CHAPTER 18 ELECTRIC FORCES AND
ELECTRIC FIELDS
PROBLEMS
______________________________________________________________________________
1.
SSM
REASONING AND SOLUTION
The charge on a single electron is
–1.60 10 –19 C . In order to give a neutral silver dollar a charge of +2.4  C, we must
remove an amount of negative charge equal to –2.4  C. This corresponds to


1 electron

 –2.4  10–6 C   –1.60
 10 –19 C
1.5  1013 electrons


______________________________________________________________________________
2.
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 C
The net charge on the sphere is, therefore,
qnet = +8.0 C – 9.6 C = 1.6 C
______________________________________________________________________________
3. REASONING The law of conservation of electric charges states that the net electric charge
of an isolated system remains constant. Initially the plate-rod system has a net charge of 3.0
μC + 2.0 μC = 1.0 μC. After the transfer this charge is shared equally by both objects, so
that each carries a charge of 0.50 μC. Therefore, 2.5 μC of negative charge must be
transferred from the plate to the rod. To determine how many electrons this is, we will divide
this charge magnitude by the magnitude of the charge on a single electron.
SOLUTION The magnitude of the charge on an electron is e, so that the number N of
electrons transferred is
N
4.
Magnitude of transferred charge
2.5 106 C

 1.6 1013
e
1.60 1019 C
REASONING AND SOLUTION Object A is negative because of an excess of electrons,
while object B is positive because of a deficiency of electrons. The mass difference between
the charged objects arises because the mass of A is greater due to the additional electrons,
764
ELECTRIC FORCES AND ELECTRIC FIELDS
while the mass of object B is less due to the loss of electrons. Since q = Ne, where e is the
magnitude of the charge on one electron, the number of excess electrons on object A is
N
q
3.0  10 –6 C

 1.9  10 13
e 1.60  10 –19 C
This corresponds to an increase in mass given by
 9.111031 kg 
17
m  1.9 10 electrons 
  1.7 10 kg
 1electron 

13

Since both objects carry charge of the same magnitude, N is also equal to the number of
electrons lost by object B. Hence, the mass of B is reduced by the amount m. If M is the
mass of either object when they are electrically neutral, then the mass of each charged object
is:
MA = M + m
while
MB = M – m
The mass difference between the charged objects is, therefore,
M = MA – MB = (M + m) – (M – m)
= 2m = 2(1.7  10–17 kg) =
3.4  10 –17 kg
From the discussion above, object A has the larger mass .
______________________________________________________________________________
5.
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.
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 .
Chapter 18 Problems
765
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
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 times the volume, as expressed by Equation 11.1. Thus, the number of
electrons is




V
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 
23
1

 10 
 1 kg   6.022  10 mol


18.0 g/mol





 3.35  10 electrons
26
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 .
______________________________________________________________________________
7.
SSM REASONING AND SOLUTION The magnitude of the force of attraction between
the charges is given by Coulomb's law (Equation 18.1): F  k q1 q2 / r 2 , where
q1 and q2 are the magnitudes of the charges and r is the separation of the charges. Let FA
766
ELECTRIC FORCES AND ELECTRIC FIELDS
and FB represent the magnitudes of the forces between the charges when the separations are
rA and rB = rA/9, respectively. Then
2
2
FB k q1 q2 / rB2  rA   rA 
2

  
  (9)  81
FA k q1 q2 / rA2  rB   rA / 9 
Therefore, we can conclude that FB  81 FA  (81)(1.5 N)= 120 N .
______________________________________________________________________________
8.
REASONING AND SOLUTION The magnitude of the electrostatic force exerted on each
proton can be obtained from Coulomb's law
F
k q1 q2
r2
8.99  109 N  m 2 / C2 1.60  10 –19 C 1.60  10 –19 C 



2
–15
3.0  10 m 
26 N
______________________________________________________________________________
9.
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
10. 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
9
2
2

6
k q1
9.0  10 N  m /C 3.5  10 C
2
q2 



(18.1)
Since q1 is positive and experiences an attractive force, the charge q2 must be negative .
______________________________________________________________________________
Chapter 18 Problems
11.
767
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
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
9
2 2
k
8.99 10 N  m /C
______________________________________________________________________________
q m
12. 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
768
ELECTRIC FORCES AND ELECTRIC FIELDS
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
r
2
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 .
______________________________________________________________________________
13.
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
k

(6.00 10 –6 kg) (4.60 103 m/s 2 ) (2.60 10 –2 m) 2
 4.56 10 –8 C
9
2 2
8.99 10 N  m /C
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/s2 )
 3.25 10–6 kg
3
2
a2
8.50 10 m/s
______________________________________________________________________________
m2 
m1a1

Chapter 18 Problems
769
14. 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  m2 / 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


______________________________________________________________________________
  tan –1 
15. 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
r2
2
or
qafter 
Fr 2
k
770
ELECTRIC FORCES AND ELECTRIC FIELDS
Using this result in the expression for N, we find that
N
Fr 2
 qbefore
k

e
 68.0 N  0.0300 m 2
 2.00 106 C
8.99 10 N  m / C
1.60 1019 C
9
2
2
 3.8 1012
16. 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
4.00 106 C  qA

 2k
cos 30.0
r2
Solving for the magnitude of the charge gives
r2
Chapter 18 Problems
qA 


