert repulsive forces of 0.020 N on each other. What is
the magnitude of the charge?
(b) The charge is
q=±
(0.020 N)(0.0150 m) 2
9.0 × 109 N m 2 /C2
= ±22 nC
The problem does not give the direction of the force. Thus, the charges could be both positive or both negative.
25.70. Solve: (a) At what distance from a 15 nC charge is the electric field strength 54,000 N/C?
(b) The distance is
r=
(9.0 × 109 N m 2 /C2 )(15 × 10−9 C)
= 0.050 m = 5.0 cm
54,000 N/C
25.71. Solve: (a) A 1.0 nC charge is placed at (0.0 cm, 2.0 cm) and another 1.0 nC charge is placed at (0 cm,
−2.0 cm). A third charge +q is placed along a line halfway between q1 and q2 such that the angle between the forces
on q due to each of the other two charges is 60°. The resultant force on q is (5.0 × 10−5 iˆ) N. What is the magnitude of
the charge q?
(b)
We have
⎛ (9.0 × 109 N m 2 /C2 )(1.0 × 10−9 C)q
⎞
F1 on 3 = ⎜
, away from q3 ⎟ = (5625 N/C)q (cos30° iˆ − sin 30° ˆj )
2
⎜
⎟
(0.020 m/ sin 30°)
⎝
⎠
ˆ
F2 on 3 = (5625 N/C)q[cos(30°) i + sin(30°) ˆj ]
Fon 3 = 2 × (5625 N/C) q cos(30°) iˆ
2(5625 N/C)q cos(30°) = 5.0 × 10−5 N ⇒ q =
(5.0 × 10−5 N)
= 5.1 nC
2(5625 N/C)cos(30°)
25.72. Model: Use the charge model.
Solve: The mass of copper in a 2.0-mm-diameter copper ball is
⎛ 4π 3 ⎞
⎡ 4π
⎤
M = ρV = ρ ⎜
r ⎟ = (8920 kg/m3 ) ⎢ (1.0 × 10−3 m)3 ⎥ = 3.736 × 10−5 kg = 0.03736 g
3
3
⎝
⎠
⎣
⎦
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25-34
Chapter 25
The number of moles in the ball is
n=
M 0.03736 g
=
= 5.884 × 10−4 mol
A 63.5 g/mol
The number of copper atoms in the ball is
N = nN A = (5.884 × 10−4 mol)(6.02 × 1023 mol−1 ) = 3.542 × 1020
Note that 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.
The number of electrons in the copper ball is thus 29 × 3.542 × 1020 = 1.03 × 1022. The number of electrons removed
from the copper ball is
50 × 10−9 C
1.60 × 10−19 C
= 3.13 × 1011
So, the fraction of electrons removed from the copper ball is
3.13 × 1011
1.03 × 1022
= 3.0 × 10−11
Assess: This is indeed a very small fraction of the available number of electrons in the copper ball.
25.73. Model: The charged balls are point charges.
Visualize:
Because of symmetry and the fact that the three balls have the same charge, the magnitude of the electric force on
each ball is the same. The other forces acting on each ball are the gravitational force on the ball and the tension force.
Solve: The force on ball 3 is the sum of the force from ball 1 and ball 2. We have
⎛ K q1 q3
⎞ Kq 2
, away from q1 ⎟ = 2 [sin(30°) iˆ + cos(30°) ˆj ]
F1 on 3 = ⎜
⎜ r2
⎟ r
13
⎝
⎠
F2 on 3 =
Fon 3 =
2 Kq 2
r2
Fe =
Kq 2
r2
cos(30°) ˆj ⇒
[ − sin(30°) iˆ + cos(30°) ˆj ]
Fon 3 =
2 Kq 2
r2
2(9.0 × 109 N m 2 /C2 )q 2 cos(30°)
(0.20 m) 2
cos(30°) = Fon 2 = Fon 1 = Fe
= (3.897 × 1011 q 2 ) N/C2
The distance l, between one of the balls and the center of the equilateral triangle, is
l cos(30°) =
0.10 m
r
= 0.10 m ⇒ l =
= 0.1155 m
2
cos(30°)
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Electric Charges and Forces
25-35
Thus, the angle made by the string with the plane containing the three balls is
cosθ =
l 0.1155 m
=
L
0.80 m
⇒ θ = 81.70°
From the free-body diagram, we have
T sinθ − mg = 0 N
− T cosθ + Fe = 0 N
mg
mg
tanθ =
=
Fe (3.897 × 1011 q 2 ) N/C2
q=
mg
11
(3.897 × 10
2
N/C ) tanθ
=
(3.0 × 10−3 kg)(9.8 N/kg)
(3.897 × 1011 N/C2 ) tan(81.70°)
