Solutions to Problems
1.
Use Coulomb’s law to calculate the magnitude of the force.
F k
2.
2
 13.47 N  13 N
2
2
2

Q1Q2
r
2

 8.988  10 N  m C
9
2
2

1.602 10
19

C 26 1.602 10 19 C
1.5 10
12
m

2
  2.7 10
3
N
Use Coulomb’s law to calculate the magnitude of the force.
F k
5.

 8.988  10 N  m
2
Use Coulomb’s law to calculate the magnitude of the force.
F k
4.
r2
 3.60 10 C
C 
 9.3 10 m 
6
9
Use the charge per electron to find the number of electrons.
 1 electron 
14
30.0  106 C 
  1.87  10 electrons
19
 1.602  10 C 

3.
Q1Q2
Q1Q2
r2

1.602 10 C 
C 
 5.0 10 m 
19
 8.988  10 N  m
9
2
2
 9.2 N
2
15
2
Use Coulomb’s law to calculate the magnitude of the force.
25  106 C 3.0  103 C
Q1Q2
9
2
2
F  k 2  8.988  10 N m C
 5.5  103 N
2

1
r
3.5  10 m







6.
Since the magnitude of the force is inversely proportional to the square of the separation
1
distance, F  2 , if the distance is multiplied by a factor of 1/8, the force will be multiplied
r
by a factor of 64.
F  64 F0  64  3.2  102 N   2.0 N
7.
Since the magnitude of the force is inversely proportional to the square of the separation
1
distance, F  2 , if the force is tripled, the distance has been reduced by a factor of 3 .
r
r0
8.45 cm
r

 4.88 cm
3
3
8.
Use the charge per electron and the mass per electron.
1 electron
 42  10 C   1.602
 10

14
14
  2.622  10  2.6  10 electrons
C
31
 2.622 1014 e   9.11110e kg   2.4 1016 kg


6
19
9.
Convert the kg of H2O to moles, then to atoms, then to electrons. Oxygen has 8 electrons
per atom,
and hydrogen has 1 electron per atom.
 1 mole H 2 O   6.02  1023 molec.   10 e   1.602  10 19 C 
  1 molec.  


2
1 mole
e

 1.8  10 kg 


1.0 kg H 2 O  1.0 kg H 2 O  
 5.4  107 C
10.
Take the ratio of the electric force divided by the gravitational force.
QQ
2
k 12 2
8.988  109 N  m 2 C2 1.602  1019 C
FE
kQ1Q2
r



 2.3  1039
m1m2 Gm1m2
FG
6.67  1011 N  m 2 kg 2 9.11  1031 kg 1.67  1027 kg
G 2
r







The electric force is about 2.3 1039 times stronger than the gravitational force for the
given scenario.
11.
(a)
between the
Let one of the charges be q , and then the other charge is QT  q . The force
charges is FE  k
then FE 
k
r2

q  QT  q 
r2
 q  q 2 
q
,
 2  qQT  q   2 Q 
    . If we let x 


QT
r
r
 QT  QT  
k
2
k
2
T

QT2 x  x 2 , where 0  x  1 . A graph of f  x   x  x 2 between the
limits of 0 and 1 shows that the maximum occurs at x  0.5 , or q  0.5 QT . Both
charges are half of the total, and the actual maximized force is FE  0.25
k
2
QT2 .
r
(b)
If one of the charges has all of the charge, and the other has no charge, then the
force between
them will be 0, which is the minimum possible force.
12. Let the right be the positive direction on the line of charges. Use the fact that like charges
repel and unlike charges attract to determine the direction of the forces. In the following
9
2
2
expressions, k  8.988 10 N  m C .
 75 C  48 C   75 C  85 C 
k
 147.2 N  1.5  10 2 N
2
2
 0.35 m 
 0.70 m 
 75 C  48 C   48 C  85 C 
F48  k
k
 563.5 N  5.6  10 2 N
2
2
 0.35 m 
 0.35 m 
 85 C  75 C   85 C  48 C 
F85   k
k
 416.3 N  4.2  10 2 N
2
2
 0.70 m 
 0.35 m 
F75   k
13. The forces on each charge lie along a line connecting the charges. Let the
variable d represent the length of a side of the triangle, and let the variable Q
represent the charge at each corner. Since the triangle is equilateral, each angle is
60o.
Q2
Q2
Q2
F12  k 2  F12 x  k 2 cos 60o , F12 y  k 2 sin 60 o
d
d
d
F13  k
Q
2
d
2
 F13 x   k
F1x  F12 x  F13 x  0
F1 
F  F  3k
2
1x
2
1y
Q2
d2
Q
2
d
2
cos 60o , F13 y  k
F1 y  F12 y  F13 y  2k