771
 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 .
17. REASONING The unknown charge q must be
positive. To see why, consider the unknown charge
FO
at the upper right corner in the drawing at the right –0.70 C
FL +q
(the unknown charge at the lower left corner could
also be used). Three forces act on this charge: (1)
FO is the repulsive force due to the other unknown
FB
charge on the opposite corner, (2) FB is the attractive
d
force due to the negative charge at the lower right
corner, and (3) FL is the attractive force due to the
negative charge at the upper left corner. These three
d
forces add to give a net force of zero. The unknown
charge can not be negative, because then FB and FL
+q
–0.70 C
would have directions opposite to
those shown in the shown in the drawing, and the sum of FO, FB, and FL could not be zero.
We note that the magnitudes of FB and FL are equal according to Coulomb’s law
(Equation 18.1), since the sides of the square have equal lengths and the charge magnitudes
are q and 0.70 C in each case. We also note that the directions of FB and FL are
perpendicular. Thus, the resultant of FB and FL is given by the Pythagorean theorem and
points along the diagonal of the square, directly opposite to the direction of FO. Since the
vector sum of FO, FB, and FL is zero, the magnitude of the resultant of FB and FL must equal
the magnitude of FO, and it is with this fact in mind that we begin our solution.
SOLUTION Using the Pythagorean to express the magnitude of the resultant of FB and FL,
which is equal to FO, we have
FB2  FL2  FO
Coulomb’s law indicates that
(1)
772
ELECTRIC FORCES AND ELECTRIC FIELDS
FB  FL 


k 0.70  10 –6 C q
FO 
and
d2
kq

2
2d

2

kq
2
2d 2
where we have used d for the length of a side of the square and the fact that the diagonal of
the square has a length of d 2  d 2  2 d . Substituting these expressions for FO, FB,
and FL into Equation (1), we find



2

2




2
2
 k 0.70  10 –6 C q   k 0.70  10 –6 C q 
 k 0.70  10 –6 C q 
kq

 
  2
 

 



d2
d2
d2
2d 2

 



Simplifying this result shows that


2 k 0.70  10 –6 C q
d
2

kq
2d
2
2
or
q  2 2 0.70  10 –6 C  2.0 C
As discussed in the REASONING, the algebraic sign of the charge is positive .
______________________________________________________________________________
18. 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
where the distance is specified in the drawing:
q1
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
Chapter 18 Problems
773
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 /s2

3
m 1.50  10 kg
where the plus sign indicates that the acceleration is along the +y axis .
______________________________________________________________________________
19. SSM REASONING Consider the drawing
qU
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 force
q2 = +3.0 C
on the charge qA. As the drawing shows,

these forces have magnitudes FA1 (vertically
downward) and FA2 (horizontally to the left).
FA1
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.
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
774
ELECTRIC FORCES AND ELECTRIC FIELDS
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  d 2
FA2 
and
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  , which must be equal to FA2, so that we have
k qA qU
 4d  2  d 2
cos 
k qA q2
or
 4d  2
qU
17
cos 
q2
16
The drawing in the REASONING, reveals that cos   4d  /
 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 .
20. 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
Chapter 18 Problems
775
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:
 F  C

 F  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
21. REASONING This is a problem that deals with motion in a circle of radius r. As Chapter
5 discusses, a centripetal force acts on the plane to keep it on its circular path. The
centripetal force Fc is the name given to the net force that acts on the plane in the radial
direction and points toward the center of the circle. When there are no electric charges
present, only the tension in the guideline supplies this force, and it has a value Tmax at the
moment the line breaks. However, when there is a charge of +q on the plane and a charge of
–q on the guideline at the center of the circle, there are two contributions to the centripetal
force. One is the electrostatic force of attraction between the charges and, since the charges
have the same magnitude, its magnitude F is given by Coulomb’s law (Equation 18.1) as
2
F  k q / r 2 . The other is the tension Tmax, which is characteristic of the rope and has the
same value as when no charges are present. Whether or not charges are present, the
centripetal force is equal to the mass m times the centripetal acceleration, according to
Newton’s second law and stated in Equation 5.3, Fc = mv2/r. In this expression v is the
speed of the plane. Since we are given information about the plane’s kinetic energy, we will
use the definition of kinetic energy, which is KE = mv2/2, according to Equation 6.2.
SOLUTION From the definition of kinetic energy, we see that mv2 = 2(KE), so that
Equation 5.3 for the centripetal force becomes
Fc 
mv 2 2  KE 

r
r
Applying this result to the situations with and without the charges, we get
776
ELECTRIC FORCES AND ELECTRIC FIELDS
Tmax 
kq
r2
2

2  KE charged
1
r
Tmax

2  KE uncharged
r
 2
Centripetal
force
Centripetal
force
Subtracting Equation (2) from Equation (1) eliminates Tmax and gives
kq
r2
2
2  KE charged –  KE uncharged 

 
r
Solving for q gives
2r  KE charged –  KE uncharged 

  2  3.0 m  51.8 J – 50.0 J   3.5  10 –5 C
q 
k
8.99  109 N  m 2 / C2
______________________________________________________________________________
22. REASONING AND SOLUTION Assume that before the objects are touched that the left
object has a negative charge of magnitude q1 and the right object has a positive charge of
magnitude q2 . The force between them then has a magnitude of
F
k q1 q2
r2
After touching the charge on each object is the same and of magnitude q2  q1 / 2 . The
magnitude of the force between the objects is now
F

