CHAPTER 18 ELECTRIC FORCES AND
ELECTRIC FIELDS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. 1.9 × 1013
2. (b) Suppose that A is positive and B is negative. Since C and A also attract each other, C
must be negative. Thus, B and C repel each other, because they have like charges (both
negative). Suppose, however, that A is negative and B is positive. Since C and A also attract
each other, C must be positive. Again we conclude that B and C repeal each other, because
they have like charges (both positive).
3. (a) The ball is electrically neutral (net charge equals zero). However, it is made from a
conducting material, so it contains electrons that are free to move. The rod attracts some of
these (negative) electrons to the side of the ball nearest the rod, leaving the opposite side of
the ball positively charged. Since the negative side of the ball is closer to the positive rod
than the positive side, a net attractive force arises.
4. (d) The fact that the positive rod repels one object indicates that that object carries a net
positive charge. The fact that the rod repels the other object indicates that that object carries
a net negative charge. Since both objects are identical and made from conducting material,
they share the combined net charges equally after they are touched together. Since the rod
repels each object after they are touched, each object must then carry a net positive charge.
But the net electric charge of any isolated system is conserved, so the total net charge
initially must also have been positive. This means that the initial positive charge had the
greater magnitude.
5. (c) This distribution is not possible because of the law of conservation of electric charge.
The total charge on the three objects here is 98 q , whereas only q was present initially.
6. (c) This is an example of charging by induction. The negatively charged rod repels free
electrons in the metal. These electrons move through the point of contact and into the sphere
farthest away from the rod, giving it an induced charge of −q. The sphere nearest the rod
acquires an induced charge of +q. As long as the rod is kept in place while the spheres are
separated, these induced charges cannot recombine and remain on the spheres.
7. (b) Coulomb’s law states that the magnitude of the force is given by F = k
q1 q2
. Doubling
r2
the magnitude of each charge as in A would increase the numerator by a factor of four, but
this is offset by the change in separation, which increases the denominator by a factor of
22 = 4. Doubling the magnitude of only one charge as in D would increase the numerator by
Chapter 18 Answers to Focus on Concepts Questions
943
a factor of two, but this is offset by the change in separation, which increases the
denominator by a factor of
( 2)
2
= 2.
8. (e) Coulomb’s law states that the magnitude of the force is given by F = k
q1 q2
. The force
r2
is directed along the line between the charges and is an attraction for unlike charges and a
repulsion for like charges. Charge B is attracted by charge A with a force of magnitude
q q
k 2 and repelled by charge C with a force of the same magnitude. Since both forces
d
q q
point to the left, the net force acting on B has a magnitude of 2k 2 . Charge A is attracted
d
q q
q q
. Since both
by charge B with a force of k 2 and also by charge C with a force of k
2
d
( 2d )
forces point to the right, the net force acting on A has a magnitude of (1.25 ) k
C is pushed to the right by B with a force of k
of k
q q
( 2d )
2
q q
d2
q q
d2
. Charge
and pulled to the left by A with a force
. Since these two forces have different directions, the net force acting on C has a
magnitude of ( 0.75 ) k
q q
d2
.
9. (b) According to Coulomb’s law, the magnitude of the force that any one of the point charges
q q
exerts on another point charge is given by F = k 2 , where d is the length of each side of
d
the triangle. The charge at B experiences a repulsive force from the charge at A and an
attractive force from the charge at C. Both forces have vertical components, but one points
in the +y direction and the other in the −y direction. These vertical components have equal
magnitudes and cancel, leaving a resultant that is parallel to the x axis.
10. 8.5 µC
11. (e) According to Equation 18.2, the force exerted on a charge by an electric field is
proportional to the magnitude of the charge. Since the negative charge has twice the
magnitude of the positive charge, the negative charge experiences twice the force.
Furthermore, the direction of the force on the positive charge is in the same direction as the
field, so that we can conclude that the field points due west. The force on the negative
charge points opposite to the field and, therefore, points due east.
944
ELECTRIC FORCES AND ELECTRIC FIELDS
12. (c) The electric field created by a point charge has a magnitude E =
kq
and is inversely
r2
proportional to the square of the distance r. If r doubles, the charge magnitude must increase
by a factor of 22 = 4 to keep the field the same.
13. (b) To the left of the positive charge the two contributions to the total field have opposite
directions. There is a spot in this region at which the field from the smaller, but closer,
positive charge exactly offsets the field from the greater, but more distant, negative charge.
14. (e) Consider the charges on opposite corners. In all of the arrangements these are like
charges. This means that the two field contributions created at the center of the square point
in opposite directions and, therefore, cancel. Thus, only the charge opposite the empty
corner determines the magnitude of the net field at the center of the square. Since the point
charges all have the same magnitude, the net field there has the same magnitude in each
arrangement.
15. 1.8 × 10−6 C/m2
16. (c) The tangent to the field line gives the direction of the electric field at a point. At A the
tangent points due south, at B southeast, and at C due east.
17. (a) The electric field has a greater magnitude where the field lines are closer together. They
are closest together at B and farthest apart at A. Therefore, the field has the greatest
magnitude at B and the smallest magnitude at A.
18. (d) In a conductor electric charges can readily move in response to an electric field. In A,
B, and C the electric charges experience an electric field and, hence, a force from
neighboring charges and will move outward, away from each other. They will rearrange
themselves so that the electric field within the metal is zero at equilibrium. This means that
they will reside on the outermost surface. Thus, only D could represent charges in
equilibrium.
19. 1.3 N·m2/C
20. 0.45 N·m2/C
Chapter 18 Problems
945
CHAPTER 18 ELECTRIC FORCES AND
ELECTRIC FIELDS
PROBLEMS
______________________________________________________________________________
1.
SSM REASONING AND SOLUTION The total charge of the electrons is
q = N(–e) = (6.0 × 1013)(–1.60 × 10–19 C)
q = – 9.6 × 10–6 C = –9.6 μC
The net charge on the sphere is, therefore,
qnet = +8.0 mC – 9.6 mC = -1.6 mC
______________________________________________________________________________
2.
REASONING The charge of a single proton is +e, and the charge of a single electron is –e,
where e = 1.60×10−19 C. The net charge of the ionized atom is the sum of the charges of its
constituent protons and electrons.
SOLUTION The ionized atom has 26 protons and 7 electrons, so its net electric charge q is
q = 26 ( +e ) + 7 ( −e ) = +19e = +19 (1.60 ×10−19 C ) = +3.04 ×10−18 C
3.
REASONING
a. Since the objects are metallic and identical, the charges on each combine and produce a
net charge that is shared equally by each object. Thus, each object ends up with one-fourth
of the net charge.
b. The number of electrons (or protons) that make up the final charge on each object is
equal to the final charge divided by the charge of an electron (or proton).
SOLUTION
a. The net charge is the algebraic sum of the individual charges. The charge q on each
object after contact and separation is one-fourth the net charge, or
q=
1
4
(1.6 μ C + 6.2 μ C – 4.8 μ C – 9.4 μ C ) =
–1.6 μ C
946
ELECTRIC FORCES AND ELECTRIC FIELDS
b. Since the charge on each object is negative, the charge is comprised of electrons. The
number of electrons on each object is the charge q divided by the charge −e of a single
electron:
q
−1.6 × 10−6 C
=
= 1.0 × 1013
Number of electrons =
−
19
−e −1.60 × 10
C
____________________________________________________________________________________________
4.
REASONING When N electrons, each carrying a charge −e = −1.6×10−19 C, are transferred
from the plate to the rod, the system consisting of the plate and the rod is isolated.
Therefore, the total charge q1i+ q2i of the system is unchanged by the process, where q1i is
the initial charge of the plate and q2i is the initial charge of the rod. At the end, the rod and
the plate each have the same final charge q1f = q2f. Therefore, each must have a charge equal
to half the total charge of the system: q1f = q2f = 12 ( q1i + q2i ) .
SOLUTION The final charge q2f on the rod is equal to its initial charge q2i plus the charge
transferred to it, which is equal to the product of the number N of electrons transferred and
the charge −e of each electron. Therefore,
q2f = q2i + N (−e) = q2i − Ne
(1)
Since the final charge on the rod is equal to half the total initial charge of the system, we can
substitute q2f = 12 ( q1i + q2i ) into Equation (1) and solve for N:
1
2
( q1i + q2i ) = q2i − Ne
or
Ne = q2i − 12 ( q1i + q2i ) =
1
2
( q2i − q1i )
or
N=
( q2i − q1i )
2e
Therefore, the number of electrons that must be transferred to the rod is
N=
5.
( q2i − q1i ) = +2.0 ×10−6 C − ( −3.0 ×10−6 C ) = 1.6 ×1013
2e
2 (1.6 × 10−19 C )
SSM REASONING Identical conducting spheres equalize their charge upon touching.
When spheres A and B touch, an amount of charge +q, flows from A and instantaneously
neutralizes the –q charge on B leaving B momentarily neutral. Then, the remaining amount
of charge, equal to +4q, is equally split between A and B, leaving A and B each with equal
amounts of charge +2q. Sphere C is initially neutral, so when A and C touch, the +2q on A
splits equally to give +q on A and +q on C. When B and C touch, the +2q on B and the +q
on C combine to give a total charge of +3q, which is then equally divided between the
spheres B and C; thus, B and C are each left with an amount of charge +1.5q.
Chapter 18 Problems
947
SOLUTION Taking note of the initial values given in the problem statement, and
summarizing the final results determined in the REASONING above, we conclude the
following:
a. Sphere C ends up with an amount of charge equal to +1.5q .
b. The charges on the three spheres before they were touched, are, according to the problem
statement, +5q on sphere A, –q on sphere B, and zero charge on sphere C. Thus, the total
charge on the spheres is +5q – q + 0 = +4q .
c. The charges on the spheres after they are touched are +q on sphere A, +1.5q on sphere B,
and +1.5q on sphere C. Thus, the total charge on the spheres is + q + 1.5q + 1.5q = +4q .
______________________________________________________________________________
6.
REASONING The conservation of electric charge states that, during any process, the net
electric charge of an isolated system remains constant (is conserved). Therefore, the net
charge (q1 + q2) on the two spheres before they touch is the same as the net charge after they
touch. When the two identical metal spheres touch, the net charge will spread out equally
over both of them. When the spheres are separated, the charge on each is the same.
SOLUTION
a. Since the final charge on each sphere is +5.0 μC, the final net charge on both spheres is
2(+5.0 μC) = +10.0 μC. The initial net charge must also be +10.0 μC. The only spheres
whose net charge is +10.0 μC are
B (qB = − 2.0 μ C) and D (qD = +12.0 μ C)
b. Since the final charge on each sphere is +3.0 μC, the final net charge on the three spheres
is 3(+3.0 μC) = +9.0 μC. The initial net charge must also be +9.0 μC. The only spheres
whose net charge is +9.0 μC are
A (qA = − 8.0 μ C), C (qC = +5.0 μ C) and D (qD = +12.0 μ C)
c. Since the final charge on a given sphere in part (b) is +3.0 μC, we would have to add
−3.0 μC to make it electrically neutral. Since the charge on an electron is −1.6 × 10−19 C,
the number of electrons that would have to be added is
−3.0 × 10−6 C
= 1.9 × 1013
−1.6 × 10−19 C
______________________________________________________________________________
Number of electrons =
948
7.
ELECTRIC FORCES AND ELECTRIC FIELDS
REASONING
a. The number N of electrons is 10 times the number of water molecules in 1 liter of water.
The number of water molecules is equal to the number n of moles of water molecules times
Avogadro’s number NA: N = 10 n NA .
b. The net charge of all the electrons is equal to the number of electrons times the change on
one electron.
SOLUTION
a. The number N of water molecules is equal to 10 n NA , where n is the number of moles of
water molecules and NA is Avogadro’s number. The number of moles is equal to the mass
m of 1 liter of water divided by the mass per mole of water. The mass of water is equal to
its density ρ times the volume, as expressed by Equation 11.1. Thus, the number of
electrons is
⎛
⎞
⎛
⎞
ρV
m
N = 10 n NA = 10 ⎜
⎟ NA = 10 ⎜
⎟ NA
⎝ 18.0 g/mol ⎠
⎝ 18.0 g/mol ⎠
⎡
−3
3
3 ⎛ 1000 g ⎞ ⎤
⎢ 1000 kg/m 1.00 × 10 m ⎜
−1
23
⎟⎥
= 10 ⎢
⎝ 1 kg ⎠ ⎥ 6.022 × 10 mol
⎢⎣
⎥⎦
18.0 g/mol
(
)(
)
(
)
= 3.35 × 1026 electrons
b. The net charge Q of all the electrons is equal to the number of electrons times the change
(
)(
)
on one electron: Q = 3.35 × 1026 −1.60 × 10−19 C = −5.36 × 107 C .
______________________________________________________________________________
8.
REASONING The magnitude F of the forces that point charges q1 and q2 exert on each
other varies with the distance r separating them according to F = k
q1 q2
(Equation 18.1),
r2
We note that both charges are given in units of
where k = 8.99×109 N·m2/C2.
microcoulombs (μC), rather than the base SI units of coulombs (C). We will replace the
prefix μ with 10−6 when calculating the distance r from Equation 18.1.
SOLUTION Solving F = k
r2 = k
q1 q2
r2
q1 q2
F
(Equation 18.1) for the distance r, we obtain
or
r= k
q1 q2
F
Chapter 18 Problems
949
Therefore, when the force magnitude F is 0.66 N, the distance between the charges must be
r= k
9.
q1 q2
F
=
(8.99 × 10
9
2
N ⋅ m /C
2
)
(8.4 × 10−6 C )( 5.6 ×10−6 C ) = 0.80 m
0.66 N
SSM REASONING The number N of excess electrons on one of the objects is equal to
the charge q on it divided by the charge of an electron (−e), or N = q/(−e). Since the charge
on the object is negative, we can write q = − q , where q is the magnitude of the charge.
The magnitude of the charge can be found from Coulomb’s law (Equation 18.1), which
states that the magnitude F of the electrostatic force exerted on each object is given by
F = k q q / r 2 , where r is the distance between them.