Q
2
d
2
d2

d
Q2
Q3
d
sin 60o  3k
2
F12
Q1
d
sin 60o
2
 3 8.988  10 N  m C
9
Q
2
F13
Q2
d2
11.0 10 C 
6
 0.150 m 2
2
 83.7 N
The direction of F1 is in the y-direction . Also notice that it lies along the bisector of the
opposite side of the triangle. Thus the force on the lower left charge is of magnitude
83.7 N , and will point 30o below the  x axis . Finally, the force on the lower right
charge is of magnitude 83.7 N , and will point 30o below the  x axis .
14. Determine the force on the upper right charge, and then use the
symmetry of the configuration to determine the force on the other three
charges. The force at the upper right corner of the square is the vector
sum of the forces due to the other three charges. Let the variable d
represent the 0.100 m length of a side of the square, and let the variable
Q represent the 6.00 mC charge at each corner.
F41  k
F42  k
F43  k
Q2
d2
Q2
2d 2
Q2
 F41 x  k
Q2
d2
 F42 x  k
Q1
, F41 y  0
2d 2
cos45o  k
 F43 x  0 , F43 y  k
2Q 2
4d 2
, F42 y  k
Q4
F41
d
Q2
Q2
F42
F43
2Q 2
4d 2
Q2
d2
d2
Add the x and y components together to find the total force, noting that F4 x  F4 y .
Q3
F4 x  F41x  F42 x  F43 x  k
F4  F  F  k
2
4x
2
4y
Q2 
4d
2
0 k
Q2 
2
1 
  F4 y
d 
4 
2
Q2 
1
1 
 2 k 2  2 
d 
4 
d 
2
2

F4 y
d
2
2Q 2
k
2
 8.988  109 N  m 2 C 2
  tan 1
Q2

 6.00 10 C 
3
 0.100 m 
2
2
1

7
 2    6.19 10 N
2

 45o above the x-direction.
F4 x
For each charge, the net force will be the magnitude determined above, and will lie along the
line from the center of the square out towards the charge.
15. Determine the force on the upper right charge, and then the symmetry of the configuration
says that the force on the lower left charge is the opposite of the force on the upper right
charge. Likewise, determine the force on the lower right charge, and then the symmetry of
the configuration says that the force on the upper left charge is the opposite of the force on
the lower right charge.
The force at the upper right corner of the square is the vector sum of the
forces due to the other three charges. Let the variable d represent the
0.100 m length of a side of the square, and let the variable Q represent
the 6.00 mC charge at each corner.
Q1
F41
F42
Q4
F43
d
F41  k
F42  k
F43  k
Q2
d2
Q2
2d 2
Q2
 F41 x   k
 F42 x  k
Q2
d2
Q2
2d 2
, F41 y  0
Q2
2Q 2
cos45o  k
 F43 x  0 , F43 y   k
4d 2
, F42 y  k
2Q 2
4d 2
Q2
d2
d2
Add the x and y components together to find the total force, noting that F4 x  F4 y .
F4 x  F41x  F42 x  F43 x  k
F4  F42x  F42y  k
d2
Q2
d
k
 0.64645
2
2Q 2
4d 2
0 k
2k
Q2
d2