k q2  q1 / 2

2
r2
Equating the equations and rearranging gives
2
2
q2  6 q1 q2  q1  0
The solutions to this quadratic equation are
q1 = 5.58 µC, q2 = 0.957 µC
and
q1 = 0.957 µC, q2 = 5.58 µC
The charge, q1, was assumed negative, so the possible solutions are
–5.58 C on left, + 0.957 C on right
and –0.957 C on left, + 5.58 C on right
Chapter 18 Problems
777
______________________________________________________________________________
23.
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  with respect to the vertical as shown in the
drawing in the problem statement. We can analyze the forces to determine the desired
quantities  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   k q1 q2 / r 2
T cos   mg
Dividing the first equation by the second yields
k q1 q2 / r 2
T sin 
 tan  
T cos
mg
Solving for , we find that
 k q1 q2
 mgr 2

  tan –1 
 tan




9
–1  (8.99 10
N  m 2 /C2 )(0.600 10 –6 C)(0.900 10 –6 C) 
  15.4
(8.00 10 –2 kg)(9.80 m/s 2 )(0.150 m) 2



b. Since T cos  mg , the tension can be obtained as follows:
mg
(8.00 102 kg) (9.80 m/s 2 )

 0.813 N
cos 
cos 15.4
______________________________________________________________________________
T
24. 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
778
ELECTRIC FORCES AND ELECTRIC FIELDS
F
kq
r2
2

8.99  109 N  m 2 / C2  2.0  106 C 

2
 0.30 m 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
The magnitude of the force due to the diagonal charge that is located at a distance of r 2 is
Fdiagonal 
kq

2
kq
2


2
2
2
r
r 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
______________________________________________________________________________
25. 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 .
______________________________________________________________________________
Chapter 18 Problems
779
26. REASONING AND SOLUTION The
electric field lines must originate on the
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
the electric field lines that meet these
criteria.
______________________________________________________________________________
27.
SSM WWW 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 .
______________________________________________________________________________
28. 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
r12
and
E2 
kq
r22
780
ELECTRIC FORCES AND ELECTRIC FIELDS
Dividing the expression for E1 by the expression for E2 gives
E1 k q / r12 r22


E2 k q / r22 r12
Solving for the ratio r2/r1 gives
r2

r1
E1
248 N/C

 1.37
E2
132 N/C
As expected, this ratio is greater than one.
29. 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 field 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.
Chapter 18 Problems
781
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
  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  m 2 /C2 8.5  106 C  8.99  109 N  m 2 /C 2  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.
______________________________________________________________________________
30. 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+
x
2
or
q
 3.0 m  x 
2

q+
x2
782
ELECTRIC FORCES AND ELECTRIC FIELDS
Solving for the ratio of the charge magnitudes gives
16.0 μC  3.0 m  x 


4.0 μC
q+
x2
q
2
or
2
3.0 m  x 

4.0 
x2
Suppressing the units for convenience and rearranging this result gives
4.0x2   3.0  x 
2
or
4.0x2  9.0  6.0 x  x2
or
3x2  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 would be 0 N .
______________________________________________________________________________
31.
SSM REASONING Before the 3.0-C point charge q is introduced into the region, the
4
region contains a uniform electric field E of magnitude 1.6 10 N/C . After the 3.0-C
charge is introduced into the region, the net electric field changes. In addition to the
uniform electric field E , the region will also contain the electric field E q due to the point
charge q. The field at any point in the region is the vector sum of E and E q . The field E q
is radial as discussed in the text, and its magnitude at any distance r from the charge q is
given by Equation 18.3, Eq  k q / r 2 . There will be one point P in the region where the net
electric field E net is zero. This point is located where the field E has the same magnitude
and points in the direction opposite to the field E q . We will use this reasoning to find the
distance r0 from the charge q to the point P.
SOLUTION Let us assume that the field E points to the right and that the charge q is
negative (the problem is done the same way if q is positive, although then the relative
positions of P and q will be reversed). Since q is negative, its electric field is radially
inward (i.e., toward q); therefore, in order for the field E q to point in the opposite direction
to E , the charge q will have to be to the left of the point P where E net is zero, as shown in
Chapter 18 Problems
783
the drawing at the right. Using Equation
18.3, Eq  k q / r02 , and solving for the
distance r0, we find
r0  k q / Eq .
Since the magnitude Eq must be equal
to the magnitude of E at the point P, we have
kq
(8.99 109 Nm2 /C2 ) (3.0 10 –6 C)

 1.3 m
E
1.6 104 N/C
______________________________________________________________________________
r0 
32. 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
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 
  tan 
  53.1
3
 6.00 10 N/C 
 Ex 
______________________________________________________________________________
  tan 1 
33.
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.
784
ELECTRIC FORCES AND ELECTRIC FIELDS
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

 4.04 10–12 C   1.601 proton

–19
10
C
2.53 107 protons
______________________________________________________________________________
34. REASONING AND SOLUTION The figure at the right shows
the configuration given in text Figure 18.21a. 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
1 1
4 4
+q
- q
2
3
+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.
785
Chapter 18 Problems
The distance r, from any of the charges to the center
of the rectangle, can be found using the Pythagorean
theorem:
1 1
d
5.00 cm
d  (3.00 cm)2 +(5.00 cm)2  5.83 cm
Therefore, r 
2

4 4
d
 2.92 cm  2.92  102 m
2
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  m2 /C2 )(8.60  1012 C)

 1.81  102 N/C
2
2
(2.92  10 m)
The figure at the right shows the configuration given in text
Figure 18.21b. 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
- q
- q
1 1
2
E 13
E 24
C
4 4
+q
3
+q
Figure 3
E24 = E13 = 1.81  102 N/C
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 
From Figure 2, we have that
sin 
and
5.00 cm 5.00 cm