SOLUTION The number N of excess electrons on one of the objects is
N=
q
q −q
=
=
e
−e −e
(1)
To find the magnitude of the charge, we solve Coulomb’s law, F = k q q / r 2 , for q :
q =
F r2
k
Substituting this result into Equation (1) gives
( 4.55 ×10 N )(1.80 ×10 m )
F r2
q
k =
8.99 ×109 N ⋅ m 2 / C2
N= =
= 8
e
e
1.60 × 10−19 C
______________________________________________________________________________
−21
−3
2
10. REASONING The gravitational force is an attractive force. To neutralize this force, the
electrical force must be a repulsive force. Therefore, the charges must both be positive or
both negative. Newton’s law of gravitation, Equation 4.3, states that the gravitational force
depends inversely on the square of the distance between the earth and the moon. Coulomb’s
law, Equation 18.1 states that the electrical force also depends inversely on the square of the
distance. When these two forces are added together to give a zero net force, the distance
can be algebraically eliminated. Thus, we do not need to know the distance between the
two bodies.
950
ELECTRIC FORCES AND ELECTRIC FIELDS
SOLUTION Since the repulsive electrical force neutralizes the attractive gravitational
force, the magnitudes of the two forces are equal:
kq q
=
2
r Electrical
force,
Equation 18.1
GM e M m
r2 Gravitational
force,
Equation 4.3
Solving this equation for the magnitude q of the charge on either body, we find
2
⎛
−11 N ⋅ m ⎞
×
6.67
10
5.98 × 1024 kg 7.35 × 1022 kg
⎜
2 ⎟
GM e M m
kg ⎠
= ⎝
= 5.71 × 1013 C
q =
2
k
N⋅m
8.99 × 109
C2
______________________________________________________________________________
(
)(
)
11. SSM WWW REASONING Initially, the two spheres are neutral. Since negative
charge is removed from the sphere which loses electrons, it then carries a net positive
charge. Furthermore, the neutral sphere to which the electrons are added is then negatively
charged. Once the charge is transferred, there exists an electrostatic force on each of the
two spheres, the magnitude of which is given by Coulomb's law (Equation 18.1),
F = k q1 q2 / r 2 .
SOLUTION
a. Since each electron carries a charge of −1.60 ×10−19 C , the amount of negative charge
removed from the first sphere is
⎛ 1.60 × 10−19 C ⎞
−6
3.0 ×1013 electrons ⎜⎜
⎟⎟ = 4.8 × 10 C
1
electron
⎝
⎠
(
)
Thus, the first sphere carries a charge +4.8 × 10–6 C, while the second sphere carries a
charge -4.8 × 10–6 C. The magnitude of the electrostatic force that acts on each sphere is,
therefore,
F=
k q1 q2
r2
8.99 × 109 N ⋅ m 2 /C2 )( 4.8 × 10−6 C )
(
=
( 0.50 m )2
2
= 0.83 N
b. Since the spheres carry charges of opposite sign, the force is attractive .
______________________________________________________________________________
Chapter 18 Problems
12. REASONING Let F2 and F1 represent the forces exerted on the
charge q at the origin by the point charges q1 and q2, respectively.
According to Equation 18.1, the magnitudes of these forces are
given by
F1 = k
q1 q
F2 = k
and
r12
q2 q
y
q2
q1 = −25 μC
(1)
r22
951
F1
where r1 is the distance between q1 and q, r2 is the distance
q = +8.4 μC
between q2 and q, and k = 8.99×109 N·m2/C2. The directions of
the forces are determined by the signs of each charge-pair. The F2
sign of q1 is opposite that of q, so F1 is an attractive force,
pointing in the positive y direction. The signs of q2 and q are both
positive, so F2 is a repulsive force, pointing in the negative y direction (see the drawing).
Because the net force F = F1 + F2 acting on q points in the positive y direction, the force F1
must have a greater magnitude than the force F2. Therefore, the magnitude F of the net
electric force acting on q is equal to the magnitude of the attractive force F1 minus that of
the repulsive force F2:
F = F1 − F2
(2)
SOLUTION Substituting Equations (1) into Equation (2) yields
F = F1 − F2 = k
q1 q
r12
−k
q2 q
(3)
r22
Solving Equation (3) for |q2|, we obtain
k
q2 q
r22
=k
q1 q
r12
−F
or
⎛q
F
q2 = r22 ⎜ 12 −
⎜r
kq
⎝ 1
⎞
⎟⎟
⎠
Substituting the given values, we find that
⎡ −25 × 10 −6 C
⎤
27 N
⎥ = 1.8 × 10 −5 C
q2 = ( 0.34 m )2 ⎢
−
2
−
9
2
2
6
⎢ ( 0.22 m )
8.99 × 10 N ⋅ m /C +8.4 × 10 C ⎥
⎣
⎦
(
)
952
ELECTRIC FORCES AND ELECTRIC FIELDS
y
13. SSM REASONING AND SOLUTION The net
electrostatic force on charge 3 at x = +3.0 m is the
1
+3.0 m
+18 μC
vector sum of the forces on charge 3 due to the other
two charges, 1 and 2. According to Coulomb's law
(Equation 18.1), the magnitude of the force on charge θ = 45°
+45 μC
3 due to charge 1 is
–12 μC
k q1 q3
3
0 2
F13 =
+3.0 m
2
r13
x
Figure 1
where the distance between charges 1 and 3 is r13.
According to the Pythagorean theorem, r132 = x 2 + y 2 . Therefore,
F13
8.99 ×109 N ⋅ m 2 / C2 )(18 ×10−6 C )( 45 × 10−6 C )
(
=
= 0.405 N
( 3.0 m )2 + ( 3.0 m )2
Charges 1 and 3 are equidistant from the origin, so that θ = 45° (see Figure 1). Since
charges 1 and 3 are both positive, the force on charge 3 due to charge 1 is repulsive and
along the line that connects them, as shown in Figure 2. The components of F13 are:
F13 x = F13 cos 45° = 0.286 N
and
F13 y = – F13 sin 45° = –0.286 N
The second force on charge 3 is the attractive force
(opposite signs) due to its interaction with charge 2
located at the origin. The magnitude of the force on
charge 3 due to charge 2 is, according to Coulomb's law ,
F23 =
k q2 q3
2
r23
=
k q2 q3
x
y
1
45°
2
2
0
8.99 ×109 N ⋅ m 2 / C2 )(12 ×10−6 C )( 45 ×10−6 C )
(
=
( 3.0 m )2
F23
Figure 2
3
45°
x
F13
F
13
= 0.539 N
Since charges 2 and 3 have opposite signs, they attract each other, and charge 3 experiences
a force to the left as shown in Figure 2. Taking up and to the right as the positive directions,
we have
F3 x = F13 x + F23 x = +0.286 N − 0.539 N = −0.253 N
F3 y = F13 y = −0.286 N
Chapter 18 Problems
Using the Pythagorean theorem, we find the magnitude of F3 to
be
F3 =
F32x
+ F32y
2
2
= (−0.253 N) + (−0.286 N) = 0.38 N
953
0.253 N
φ
0.286 N
F3
The direction of F3 relative to the –x axis is specified by the
angle f (see Figure 3), where
Figure 3
N⎞
φ = tan ⎜
⎟ = 49° below the − x axis
⎝ 0.253 N ⎠
______________________________________________________________________________
−1 ⎛ 0.286
14. REASONING The electrical force that each charge exerts on charge 2 is shown in the
following drawings. F21 is the force exerted on 2 by 1, and F23 is the force exerted on 2
by 3. Each force has the same magnitude, because the charges have the same magnitude
and the distances are equal.
−q
F21
+q
1
F23
2
+q
+q
F23 +q
3
1
2
(a)
F21
+q
+q
3
1
F21 −q
2
F23
(b)
3
+q
(c)
The net electric force F that acts on charge 2 is shown in the following diagrams.
F21
F23
F21
F23
F
F=0N
(a)
(b)
F23
F
F21
(c)
It can be seen from the diagrams that the largest electric force occurs in (a), followed by (c),
and then by (b).
SOLUTION The magnitude F21 of the force exerted on 2 by 1 is the same as the
magnitude F23 of the force exerted on 2 by 3, since the magnitudes of the charges are the
same and the distances are the same. Coulomb’s law gives the magnitudes as
954
ELECTRIC FORCES AND ELECTRIC FIELDS
F21 = F23 =
kq q
r2
8.99 × 109 N ⋅ m 2 /C2 )( 8.6 × 10−6 C )( 8.6 × 10−6 C )
(
=
= 4.6 × 104 N
2
−3
(3.8 × 10 m )
In part (a) of the drawing showing the net electric force acting on charge 2, both F21 and F23
point to the left, so the net force has a magnitude of
(
)
F = 2 F12 = 2 4.6 × 104 N = 9.2 × 104 N
In part (b) of the drawing showing the net electric force acting on charge 2, F21 and F23
point in opposite directions, so the net force has a magnitude of 0 N .
In part (c) showing the net electric force acting on charge 2, the magnitude of the net force
can be obtained from the Pythagorean theorem:
( 4.6 × 104 N ) + ( 4.6 × 104 N )
2
2
2
F = F21
+ F23
=
2
= 6.5 × 104 N
______________________________________________________________________________
15. SSM REASONING AND SOLUTION
a. Since the gravitational force between the spheres is one of attraction and the electrostatic
force must balance it, the electric force must be one of repulsion. Therefore, the charges
must have the same algebraic signs, both positive or both negative .
b. There are two forces that act on each sphere; they are the gravitational attraction FG of
one sphere for the other, and the repulsive electric force FE of one sphere on the other.
From the problem statement, we know that these two forces balance each other, so that
FG = FE. The magnitude of FG is given by Newton's law of gravitation (Equation 4.3:
FG = Gm1m2 / r 2 ), while the magnitude of FE is given by Coulomb's law (Equation 18.1:
FE = k q1 q2 / r 2 ). Therefore, we have
Gm1m2
r
2
=
k q1 q2
r
2
or
Gm2 = k q
2
Chapter 18 Problems
955
since the spheres have the same mass m and carry charges of the same magnitude q .
Solving for q , we find
G
6.67 ×10 –11 N ⋅ m 2 /kg 2
= (2.0 ×10 –6 kg)
= 1.7 × 10 –16 C
k
8.99 × 109 N ⋅ m 2 /C2
______________________________________________________________________________
q =m
16. REASONING AND SOLUTION The electrostatic forces decreases with the square of the
distance separating the charges. If this distance is increased by a factor of 5 then the force
will decrease by a factor of 25. The new force is, then,
3.5 N
= 0.14 N
25
______________________________________________________________________________
F=
17. REASONING The electrons transferred increase the magnitudes of the positive and negative
charges from 2.00 μC to a greater value. We can calculate the number N of electrons by
dividing the change in the magnitude of the charges by the magnitude e of the charge on an
electron. The greater charge that exists after the transfer can be obtained from Coulomb’s law
and the value given for the magnitude of the electrostatic force.
SOLUTION The number N of electrons transferred is
N=
qafter − qbefore
e
where qafter and qbefore are the magnitudes of the charges after and before the transfer of
electrons occurs. To obtain qafter , we apply Coulomb’s law with a value of 68.0 N for the
electrostatic force:
F =k
qafter
2
qafter =
or
r2
Fr 2
k
Using this result in the expression for N, we find that
N=
Fr 2
− qbefore
k
=
e
( 68.0 N )( 0.0300 m )2
9
2
2
− 2.00 × 10−6 C
8.99 ×10 N ⋅ m / C
1.60 ×10−19 C
= 3.8 × 1012
956
ELECTRIC FORCES AND ELECTRIC FIELDS
18. REASONING AND SOLUTION Calculate the magnitude of each force acting on the
center charge. Using Coulomb’s law, we can write
F43 =
k q4 q3
2
r43
(
8.99 × 109 N ⋅ m 2 / C2 )( 4.00 × 10−6 C )( 3.00 × 10−6 C )
=
( 0.100 m )2
= 10.8 N (toward the south)
F53 =
k q5 q3
2
r53
=
(8.99 × 109 N ⋅ m2 / C2 )( 5.00 × 10−6 C )( 3.00 × 10−6 C )
( 0.100 m )2
= 13.5 N (toward the east)
Adding F43 and F53 as vectors, we have
2
2
F = F43
+ F53
=
(10.8 N )2 + (13.5 N )2
= 17.3 N
⎛ F43 ⎞
–1 ⎛ 10.8 N ⎞
⎟⎟ = tan ⎜
⎟ = 38.7° S of E
F
13.5
N
⎝
⎠
⎝ 53 ⎠
______________________________________________________________________________
θ = tan –1 ⎜⎜
19. REASONING According to Newton’s second law, the centripetal acceleration experienced
by the orbiting electron is equal to the centripetal force divided by the electron’s mass.
Recall from Section 5.3 that the centripetal force Fc is the name given to the net force
required to keep an object on a circular path of radius r. For an electron orbiting about two
protons, the centripetal force is provided almost exclusively by the electrostatic force of
attraction between the electron and the protons. This force points toward the center of the
circle and its magnitude is given by Coulomb’s law.
SOLUTION The magnitude ac of the centripetal acceleration is equal to the magnitude Fc
of the centripetal force divided by the electron’s mass: ac = Fc / m (Equation 5.3). The
centripetal force is provided almost entirely by the electrostatic force, so Fc = F, where F is
the magnitude of the electrostatic force of attraction between the electron and the two
protons, Thus, ac = F / m . The magnitude of the electrostatic force is given by Coulomb’s
law, F = k q1 q2 / r 2 (Equation 18.1), where q1 = −e and q2 = +2e are the magnitudes
of the charges, r is the radius of the orbit, and k = 8.99 ×109 N ⋅ m2 / C2 . Substituting this
expression for F into ac = F / m , and using m = 9.11 × 10−31 kg for the mass of the electron,
we find that
Chapter 18 Problems
ac =
=
F
=
m
k
957
−e +2e
k −e +2e
r2
=
m
m r2
( 8.99 ×109 N ⋅ m 2 / C2 ) −1.60
( 9.11×10
−31
× 10−19 C +2 × 1.60 × 10−19 C
kg )( 2.65 × 10
−11
m)
2
= 7.19 × 1023 m/s 2
____________________________________________________________________________________________
20. REASONING
The drawing at the right shows the forces that act
on the charges at each corner. For example, FAB is
the force exerted on the charge at corner A by the
charge at corner B. The directions of the forces
are consistent with the fact that like charges repel
and unlike charges attract.