9
F4 y
F4 x
2
Q2 
2
Q2

1



0.64645
k
 F4 y


d2 
4 
d2
 0.9142 
 6.00 10 C   0.9142  2.96 10 N
C 
2
3
 8.988  10 N  m
  tan 1
Q2
2
7
 0.100 m 
2
 225o from the x-direction, or exactly towards the center of the square.
For each charge, the net force will be the magnitude of 2.96 107 N and each net force will
lie along the line from the charge inwards towards the center of the square.
Q3
16.
Take the lower left hand corner of the square to be the origin of coordinates. Each charge
will have a
horizontal force on it due to one charge, a vertical force on it due to one charge, and a
diagonal force on it due to one charge. Find the components of each force, add the
components, find the magnitude of the net force, and the direction of the net force. At the
conclusion of the problem is a diagram showing the net force on each of the two charges.
Q2
kQ 2
 2Q  Q  2Q  4Q 
o
(a)
2Q : F2 Qx  k

k
cos
45

k
2

2
2

4.8284
l2
2l 2
l2
l2
Q2
kQ 2
 2Q  3Q   2Q  4Q 
o
F2 Qy  k

k
sin
45

k
6

2
2

8.8284
l2
2l 2
l2
l2


F2 Q  F22Qx  F22Qy  10.1
3Q : F3Qx  k
(b)
F3Qy  k
kQ 2
l
 3Q  4Q 
l2
 3Q  2Q 
l2
k
k
 3Q  Q
 3Q  Q
2l 2
F3Q  F32Qx  F32Qy  14.8
2l 2
F2 x
8.8284
4.8284
 61o
Q 
3
kQ 2

cos 45  k 2 12 
2   13.0607 2
l 
4
l

2Q
Q
 tan 1
o
Q2 
3
kQ 2

6

2


7.0607


l2 
4
l2

 3Q  tan 1
2

2
sin 45o  k
kQ 2
l
F2 y
 2 Q  tan 1
2

F3 y
F3 x
 tan 1
7.0607
13.0607
 332o
F2Q
l
3Q F3Q
4Q
F13
17. The forces on each charge lie along a line connecting the charges. Let
the variable d represent the length of a side of the triangle. Since the
triangle is equilateral, each angle is 60o. First calculate the magnitude
of each individual force.
F12  k
Q1Q2
d2

 8.988  109 N  m 2 C 2

 0.1997 N  F21
F13  k
Q1Q3
d2

 8.988  10 N  m C
 0.1498 N  F31
9
2
2

 4.0 10 C 8.0 10 C 
6
6
1.20 m 
F23
6
F21
6
2
Q1
d
Q2
2
 4.0  10 C  6.0 10 C 
1.20 m 
d
F12
Q3
d
F32
F31
F23  k
Q2Q3
d2

 8.988  109 N  m 2 C 2

8.0 10 C  6.0 10 C   0.2996 N  F
6
6
1.20 m 
32
2
Now calculate the net force on each charge and the direction of that net force, using
components.
F1x  F12 x  F13 x    0.1997 N  cos 60o   0.1498 N  cos 60o  2.495  102 N
F1 y  F12 y  F13 y    0.1997 N  sin 60o   0.1498 N  sin 60o  3.027  10 1 N
1  tan
F1  F  F  0.30 N
2
1x
2
1y
1
F1 y
 tan
F1x
3.027  101 N
1
 265o
2
2.495  10 N
F2 x  F21x  F23 x   0.1997 N  cos 60   0.2996 N   1.998  101 N
o
F2 y  F21 y  F23 y   0.1997 N  sin 60o  0  1.729  101 N
 2  tan 1
F2  F22x  F22y  0.26 N
F2 y
F2 x
 tan 1
1.729  101 N
1
1.998  10 N
 139o
F3 x  F31x  F32 x    0.1498 N  cos 60   0.2996 N   2.247  101 N
o
F3 y  F31 y  F32 y   0.1498 N  sin 60o  0  1.297  101 N
3  tan 1
F3  F32x  F32y  0.26 N
F3 y
F3 x
 tan 1
1.297  101 N
2.247  101 N
 30o
18.
Since the force is repulsive, both charges must be the same sign. Since the total charge is
positive,
both charges must be positive. Let the total charge be Q. Then if one charge is of magnitude
q, then the other charge must be of magnitude Q  q . Write a Coulomb’s law expression for
one of the charges.
q Q  q 
Fr 2
2
F k