 0.858
d
5.83 cm


EC  2  E13  sin   2 1.81102 N/C  0.858  3.11102 N/C
786
ELECTRIC FORCES AND ELECTRIC FIELDS
______________________________________________________________________________
35. 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 .
______________________________________________________________________________
36. REASONING The magnitude of the electric field between the plates of a parallel plate
capacitor is given by Equation 18.4 as E 

, where σ is the charge density for each plate
0
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


 0  0 r 2
Chapter 18 Problems
787
Applying this result to each capacitor gives
E1 
q
and
 0 r12
First capacitor
E2 
q
 0 r22
Second capacitor
Dividing the expression for E1 by the expression for E2 gives




2
E1 q /  0 r1
r2

 22
E2 q /   r 2
r1
0 2
Solving for the ratio r2/r1 gives
r2

r1
E1
2.2 105 N/C

 0.76
E2
3.8 105 N/C
As expected, this ratio is less than one.
37. REASONING AND SOLUTION The average force F on the proton can be determined
from the impulse-momentum theorem (Equation 7.4):
F t = (mv)
Therefore, the magnitude of the force is
F
mv  mv0
t
5.0 1023 kg  m/s   1.5 1023 kg  m/s 


 5.6  1018 N
6.3 106 s
From the definition of electric field, E =
F
, we find that the magnitude of the field is
q0
F
5.6 1018 N

 35 N/C
q0 1.60 1019 C
______________________________________________________________________________
E
38. REASONING AND SOLUTION From kinematics, vy2 = v0y2 + 2ayy. Since the electron
starts from rest, v0y = 0 m/s. The acceleration of the proton is given by
788
ELECTRIC FORCES AND ELECTRIC FIELDS
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 = /0, where  is the magnitude of the charge per unit area on
each plate. Thus, ay = e/(m0). Combining this expression for a with the kinematics
equation we have
 e 
v 2y  2 
y
 m 
 0
Solving for vy gives





2 1.60 1019 C 1.8 107 C/m2 1.5 102 m
2e y
vy 

 1.0 107 m/s

31

12
2
2
m 0
9.1110 kg 8.85 10
C / Nm 


______________________________________________________________________________
39.



E2
SSM WWW REASONING
The
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
Eq
d 5
–q
E1

d
+q2
2d
According to the Pythagorean theorem, the distance from the charge –q to the empty corner
along the diagonal is given by (2d ) 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:
E1 
Eq 
kq
k q1

 2d  2
r
2

k q1
4d 2
kq
d 5 
2
and

kq
5d 2
E2 
k q2
d2
Chapter 18 Problems
789
The angle  that the diagonal makes with the horizontal is   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
d
2
–
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
______________________________________________________________________________
40. REASONING The proton (charge = +e = 1.60  10–19 C and mass = m = 1.67  10–27 kg)
moves in the direction of the electric field because of the force that the field applies to the
proton. This force does work and thereby changes the proton’s kinetic energy. According to
the work-energy theorem (Equation 6.3), the work done causes the proton’s kinetic energy
to change. Kinetic energy is mv2/2, where v is the speed. The work-energy theorem will
involve the final speed of the proton and we will use it to obtain that speed.
SOLUTION According to Equation 6.3, the work-energy theorem is W = KEf – KE0, where
W is the work done by the net external force that acts on the proton and KEf and KE0 are,
respectively, the final and initial kinetic energies. The force applied by the electric field E is
the only force acting, so it is the net force and, according to Equation 18.2, has a magnitude
of F = Ee. The direction of the force is in the same direction as the electric field, since the
proton has a positive charge. Since the motion is in the direction of the field and the force,
the work done by the force is given by Equation 6.1 as W = Fs = Ees, where s is the distance
traveled. Thus, the work-energy theorem can be stated as follows:
W  KE f – KE 0
or
Solving this result for the final speed vf, we find
1
1
2
2
Ees  mv 2f – mv 02
790
ELECTRIC FORCES AND ELECTRIC FIELDS
vf 

2Ees
 v 20
m



2 2.3  10 3 N/C 1.60  10 –19 C 2.0  10 –3 m
1.67  10
–27
kg

2.5  10 4
m/s
2
 3.9  10 4 m/s
______________________________________________________________________________
41. 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.
E2
60.0º
30.0º
E2 sin 60.0º
E1
E1
30.0º
60.0º
sin
30.0º
q2
SOLUTION Setting the magnitudes of the
horizontal components of the fields equal gives
q1
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
r
2
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
42. REASONING AND SOLUTION From two-dimensional kinematics, taking the entry point
as the origin, we have
791
Chapter 18 Problems
1
x = v0x t + 2 axt2
(1)
1
ayt2
2
(2)
y = v0y t +
There is no acceleration in the x-direction, so ax = 0 m/s2. Initially, the electron travels in
the +x direction, so v0y = 0 m/s. Solving Equation (1) for t and substituting into
Equation (2) gives:
1  x
y  ay 
2  v0 x



2
(3)
From Newton's second law and the definition of electric field we have
ay 
F eE

m m
Thus, Equation (3) becomes
1  eE   x
y   
2  m   v0 x



2
where e is the magnitude of the electron’s charge and m is its mass. Solving for E gives



2
2 9.111031 kg 1.50 103 m  7.00 106 m/s 2
2my  v0 x 
3
E

  2.09 10 N/C

 