Coulomb’s law
indicates that all of the forces shown have the
2
FBC
+
FBA
FAB
2
same magnitude, namely, F = k q / L , where q
is the magnitude of each of the charges and L is the
length of each side of the equilateral triangle. The
magnitude is the same for each force, because q
B
A
+
-
FAC
C
FCA
FCB
and L are the same for each pair of charges.
The net force acting at each corner is the sum of the two force vectors shown in the drawing,
and the net force is greatest at corner A. This is because the angle between the two vectors at
A is 60º. With the angle less than 90º, the two vectors partially reinforce one another. In
comparison, the angles between the vectors at corners B and C are both 120º, which means
that the vectors at those corners partially offset one another.
The net forces acting at corners B and C have the same magnitude, since the magnitudes of
the individual vectors are the same and the angles between the vectors at both B and C are
the same (120º). Thus, vector addition by either the tail-to-head method (see Section 1.6) or
the component method (see Section 1.8) will give resultant vectors that have different
directions but the same magnitude. The magnitude of the net force is the smallest at these
two corners.
958
ELECTRIC FORCES AND ELECTRIC FIELDS
SOLUTION
As pointed out in the
REASONING, the magnitude of any
+y
2
individual force vector is F = k q / L2 .
With this in mind, we apply the component
method for vector addition to the forces at
corner A, which are shown in the drawing at
the right, together with the appropriate
components. The x component Σ Fx and the
FAB
FAB sin 60.0º
FAC
60.0
+x
A
y component Σ Fy of the net force are
FAB cos 60.0
( Σ Fx )A = FAB cos 60.0° + FAC = F ( cos 60.0° + 1)
( Σ Fy )A = FAB sin 60.0° = F sin 60.0°
where we have used the fact that FAB = FAC = F . The Pythagorean theorem indicates that
the magnitude of the net force at corner A is
( Σ F )A = ( Σ Fx )A + ( Σ Fy )A
2
=F
( cos 60.0° + 1)
(
2
2
= F 2 ( cos 60.0° + 1) + ( F sin 60.0° )
2
+ ( sin 60.0° ) = k
= 8.99 × 109 N ⋅ m 2 / C2
2
)
(5.0 ×10−6 C )
( 0.030 m )
2
2
q
2
L
2
( cos 60.0° + 1)2 + ( sin 60.0° )2
2
( cos 60.0° + 1)2 + ( sin 60.0° )2
= 430 N
We now apply the component method
for vector addition to the forces at
corner B. These forces, together with
the appropriate components are shown
in the drawing at the right. We note
immediately that the two vertical
components cancel, since they have
opposite directions. The two horizontal
components, in contrast, reinforce since
they have the same direction. Thus, we
have the following components for the
net force at corner B:
FBC
+y
60.0º
FBC cos 60.0º
B
FBA cos 60.0
FBA
+x
Chapter 18 Problems
959
( Σ Fx )B = − FBC cos 60.0° − FBA cos 60.0° = −2 F cos 60.0°
( Σ Fy )B = 0
where we have used the fact that FBC = FBA = F . The Pythagorean theorem indicates that
the magnitude of the net force at corner B is
( Σ F )B = ( Σ Fx )B + ( Σ Fy )B = ( −2 F cos 60.0° )2 + ( 0 )2 = 2 F cos 60.0°
2
= 2k
2
q
2
L
2
(
cos 60.0° = 2 8.99 × 109 N ⋅ m 2 / C2
)
(5.0 ×10−6 C )
( 0.030 m )2
2
cos 60.0°
= 250 N
As discussed in the REASONING, the magnitude of the net force acting on the charge at
corner C is the same as that acting on the charge at corner B, so ( ΣF )C = 250 N .
21. REASONING
a. There are two electrostatic forces that act on q1; that due to q2 and that due to q3. The
magnitudes of these forces can be found by using Coulomb’s law. The magnitude and
direction of the net force that acts on q1 can be determined by using the method of vector
components.
b. According to Newton’s second law, Equation 4.2b, the acceleration of q1 is equal to the
net force divided by its mass. However, there is only one force acting on it, so this force is
the net force.
q2
+y
SOLUTION
F13
F12
a. The magnitude F12 of the force exerted on q1
1.30 m
23.0°
23.0°
by q2 is given by Coulomb’s law, Equation 18.1,
+x
q1
where the distance is specified in the drawing:
1.30 m
q3
F12 =
k q1 q2
r122
8.99 × 109 N ⋅ m 2 /C2 )( 8.00 × 10−6 C )( 5.00 × 10−6 C )
(
=
= 0.213 N
(1.30 m )2
960
ELECTRIC FORCES AND ELECTRIC FIELDS
Since the magnitudes of the charges and the distances are the same, the magnitude of F13 is
the same as the magnitude of F12, or F13 = 0.213 N. From the drawing it can be seen that the
x-components of the two forces cancel, so we need only to calculate the y components of the
forces.
Force
y component
F12
+F12 sin 23.0° = +(0.213 N) sin 23.0° = +0.0832 N
F13
+F13 sin 23.0° = +(0.213 N) sin 23.0° = +0.0832 N
F
Fy = +0.166 N
Thus, the net force is F = + 0.166 N (directed along the +y axis) .
b. According to Newton’s second law, Equation 4.2b, the acceleration of q1 is equal to the
net force divided by its mass. However, there is only one force acting on it, so this force is
the net force:
F
+0.166 N
a= =
= +111 m /s 2
3
−
m 1.50 × 10 kg
where the plus sign indicates that the acceleration is along the +y axis .
______________________________________________________________________________
22. REASONING We will use Coulomb’s law to calculate the force that any one charge exerts
on another charge. Note that in such calculations there are three separations to consider.
Some of the charges are a distance d apart, some a distance 2d, and some a distance 3d. The
greater the distance, the smaller the force. The net force acting on any one charge is the
vector sum of three forces. In the following drawing we represent each of those forces by an
arrow. These arrows are not drawn to scale and are meant only to “symbolize” the three
different force magnitudes that result from the three different distances used in Coulomb’s
law. In the drawing the directions are determined by the facts that like charges repel and
unlike charges attract. By examining the drawing we will be able to identify the greatest and
the smallest net force.
+
A
d
+
B
d
+
C
d
D
The greatest net force occurs for charge C, because all three force contributions point in the
same direction and two of the three have the greatest magnitude, while the third has the next
Chapter 18 Problems
961
greatest magnitude. The smallest net force occurs for charge B, because two of the three
force contributions cancel.
SOLUTION Using Coulomb’s law for each contribution to the net force, we calculate the
ratio of the greatest to the smallest net force as follows:
( ΣF ) C
=
( ΣF ) B
k
k
q
2
d2
q
+k
2
d2
−k
q
2
d2
q
+k
2
d2
+k
q
2
( 2d ) 2
q
2
=
1 + 1 + 14
1
4
= 9.0
( 2d ) 2
23. REASONING The kinetic energy of the orbiting electron is KE = 12 mv 2 (Equation 6.2),
where m and v are its mass and speed, respectively. We can obtain the speed by noting that
the electron experiences a centripetal force whose magnitude Fc is given by Fc = mv 2 / r
(Equation 5.3), where r is the radius of the orbit. The centripetal force is provided almost
exclusively by the electrostatic force of attraction F between the electron and the protons, so
Fc = F. The electrostatic force points toward the center of the circle and its magnitude is
given by Coulomb’s law as F = k q1 q2 / r 2 (Equation 18.1), where q1 and q2 are the
magnitudes of the charges.
SOLUTION The kinetic energy of the electron is
KE = 12 mv 2
(6.2)
Solving the centripetal-force expression, Fc = mv 2 / r (Equation 5.3), for the speed v, and
substituting the result into Equation 6.2 gives
KE = 12 mv 2 =
1
2
⎛ rF ⎞
m ⎜ c ⎟ = 12 r Fc
⎝ m ⎠
(1)
The centripetal force is provided almost entirely by the electrostatic force, so Fc = F, where
F is the magnitude of the electrostatic force of attraction between the electron and the three
protons. This force is given by Coulomb’s law, F = k q1 q2 / r 2 (Equation 18.1).
Substituting Coulomb’s law into Equation 1 yields
⎛ k q q ⎞ k q1 q2
KE = 12 r Fc = 12 r F = 12 r ⎜ 12 2 ⎟ =
2r
⎝ r
⎠
962
ELECTRIC FORCES AND ELECTRIC FIELDS
Setting q1 = −e and q2 = +3e , we have
KE =
=
k −e +3e
2r
(8.99 ×109 N ⋅ m 2 / C2 ) −1.60 × 10−19 C
2(1.76 ×10−11 m )
+3 × 1.60 × 10−19 C
= 1.96 × 10−17 J
______________________________________________________________________________
24. REASONING When the airplane and the other end of the guideline carry point charges +q
and −q, the airplane is subject to an attractive electric force of magnitude
q q
q −q
q2
(Equation 18.1), where k = 8.99 ×109 N ⋅ m2 / C2 and r is the
F =k 1 22 =k
=
k
2
2
r
r
r
mv22
length of the guideline. This electric force provides part of the centripetal force Fc2 =
r
(Equation 5.3) necessary to keep the airplane (mass = m) flying along its circular path at the
higher speed v2 (which is associated with the greater kinetic energy KE2). The remainder of
the centripetal force is provided by the maximum tension Tmax in the guideline, so we have
that
mv22
F + Tmax = Fc2 =
(1)
r
When the airplane is neutral and flying at the slower speed v1 (which is associated with the
smaller kinetic energy KE1), there is no electrical force, so the centripetal force Fc1 =
mv12
r
acting on the airplane is due solely to the maximum tension Tmax in the guideline:
Tmax = Fc1 =
mv12
r
(2)
We note that the kinetic energy of the airplane is given by KE = 12 mv 2 (Equation 6.2), so
that the quantities in the numerators of Equations (1) and (2) are proportional to the kinetic
energies KE2 and KE1 of the airplane:
mv22 = 2 ( KE 2 )
SOLUTION Solving F = k
sides, we obtain
and
mv12 = 2 ( KE1 )
(3)
q2
(Equation 18.1) for q2 and taking the square root of both
2
r
Chapter 18 Problems
q2 =
Fr 2
k
Fr 2
k
q=
or
963
(4)
Substituting Equations (3) into Equations (1) and (2) yields
F + Tmax =
2 ( KE 2 )
r
Tmax =
and
2 ( KE1 )
(5)
r
Substituting the second of Equations (5) into the first of Equations (5), we obtain
F+
2 ( KE1 )
r
=
2 ( KE 2 )
r
or
F=
2 ( KE 2 )
r
−
2 ( KE1 )
r
=
2 ( KE 2 − KE1 )
r
(6)
Substituting Equation (6) into Equation (4), we find that
q=
Fr 2
=
k
2 ( KE 2 − KE1 ) ⎛ r 2 ⎞
⎜
⎟=
⎝ k ⎠
r
2 ( KE 2 − KE1 ) r
k
Therefore, the magnitude q of the charge on the airplane is
q=
25.
2 ( 51.8 J − 50.0 J )( 3.0 m )
= 3.5 × 10 −5 C
9
2 2
8.99 × 10 N ⋅ m /C
qU
SSM REASONING Consider the drawing
q1
FAU
at the right. It is given that the charges qA,
d
FA2
q1, and q2 are each positive. Therefore, the
4d
qA
charges q1 and q2 each exert a repulsive
q2 = +3.0 mC
force on the charge qA. As the drawing
q
shows, these forces have magnitudes FA1
(vertically downward) and FA2 (horizontally
FA1
to the left). The unknown charge placed at
the empty corner of the rectangle is qU, and it exerts a force on qA that has a magnitude FAU.
In order that the net force acting on qA point in the vertical direction, the horizontal
component of FAU must cancel out the horizontal force FA2. Therefore, FAU must point as
shown in the drawing, which means that it is an attractive force and qU must be negative,
since qA is positive.
964
ELECTRIC FORCES AND ELECTRIC FIELDS
SOLUTION The basis for our solution is the fact that the horizontal component of FAU
must cancel out the horizontal force FA2. The magnitudes of these forces can be expressed
using Coulomb’s law F = k q q′ / r 2 , where r is the distance between the charges q and q′ .
Thus, we have
FAU =
k qA qU
( 4d )
2
FA2 =
and
+ d2
k qA q2
( 4d ) 2
where we have used the fact that the distance between the charges qA and qU is the diagonal
of the rectangle, which is
( 4d )2 + d 2
according to the Pythagorean theorem, and the fact
that the distance between the charges qA and q2 is 4d. The horizontal component of FAU is
FAU cos θ , which must be equal to FA2, so that we have
k qA qU
( 4d ) 2 + d 2
cosθ =
k qA q2
qU
or
( 4d )2
17
cosθ =
q2
16
The drawing in the REASONING, reveals that cos θ = ( 4 d ) /
( 4d ) 2 + d 2
= 4 / 17 .
Therefore, we find that
qU ⎛ 4 ⎞ q2
=
17 ⎜⎝ 17 ⎟⎠ 16
or
qU =
(
)
17 17
17 17
q2 =
3.0 × 10−6 C = 3.3 × 10−6 C
64
64
As discussed in the REASONING, the algebraic sign of the charge qU is negative .
26. REASONING AND SOLUTION
a. To find the charge on each ball we first need to determine the electric force acting on
each ball. This can be done by noting that each thread makes an angle of 18° with respect to
the vertical.
