Physics 107 HOMEWORK ASSIGNMENT #14
Cutnell & Johnson, 7th edition
Chapter 17: Problem 44, 60
Chapter 18: Problems 14, 18, 28
**44 A tube, open at only one end, is cut into two shorter (nonequal) lengths. The piece that is
open at both ends has a fundamental frequency of 425 Hz, while the piece open only at one end
has a fundamental frequency of 675 Hz. What is the fundamental frequency of the original tube?
60 Concept Questions A copper block is suspended in air from a wire in
Part 1 of the drawing. A container of mercury is then raised up around the
block as in Part 2. (a) The fundamental frequency of the wire is given by
Equation 17.3 with n = 1: f1 = v/(2L). How is the speed v at which individual
waves travel on the wire related to the tension in the wire? (b) Is the tension
in the wire in Part 2 less than, greater than, or equal to the tension in Part 1 ?
(c) Is the fundamental frequency of the wire in Part 2 less than, greater than,
or equal to the fundamental frequency in Part 1? Justify each of your
answers.
14 Two tiny conducting spheres are identical and carry charges of -20 µC and +50 µC. They are
separated by a distance of 2.50 cm. (a) What is the magnitude of the force that each sphere
experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and
then separated to a distance of 2.50 cm. Determine the magnitude of the force that each sphere
now experiences, and state whether the force is attractive or repulsive.
*18 The drawing shows an equilateral triangle, each side of which has a length
of 2.00 cm. Point charges are fixed to each corner, as shown. The 4.00 µC
charge experiences a net force due to the charges qA and qB. This net force
points vertically downward and has a magnitude of 405 N. Determine the
magnitudes and algebraic signs of the charges qA and qB.
28 Four point charges have the same magnitude of 2.4 x 10-12 C and are fixed to the corners of a
square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine
the magnitude of the net electric field that exists at the center of the square.
44. REASONING AND SOLUTION The original tube has a fundamental given by f = v/(4L),
so that its length is L = v/(4f ). The cut tube that has one end closed has a length of Lc = v/(4fc),
while the cut tube that has both ends open has a length Lo = v/(2fo).
We know that L = Lc + Lo. Substituting the expressions for the lengths and solving for f
gives
fo fc
 425 Hz  675 Hz   162 Hz
f 

2 f c  f o 2  675 Hz   425 Hz
60. CONCEPT QUESTIONS a. The speed v at which individual waves travel on the wire
F
related to the tension F according to Equation 16.2: v 
, where m/L is the mass per
m/ L
unit length of the wire.
b. The tension in the wire in Part 2 less than the tension in Part 1. The reason is related to
Archimedes’ principle (Equation 11.6). This principle indicates that when an object is
immersed in a fluid, the fluid exerts an upward buoyant force on the object. In Part 2 the
upward buoyant force from the mercury supports part of the block’s weight, thus reducing
the amount of the weight that the wire must support. As a result, the tension in the wire is
less than in Part 1.
c. Since the tension F is less in Part 2, the speed v is also less. The fundamental frequency
of the wire is given by Equation 17.3 with n = 1: f1 = v/(2L). Since v is less, the
fundamental frequency of the wire is less in Part 2 than in Part 1.
SOLUTION Using Equations 17.3 and 16.2, we can obtain the fundamental frequency of
the wire as follows:
v
1
F
(1)
f1 

2L 2L m / L
In Part 1 of the drawing, the tension F balances the weight of the block, keeping it from
falling. The weight of the block is its mass times the acceleration due to gravity. The mass,
according to Equation 11.1 is the density copper times the volume V of the block. Thus, the
tension in Part 1 is
Part 1 tension
F   mass  g  copperVg
In Part 2 of the drawing, the tension is reduced from this amount by the amount of the
upward buoyant force. According to Archimedes’ principle, the buoyant force is the weight
of the liquid mercury displaced by the block. Since half of the block’s volume is immersed,
the volume of mercury displaced is V/2. The weight of this mercury is the mass times the
acceleration due to gravity. Once again, according to Equation 11.1, the mass is the density
mercury times the volume, which is V/2. Thus, the tension in Part 2 is
Part 2 tension
F  copperVg   mercury V / 2  g
With these two values for the tension we can apply Equation (1) to both parts of the drawing
and obtain
f1 
Part 1
f1 
Part 2
1
2L
1
2L
copperVg
m/ L
copperVg  mercury V / 2  g
m/ L
Dividing the Part 2 by the Part 1 result, gives
f1, Part 2
f1, Part 1
1
 2L
copperVg  mercury V / 2  g
1
2L

m/ L
copperVg

copper  12 mercury
copper
m/ L

8890 kg/m3  12 13 600 kg/m3
8890 kg/m3

0.485
As expected, the fundamental frequency is less in Part 2 than Part 1.
14. 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:
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 
Because the charges have opposite signs, the force is attractive .
1.44  104 N
b. The net charge on the spheres is 20.0  C + 50.0  C = +30.0  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  C. Since the spheres are identical, the charge on each after
being separated is one-half the net charge, so q1  q2   15.0  C . The electrostatic force
that acts on each sphere is now
F
k q1 q2
r2
8.99  109 N  m2 /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 .
18. REASONING The unknown charges can
+4.00 μC
be determined using Coulomb’s law to
express the electrostatic force that each
30.0º
F cos 30.0º
unknown charge exerts on the 4.00 μC
charge. In applying this law, we will use
F
the fact that the net force points downward
in the drawing. This tells us that the
F sin 30.0º
unknown charges are both negative and
have the same magnitude, as can be
understood with the help of the free-body
qA
qB
diagram for the 4.00 μC charge that is
shown at the right. The diagram shows
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


F  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 


 F  r 2

2k 4.00 106 C cos 30.0

 405 N  0.0200 m 2
2 8.99 10 N  m / C
9
2
2
 4.00 10
6

C cos 30.0
 2.60 106 C
Thus, we have qA  qB  2.60 106 C .
28. 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
E  E A  EB  EC  ED  E A  EB  EC  EB  E A  EC  2E A
+ D
Using Equation 18.3, we find that the magnitude of the net field is
 E  2 EA  2
kq
r2
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  12 L2  L2 . With this substitution for r, the
magnitude of the net field becomes
E  2

kq
1
2
L2  L2

2

4k q
L2



4 8.99 109 N  m 2 / C2 2.4 1012 C
 0.040 m 2
  54 N/C