19

2
e  x 
1.60 10
C
 2.00 10 m 
______________________________________________________________________________
43. 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,
Fe  E qball
(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
792
ELECTRIC FORCES AND ELECTRIC FIELDS
E
q
0 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
Fe  mg tan 30.0 
q qball
0 A
Solving for q yields
q
 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
______________________________________________________________________________
44. REASONING AND SOLUTION The electric field due to a parallel plate capacitor is given
by E = /o (Equation 18.4).
a. The induced charge density  is, therefore,
 = oE = [8.85  10–12 C2/(N.m2)](480 N/C) = 4.2 × 109 C/m2
b. The area of one face of the circular coin is
A =  r2 =  (0.019 m)2
The total charge on each face of the coin is
q = A = (4.2 × 109 C/m2)[ (0.019 m)2] = 4.8 × 1012 C
______________________________________________________________________________
45.
SSM REASONING 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.
Chapter 18 Problems
793
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 . Therefore, E  E–  k q / 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  +E – cos   2  2


 cos 

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
Since EM/EP = 9.0, we have
EM
2k q / d 2

EP 2k q  cos   /
2

1
 cos   d 2 /
2
 9.0
But from Figure 1, we can see that d/ = cos . Thus, it follows that
1
 9.0
cos3
or
cos   3 1/ 9.0  0.48
The value for  is, then,   cos –1  0.48  61 .
______________________________________________________________________________
46. REASONING AND SOLUTION The maximum possible flux occurs when the electric
field is parallel to the normal of the rectangular surface (that is, when the angle between the
direction of the field and the direction of the normal is zero). Then
E = (E cos)A = (580 N/C)(cos 0°)(0.16 m)(0.38 m) =
35 N  m 2 / C
794
ELECTRIC FORCES AND ELECTRIC FIELDS
______________________________________________________________________________
47.
SSM REASONING As discussed in Section 18.9, the electric flux  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, 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
 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
 E 2  (E cos 55) A2  (250 N/C)(cos 55)(3.2 m2 ) 
460 N m 2 /C
______________________________________________________________________________
48. REASONING AND SOLUTION Gauss' Law is given by text Equation 18.7: E =
Q
0
,
where Q is the net charge enclosed by the Gaussian surface.
a.  E 
3.5  10 – 6 C
 4.0  10 5 N m 2 /C
12
2
2
8.85  10
C /(N  m )
b.  E 
2.3  10 – 6 C
 –2.6  10 5 N m 2 /C
12
2
2
8.85  10
C /(N  m )
(3.5  10 – 6 C)  (2.3  10 – 6 C)
 1.4  10 5 N  m 2 /C
12
2
2
8.85  10
C /(N  m )
______________________________________________________________________________
c.  E 
49. REASONING AND SOLUTION
a. In all three cases, the net charge enclosed by the surface is the same, because the net
charge enclosed by each surface is the same; therefore, by Gauss' Law, the electric flux
through the surfaces described in parts (a) through (c) is the same:
E 
Q
0

2.0  10 – 6 C
8.85  10
b.  E  2.3 10 5 N  m 2 /C
12
C /(N  m )
2
2
 2.3  10 5 N m 2 /C
Chapter 18 Problems
795
c.  E  2.3 10 5 N  m 2 /C
______________________________________________________________________________
50. REASONING AND SOLUTION Since the electric field is uniform, its magnitude and
direction are the same at each point on the wall. The angle  between the electric field and
the normal to the wall is 35°. Therefore, the electric flux is
E = (E cos ) A = (150 N/C)(cos 35°)[(5.9 m)(2.5 m)] = 1.8 10 3 N  m 2 /C
______________________________________________________________________________
51.
SSM REASONING The electric flux through each
face of the cube is given by  E  ( E cos  ) 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  is the
angle between the electric field and the outward
normal of that face. We can use this expression to
calculate the electric flux  E through each of the six
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  = 180°. Therefore,
 E bottom  1500 N/C cos 180° 0.20 m2 
6.0  101 N  m2 /C
On the top face of the cube, the outward normal points parallel to the +y axis, and  = 0.0°.
The electric flux is, therefore,
(E )top  (1500 N/C)(cos 0.0)(0.20 m)2  +6.0 101N.m2 /C
On each of the other four faces, the outward normals are perpendicular to the direction of
the electric field, so  = 90°. So for each of the four side faces,
( E )sides  (1500 N/C)(cos 90)(0.20 m) 2  0 N  m 2 / C
b. The total flux through the cube is
(E )
total
Therefore,
 ( E ) top  ( E )bottom  ( E )side 1  ( E )side 2  ( E )side 3  (E )side 4
796
ELECTRIC FORCES AND ELECTRIC FIELDS
( E ) total  (  6.0 101N.m 2 /C)  (–6.0 101N.m 2 /C)  0  0  0  0  0 N  m 2 / C
______________________________________________________________________________
52. REASONING AND SOLUTION Since both charge distributions are uniformly spread over
concentric spherical shells, the electric field possesses spherical symmetry. Gauss' law can
be used to determine the magnitude of the electric field, provided we choose spherical
Gaussian surfaces (concentric with the spherical shells) to evaluate the electric flux. To find
the magnitude of the electric field at any distance r from the center of the spherical shells,
we construct a spherical Gaussian surface of radius r. The electric flux through this
Gaussian surface is
E = (E cos)A
Because the charge distributions have spherical symmetry, we expect the electric field to be
directed radially. That is, the electric field is everywhere perpendicular to the Gaussian
surface. Thus, for any surface element,  will be 0 or 180°. Furthermore, since the charge
distribution possesses spherical symmetry, we expect the electric field to be uniform in
magnitude over any sphere concentric with the shells. Thus, E is constant over any
Gaussian surface concentric with the shells. Then, (E cos ) can be factored out of the
summation.
E = (E cos)A = (E cos)A
where A is the sum of the area elements that make up the Gaussian surface. This sum
must equal the surface area of the Gaussian surface or
E = (E cos)A = (E cos)4r2)
where r is the radius of the Gaussian surface. From Gauss' law this becomes
(E cos)r2) =
Q
0
where Q is the net charge enclosed by the Gaussian surface.
(1)
Chapter 18 Problems
a. r = 0.20 m
797
Gaussian surface
The Gaussian surface encloses both shells. The net
charge enclosed is
(+5.1 x 10–6 C) + (–1.6 x 10–6 C) = + 3.5 x 10–6 C
r1
Since the net charge is positive, E will be radially
outward for all points on the Gaussian surface, and
 = 0.0° for all elements on the Gaussian surface.