Fe = mg tan 18° = (8.0 × 10–4 kg)(9.80 m/s2) tan 18° = 2.547 × 10–3 N
We also know that
Fe =
k q1 q2
r2
where r = 2(0.25 m) sin 18° = 0.1545 m. Now
=
kq
r2
2
Chapter 18 Problems
Fe
q =r
k
= ( 0.1545 m )
965
2.547 × 10 –3 N
= 8.2 × 10 –8 C
8.99 × 109 N ⋅ m 2 / C2
b. The tension is due to the combination of the weight of the ball and the electric force, the
two being perpendicular to one another. The tension is therefore,
( mg )2 + Fe2
(
)(
)
2
(
)
2
⎡ 8.0 × 10 –4 kg 9.80 m/s 2 ⎤ + 2.547 × 10 –3 N = 8.2 × 10 –3 N
⎣
⎦
______________________________________________________________________________
T=
=
27. SSM REASONING The charged insulator experiences an electric force due to the
presence of the charged sphere shown in the drawing in the text. The forces acting on the
insulator are the downward force of gravity (i.e., its weight, W = mg ), the electrostatic force
F = k q1 q2 / r 2 (see Coulomb's law, Equation 18.1) pulling to the right, and the tension T
in the wire pulling up and to the left at an angle θ with respect to the vertical as shown in the
drawing in the problem statement. We can analyze the forces to determine the desired
quantities q and T.
SOLUTION.
a. We can see from the diagram given with the problem statement that
Tx = F
which gives
and
Ty = W
which gives
T sin θ = k q1 q2 / r 2
T cos θ = mg
Dividing the first equation by the second yields
k q1 q2 / r 2
T sin θ
= tan θ =
T cos θ
mg
Solving for q, we find that
⎛ k q1 q2
⎜ mgr 2
⎝
θ = tan –1 ⎜
⎞
⎟
⎟
⎠
⎡ (8.99 ×109 N ⋅ m 2 /C 2 )(0.600 × 10 –6 C)(0.900 × 10 –6 C) ⎤
= tan –1 ⎢
⎥ = 15.4°
(8.00 × 10 –2 kg)(9.80 m/s 2 )(0.150 m) 2
⎣
⎦
966
ELECTRIC FORCES AND ELECTRIC FIELDS
b. Since T cos θ = mg , the tension can be obtained as follows:
mg
(8.00 ×10−2 kg) (9.80 m/s 2 )
=
= 0.813 N
cosθ
cos 15.4°
______________________________________________________________________________
T=
28. REASONING AND SOLUTION Since the objects attract each other initially, one object
has a negative charge, and the other object has a positive charge. Assume that the negative
charge has a magnitude of q1 and that the positive charge has a magnitude of q2 .
Assume also that q2 is greater than q1 . The magnitude F of the initial attractive force
between the objects is
F=
k q1 q2
(1)
r2
After the objects are brought into contact and returned to their initial positions, the charge
on each object is the same and has a magnitude of q2 − q1 / 2 . The magnitude F of the
(
)
force between the objects is now
(
)
k ⎡ q2 − q1 / 2 ⎤
⎦
F= ⎣
2
r
2
It is given that F is the same in Equations (1) and (2).
Equations (1) and (2) and rearrange the result to find that
2
(2)
Therefore, we can equate
2
q1 − 6 q2 q1 + q2 = 0
(3)
The solutions to this quadratic equation are
q1 = 5.828 q2
and
q1 = 0.1716 q2
(4)
Since we have assumed that q2 is greater than q1 , we must choose the solution
q1 = 0.1716 q2 . Substituting this result into Equation (1) and using the given values of
F = 1.20 N and r = 0.200 m, we find that
q1 = 0.957 μ C
and
q2 = 5.58 μ C
(5)
Note that we need not have assumed that q2 is greater than q1 . We could have assumed
that q2 is less than q1 . Had we done so, we would have found that
Chapter 18 Problems
q1 = 5.58 μ C
q2 = 0.957 μ C
and
967
(6)
Considering Equations (5) and (6) and remembering that q1 is the negative charge, we
conclude that the two possible solutions to this problem are
q1 = −0.957 μ C
and
q2 = +5.58 μ C
or
q1 = −5.58 μ C
and
q2 = +0.957 μ C
______________________________________________________________________________
29. SOLUTION Knowing the electric field at a spot allows us to calculate the force that acts on
a charge placed at that spot, without knowing the nature of the object producing the field.
This is possible because the electric field is defined as E = F/q0, according to Equation 18.2.
This equation can be solved directly for the force F, if the field E and charge q0 are known.
SOLUTION Using Equation 18.2, we find that the force has a magnitude of
(
)
F = E q0 = ( 260 000 N/C ) 7.0 × 10 –6 C = 1.8 N
If the charge were positive, the direction of the force would be due west, the same as the
direction of the field. But the charge is negative, so the force points in the opposite
direction or due east. Thus, the force on the charge is 1.8 N due east .
______________________________________________________________________________
30. REASONING
a. The magnitude of the electric field is obtained by dividing the magnitude of the force
(obtained from the meter) by the magnitude of the charge. Since the charge is positive, the
direction of the electric field is the same as the direction of the force.
b. As in part (a), the magnitude of the electric field is obtained by dividing the magnitude of
the force by the magnitude of the charge. Since the charge is negative, however, the
direction of the force (as indicated by the meter) is opposite to the direction of the electric
field. Thus, the direction of the electric field is opposite to that of the force.
SOLUTION
a. According to Equation 18.2, the magnitude of the electric field is
E=
F 40.0 μ N
=
= 2.0 N /C
q 20.0 μ C
As mentioned in the REASONING, the direction of the electric field is the same as the
direction of the force, or due east .
968
ELECTRIC FORCES AND ELECTRIC FIELDS
b. The magnitude of the electric field is
E=
F 20.0 μ N
=
= 2.0 N /C
q 10.0 μ C
Since the charge is negative, the direction of the electric field is opposite to the direction of
the force, or due east . Thus, the electric fields in parts (a) and (b) are the same.
______________________________________________________________________________
31. SSM WWW REASONING The electric field created by a point charge is inversely
proportional to the square of the distance from the charge, according to Equation 18.3.
Therefore, we expect the distance r2 to be greater than the distance r1, since the field is
smaller at r2 than at r1. The ratio r2/r1, then, should be greater than one.
SOLUTION Applying Equation 18.3 to each position relative to the charge, we have
E1 =
kq
and
r12
E2 =
kq
r22
Dividing the expression for E1 by the expression for E2 gives
E1
E2
=
k q / r12
k q / r22
=
r22
r12
Solving for the ratio r2/r1 gives
r2
r1
=
E1
E2
=
As expected, this ratio is greater than one.
248 N/C
= 1.37
132 N/C
Chapter 18 Problems
969
32. REASONING AND SOLUTION The
electric field lines must originate on the -4 q
+5 q
positive charges and terminate on the
negative charges. They cannot cross
one another. Furthermore, the number
of field lines beginning or ending on
any charge must be proportional to the
magnitude of the charge. If 10 electric
field lines leave the +5q charge, then
six lines must originate from the +3q
charge, and eight lines must end on
each –4q charge. The drawing shows
-4 q
the electric field lines that meet these +3 q
criteria.
______________________________________________________________________________
33. REASONING Each charge creates an electric field at the center of the square, and the four
fields must be added as vectors to obtain the net field. Since the charges all have the same
magnitude and since each corner is equidistant from the center of the square, the magnitudes
kq
of the four individual fields are identical. Each is given by Equation 18.3 as E = 2 . The
r
directions of the various contributions are not the same, however. The field created by a
positive charge points away from the charge, while the field created by a negative charge
points toward the charge.
SOLUTION The drawing at the right shows
each of the field contributions at the center of
the square (see black dot). Each is directed
along a diagonal of the square. Note that ED
and EB point in opposite directions and,
therefore, cancel, since they have the same
magnitude. In contrast EA and EC point in
the same direction toward corner A and,
therefore, combine to give a net field that is
twice the magnitude of EA or EC. In other
words, the net field at the center of the square
is given by the following vector equation:
+ C
B+
ED
EA
EC
EB
A -
Σ E = E A + E B + EC + E D = E A + E B + E C − E B = E A + E C = 2 E A
Using Equation 18.3, we find that the magnitude of the net field is
Σ E = 2 EA = 2
kq
r2
+ D
970
ELECTRIC FORCES AND ELECTRIC FIELDS
In this result r is the distance from a corner to the center of the square, which is one half of
the diagonal distance d. Using L for the length of a side of the square and taking advantage
of the Pythagorean theorem, we have r = 12 d =
L2 + L2 . With this substitution for r, the
1
2
magnitude of the net field becomes
ΣE = 2
(
kq
1
2
L2 + L2
)
2
=
4k q
L2
=
(
)(
4 8.99 × 109 N ⋅ m 2 / C 2 2.4 × 10−12 C
( 0.040 m )2
) = 54 N/C
34. REASONING
Part (a) of the drawing given in the text. The electric field produced by a charge points
away from a positive charge and toward a negative charge. Therefore, the electric field E+2
produced by the +2.0 μC charge points away from it, and the electric fields E−3 and E−5
produced by the −3.0 μC and −5.0 μC charges point toward them (see the left-hand side of
the following drawing). The magnitude of the electric field produced by a point charge is
given by Equation 18.3 as E = k q /r2. Since the distance from each charge to the origin is
the same, the magnitude of the electric field is proportional only to the magnitude q of the
charge. Thus, the x component Ex of the net electric field is proportional to
5.0 μC (2.0 μC + 3.0 μC). Since only one of the charges produces an electric field in the y
direction, the y component Ey of the net electric field is proportional to the magnitude of this
charge, or 5.0 μC. Thus, the x and y components are equal, as indicated at the right-hand
side of the following drawing, where the net electric field E is also shown.
−5.0 μ C
E−5
Ey
E+2
+2.0 μ C
O
E−3
E
−3.0 μ C
Ex
Part (b) of the drawing given in the text. Using the same arguments as earlier, we find
that the electric fields produced by the four charges are shown at the left-hand side of the
following drawing. These fields also produce the same net electric field E as before, as
indicated at the right-hand side of the following drawing.
Chapter 18 Problems
E+6
971
+1.0 μ C
E
Ey
E−1
+4.0 μ C
−1.0 μ C
Ex
E+4
E+1
+6.0 μ C
SOLUTION
Part (a) of the drawing given in the text. The net electric field in the x direction is
Ex
8.99 × 109 N ⋅ m 2 /C2 )( 2.0 × 10−6 C ) ( 8.99 × 109 N ⋅ m 2 /C2 )( 3.0 × 10−6 C )
(
=
+
( 0.061 m )2
( 0.061 m )2
= 1.2 × 107 N /C
The net electric field in the y direction is
Ey
8.99 × 109 N ⋅ m 2 /C2 )( 5.0 × 10−6 C )
(
=
= 1.2 × 107 N /C
( 0.061 m )2
The magnitude of the net electric field is
E = Ex2 + E y2 =
(1.2 × 107 N /C ) + (1.2 × 107 N /C)
2
2
= 1.7 × 107 N /C
Part (b) of the drawing given in the text. The magnitude of the net electric field is the
same as determined for part (a); E = 1.7 × 107 N/C .
______________________________________________________________________________
35. REASONING AND SOLUTION
a. In order for the field to be zero, the point cannot be between the two charges. Instead, it
must be located on the line between the two charges on the side of the positive charge and
away from the negative charge. If x is the distance from the positive charge to the point in
question, then the negative charge is at a distance (3.0 m + x) meters from this point. For
the field to be zero here we have
k q−
( 3.0 m + x )2
=
k q+
x2
or
q−
( 3.0 m + x )2
=
q+
x2
972
ELECTRIC FORCES AND ELECTRIC FIELDS
Solving for the ratio of the charge magnitudes gives
16.0 μ C ( 3.0 m + x )
=
=
4.0 μ C
q+
x2
q−
2
3.0 m + x )
(
4.0 =
2
or
x2
Suppressing the units for convenience and rearranging this result gives
4.0x 2 = ( 3.0 + x )
2
or
4.0x 2 = 9.0 + 6.0 x + x 2
or
3x 2 − 6.0 x − 9.0 = 0
Solving this quadratic equation for x with the aid of the quadratic formula (see Appendix
C.4) shows that
x = 3.0 m or x = -1.0 m
We choose the positive value for x, since the negative value would locate the zero-field spot
between the two charges, where it can not be (see above). Thus, we have x = 3.0 m .
b. Since the field is zero at this point, the force acting on a charge at that point is 0 N .
______________________________________________________________________________
36. REASONING
a. The magnitude E of the electric field is given by E = σ / ε 0 (Equation 18.4), where σ is
the charge density (or charge per unit area) and ε 0 is the permittivity of free space.
b. The magnitude F of the electric force that would be exerted on a K+ ion placed inside the
membrane is the product of the magnitude q0 of the charge and the magnitude E of the
electric field (see Equation 18.2), or F = q0 E .
SOLUTION
a. The magnitude of the electric field is
σ
7.1×10−6 C/m 2
E=
=
= 8.0 ×105 N/C
−
12
2
2
ε 0 8.85 ×10 C / N ⋅ m
(
)
b. The magnitude F of the force exerted on a K+ ion (q0 = +e) is
F = q0 E = e E = 1.60 × 10−19 C ( 8.0 × 105 N/C ) = 1.3 × 10−13 N
______________________________________________________________________________
Chapter 18 Problems
973
37. SSM REASONING
a. The drawing shows the two point charges q1 and q2. Point A is located at x = 0 cm, and
point B is at x = +6.0 cm.
A
E1
3.0 cm
3.0 cm
B
3.0 cm
q2
q1
E2
Since q1 is positive, the electric field points away from it. At point A, the electric field E1
points to the left, in the −x direction. Since q2 is negative, the electric field points toward it.
At point A, the electric field E2 points to the right, in the +x direction. The net electric field
is E = −E1 + E2. We can use Equation 18.3, E = k q / r 2 , to find the magnitude of the
electric field due to each point charge.
b. The drawing shows the electric fields produced by the charges q1 and q2 at point B,
which is located at x = +6.0 cm.
A
3.0 cm
3.0 cm
q1
B
3.0 cm
q2
E1
E2
Since q1 is positive, the electric field points away from it. At point B, the electric field
points to the right, in the +x direction. Since q2 is negative, the electric field points toward
it. At point B, the electric field points to the right, in the +x direction. The net electric field
is E = +E1 + E2.