q

Qq

0 
r2
k
Q Q 
2
q
2
4

4 Fr 2
k
 
560  10 6 C 


2
560  10 6 C 
4  22.8 N 1.10 m 
8.988 10
9
2
N  m 2 C2

2
6
 5.54  10 C , 5.54  10 C
6
Q  q  5.54  10 C , 5.54  10 4 C
19. The negative charges will repel each other, and so the third charge
Q
Q0
3Q0
must put an opposite force on each of the original charges.
x
l–x
Consideration of the various possible configurations leads to the
conclusion that the third charge must be positive and must be between
l
the other two charges. See the diagram for the definition of variables.
For each negative charge, equate the magnitudes of the two forces on the charge. Also note
that 0  x  l .
left: k
k
k
Q0Q
x
2
Q0Q
x
2
Q0Q
x2
k
k
k
3Q02
l2
3Q0Q
l  x
3Q02
l
2
2
3Q0Q
right: k
l  x
l
 x
3 1
 Q  3Q0
x2
l
2
2
3Q02
k

l2
 0.366l
 Q0

3

3 1
2
 0.402Q0
Thus the charge should be of magnitude 0.40 Q0 , and a distance
0.37 l from  Q0 towards  3Q0 .
20. Assume that the negative charge is d = 18.5 cm to the right of the
Q1
Q2
Q
positive charge, on the x-axis. To experience no net force, the
4.7 C
–3.5 C
third charge Q must be closer to the smaller magnitude charge
–
+
x
d
(the negative charge). The third charge cannot be between the
charges, because it would experience a force from each charge in
the same direction, and so the net force could not be zero. And the third charge must be on
the line joining the other two charges, so that the two forces on the third charge are along the
same line. See the diagram. Equate the magnitudes of the two forces on the third charge,
and solve for x > 0.
F1  F2
xd

 k
Q1 Q
d  x
Q2
Q1 
Q2

2
Q2 Q
k
x2
 18.5cm 

 xd
Q2

Q1 
Q2
3.5  106 C
4.7  106 C  3.5  106 C


 116 cm
21.
(a)
If the force is repulsive, both charges must be positive since the total charge is
positive. Call the
total charge Q.
kQ  Q  Q 
kQ Q
Fd 2
Q1  Q2  Q
F  12 2  1 2 1
 Q12  QQ1 
0
d
d
k
Q  Q2  4
Q1 
2
Fd 2
k
Q  Q2  4


k
2
 90.0 10 C    90.0  10 C 
6
Fd 2
6
2
4
12.0N 1.06 m 
8.988 10
9
2
N  m 2 C2

2
 69.9  10 6 C , 22.1  10 6 C
(b)
If the force is attractive, then the charges are of opposite sign. The value used for
F must then
be negative. Other than that, the solution method is the same as for part (a).
Q1  Q2  Q
F
Q  Q2  4
Q1 
kQ1Q2
d2
Fd 2
k
2

kQ1  Q  Q1 
d2
Q  Q2  4

 Q12  QQ1 
Fd 2
k
0
Fd 2
k
2
 90.0 10 C    90.0  10 C 
6
6

2
4
 12.0N 1.06 m 
8.988 10
9
2
N  m 2 C2

2
 104.4  10 6 C ,  14.4  10 6 C
22.
The spheres can be treated as point charges since they are spherical, and so Coulomb’s
law may be
used to relate the amount of charge to the force of attraction. Each sphere will have a
magnitude Q
of charge, since that amount was removed from one sphere and added to the other, being
initially uncharged.
F k
Q1Q2
r2
k
Q2
r2
 Qr
F
k
  0.12 m 
1.7  102 N
8.988  109 N  m 2 C2
 1 electron 
12
  1.0  10 electrons
19
 1.602  10 C 
 1.650  107 C 
23.
Use Eq. 16–3 to calculate the force.
F
E
 F  qE   1.602  1019 C   2360 N C east   3.78  10 16 N west
q
24.
Use Eq. 16–3 to calculate the electric field.
F 3.75 1014 N south
E 
 2.34  105 N C south
19
q
1.602 10 C
25.
Use Eq. 16–3 to calculate the electric field.
F 8.4 N down
E 
 9.5  105 N C up
q 8.8  106 C
26.
Use Eq. 16–4a to calculate the electric field due to a point charge.
Q
33.0  106 C
E  k 2  8.988  109 N  m 2 C 2
 7.42  106 N C up
2
1
r
2.00  10 m




