Physics 11
HW #11 Solutions
Chapter 18
Focus on Concepts: 4, 8, 13, 17
Problems: 2, 8, 10, 45, 47, 66, 67, 74
Focus On Concepts 18-4
(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.
Focus On Concepts 18-8
(e) Coulomb’s law states that the magnitude of the force is given by F = k
q1 q2
. The
r2
force 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
q q
magnitude k 2 and repelled by charge C with a force of the same magnitude. Since
d
q q
both forces point to the left, the net force acting on B has a magnitude of 2k 2 .
d
q q
Charge A is attracted by charge B with a force of k 2 and also by charge C with a
d
q q
. Since both forces point to the right, the net force acting on A has a
force of k
2
( 2d )
magnitude of (1.25 ) k
q q
d
2
. Charge C is pushed to the right by B with a force of k
and pulled to the left by A with a force of k
q q
( 2d )
2
q q
d2
. Since these two forces have different
directions, the net force acting on C has a magnitude of ( 0.75 ) k
q q
.
d2
Focus On Concepts 18-13
(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.
Focus On Concepts 18-17
(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.
Problem 18-2
REASONING Since the object has a charge of −2.0 μC to begin with, it becomes neutral
when that charge is removed. To create a charge of +3.0 μC on the object, an additional
charge of −3.0 μC must be removed. Thus, the total charge that must be removed from the
object is −5.0 μC. The number of electrons that correspond to this charge can be obtained
by dividing the charge by the charge on a single electron, which is −e =−1.60 × 10−19 C .
SOLUTION The number N of electrons corresponding to −5.0 μC is
=
N
−5.0 × 10−6 C
= 3.1 × 1013
−19
−1.60 × 10
C
Problem 18-8
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
F r2
q2
=
=
k q1
( 3.4 N )( 0.26 m )2
(8.99 × 10
9
N ⋅ m /C
2
2
)(
=
3.5 × 10 C
−6
)
7.3 × 10−6 C
(18.1)
Since q1 is positive and experiences an attractive force, the charge q2 must be negative .
Problem 18-10
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.
SOLUTION
a. The magnitude of the force that each sphere experiences is given by Coulomb’s law as:
k q1 q2
F
=
=
r2
C)
(8.99 × 109 N ⋅ m2 /C2 )( 20.0 × 10−6 C )(50.0 × 10−6=
2
( 2.50 × 10−2 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
k q1 q2
=
=
F
r2
6
C)
(8.99 × 109 N ⋅ m2 /C2 )(15.0 × 10−6 C )(15.0 × 10−=
2
( 2.50 × 10−2 m )
3.24 × 103 N
Since the charges now have the same signs, the force is repulsive .
Problem 18-45
SSM 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
−k q
kq
E=
− E3 + E4 = 2 3 + 24
r3
r4
(
)(
2
(5.0 × 10−2 m )
− 8.99 × 109 N ⋅ m 2 /C2 3.0 × 10−6 C
=+3.9 × 106 N/C
) + (8.99 × 109 N ⋅ m2 /C2 )(8.0 × 10−6 C )
2
( 7.0 × 10−2 m )
The plus sign indicates that the net electric field points along the +y direction .
Problem 18-47
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 xt + 12 ax 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 
(1)
x=
v0 xt + 12 axt 2 =
v0 xt + 12  x  t 2
m
Since the particle is moving in a uniform electric field Ex, it experiences an electrostatic
force Fx given by Fx = q0 E x (Equation 18.2), where q0 is the charge. Substituting this
expression for Fx into Equation (1) gives
F
x=
v0 xt + 12  x
m
 2
q E
v0 xt + 12  0 x
t =

 m
 2
t

−6 ) (
(
2
)
( 0 m/s ) (1.6 ×10−2 s ) + 1  +12 ×10 C +480 N/C  (1.6 ×10−2 s ) =+1.9 ×10−2 m
=
2
3.8 ×10−5 kg


Problem 18-66
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.
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
The charge on a proton is 1.60 × 10–19 C, so the excess number of protons is

2.53 ×107 protons
( 4.04 ×10–12 C )  1.601 ×proton
=
–19
10
C
Problem 18-67
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.
E1
A
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:
−k q
kq
− E1 + E2 = 2 1 + 22
E=
r1
r2
(
)(
2
(3.0 × 10−2 m )
− 8.99 × 109 N ⋅ m 2 /C2 8.5 × 10−6 C
) + (8.99 × 109 N ⋅ m2 /C2 )( 21 × 10−6 C )
2
(9.0 × 10−2 m )
=−6.2 × 107 N/C
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 ⋅ m2 /C2 )(8.5 × 10−6 C ) + (8.99 × 109 N ⋅ m2 /C2 )( 21 × 10−6 C )
2
2
(3.0 × 10−2 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.
Problem 18-74
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
d
+
B
+
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
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
d
+k
2
2
−k
q
2
d2
q
d
+k
2
2
+k
q
2
2
2d )
(=
q
2
( 2d )
1 + 1 + 14
=
9.0
1
4
2