SOLUTION
a. The net electric field at the origin (point A) is E = −E1 + E2:
E = − E1 + E2 =
=
(
−k q1
r12
+
k q2
r22
)(
2
(3.0 × 10−2 m )
− 8.99 × 109 N ⋅ m 2 /C2 8.5 × 10−6 C
= −6.2 × 107 N/C
) + (8.99 × 109 N ⋅ m2 /C2 )( 21 × 10−6 C )
2
(9.0 × 10−2 m )
974
ELECTRIC FORCES AND ELECTRIC FIELDS
The minus sign tells us that the net electric field points along the −x axis.
b. The net electric field at x = +6.0 cm (point B) is E = E1 + E2:
E = E1 + E2 =
k q1
r12
+
k q2
r22
8.99 × 109 N ⋅ m 2 /C2 )( 8.5 × 10−6 C ) ( 8.99 × 109 N ⋅ m 2 /C2 )( 21 × 10−6 C )
(
=
+
2
2
−2
(3.0 × 10 m )
(3.0 × 10−2 m )
= +2.9 × 108 N/C
The plus sign tells us that the net electric field points along the +x axis.
______________________________________________________________________________
38. REASONING At every position in space, the
net electric field E is the vector sum of the
external electric field Eext and the electric
field Epoint created by the point charge at the
origin: E = Eext + Epoint. The external electric
field is uniform, which means that it has the
same magnitude Eext = 4500 N/C at all
locations and that it always points in the
positive x direction (see the drawing). The
magnitude Epoint of the electric field due to
y
Eext
Epoint
Epoint
Epoint
Eext
the point charge at the origin is given by Epoint =
Eext
x
kq
(Equation 18.3), where r is the
r2
distance between the origin and the location where the electric field is to be evaluated, q is
the charge at the origin, and k = 8.99×109 N·m2/C2. All three locations given in the problem
are a distance r = 0.15 m from the origin, so we have that
Epoint =
kq
r2
8.99 × 109 N ⋅ m 2 /C 2 ) −8.0 × 10 −9 C
(
=
= 3200 N/C
( 0.15 m )2
The direction of the electric field Epoint varies from location to location, but because the
charge q is negative, Epoint is always directed towards the origin (see the drawing).
Chapter 18 Problems
975
SOLUTION
a. At x = −0.15 m, the electric field of the point charge and the external electric field both
point in the positive x direction (see the drawing), so the magnitude E of the net electric
field is the sum of the magnitudes of the individual electric fields:
E = Eext + Epoint = 4500 N/C + 3200 N/C = 7700 N/C
b. At x = +0.15 m, the electric field of the point charge is opposite the external electric field
(see the drawing). Therefore, the magnitude E of the net electric field is the difference
between the magnitudes of the individual electric fields:
E = Eext − Epoint = 4500 N/C − 3200 N/C = 1300 N/C
c. At y = +0.15 m, the electric fields Epoint and Eext are perpendicular (see the drawing).
This makes them, in effect, the x-component (Eext) and y-component (Epoint) of the net
electric field E. The magnitude E of the net electric field, then, is given by the Pythagorean
theorem (Equation 1.7):
2
2
E = Eext
+ Epoint
= ( 4500 N/C )2 + ( 3200 N/C )2 = 5500 N/C
39. SSM REASONING Since the charged droplet (charge = q) is suspended motionless in
the electric field E, the net force on the droplet must be zero. There are two forces that act
on the droplet, the force of gravity W = mg , and the electric force F = qE due to the electric
field. Since the net force on the droplet is zero, we conclude that mg = q E . We can use
this reasoning to determine the sign and the magnitude of the charge on the droplet.
SOLUTION
a. Since the net force on the droplet is zero, and the weight of magnitude W
points downward, the electric force of magnitude F = q E must point
upward. Since the electric field points upward, the excess charge on the
droplet must be positive in order for the force F to point upward.
F
mg
b. Using the expression mg = q E , we find that the magnitude of the excess charge on the
droplet is
mg (3.50 × 10 –9 kg)(9.80 m/s 2 )
q =
=
= 4.04 × 10 –12 C
E
8480 N/C
976
ELECTRIC FORCES AND ELECTRIC FIELDS
The charge on a proton is 1.60 × 10–19 C, so the excess number of protons is
⎞
( 4.04 ×10–12 C ) ⎛⎜⎝ 1.601 ×proton
⎟=
–19
10
C⎠
2.53 ×107 protons
______________________________________________________________________________
40. REASONING Since the proton and the electron have the same charge magnitude e, the
electric force that each experiences has the same magnitude. The directions are different,
however. The proton, being positive, experiences a force in the same direction as the electric
field (due east). The electron, being negative, experiences a force in the opposite direction
(due west).
Newton’s second law indicates that the direction of the acceleration is the same as the
direction of the net force, which, in this case, is the electric force. The proton’s acceleration
is in the same direction (due east) as the electric field. The electron’s acceleration is in the
opposite direction (due west) as the electric field.
Newton’s second law indicates that the magnitude of the acceleration is equal to the
magnitude of the electric force divided by the mass. Although the proton and electron
experience the same force magnitude, they have different masses. Thus, they have
accelerations of different magnitudes.
SOLUTION According to Newton’s second law, Equation 4.2, the acceleration a of an
object is equal to the net force divided by the object’s mass m. In this situation there is only
one force, the electric force F, so it is the net force. According to Equation 18.2, the
magnitude of the electric force is equal to the product of the magnitude of the charge and the
magnitude of the electric field, or F = q0 E. Thus, the magnitude of the acceleration can be
written as
a=
F q0 E
=
m
m
The magnitude of the acceleration of the proton is
a=
q0 E
m
1.60 × 10−19 C )( 8.0 × 104 N /C )
(
=
=
1.67 × 10
−27
kg
7.7 × 1012 m /s 2
The magnitude of the acceleration of the electron is
a=
q0 E
1.60 × 10−19 C )( 8.0 × 104 N /C )
(
=
=
−31
1.4 × 1016 m /s 2
m
9.11 × 10 kg
______________________________________________________________________________
977
Chapter 18 Problems
41. REASONING AND SOLUTION Figure 1 at the right shows
the configuration given in text Figure 18.20a. The electric field
at the center of the rectangle is the resultant of the electric fields
at the center due to each of the four charges. As discussed in
Conceptual Example 11, the magnitudes of the electric field at
the center due to each of the four charges are equal. However,
the fields produced by the charges in corners 1 and 3 are in
opposite directions. Since they have the same magnitudes, they
combine to give zero resultant.
- q
+q
1 1
2
4 4
3
+q
+q
The fields produced by the charges in corners 2 and 4 point in
Figure 1
the same direction (toward corner 2). Thus, EC = EC2 + EC4,
where EC is the magnitude of the electric field at the center of the rectangle, and EC2 and
EC4 are the magnitudes of the electric field at the center due to the charges in corners 2 and
4 respectively. Since both EC2 and EC4 have the same magnitude, we have EC = 2 EC2.
The distance r, from any of the charges to the center
of the rectangle, can be found using the Pythagorean
theorem (see Figure 2 at the right):
1 1
d
5.00 cm
d = (3.00 cm) 2 +(5.00 cm)2 = 5.83 cm
d
Therefore, r = = 2.92 cm = 2.92 × 10−2 m
2
2
θ
4 4
3
3.00 cm
Figure 2
The electric field at the center has a magnitude of
EC = 2 EC 2 =
2k q2
r2
=
2(8.99 × 109 N ⋅ m 2 /C2 )(8.60 × 10−12 C)
= 1.81 × 102 N/C
−2
2
(2.92 × 10 m)
Figure 3 at the right shows the configuration given in text
Figure 18.20b. All four charges contribute a non-zero
component to the electric field at the center of the rectangle.
As discussed in Conceptual Example 11, the contribution
from the charges in corners 2 and 4 point toward corner 2 and
the contribution from the charges in corners 1 and 3 point
toward corner 1.
Notice also, the magnitudes of E24 and E13 are equal, and,
from the first part of this problem, we know that
E24 = E13 = 1.81 × 102 N/C
- q
- q
1 1
2
E 13
E 24
C
4 4
+q
Figure 3
3
+q
978
ELECTRIC FORCES AND ELECTRIC FIELDS
The electric field at the center of the rectangle is the vector sum of E24 and E13. The x
components of E24 and E13 are equal in magnitude and opposite in direction; hence
(E13)x – (E24)x = 0
Therefore,
EC = ( E13 ) y + ( E24 ) y = 2( E13 ) y = 2( E13 ) sin θ
From Figure 2, we have that
sinθ =
and
5.00 cm 5.00 cm
=
= 0.858
d
5.83 cm
(
)
EC = 2 ( E13 ) sin θ = 2 1.81× 102 N/C ( 0.858 ) = 3.11× 102 N/C
______________________________________________________________________________
42. REASONING AND SOLUTION The magnitude of the force on q1 due to q2 is given by
Coulomb's law:
k q1 q2
F12 =
(1)
r12 2
The magnitude of the force on q1 due to the electric field of the capacitor is given by
⎛σ ⎞
F1C = q1 EC = q1 ⎜ ⎟
⎜ε ⎟
⎝ 0⎠
(2)
Equating the right hand sides of Equations (1) and (2) above gives
k q1 q2
r12 2
⎛σ ⎞
= q1 ⎜ ⎟
⎜ε ⎟
⎝ 0⎠
Solving for r12 gives
r12 =
ε 0 k q2
σ
[8.85 ×10−12 C 2 /(N ⋅ m 2 )](8.99 ×109 N ⋅ m 2 /C2 )(5.00 ×10− 6 C)
= 5.53 ×10 –2 m
−4
2
(1.30 ×10 C/m )
______________________________________________________________________________
=
Chapter 18 Problems
979
43. REASONING The magnitude E of the electric field is the magnitude F of the electric force
exerted on a small test charge divided by the magnitude of the charge: E = F/ q . According
to Newton’s second law, Equation 4.2, the net force acting on an object is equal to its mass
m times its acceleration a. Since there is only one force acting on the object, it is the net
force. Thus, the magnitude of the electric field can be written as
E=
F ma
=
q
q
The acceleration is related to the initial and final velocities, v0 and v, and the time t through
v − v0
Equation 2.4, as a =
. Substituting this expression for a into the one above for E gives
t
⎛ v − v0 ⎞
m⎜
t ⎟⎠ m ( v − v0 )
ma
⎝
E=
=
=
q
q
qt
SOLUTION The magnitude E of the electric field is
E=
m ( v − v0 )
qt
9.0 × 10−5 kg )( 2.0 × 103 m /s − 0 m /s )
(
=
=
( 7.5 × 10−6 C ) ( 0.96 s )
2.5 × 104 N /C
______________________________________________________________________________
44. REASONING The external electric field E exerts a force FE = qE
(Equation 18.2) on the sphere, where q = +6.6 μC is the net charge
FE
of the sphere. The external electric field E is directed upward, so
the force FE it exerts on the positively charged sphere is also
q = +6.6 μC
directed upward (see the free-body diagram). Balancing this
upward force are two downward forces: the weight mg of the
sphere (where m is the mass of the sphere and g is the acceleration mg
Fs
due to gravity) and the force Fs exerted on the sphere by the spring
(see the free-body diagram). We know that the spring exerts a
downward force on the sphere because the equilibrium length
Free-body diagram
L = 0.059 m of the spring is shorter than its unstrained length
L0 = 0.074 m. The magnitude Fs of the spring force is given by Fs = kx (Equation 10.2,
without the minus sign), where k is the spring constant of the spring and
x = L0 − L = 0.074 m – 0.059 m = 0.015 m is the distance by which the spring has been
compressed.
980
ELECTRIC FORCES AND ELECTRIC FIELDS
SOLUTION Solving FE = qE (Equation 18.2) for the magnitude E of the external electric
field, we find that
F
E= E
(1)
q
The sphere is in equilibrium, so the upward force (FE) must exactly balance the two
downward forces (mg and Fs). Therefore, the magnitudes of the three forces are related by
FE = mg + Fs
(2)
Substituting Equation (2) into Equation (1) yields
E=
mg + Fs
(3)
q
Lastly, substituting Fs = kx (Equation 10.2, without the minus sign) into Equation (3), we
obtain the desired electric field magnitude:
(
)(
)
5.1× 10 −3 kg 9.80 m/s 2 + ( 2.4 N/m )( 0.015 m )
mg + kx
=
= 1.3 × 10 4 N/C
E=
−
6
q
6.6 × 10 C
E2
The
45. SSM WWW REASONING
drawing shows the arrangement of the
three charges. Let E q represent the
electric field at the empty corner due to
the –q charge. Furthermore, let E1 and E2
be the electric fields at the empty corner
due to charges +q1 and +q2, respectively.
+q1
d
Eq
d 5
–q
E1
θ
+q2
2d
According to the Pythagorean theorem, the distance from the charge –q to the empty corner
along the diagonal is given by (2 d ) 2 + d 2 = 5d 2 = d 5 . The magnitude of each
electric field is given by Equation 18.3, E = k q / r 2 . Thus, the magnitudes of each of the
electric fields at the empty corner are given as follows:
Eq =
kq
r2
=
kq
(d 5 )
2
=
kq
5d 2
Chapter 18 Problems
E1 =
k q1
( 2d ) 2
=
k q1
and
4d 2
E2 =
981
k q2
d2
The angle q that the diagonal makes with the horizontal is θ = tan −1 (d / 2d ) = 26.57° . Since
the net electric field Enet at the empty corner is zero, the horizontal component of the net
field must be zero, and we have
E1 – Eq cos 26.57° = 0
k q1
or
4d
2
–
k q cos 26.57°
5d 2
=0
Similarly, the vertical component of the net field must be zero, and we have
E2 – Eq sin 26.57° = 0
or
k q2
d2
–
k q sin 26.57°
5d 2
=0
These last two expressions can be solved for the charge magnitudes q1 and q2 .