Note that the electric field points away from the positive charge.
27.
Assuming the electric force is the only force on the electron, then Newton’s 2nd law may
be used to
find the acceleration.
1.602  1019 C
q
Fnet  ma  qE  a  E 
750 N C  1.32  1014 m s 2
m
9.11 1031 kg




Since the charge is negative, the direction of the acceleration is opposite to the field .
28. The electric field due to the negative charge will point
Q1  0
toward the negative charge, and the electric field due
to the positive charge will point away from the
positive charge. Thus both fields point in the same
d 2
direction, towards the negative charge, and so can be
added.
Q
Q
Q1
Q2
4k
E  E1  E2  k 21  k 22  k
k
 2  Q1  Q2 
2
2
r1
r2
 d / 2
 d / 2 d


4 8.988  109 N  m 2 C2
8.0 10 m 
2
2

8.0 10
6
E1
E2

C  7.0  106 C  8.4  107 N C
The direction is towards the negative charge .
29.
30.
Assuming the electric force is the only force on the electron, then Newton’s 2nd law may
be used to
find the electric field strength.
Fnet  ma  qE  E 
ma
q
1.67 10

27

kg 1  106
1.602 10
19
 9.80 m s   0.102 N C  0.1N C
C
2
31.
Since the electron accelerates from rest towards the north, the net force on it must be to
the north.
Assuming the electric force is the only force on the electron, then Newton’s 2nd law may be
used to
find the electric field.
Q2  0
m
Fnet  ma  qE  E 
q
 9.1110
a
31
 1.602 10
kg
19

C

115 m s
2

north  6.54  10 10 N C south
32.
The field due to the negative charge will point
towards
the negative charge, and the field due to the positive
charge will point towards the negative charge. Thus
the magnitudes of the two fields can be added together
to find the charges.
Enet  2 EQ  2k
Q
 d / 2
2

8kQ
d2
 Q
Ed 2
8k

d 2
E  E1  E2  k
Q1
2
d 2
k
Q2
2
d 2

 8.988  109 N  m 2 C2
34.
of
k
EQ
 745 N C  1.60  101 m 

8 8.988  109 N  m 2 C 2
33. The field at the center due to the two negative charges on opposite
corners (lower right and upper left in the diagram) will cancel
each other, and so only the positive charge and the opposite
negative charge need to be considered. The field due to the
negative charge will point directly toward it, and the field due to
the positive charge will point directly away from it. Accordingly,
the two fields are in the same direction and can be added
algebraically.
Q
EQ
Q

 2.65  10 10 C
Q2  27.0 C
Q2
E1
d
E2
Q2
Q1  45.0 C
Q1  Q2
d2 2
 47.0  27.0   106 C


2
 0.525 m  2
4.70  106 N C at 45o
The field at the upper right corner of the square is the vector sum
E3
the fields due to the other three charges. Let the variable d represent Q1
the 1.0 m length of a side of the square, and let the variable Q
represent the charge at each of the three occupied corners.
d
E1  k
E2  k
E3  k
Q
d
 E1x  k
2
Q
2d
Q
2
2
Q
d2
 E2 x  k
, E1 y  0
Q2
Q
2d
2
cos45o  k
 E3 x  0 , E1 y  k
2Q
4d
2
, E2 y  k
2Q
4d 2
Q
d
d2
Add the x and y components together to find the total electric field, noting that Ex  E y .
E2
E1
Q3
Q
Ex  E1x  E2 x  E3 x  k
E  Ex2  E y2  k
Ex
4d
2
0 k
Q
2
1 
  Ey
d 
4 
2
2
Q
1
1