Q
 0 4 r 2


8.85 1012

r2
r1 = 0.050 m
Solving Equation (1) for E gives
E
r
r2 = 0.15 m
3.5 106 C
 7.9 105 N/C
2
2
2 
C / Nm
4  0.20 m 




The direction of E is radially outward , because the net charge within the Gaussian surface
is positive.
b. r = 0.10 m
Gaussian surface
The Gaussian surface encloses only the inner shell.
The net charge enclosed is
r1
Q = –1.6 x 10–6 C
Since the net charge is negative, E will be radially
inward for all points on the Gaussian surface, and  =
180° for all elements on the Gaussian surface.
r
r2
r
1 = 0.050 m
r
2 = 0.15 m
Solving Equation (1) for E gives
1.6 106 C 

E

 1.4 106 N/C
2
2

12
2
2
8.85 10
 0  4 r 
C /  N  m    4  0.10 m  



Q
The direction of E is radially inward , because the net charge within the Gaussian surface
is negative.
c. r = 0.025 m
The net charge enclosed by the Gaussian surface is zero. This implies that the net electric
flux is zero, so the electric field is either a constant or zero everywhere within the Gaussian
surface. However, an electric field does not exist within the Gaussian surface, because
798
ELECTRIC FORCES AND ELECTRIC FIELDS
there are only negative charges on the shell of radius r1, so electric field lines cannot
originate from any place on this shell. Thus, E = 0 N/C in this region.
______________________________________________________________________________
53. 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
  E cos    A 
(18.7)
0
Electric flux,  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 between E
E
and the normal is 0º
symmetric. Consequently, the electric
field is directed radially outward, and for
each element of area  A is perpendicular
to the surface. This means that the
angle
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:
  E cos    A    E cos 0  A  E   A
The term  A is the entire area of the spherical Gaussian surface or 4πr2. With this
substitution, the electric flux becomes
  E cos    A  E   A  E 4 r 2
(1)


Now consider the net charge Q within the Gaussian surface. This charge is the charge density
times the volume 43  r 3 encompassed by that surface:
Q
q

4  R3
3
 43  r 3   qrR3
Charge
density
Volume of
Gaussian
surface
3
Substituting Equations (1) and (2) into Equation 18.7 gives


E 4 r 2 
Rearranging this result shows that
qr 3 / R3
0
(2)
Chapter 18 Problems
E
qr 3 / R 3
 4 r 2   0

799
qr
4 0 R 3
54. 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.
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/0 (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  = cos 90° = 0. Thus, there is no flux through these two surfaces:
1 = 3 = 0 N  m 2 /C .
Surface 2 – the curved wall – is everywhere perpendicular to the electric field E, so
cos  = 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:
 2   E cos 0  A  E  A
The area A of this surface is just the circumference 2 r of the cylinder times its length L:
A = (2 r)L. The electric flux through the entire cylinder is, then,
 E  1   2  3  0  E 2  rL  0  E2  rL 
800
ELECTRIC FORCES AND ELECTRIC FIELDS
Following Gauss’ law, we set E equal to Q/0, where Q is the net charge inside the
Gaussian cylinder: E(2 rL) = Q/0. The ratio Q/L is the charge per unit length of the wire
and is known as the linear charge density :  = Q/L. Solving for E, we find that
E
Q/ L


2  0 r 2  0 r
______________________________________________________________________________
55. 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.
______________________________________________________________________________
56.
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,
Chapter 18 Problems
F
k q1 q2
r2
8.99 109 N  m 2 /C2  4.8 106 C 


801
2
 0.50 m 2
 0.83 N
b. Since the spheres carry charges of opposite sign, the force is attractive .
______________________________________________________________________________
57. 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 .
______________________________________________________________________________
58. 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
Using Equation 18.3, we find that the magnitude of the net field is
 E  2 EA  2
kq
r2
+ D
802
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  12 L2  L2 . With this substitution for r, the
magnitude of the net field becomes
E  2
59.