SOLUTION Solving the last two expressions for q1 and q2 , we find that
4
5
q1 = q cos 26.57° = 0.716 q
1
5
q2 = q sin 26.57° = 0.0895 q
______________________________________________________________________________
46. REASONING AND SOLUTION From kinematics, vy2 = v0y2 + 2ayy. Since the electron
starts from rest, v0y = 0 m/s. The acceleration of the electron is given by
ay =
F eE
=
m m
where e and m are the electron's charge magnitude and mass, respectively, and E is the
magnitude of the electric field. The magnitude of the electric field between the plates of a
parallel plate capacitor is E = s/ε0, where s is the magnitude of the charge per unit area on
each plate. Thus, ay = es/(mε0). Combining this expression for a with the kinematics
equation we have
⎛ eσ ⎞
v 2y = 2 ⎜
y
⎜ mε ⎟⎟
⎝ 0⎠
982
ELECTRIC FORCES AND ELECTRIC FIELDS
Solving for vy gives
(
)(
)
)(
)
2 1.60 ×10−19 C 1.8 × 10−7 C/m 2 1.5 ×10−2 m
2eσ y
=
= 1.0 × 107 m/s
vy =
2 ⎤
−31
−12 2
mε 0
⎡
9.11× 10
kg 8.85 ×10
C / N⋅m
⎣
⎦
______________________________________________________________________________
(
47. REASONING The fact that the net electric
field points upward along the vertical axis
holds the key to this problem. The drawing
at the right shows the fields from each
charge, together with the horizontal
components of each. The reason that the
net field points upward is that these
horizontal components point in opposite
directions and cancel. Since they cancel,
they must have equal magnitudes, a fact
that will quickly lead us to a solution.
(
)
60.0º
30.0º
E1
E2
E2 sin 60.0º
E1 sin 30.0º
30.0º
60.0º
q2
q1
SOLUTION Setting the magnitudes of the
horizontal components of the fields equal gives
E2 sin 60.0° = E1 sin 30.0°
The magnitude of the electric field created by a point charge is given by Equation 18.3.
Using this expression for E1 and E2 and noting that each point charge is the same distance r
from the center of the circle, we obtain
k q2
r2
sin 60.0° =
k q1
r2
sin 30.0°
or
q2 sin 60.0° = q1 sin 30.0°
Solving for the ratio of the charge magnitudes gives
q2
q1
=
sin 30.0°
= 0.577
sin 60.0°
48. REASONING
a. The drawing at the right shows the
electric fields at point P due to the two
charges in the case that the second
charge is positive. The presence of the
q2
+
+q1
d
P
d
E2
E1
Chapter 18 Problems
983
second charge causes the magnitude of the net field at P to be twice as great as it is when
only the first charge is present. Since both fields have the same direction, the magnitude of
E2 must, then, be the same as the magnitude of E1. But the second charge is further away
from point P than is the first charge, and more distant charges create weaker fields. To offset
the weakness that comes from the greater distance, the second charge must have a greater
magnitude than that of the first charge.
b. The drawing at the right shows the
q2
+q1
E2
P E1
electric fields at point P due to the two
d
d
charges in the case that the second
charge is negative. The presence of the
second charge causes the magnitude of the net field at P to be twice as great as it is when
only the first charge is present. Since the fields now have opposite directions, the magnitude
of E2 must be greater than the magnitude of E1. This is necessary so that E2 can offset E1
and still lead to a net field with twice the magnitude as E1. To create this greater field E2, the
second charge must now have a greater magnitude than it did in question (a).
SOLUTION
a. The magnitudes of the field contributions of each charge are given according to
kq
Equation 18.3 as E = 2 . With q2 present, the magnitude of the net field at P is twice what
r
it is when only q1 is present. Using Equation 18.3, we can express this fact as follows:
k q1
d2
+
k q2
( 2d ) 2
=2
k q1
d2
or
k q2
( 2d )2
=
k q1
d2
Solving for q2 gives
q2 = 4 q1 = 4 ( 0.50 μ C ) = 2.0 μ C
Thus, the second charge is q2 = +2.0 μC .
b. Now that the second charge is negative, we have
k q2
( 2d ) 2
−
k q1
d2
=2
k q1
d2
or
k q2
( 2d ) 2
=3
Solving for q2 gives
q2 = 12 q1 = 12 ( 0.50 μ C ) = 6.0 μ C
Thus, the second charge is q2 = -6.0 μC .
k q1
d2
984
ELECTRIC FORCES AND ELECTRIC FIELDS
49. REASONING Since we know the initial velocity and time, we can determine the particle’s
displacement from an equation of kinematics, provided its acceleration can be determined.
The acceleration is given by Newton’s second law as the net force acting on the particle
divided by its mass. The net force is the electrostatic force, since the particle is moving in an
electric field. The electrostatic force depends on the particle’s charge and the electric field,
both of which are known.
SOLUTION To obtain the displacement x of the particle we employ Equation 3.5a from the
equations of kinematics: x = v0 x t + 12 a x t 2 . We use this equation because two of the
variables, the initial velocity v0x and the time t, are known. The initial velocity is zero, since
the particle is released from rest. The acceleration ax can be found from Newton’s second
law, as given by Equation 4.2a, as the net force ΣFx acting on the particle divided by its
mass m: ax = ΣFx / m . Only the electrostatic force Fx acts on the proton, so it is the net
force. Setting ΣFx = Fx in Newton’s second law gives ax = Fx / m . Substituting this result
into Equation 3.5a, we have that
⎛F ⎞
x = v0 x t + 12 axt 2 = v0 x t + 12 ⎜ x ⎟ t 2
(1)
⎝m⎠
Since the particle is moving in a uniform electric field Ex, it experiences an electrostatic
force Fx given by Fx = q0 Ex (Equation 18.2), where q0 is the charge. Substituting this
expression for Fx into Equation (1) gives
⎛F ⎞
⎛q E ⎞
x = v0 xt + 12 ⎜ x ⎟ t 2 = v0 x t + 12 ⎜ 0 x ⎟ t 2
⎝m⎠
⎝ m ⎠
⎡ ( +12 × 10−6 C ) ( +480 N/C ) ⎤
−2 ) 2
(
= ( 0 m/s ) (1.6 × 10−2 s ) + 12 ⎢
×
1.6
10
s = +1.9 × 10−2 m
⎥
−5
3.8
10
kg
×
⎣⎢
⎦⎥
______________________________________________________________________________
50. REASONING The following drawing shows the two particles in the electric field Ex. They
are separated by a distance d. If the particles are to stay at the same distance from each other
after being released, they must have the same acceleration, so ax, 1 = ax, 2. According to
Newton’s second law (Equation 4.2a), the acceleration ax of each particle is equal to the net
force ΣFx acting on it divided by its mass m, or ax = ΣFx / m .
Chapter 18 Problems
d
Ex
q1 = −7.0 μ C
985
+x
q2 = +18 μ C
m1 = 1.4 × 10−5 kg
m2 = 2.6 × 10−5 kg
SOLUTION The net force acting on each particle and its resulting acceleration are:
q1: The charge q1 experiences a force q1Ex due to the electric field (see Equation 18.2).
The charge also experiences an attractive force in the +x direction due to the presence
of q2. This force is given by Coulomb’s law as + k q1 q2 / d 2 (see Equation 18.1). The
net force acting on q1 is
ΣFx, 1 = q1Ex + k
q1 q2
d2
The acceleration of q1 is
a x,1 =
ΣFx, 1
m1
=
q1Ex + k
q1 q2
d2
m1
q2: The charge q2 experiences a force q2 E x due to the electric field. It also experiences an
attractive force in the −x direction due to the presence of q1. This force is given by
Coulomb’s law as − k q1 q2 / d 2 . The net force acting on q2 is
ΣFx, 2 = q2 Ex − k
q1 q2
d2
The acceleration of q2 is
ax, 2 =
ΣFx, 2
m2
=
q2 E x − k
q1 q2
d2
m2
Setting ax,1 = ax,2 gives
q1Ex + k
m1
q1 q2
d
2
=
q2 Ex − k
m2
q1 q2
d2
986
ELECTRIC FORCES AND ELECTRIC FIELDS
Solving this expression for d, we find that
1 ⎞
⎛ 1
k q1 q2 ⎜ +
⎟
⎝ m1 m2 ⎠
d=
q ⎞
⎛q
Ex ⎜ 2 − 1 ⎟
⎝ m2 m1 ⎠
2
⎛
1
1
⎛
⎞
9 N⋅m ⎞
−7.0 × 10−6 C +18 × 10−6 C ⎜
+
⎜ 8.99 ×10
−5
−5
2 ⎟
C ⎠
1.4 ×10 kg 2.6 ×10 kg ⎟⎠
⎝
⎝
=
= 6.5 m
⎛ +18 × 10−6 C −7.0 ×10−6 C ⎞
( +2500 N/C ) ⎜
−
⎟
−5
×
2.6
10
kg
1.4 × 10−5 kg ⎠
⎝
______________________________________________________________________________
51. SSM REASONING AND SOLUTION The net electric field at point P in Figure 1 is the
vector sum of the fields E+ and E–, which are due, respectively, to the charges +q and –q.
These fields are shown in Figure 2.
E+
P
P α
α
+q
a
E–
l
l
a
M
2d
Figure 1
–q
Figure 2
According to Equation 18.3, the magnitudes of the fields E+ and E– are the same, since the
triangle is an isosceles triangle with equal sides of length A. Therefore, E+ = E– = k q / A 2 .
The vertical components of these two fields cancel, while the horizontal components
reinforce, leading to a total field at point P that is horizontal and has a magnitude of
⎛k q ⎞
EP = E+ cos α +E – cos α = 2 ⎜ 2 ⎟ cos α
⎝ A ⎠
At point M in Figure 1, both E+ and E– are horizontal and point to the right. Again using
Equation 18.3, we find
k q k q 2k q
EM = E+ +E – = 2 + 2 = 2
d
d
d
Chapter 18 Problems
987
Since EM/EP = 9.0, we have
EM
EP
=
2k q / d 2
2k q ( cos α ) / A 2
=
1
= 9.0
( cos α ) d 2 / A 2
But from Figure 1, we can see that d/A = cos α. Thus, it follows that
1
= 9.0
cos3α
cos α = 3 1/ 9.0 = 0.48
or
The value for α is, then, α = cos –1 ( 0.48 ) = 61° .
______________________________________________________________________________
52. REASONING The net charge q carried by the particle determines the magnitude F of the
electrical force that the external electric field of magnitude E exerts on the particle, via
F = qE (Equation 18.2). The electric field is horizontal, so the electric force acting on the
particle is also horizontal. Taking east as the positive x direction, the electric force gives the
particle a horizontal component of acceleration: ax. Newton’s second law holds that the
particle’s horizontal acceleration is ax =
F
(Equation 4.2a), where m is the mass of the
m
particle. Combining Equation 18.2 and Equation 4.2a, we obtain
ax =
F qE
=
m m
(1)
We will use Equation (1) to determine q. Note that the algebraic sign of q is the same as
that of ax, which we will obtain by using kinematics. The particle’s horizontal acceleration
is related to its horizontal displacement x by x = v0 xt + 12 axt 2 (Equation 3.5a), where v0x is
the horizontal component of its initial velocity v0 and t is the elapsed time. The particle’s
initial velocity is entirely horizontal, so that v0x = v0, and Equation 3.5a becomes
x = v0t + 12 axt 2
(2)
In order to determine the elapsed time t, we consider the vertical component of the particle’s
projectile motion. Because the particle is launched horizontally, the vertical component v0y
of its initial velocity is zero, and the vertical component ay of its acceleration is that due to
gravity: ay = −9.80 m/s2, where we have taken upward as the positive y direction.
Therefore, from y = v0 yt + 12 a yt 2 (Equation 3.5b), we have that
988
ELECTRIC FORCES AND ELECTRIC FIELDS
y = ( 0 m/s ) t + 12 ayt 2 = 12 ayt 2
(3)
where y = −6.71×10−3 m is the particle’s vertical displacement. The value of y is negative
because the particle moves downward.
SOLUTION Solving Equation (1) for q, we obtain
q=
max
(4)
E
Solving Equation (2) for ax yields
1 a t2
2 x
= x − v0t
ax =
or
2 ( x − v0t )
(5)
t2
Substituting Equation (5) into Equation (4), we find that
q=
2m ( x − v0t )
(6)
Et 2
Solving Equation (3) for t yields
2y
t =
ay
2
or
(
)
2 −6.71× 10−3 m
2y
t=
=
= 0.0370 s
ay
−9.80 m/s 2
Therefore, from Equation (6), the charge carried by the particle is
q=
2 m ( x − v0t )
Et
2
=
(
)
2 1.50 × 10 −6 kg ⎡⎣ 0.160 m − ( 8.80 m/s )( 0.0370 s ) ⎤⎦
( 925 N/C )( 0.0370 s )
2
= −3.92 × 10 −7 C
53. REASONING AND SOLUTION Since the thread makes an angle of 30.0° with the
vertical, it can be seen that the electric force on the ball, Fe, and the gravitational force, mg,
are related by
Fe = mg tan 30.0°
The force Fe is due to the charged ball being in the electric field of the parallel plate
capacitor. That is,
Chapter 18 Problems
Fe = E qball
989
(1)
where qball is the magnitude of the ball's charge and E is the magnitude of the field due to
the plates. According to Equation 18.4 E is
E=
q
ε0 A
(18.4)
where q is the magnitude of the charge on each plate and A is the area of each plate.
Substituting Equation 18.4 into Equation (1) gives
q qball
Fe = mg tan 30.0° =
ε0 A
Solving for q yields
q=
ε 0 Amg tan 30.0°
qball
(
)(
)(
)(
)
⎡8.85 × 10 –12 C2 / N ⋅ m 2 ⎤ 0.0150 m 2 6.50 × 10 –3 kg 9.80 m/s 2 tan 30.0°
⎦
=⎣
0.150 × 10 –6 C
= 3.25 × 10 –8 C
______________________________________________________________________________
54. REASONING AND SOLUTION Gauss' Law is given by text Equation 18.7: ΦE =
Q
ε0
,
where Q is the net charge enclosed by the Gaussian surface.
a. Φ E =
b. Φ E
3.5 × 10 – 6 C
= 4.0 × 10 5 N ⋅ m 2 /C
8.85 × 10 −12 C 2 /(N ⋅ m 2 )
−2.3 × 10 – 6 C
=
= –2.6 × 10 5 N ⋅ m 2 /C
−12
2
2
C /(N ⋅ m )
8.85 × 10
(3.5 × 10 – 6 C) + (−2.3 × 10 – 6 C)
= 1.4 × 10 5 N ⋅ m 2 /C
c. Φ E =
−12
2
2
C /(N ⋅ m )
8.85 × 10
______________________________________________________________________________
990
ELECTRIC FORCES AND ELECTRIC FIELDS
55. SSM REASONING As discussed in Section 18.9, the electric flux Φ E through a surface
is equal to the component of the electric field that is normal to the surface multiplied by the
area of the surface, Φ E = E ⊥ A , where E⊥ is the component of E that is normal to the
surface of area A. We can use this expression and the figure in the text to determine the flux
through the two surfaces.