= 1.05 × 10−7 C ≈ 0.11 μ C
25.74. Model: The charged spheres are point charges.
Visualize:
The figure shows the free-body diagram of the forces on the sphere with the negative charge that is shown in
Figure CP25.74. The force FE is due to the electric field. The force Fe is the attractive force between the positive
and the negative spheres. The tension in the string and the gravitational force are the remaining two forces on the
spheres.
Solve: The two electrical forces are calculated as follows:
FE = q E = (100 × 10−9 C)(1.00 × 105 N/C) = 1.00 × 10−2 N
Fe =
K q1 q2
r2
=
(9.0 × 109 N m 2 /C2 )(100 × 10−9 C) 2
r2
=
9.0 × 10−5 N m 2 /C
r2
From the geometry of Figure CP25.74,
r = 2 L sin(10°) = 2(0.50 m)sin(10°) = 0.174 E ⇒
Fe =
9.0 × 10−5 N m 2 /C
(0.174 m) 2
= 3.0 × 10−3 N
From the free-body diagram,
T cos10° = mg
T sin10° + Fe = FE
Rearranging and dividing the two equations gives
sin(10°)
F − Fe
= tan(10°) = E
cos(10°)
mg
m=
1.00 × 10−2 N − 0.30 × 10−2 N
FE − Fe
=
= 0.41 × 10−2 kg = 4.1 g
(9.8 N/kg) tan10°
g tan(10°)
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25-36
Chapter 25
25.75. Model: The charges are point charges.
Visualize:
Solve: The forces on the −1.0 nC charge lie along the line connecting the pairs of charges. Since the 10 nC charge is
positive, F10 points as shown. The angle formed by the dashed lines where they meet is 90°, so Fq must point
towards q in order for F = F10 + Fq to hold.
Therefore
F10
cos(30°)
The distance between the 10 nC and −1.0 nC charges is r = 5.0sin 30° cm = 2.5 cm. Thus
F=
F10 = (9.0 × 109 N m 2 /C2 )
(10 × 10−9 C)(1.0 × 10−9 C)
(2.5 × 10−2 m)2
= 1.44 × 10−4 N
Thus,
F=
1.44 × 10−4 N
= 1.7 × 10−4 N
cos(30°)
25.76. Model: Charge Q and the dipole charges (q and −q) are point charges.
Visualize: Please refer to Figure CP25.76.
Solve: (a) The force on the dipole is the vector sum of the force on q and −q. We have
⎛ KQ q
⎞
KQq
, away from Q ⎟⎟ =
(−iˆ)
FQ on + q = ⎜⎜
2
2
+
+
(
/2)
(
/2)
r
s
r
s
⎝
⎠
FQ on − q =
KQq
(r − s/2)
2
( +iˆ) ⇒
⎛
⎞
1
1
ˆ
−
Fnet = KQq ⎜⎜
2
2⎟
⎟i
(
/2)
(
/2)
−
+
r
s
r
s
⎝
⎠
(b) The net force Fnet is toward the charge Q, because the attractive force due to Q on the negative charge of the
dipole is more than the repulsive force of Q on the positive charge.
(c) Using the binomial approximation, we get
s ⎞
⎛
(r ± s/2) −2 = r −2 ⎜1 ± ⎟
2
r⎠
⎝
−2
⎛ 2s
≅ r −2 ⎜1 ∓
+
⎝ 2r
⎞
⎟ ⇒
⎠
Fnet =
KQq ⎡⎛ 2s ⎞ ⎛ 2s ⎞ ⎤ 2 KQqs
⎢⎜ 1 + ⎟ − ⎜ 1 − ⎟ ⎥ =
r 2 ⎣⎝ 2 r ⎠ ⎝ 2 r ⎠ ⎦
r3
(d) Coulomb’s law applies only to the force between two point charges. A dipole is not a point charge, so there’s no
reason that the force between a dipole and a point charge should be an inverse-square force.
Assess: Note that when s → 0 m, Fnet → 0 N. In this limit the dipole is a point with zero charge.
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