kq
1
2
L2  L2

2

4k q
2
L



4 8.99 109 N  m 2 / C2 2.4 1012 C
 0.040 m 
2
  54 N/C
SSM REASONING AND SOLUTION The force F exerted on a charge q0 placed in an
electric field E can be determined from Equation 18.2, the definition of electric field
( E  F / q0 ). Writing this in terms of magnitudes, and taking due east as the positive
direction, we have, solving for the magnitude F of the force,
F  q0 E  (3.0 10–5 C)(15 000 N/C)= 0.45 N
Since the charge is positive, the direction of the force is the same as the direction of the
electric field, namely, due east .
______________________________________________________________________________
60. REASONING The drawing at the right
shows the set-up. The force on the +q
charge at the origin due to the other +q
charge is given by Coulomb’s law
(Equation 18.1), as is the force due to the
+2q charge. These two forces point to the
left, since each is repulsive. The sum of
the two is twice the force on the +q charge at the origin due to the other +q charge alone.
SOLUTION Applying Coulomb’s law, we have
kq q
 0.50 m 2
Force due to +q
charge at x 0.50 m

k 2q q
 d 2
Force due to +2q
charge at x  d

2
kq q
 0.50 m 2
Twice the force due to
+ q charge at x 0.50 m
Rearranging this result and solving for d give
k 2q q
d 
2

kq q
 0.50 m 
2
or
d 2  2  0.50 m 
2
or
d  0.71 m
Chapter 18 Problems
803
We reject the negative root, because a negative value for d would locate the +2q charge to
the left of the origin. Then, the two forces acting on the charge at the origin would have
different directions, contrary to the statement of the problem. Therefore, the +2q charge is
located at a position of x  0.71 m .
______________________________________________________________________________
61.
SSM REASONING AND SOLUTION The net
electrostatic force on charge 3 at x  3.0 m is the
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
3 due to charge 1 is
k q1 q3
F13 
r132
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   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:
F13x  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
x2
8.99 109 N  m 2 / C2 12 106 C  45 106 C 


 3.0 m 
 0.539 N
2
Figure 2
F13
804
ELECTRIC FORCES AND ELECTRIC FIELDS
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
Using the Pythagorean theorem, we find the magnitude of F3 to
be
F3  F32x  F32y  (0.253 N)2  (0.286 N)2  0.38 N
The direction of F3 relative to the –x axis is specified by the
angle , where
Figure 3
N
  tan 
  49 below the  x axis
 0.253 N 
______________________________________________________________________________
1  0.286
+z
62. REASONING The drawing at the right shows the setup. Here, the electric field E points along the +y axis
–F
and applies a force of +F to the +q charge and a force of
–q
–F to the –q charge, where q = 8.0 C denotes the
magnitude of each charge. Each force has the same
+y
magnitude of F = E q , according to Equation 18.2.
E
The torque is measured as discussed in Section 9.1.
+q
According to Equation 9.1, the torque produced by each
force has a magnitude given by the magnitude of the
+F
force times the lever arm, which is the perpendicular
+x
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
 = (E q )(L/2) for each torque, we find that the magnitude of the net torque is
L
L



  E q    E q    E q L  5.0  103 N/C 8.0  10 –6 C  4.0 m   0.16 N  m
2
2
Chapter 18 Problems
805
______________________________________________________________________________
63. 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
 
F1C  q1 EC  q1  
 
 0
(2)
Equating the right hand sides of Equations (1) and (2) above gives
k q1 q2
r12
2
 
 q1  
 
 0
Solving for r12 gives
r12 =
 0 k q2

[8.85 ×1012 C2 /(N  m 2 )](8.99 ×109 N  m 2 /C 2 )(5.00 ×10  6 C)
= 5.53 ×10 –2 m

2
(1.30 ×10 C/m )
______________________________________________________________________________
=
64. REASONING AND SOLUTION
a. Since the spring is stretched, the electric force must be a repulsion. Therefore, the
charges must be the same polarity (both positive or both negative) .
b. The force needed to stretch the spring is F = kspring x, which is provided by the electric
force given by Coulomb’s law.
kq
r2
q 
kspring xr 2

2
 kspring x
 220 N/m  0.020 m  0.34 m 2
 7.5  10 –6 C
8.99  10 N  m / C
______________________________________________________________________________
k
9
2
2
65. 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
806
ELECTRIC FORCES AND ELECTRIC FIELDS
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
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 F = 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 .
66. REASONING The magnitude E of the electric field is defined as 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
Chapter 18 Problems
E
m  v  v0 
qt
9.0  105 kg  2.0  103 m /s  0 m /s 



6
7.5

10
C
0.96
s




807
2.5  104 N /C
______________________________________________________________________________
67.
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  m2 /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:
F  kx . Therefore, solving for k, we find
F
2.30 N

 92.0 N/m
x 0.0250 m
______________________________________________________________________________
k
68. 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

kq
2
r2
where r = 2(0.25 m) sin 18° = 0.1545 m. Now
q r
Fe
k
  0.1545 m 
2.547  10 –3 N
 8.2  10 –8 C
9
2
2
8.99  10 N  m / C
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,
T
 mg 2  Fe2




2

 8.0  10 –4 kg 9.80 m/s2   2.547  10 –3 N



2
 8.2  10 –3 N
808
ELECTRIC FORCES AND ELECTRIC FIELDS
______________________________________________________________________________
69. CONCEPT QUESTIONS
a. 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.
b. When the two identical 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 C, the final net charge on both spheres is
2(+5.0 C) = +10.0 C. The initial net charge must also be +10.0 C. The only spheres
whose net charge is +10.0 C are
B (qB =  2.0  C) and D (qD = +12.0  C) .
b. Since the final charge on each sphere is +3.0 C, the final net charge on the three spheres
is 3(+3.0 C) = +9.0 C. The initial net charge must also be +9.0 C. The only spheres
whose net charge is +9.0 C are
A (qA =  8.0  C), C (qC = +5.0  C) and D (qD = +12.0  C) .
c. Since the final charge on a given sphere in part (b) is +3.0 C, we would have to add
3.0 C 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
19
1.6  10 C
______________________________________________________________________________
Number of electrons 
70. CONCEPT QUESTIONS
a. The electrical force that each charge exerts on the middle charge is shown in the drawing
below. 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
(a)
+q
+q
F23 +q
3
1
2
F21
+q
+q
3
1
F21 q
2
(b)
3
(c)
F23
+q
Chapter 18 Problems
809
b. The net electric force F that acts on 2 is shown in the diagrams below.
F21
F23
F21
F23
F
F=0N
(a)
(b)
F23
F21
F
(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
F21  F23 
kq q
r2
8.99  109 N  m2 /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, both F21 and F23 point to the left, so the net force has a magnitude
of