SOLUTION
a. The flux through surface 1 is
(Φ E )1 = ( E cos 35° ) A1 = (250 N/C)(cos 35° )(1.7 m 2 ) =
350 N ⋅ m 2 /C
b. Similarly, the flux through surface 2 is
(Φ E )2 = ( E cos 55° ) A2 = (250 N/C)(cos 55° )(3.2 m2 ) =
460 N ⋅ m 2 /C
______________________________________________________________________________
56. REASONING The electric flux ΦE through the circular surface is determined by the angle
φ between the electric field and the normal to the surface, as well as the magnitude E of the
electric field and the area A of the surface:
Φ E = ( E cos φ ) A
(18.6)
We will use Equation 18.6 to determine the angle φ.
SOLUTION Solving Equation 18.6 for the angle φ, we obtain
cos φ =
ΦE
EA
⎛ ΦE ⎞
⎟
⎝ EA ⎠
φ = cos −1 ⎜
or
(1)
The surface is circular, with a radius r, so its area is A = π r 2 . Making this substitution in
Equation (1) yields
⎡
⎤
78 N ⋅ m 2 /C
⎞
−1 ⎛ Φ E ⎞
−1 ⎢
D
φ = cos ⎜
cos
cos
=
=
2 ⎥ = 58
⎜
⎟
4
2⎟
π
1.44
10
N/C
0.057
m
×
π
E
r
(
) ⎥⎦
⎝ EA ⎠
⎢⎣
⎝
⎠
−1 ⎛ Φ E
(
)
57. REASONING AND SOLUTION Since the electric field is uniform, its magnitude and
direction are the same at each point on the wall. The angle φ between the electric field and
the normal to the wall is 35°. Therefore, the electric flux is
ΦE = (E cos φ) A = (150 N/C)(cos 35°)[(5.9 m)(2.5 m)] =
1.8 × 10 3 N ⋅ m 2 /C
Chapter 18 Problems
991
______________________________________________________________________________
58. REASONING The charge Q inside the rectangular box is related to the electric flux ΦE that
passes through the surfaces of the box by Gauss’ law, Q = ε 0Φ E (Equation 18.7), where ε0
is the permittivity of free space. The electric flux is the algebraic sum of the flux through
each of the six surfaces.
SOLUTION The charge inside the box is
Q = ε 0 Φ E = ε 0 ( Φ1 + Φ 2 + Φ 3 + Φ 4 + Φ 5 + Φ 6 )
⎡
C2 ⎤ ⎛
N ⋅ m2
N ⋅ m2
N ⋅ m2
+1500
2200
4600
= ⎢8.85 × 10−12
+
+
⎜
⎥
C
C
C
N ⋅ m2 ⎦ ⎝
⎣
− 1800
N ⋅ m2
N ⋅ m2
N ⋅ m2 ⎞
− 3500
− 5400
⎟
C
C
C ⎠
= −2.1 × 10−8 C
______________________________________________________________________________
59. SSM REASONING The electric flux through each
y
face of the cube is given by Φ E = ( E cos φ ) A (see
Section 18.9) where E is the magnitude of the electric
field at the face, A is the area of the face, and φ is the
angle between the electric field and the outward
normal of that face. We can use this expression to
calculate the electric flux Φ E through each of the six
E
x
z
faces of the cube.
SOLUTION
a. On the bottom face of the cube, the outward normal points parallel to the –y axis, in the
opposite direction to the electric field, and φ = 180°. Therefore,
( Φ E )bottom = (1500 N/C )( cos 180° )( 0.20 m )2 =
−6.0 × 101 N ⋅ m 2 /C
On the top face of the cube, the outward normal points parallel to the +y axis, and φ = 0.0°.
The electric flux is, therefore,
(Φ E ) top = (1500 N/C)(cos 0.0°)(0.20 m)2 = +6.0 × 101N.m 2 /C
992
ELECTRIC FORCES AND ELECTRIC FIELDS
On each of the other four faces, the outward normals are perpendicular to the direction of
the electric field, so φ = 90°. So for each of the four side faces,
(Φ E )sides = (1500 N/C)(cos 90°)(0.20 m) 2 = 0 N ⋅ m 2 / C
b. The total flux through the cube is
(Φ E )
total
= (Φ E ) top + (Φ E ) bottom + (Φ E )side 1 + (Φ E )side 2 + (Φ E )side 3 + (Φ E )side 4
Therefore,
(Φ E ) total = ( + 6.0 × 101N.m 2 /C) + (–6.0 × 101N.m 2 /C) + 0 + 0 + 0 + 0 = 0 N ⋅ m 2 / C
______________________________________________________________________________
60. REASONING Gauss’ Law, Σ ( E cos φ ) ΔA =
Q
(Equation 18.7), relates the electric field
ε0
magnitude E on a Gaussian surface to the net charge Q enclosed by that surface;
Φ E = Σ ( E cos φ ) ΔA (Equation 18.6) is the electric flux through the Gaussian surface
(divided into many tiny sections of area ΔA) and ε0 is the permittivity of free space. We are
to determine the magnitude E of the electric field due to electric charges that are spread
uniformly over the surfaces of two concentric spherical shells. The electric field due to these
charges possesses spherical symmetry, so we will choose Gaussian surfaces in the shape of
spheres concentric with the shells. The radius r of each Gaussian surface will be equal to the
distance from the common center of the shells and will be the distance at which we are to
evaluate the electric field.
Because the electric field has spherical symmetry, the magnitude E of the electric field is
constant at all points on any such spherical Gaussian surface. Furthermore, the electric field
is directed either radially outward (if the net charge within the Gaussian surface is positive)
or radially inward (if the net charge within the Gaussian surface is negative). This means
that the angle φ between the electric field and the normal to any such spherical Gaussian
surface is either 0.0° or 180°. The quantity E cos φ , therefore, is constant, and may be
factored out of the summation in Equation 18.6:
Φ E = Σ ( E cos φ ) ΔA = ( E cos φ ) ΣΔA
(1)
The sum ΣΔA of all the tiny sections of area ΔA that compose a spherical Gaussian surface
is the total surface area ΣΔA = A = 4π r 2 of a sphere of radius r. Thus, Equation (1) becomes
Φ E = ( E cos φ ) ΣΔA = 4π r 2 E cos φ
(2)
Chapter 18 Problems
993
SOLUTION
a. The outer shell has a radius r2 = 0.15 m, and we are to determine the electric field at a
distance r = 0.20 m from the common center of the shells. Therefore, we will choose a
spherical Gaussian surface (radius r = 0.20 m) that encloses both shells and shares their
common center. According to Gauss’ law and Equation (2), we have that the net electric
flux ΦE through this sphere is
Σ ( E cos φ ) ΔA = 4π r 2 E cos φ =
Q
(3)
ε0
Solving Equation (3) for E yields
E=
Q
4πε 0 r 2 cos φ
(4)
Because the chosen Gaussian surface encloses both shells, the net charge Q enclosed by the
surface is Q = q1 + q2. The positive charge q2 on the outer shell has a larger magnitude than
the negative charge q1 on the inner shell, so that Q is a positive net charge. Therefore, the
electric field is directed radially outward , and the angle between the electric field and the
normal to the surface of the spherical Gaussian surface is φ = 0.0°. Therefore, Equation (4)
gives the electric field magnitude as
E=
q1 + q2
Q
−1.6 × 10 −6 C + 5.1× 10 −6 C
=
=
4πε 0 r 2 cos φ 4πε 0 r 2 cos φ 4π ⎡8.85 × 10 −12 C 2 / N ⋅ m 2 ⎤ ( 0.20 m )2 cos 0.0D
⎣
⎦
(
)
= 7.9 × 105 N/C
b. We again choose a spherical Gaussian surface concentric with the shells, this time of
radius r = 0.10 m. The radius of this sphere is greater than the radius (r1 = 0.050 m) of the
inner shell but less than the radius (r2 = 0.15 m) of the outer shell. Therefore, this Gaussian
surface is located between the two shells and encloses only the charge on the inner shell:
Q = q1. This is a negative charge, so that the electric field is directed radially inward , and
the angle between the electric field and the normal to the surface of the Gaussian sphere is
φ = 180°. From Equation (4), then, we have that
E=
q1
Q
−1.6 × 10 −6 C
=
=
4πε 0 r 2 cos φ 4πε 0 r 2 cos φ 4π ⎡8.85 × 10 −12 C 2 / N ⋅ m 2 ⎤ ( 0.10 m )2 cos180D
⎣
⎦
= 1.4 × 106 N/C
(
)
994
ELECTRIC FORCES AND ELECTRIC FIELDS
c. Choosing a spherical Gaussian surface with a radius of r = 0.025 m, we see that it is
entirely inside the inner shell (r1 = 0.050 m). Therefore, the enclosed charge is zero:
Q = 0 C. Equation (4) shows that the electric field at this distance from the common center
is zero:
E=
Q
0C
=
= 0 N/C
2
4πε 0 r cos φ 4πε 0 r 2 cos φ
61. REASONING We use a Gaussian surface that is a sphere centered within the solid sphere.
The radius r of this surface is smaller than the radius R of the solid sphere. Equation 18.7
gives Gauss’ law as follows:
Q
Σ ( E cos φ ) Δ A =
(18.7)
ε0
Electric flux, Φ E
We will deal first with the left side of this equation and evaluate the electric flux ФE. Then
we will evaluate the net charge Q within the Gaussian surface.
SOLUTION The positive charge is
Normal
spread uniformly throughout the solid
sphere and, therefore, is spherically Angle φ between E
E
symmetric. Consequently, the electric and the normal is 0º
field is directed radially outward, and for
each element of area D A is perpendicular
to the surface. This means that the
angle φ between the normal to the surface and the field is 0º, as the drawing shows.
Furthermore, the electric field has the same magnitude everywhere on the Gaussian surface.
Because of these considerations, we can write the electric flux as follows:
Σ ( E cos φ ) Δ A = Σ ( E cos 0° ) Δ A = E ( ΣΔ A)
The term ΣD A is the entire area of the spherical Gaussian surface or 4πr2. With this
substitution, the electric flux becomes
Σ ( E cos φ ) Δ A = E ( ΣΔ A) = E 4π r 2
(1)
(
)
Now consider the net charge Q within the Gaussian surface. This charge is the charge
density times the volume 43 π r 3 encompassed by that surface:
Chapter 18 Problems
Q=
q
qr 3
3
4
r
× 3π
= 3
4 π R3
R
3
(
)
995
(2)
Volume of
Gaussian
surface
Charge
density
Substituting Equations (1) and (2) into Equation 18.7 gives
(
)
E 4π r 2 =
qr 3 / R 3
ε0
Rearranging this result shows that
E=
qr 3 / R3
( 4π r ) ε 0
2
=
qr
4πε 0 R3
62. REASONING Because the charge is distributed uniformly along the straight wire, the
electric field is directed radially outward, as the following end view of the wire illustrates.
Long straight wire
Gaussian cylinder
E
Gausssian
cylinder
r
+++ + +
1
+ ++
2
3
L
End view
And because of symmetry, the magnitude of the electric field is the same at all points
equidistant from the wire. In this situation we will use a Gaussian surface that is a cylinder
concentric with the wire. The drawing shows that this cylinder is composed of three parts,
the two flat ends (1 and 3) and the curved wall (2). We will evaluate the electric flux for
this three-part surface and then set it equal to Q/ε0 (Gauss’ law) to find the magnitude of the
electric field.
SOLUTION Surfaces 1 and 3 – the flat ends of the cylinder – are parallel to the electric
field, so cos φ = cos 90° = 0. Thus, there is no flux through these two surfaces:
Φ1 = Φ3 = 0 N ⋅ m 2 /C .
Surface 2 – the curved wall – is everywhere perpendicular to the electric field E, so
cos φ = cos 0° = 1. Furthermore, the magnitude E of the electric field is the same for all
points on this surface, so it can be factored outside the summation in Equation 18.6:
996
ELECTRIC FORCES AND ELECTRIC FIELDS
Φ 2 = Σ (E cos 0° )Δ A = E Σ A
The area ΣA of this surface is just the circumference 2π r of the cylinder times its length L:
ΣA = (2π r)L. The electric flux through the entire cylinder is, then,
Φ E = Φ 1 + Φ 2 + Φ 3 = 0 + E (2 π rL ) + 0 = E (2 π rL )
Following Gauss’ law, we set ΦE equal to Q/ε0, where Q is the net charge inside the
Gaussian cylinder: E(2π rL) = Q/ε0. The ratio Q/L is the charge per unit length of the wire
and is known as the linear charge density λ: λ = Q/L. Solving for E, we find that
E=
Q/L
λ
=
2π ε 0 r 2π ε 0 r
______________________________________________________________________________
63. REASONING AND SOLUTION The electric
field lines must originate on the positive charges
and terminate on the negative charge. They
-3q
cannot cross one another. Furthermore, the
number of field lines beginning or terminating on
any charge must be proportional to the magnitude
of the charge. Thus, for every field line that
leaves the charge +q, two field lines must leave
the charge +2q. These three lines must terminate
+q
+2q
on the -3q charge. If the sketch is to have six
field lines, two of them must originate on +q, and
four of them must originate on the charge +2q.
______________________________________________________________________________
64. REASONING
a. The magnitude of the electrostatic force that acts on each sphere is given by Coulomb’s
law as F = k q1 q2 / r 2 , where q1 and q2 are the magnitudes of the charges, and r is the
distance between the centers of the spheres.
b. When the spheres are brought into contact, the net charge after contact and separation
must be equal to the net charge before contact. Since the spheres are identical, the charge
on each after being separated is one-half the net charge. Coulomb’s law can be applied
again to determine the magnitude of the electrostatic force that each sphere experiences.