 2k 2 2 
2
d 
4 
d 
2

Ey
2Q
Q
 8.988  109 N  m 2 C 2
  tan 1
d
k
2

 2.25 10 C  
6
1.00 m 
2
1
4
 2    3.87  10 N C
2

 45o from the x-direction.
35.
Choose the rightward direction to be positive. Then the field due to +Q will be positive,
and the
field due to –Q will be negative.
Ek
Q
 x  a
2
k
 1

1
4kQxa


  x  a 2  x  a 2 
2
2 2

 x  a 
Q
 x  a
 kQ 
2
The negative sign means the field points to the left .
36.
For the net field to be zero at point P, the magnitudes of the fields created by Q1 and Q2
must be
equal. Also, the distance x will be taken as positive to the left of Q1 . That is the only
region where the total field due to the two charges can be zero. Let the variable d represent
the 12 cm distance, and note that Q1  12 Q2 .
E1  E 2
xd

 k
Q1
Q1
Q2 
Q1
x
2

k
d
Q2
x  d

1
2
Q2 

2
Q2
1
2

Q2
 
d

2 1

12 cm
2 1
 29 cm
37. (a) The field due to the charge at A will point straight downward, and the
field due to the charge at B will point along the line from A to the
origin, 30o below the negative x axis.
A
Q
l
Q
l
B
l
EB
EA
EA  k
EB  k
Q
 EAx  0 , EAx   k
2
l
Q
l
Q
 EBx   k
2
2
EBy   k
Ex  EAx  EBx   k
E  Ex2  E y2 
  tan 1
Ey
Ex
l
Q
l2
3Q
2l
4l
4
l2
cos 30o   k
sin 30o   k
3Q
2l 2
Q
2l 2
3Q
E y  EAy  EBy   k
2
3k 2Q 2
Q

9k 2 Q 2
4l
12k 2Q 2

4
4l
4
2l 2
3kQ

l2
3Q
k
2l 2  tan 1 3  tan 1 3  240o
3Q
 3
 tan 1
k
2l 2
(b) Now reverse the direction of E A
EA  k
EB  k
Q
l
2
Q
l
2
 EAx  0 , EAx   k
 EBx  k
Ex  EAx  EBx  k
E  Ex2  E y2 
  tan 1
Ey
Ex
Q
l
2
3Q
2l
4l
4
k
 tan 1
k
l2
cos 30o  k
3Q
2l
2
, EBy  k
E y  EAy  EBy   k
2
3k 2Q 2
Q

k 2Q 2
4l
4
4k 2 Q 2

4l
4

Q
l
2
sin 30o  k
Q
2l 2
Q
2l 2
kQ
l2
Q
2l 2  tan 1 1  330o
3Q
 3
2l 2


38. In each case, find the vector sum of the field caused by the charge on the left E left and the

field caused by the charge on the right E right

Point A: From the symmetry of the geometry, in
calculating the electric field at point A only the vertical
components of the fields need to be considered. The
horizontal components will cancel each other.
5.0
  tan 1
 26.6o
10.0
d
 5.0 cm   10.0 cm 
2
2
 0.1118 m
E right
Eleft
A
d
Q

d

Q
EA  2
kQ
d
2

sin  2 8.988 109 N  m 2 C2
 7.0 10
6
C
 0.1118 m 
2
sin 26.6o  4.5 106 N C
Point B: Now the point is not symmetrically placed, and
so horizontal and vertical components of each individual
field need to be calculated to find the resultant electric
field.
5.0
5.0
 left  tan 1
 45o
 left  tan 1
 18.4o
5.0
15.0
d left 
d right 
 5.0 cm    5.0 cm 
2
 5.0 cm 2  15.0 cm 2
Ex   Eleft  x   E right  x  k
Q
d

 8.988  109 N  m 2 C 2
E y   Eleft  y   E right  y  k

 8.988  109 N  m 2 C 2
E right
Q
2
d right
Q

cos45o
Q
sin  right
6
  0.0707 m 

 right
 left
Q
cos  right
 7.0 10 C  
2
left
d right
d left
 0.1581m
cos left  k
2
left
d
Eleft
 0.0707 m
2
 A  90o
Q
sin left  k
2
d right