F  2F12  2 4.6  104 N  9.2  104 N
In part (b) of the drawing, F21 and F23 point in opposite directions, so the net force has a
magnitude of 0 N .
In part (c) the magnitude can be obtained from the Pythagorean theorem:
2
2
F  F21
 F23

 4.6  104 N    4.6  104 N 
2
2
 6.5  104 N
______________________________________________________________________________
71. CONCEPT QUESTIONS
a. 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.
810
ELECTRIC FORCES AND ELECTRIC FIELDS
b. 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.
SOLUTION Since the repulsive electrical force neutralizes the attractive gravitational
force, the magnitudes of the two forces are equal:
kq q

r2
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 
5.98  1024 kg 7.35  1022 kg
 6.67  10
2 
GM e M m
kg 
q 
 
 5.71  1013 C
2
k
Nm
8.99  109
C2
______________________________________________________________________________



72. CONCEPT QUESTIONS
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  N

 2.0 N /C
q 20.0  C
As mentioned in the answer to Concept Question (a), the direction of the electric field is the
same as the direction of the force, or due east .
b. The magnitude of the electric field is
Chapter 18 Problems
E
811
F 20.0  N

 2.0 N /C
q 10.0  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.
______________________________________________________________________________
73. CONCEPT QUESTION
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 C charge points away from it, and the electric fields E3 and E5
produced by the 3.0 C and 5.0 C charges point toward them (see the drawing that
follows). 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 C (2.0 C +
3.0 C). 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 C. Thus, the x and y components are equal, as indicated in the right drawing, where the
net electric field E is also shown.
5.0 C
E5
Ey
E+2
+2.0 C
E
3.0 C
Ex
E3
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 in the left drawing. These
fields also produce the same net electric field E as before, as indicated in the following
drawing.
E+6
+1.0 C
Ey
E1
+4.0 C
E
1.0 C
E+4
E+1
+6.0 C
Ex
812
ELECTRIC FORCES AND ELECTRIC FIELDS
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 /C 2 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).
______________________________________________________________________________
74. CONCEPT QUESTIONS
a. 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).
b. 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.
c. 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
Chapter 18 Problems
813
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 electron is
a
q0 E
m
1.60  1019 C 8.0  104 N /C 



9.11  10
31
kg
1.4  1016 m /s 2
The magnitude of the acceleration of the proton is
a
q0 E
1.60  1019 C  8.0  104 N /C 



27
7.7  1012 m /s 2
m
1.67  10
kg
______________________________________________________________________________
75. CONCEPT QUESTIONS
a. 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 same
2
FBC
+
FBA
FAB
magnitude, namely, F  k q / L , where q is the
magnitude of each of the charges and L is the A
length of each side of the equilateral triangle. The

magnitude is the same for each force, because q
and L are the same for each pair of charges.
2
B
+
FAC
FCA
C
FCB
b. 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.
c. 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
814
ELECTRIC FORCES AND ELECTRIC FIELDS
directions but the same magnitude. The magnitude of the net force is the smallest at these
two corners.
SOLUTION As pointed out in the answer
to Concept Question (a), the magnitude of
+y
2
any 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  Fx and the
FAB
FAB sin 60.0º
FAC
60.0
º
A
y component  Fy of the net force are
+x
FAB cos 60.0
º
  Fx A  FAB cos 60.0  FAC  F  cos 60.0  1
  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
  F A    Fx A    Fy A
2
F
2
 F 2  cos 60.0  1   F sin 60.0 
2
 cos 60.0  1   sin 60.0   k

2
 8.99 109 N  m 2 / C 2
 430 N
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
Chapter 18 Problems
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:
815
+y
FBC
60.0º
FBC cos 60.0º
+x
B
FBA cos 60.0
º
FBA
  Fx B   FBC cos 60.0  FBA cos 60.0  2 F cos 60.0
  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
  F B    Fx B    Fy B   2 F cos 60.0 2   0 2
2
 2k
2
q
2
L
2

cos 60.0  2 8.99 109 N  m 2 / C 2

 2 F cos 60.0
5.0 106 C 
 0.030 m 
2
2
cos 60.0
 250 N
As discussed in the answer to Concept Question (c), 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
 F C  250 N .
These values of 430 and 250 N for the magnitudes of the net forces at corners A and B,
respectively, are consistent with our answers to the Concept Questions.
76. CONCEPT QUESTIONS
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
816
ELECTRIC FORCES AND ELECTRIC FIELDS
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
d
2

k q2
 2d 
2
2
k q1
d
2
or
k q2
 2d 
2
=
k q1
d2
Solving for q2 gives
q2  4 q1  4  0.50  C   2.0  C
Thus, the second charge is q2 = +2.0 μC , which is consistent with our answer to Concept
Question (a).
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  C   6.0  C
k q1
d2
Chapter 18 Problems
817
Thus, the second charge is q2 = 6.0 μC , which is consistent with our answer to Concept
Question (b).