Chapter 18 Problems
997
SOLUTION
a. The magnitude of the force that each sphere experiences is given by Coulomb’s law as:
F=
k q1 q2
r
2
8.99 × 109 N ⋅ m 2 /C2 )( 20.0 × 10−6 C )( 50.0 × 10−6 C )
(
=
=
2
−2
2.50
×
10
m
(
)
1.44 × 104 N
Because the charges have opposite signs, the force is attractive .
b. The net charge on the spheres is −20.0 μ C + 50.0 μ C = +30.0 μ C. When the spheres are
brought into contact, the net charge after contact and separation must be equal to the net
charge before contact, or +30.0 μ C. Since the spheres are identical, the charge on each after
being separated is one-half the net charge, so q1 = q2 = + 15.0 μ C . The electrostatic force
that acts on each sphere is now
F=
k q1 q2
r
2
8.99 × 109 N ⋅ m 2 /C2 )(15.0 × 10−6 C )(15.0 × 10−6 C )
(
=
=
2
−2
( 2.50 × 10 m )
3.24 × 103 N
Since the charges now have the same signs, the force is repulsive .
______________________________________________________________________________
65. REASONING AND SOLUTION The +2q of charge initially on the sphere lies entirely on
the outer surface. When the +q charge is placed inside of the sphere, then a -q charge will
still be induced on the interior of the sphere. An additional +q will appear on the outer
surface, giving a net charge of +3q .
______________________________________________________________________________
66. REASONING Two forces act on the charged ball (charge q); they are the downward force
of gravity mg and the electric force F due to the presence of the charge q in the electric field
E. In order for the ball to float, these two forces must be equal in magnitude and opposite in
direction, so that the net force on the ball is zero (Newton's second law). Therefore, F must
point upward, which we will take as the positive direction. According to Equation 18.2,
F = qE. Since the charge q is negative, the electric field E must point downward, as the
product qE in the expression F = qE must be positive, since the force F points upward. The
magnitudes of the two forces must be equal, so that mg = q E . This expression can be
solved for E.
SOLUTION The magnitude of the electric field E is
E=
mg (0.012 kg)(9.80 m/s 2 )
=
= 6.5 × 103 N/C
q
18 × 10 –6 C
As discussed in the reasoning, this electric field points downward .
______________________________________________________________________________
998
ELECTRIC FORCES AND ELECTRIC FIELDS
+z
67. SSM REASONING The drawing at the right shows
the set-up. Here, the electric field E points along the +y
–F
–q
axis and applies a force of +F to the +q charge and a
force of –F to the –q charge, where q = 8.0 μC denotes
+y
the magnitude of each charge. Each force has the same
magnitude of F = E q , according to Equation 18.2.
E
+q
The torque is measured as discussed in Section 9.1.
According to Equation 9.1, the torque produced by each
+F
force has a magnitude given by the magnitude of the
+x
force times the lever arm, which is the perpendicular
distance between the point of application of the force
and the axis of rotation. In the drawing the z axis is the axis of rotation and is midway
between the ends of the rod. Thus, the lever arm for each force is half the length L of the rod
or L/2, and the magnitude of the torque produced by each force is (E q )(L/2).
SOLUTION The +F and the –F force each cause the rod to rotate in the same sense about
the z axis. Therefore, the torques from these forces reinforce one another. Using the
expression (E q )(L/2) for the magnitude of each torque, we find that the magnitude of the
net torque is
Magnitude of = E q ⎛ L ⎞ + E q ⎛ L ⎞ = E q L
⎜ ⎟
⎜ ⎟
net torque
⎝2⎠
⎝2⎠
(
)(
)
= 5.0 × 103 N/C 8.0 × 10 –6 C ( 4.0 m ) = 0.16 N ⋅ m
______________________________________________________________________________
68. REASONING The magnitude of the electrostatic force that acts on particle 1 is given by
Coulomb’s law as F = k q1 q2 / r 2 . This equation can be used to find the magnitude q2
of the charge.
SOLUTION Solving Coulomb’s law for the magnitude q2 of the charge gives
( 3.4 N )( 0.26 m )
F r2
=
= 7.3 × 10−6 C
q2 =
−6
9
2
2
k q1
8.99 × 10 N ⋅ m /C 3.5 × 10 C
2
(
)(
)
(18.1)
Since q1 is positive and experiences an attractive force, the charge q2 must be negative .
______________________________________________________________________________
Chapter 18 Problems
999
69. SSM WWW REASONING Each particle will experience an electrostatic force due to
the presence of the other charge. According to Coulomb's law (Equation 18.1), the
magnitude of the force felt by each particle can be calculated from F = k q1 q2 / r 2 , where
q1 and q2 are the respective charges on particles 1 and 2 and r is the distance between
them. According to Newton's second law, the magnitude of the force experienced by each
particle is given by F = ma , where a is the acceleration of the particle and we have assumed
that the electrostatic force is the only force acting.
SOLUTION
a. Since the two particles have identical positive charges, q1 = q2 = q , and we have, using
the data for particle 1,
kq
r2
2
= m1a1
Solving for q , we find that
q =
m1a1r 2
(6.00 × 10 –6 kg) (4.60 × 103 m/s 2 ) (2.60 × 10 –2 m) 2
=
= 4.56 × 10 –8 C
k
8.99 × 109 N ⋅ m 2 /C 2
b. Since each particle experiences a force of the same magnitude (From Newton's third law),
we can write F1 = F2, or m1a1 = m2a2. Solving this expression for the mass m2 of particle 2,
we have
(6.00 × 10 –6 kg)(4.60 × 103 m/s 2 )
= 3.25 × 10 –6 kg
3
2
a2
8.50 × 10 m/s
______________________________________________________________________________
m2 =
m1a1
=
70. REASONING AND SOLUTION The electric field is defined by Equation 18.2: E = F/q0.
Thus, the magnitude of the force exerted on a charge q in an electric field of magnitude E is
given by
F=qE
(1)
The magnitude of the electric field can be determined from its x and y components by using
the Pythagorean theorem:
E = Ex2 + E y2 =
( 6.00 ×103 N/C ) + (8.00 ×103 N/C)
2
2
= 1.00 × 104 N/C
a. From Equation (1) above, the magnitude of the force on the charge is
F = (7.50 × 10–6 C)(1.00 × 104 N/C) =
7.5 × 10 –2 N
1000
ELECTRIC FORCES AND ELECTRIC FIELDS
b. From the defining equation for the electric field, it follows that the direction of the force
on a charge is the same as the direction of the field, provided that the charge is positive.
Thus, the angle that the force makes with the x axis is given by
3
⎛ Ey ⎞
−1 ⎛ 8.00 × 10 N/C ⎞
θ = tan ⎜⎜ ⎟⎟ = tan ⎜⎜
⎟⎟ = 53.1°
3
⎝ 6.00 × 10 N/C ⎠
⎝ Ex ⎠
______________________________________________________________________________
−1
71. REASONING The two charges lying on the x axis produce no net electric field at the
coordinate origin. This is because they have identical charges, are located the same distance
from the origin, and produce electric fields that point in opposite directions. The electric
field produced by q3 at the origin points away from the charge, or along the −y direction.
The electric field produced by q4 at the origin points toward the charge, or along the +y
direction. The net electric field is, then, E = –E3 + E4, where E3 and E4 can be determined
by using Equation 18.3.
SOLUTION The net electric field at the origin is
E = − E3 + E4 =
=
(
−k q3
r32
+
k q4
r42
)(
2
(5.0 × 10−2 m )
− 8.99 × 109 N ⋅ m 2 /C2 3.0 × 10−6 C
) + (8.99 × 109 N ⋅ m2 /C2 )(8.0 × 10−6 C )
2
( 7.0 × 10−2 m )
= +3.9 × 106 N/C
The plus sign indicates that the net electric field points along the +y direction .
______________________________________________________________________________
72. REASONING The unknown charges can
be determined using Coulomb’s law to
express the electrostatic force that each
unknown charge exerts on the 4.00 μC
charge. In applying this law, we will use
the fact that the net force points downward
in the drawing. This tells us that the
unknown charges are both negative and
have the same magnitude, as can be
understood with the help of the free-body
diagram for the 4.00 μC charge that is
shown at the right. The diagram shows
+4.00 μC
30.0º
F cos 30.0º
F
F sin 30.0º
qA
qB
Chapter 18 Problems
1001
the attractive force F from each negative charge directed along the lines between the charges.
Only when each force has the same magnitude (which is the case when both unknown
charges have the same magnitude) will the resultant force point vertically downward. This
occurs because the horizontal components of the forces cancel, one pointing to the right and
the other to the left (see the diagram). The vertical components reinforce to give the
observed downward net force.
SOLUTION Since we know from the REASONING that the unknown charges have the
same magnitude, we can write Coulomb’s law as follows:
4.00 ×10−6 C ) qA
(
F =k
r2
4.00 × 10−6 C ) qB
(
=k
r2
The magnitude of the net force acting on the 4.00 μC charge, then, is the sum of the
magnitudes of the two vertical components F cos 30.0º shown in the free-body diagram:
4.00 × 10−6 C ) qA
4.00 × 10−6 C ) qB
(
(
ΣF = k
cos 30.0° + k
cos 30.0°
r2
r2
4.00 × 10−6 C ) qA
(
= 2k
cos 30.0°
r2
Solving for the magnitude of the charge gives
qA =
=
(
( ΣF ) r 2
)
2k 4.00 ×10−6 C cos 30.0°
(
( 405 N )( 0.0200 m )2
9
2
2 8.99 × 10 N ⋅ m / C
2
)( 4.00 ×10
−6
)
C cos 30.0°
= 2.60 ×10−6 C
Thus, we have qA = qB = −2.60 × 10−6 C .
73. REASONING The electric field is given by Equation 18.2 as the force F that acts on a test
charge q0, divided by q0. Although the force is not known, the acceleration and mass of the
charged object are given. Therefore, we can use Newton’s second law to determine the force
as the mass times the acceleration and then determine the magnitude of the field directly
from Equation 18.2. The force has the same direction as the acceleration. The direction of
the field, however, is in the direction opposite to that of the acceleration and force. This is
1002
ELECTRIC FORCES AND ELECTRIC FIELDS
because the object carries a negative charge, while the field has the same direction as the
force acting on a positive test charge.
SOLUTION According to Equation 18.2, the magnitude of the electric field is
E=
F
q0
According to Newton’s second law, the net force acting on an object of mass m and
acceleration a is ΣF = ma. Here, the net force is the electrostatic force F, since that force
alone acts on the object. Thus, the magnitude of the electric field is
(
)(
)
3.0 ×10−3 kg 2.5 ×103 m/s 2
F
ma
E=
=
=
= 2.2 ×105 N/C
−
6
q0
q0
34 ×10 C
The direction of this field is opposite to the direction of the acceleration. Thus, the field
points along the -x axis .
74. REASONING The magnitude of the electric field between the plates of a parallel plate
capacitor is given by Equation 18.4 as E =
σ
, where σ is the charge density for each plate
ε0
and ε0 is the permittivity of free space. It is the charge density that contains information
about the radii of the circular plates, for charge density is the charge per unit area. The area
of a circle is πr2. The second capacitor has a greater electric field, so its plates must have the
greater charge density. Since the charge on the plates is the same in each case, the plate area
and, hence, the plate radius, must be smaller for the second capacitor. As a result, we expect
that the ratio r2/r1 is less than one.
SOLUTION Using q to denote the magnitude of the charge on the capacitor plates and
A = πr2 for the area of a circle, we can use Equation 18.4 to express the magnitude of the
field between the plates of a parallel plate capacitor as follows:
E=
q
σ
=
ε 0 ε 0π r 2
Applying this result to each capacitor gives
E1 =
q
ε 0π r12
First capacitor
and
E2 =
q
ε 0π r22
Second capacitor
Chapter 18 Problems
1003
Dividing the expression for E1 by the expression for E2 gives
(
(
)
)
2
E1 q / ε 0π r1
r2
=
= 22
E2 q / ε π r 2
r1
0 2
Solving for the ratio r2/r1 gives
r2
r1
=
E1
E2
=
2.2 ×105 N/C
= 0.76
3.8 ×105 N/C
As expected, this ratio is less than one.
75. SSM REASONING AND SOLUTION Before the spheres have been charged, they exert
no forces on each other. After the spheres are charged, each sphere experiences a repulsive
force F due to the charge on the other sphere, according to Coulomb's law (Equation 18.1).
Therefore, since each sphere has the same charge, the magnitude F of this force is
F=
k q1 q2
r2
=
(8.99 × 109 N ⋅ m 2 /C2 )(1.60 × 10−6 C) 2
= 2.30 N
(0.100 m) 2
The repulsive force on each sphere compresses the spring to which it is attached. The
magnitude of this repulsive force is related to the amount of compression by
Equation 10.1: FxApplied = kx . Setting FxAppled = F and solving for k, we find that
F
2.30 N
=
= 92.0 N/m
x 0.0250 m
______________________________________________________________________________
k=
76. REASONING AND SOLUTION In order for the net force on any charge to be directed
inward toward the center of the square, the charges must be placed with alternate + and –
signs on each successive corner. The magnitude of the force on any charge due to an
adjacent charge located at a distance r is
F=
kq
r2
2
(8.99 × 109 N ⋅ m2 / C2 )( 2.0 × 10−6 C )
=
( 0.30 m )
2
2
= 0.40 N
The forces due to two adjacent charges are perpendicular to one another and produce a
resultant force that has a magnitude of
Fadjacent = 2 F 2 = 2 ( 0.40 N ) = 0.57 N
2
1004
ELECTRIC FORCES AND ELECTRIC FIELDS
The magnitude of the force due to the diagonal charge that is located at a distance of r 2 is
Fdiagonal =
kq
2
(r 2 )
2
=
kq
2
2r 2
=
0.40 N
= 0.20 N
2
since the diagonal distance is r 2 . The force Fdiagonal is directed opposite to Fadjacent (since
the diagonal charges are of the same sign). Therefore, the net force acting on any of the
charges is directed inward and has a magnitude
Fnet = Fadjacent – Fdiagonal = 0.57 N – 0.20 N = 0.37 N
______________________________________________________________________________