7.0  10 6 C 
EB  Ex2  E y2  1.2  107 N C
2

2

sin45o
  0.0707 m 
 B  tan 1
cos18.4o 
6
  6.51 10 N C
 0.1581m  
2
sin18.4o 
6
  9.69  10 N C
 0.1581m  
2
Ey
 56o
Ex
The results are consistent with Figure 16-31b. In the figure, the field at Point A points
straight up, matching the calculations. The field at Point B should be to the right and
vertical, matching the calculations. Finally, the field lines are closer together at Point B than
at Point A, indicating that the field is stronger there, matching the calculations.
39. Both charges must be of the same sign so that the electric fields created by the two charges
oppose each other, and so can add to zero. The magnitudes of the two electric fields must be
equal.
E1  E2  k
Q1
 l 3
2
k
Q2
 2l 3
2
 9Q1 
9Q2
4

Q1
Q2

1
4
40. From the diagram, we see that the x components of the two fields will cancel each other at
the point P. Thus the net electric field will be in the
Q
negative y-direction, and will be twice the ycomponent of either electric field vector.
a

x
a
Q
E Q
EQ
Enet  2 E sin   2


2kQ
kQ
x  a2
a
2

x  a2 x2  a2
2
2kQa
x
2
a
2

sin 

1/ 2
in the negative y direction
3/ 2
41.
We assume that gravity can be ignored, which is proven in part (b).
(a)
The electron will accelerate to the right. The magnitude of the acceleration can
be found from
setting the net force equal to the electric force on the electron. The acceleration is
constant, so constant acceleration relationships can be used.
Fnet  ma  q E  a 
qE
m
v 2  v02  2ax  v  2ax  2
qE
m
x
1.602 10 C 1.45 10
2
 9.1110 kg 
19

(b)
4
31
N C

1.10 10 m   7.49 10
2
6
m s
The value of the acceleration caused by the electric field is compared to g.
1.602  1019 C 1.45  10 4 N C
qE
a

 2.55  1015 m s 2
31
m
9.11  10 kg

a





2.55  1015 m s 2
 2.60  1014
g
9.80 m s 2
The acceleration due to gravity can be ignored compared to the acceleration
caused by the
electric field.
42.
(a)
The electron will experience a force in the opposite direction to the electric field.
Since the
electron is to be brought to rest, the electric field must be in the same direction as the
initial velocity of the electron, and so is to the right .
(b) Since the field is uniform, the electron will experience a constant force, and therefore
have a constant acceleration. Use constant acceleration relationships to find the field
strength.
qE
F  qE  ma  a 
E

m v 2  v02
2qx
m
  mv
v 2  v02  2ax  v02  2
qE
m
x 
 9.1110 kg  3.0 10 m s   6.4 10

2qx
2  1.602  10 C  4.0  10 m 
2
0
31
2
6
19
2
2
N C
43.
Use Gauss’s law to determine the enclosed charge.
Q
 E  encl  Qencl   E o  1.45  103 N  m 2 C 8.85  10 12 C 2 N  m 2  1.28  10 8 C
o



 


2
44. (a)  E  E A  E r 2  5.8 102 N C  1.8 101 m  59 N  m2 C



 E A   E cos 90   r  0

 
(b)  E  E A  E cos 45o  r 2  5.8 102 N C cos 45o  1.8 10 1 m
(c)  E
o

2
 42 N  m 2 C
2

45. (a) Use Gauss’s law to determine the electric flux.
Q
1.0  106 C
 E  encl 
 1.1 105 N  m 2 C
12
2
2
o
8.85  10 C N  m
(b) Since there is no charge enclosed by surface A2, E  0 .
46. (a) Assuming that there is no charge contained within the cube, then the net flux through the
cube is
0 . All of the field lines that enter the cube also leave the cube.
(b)
There are four faces that have no flux through them,
because none of the field lines pass through those faces.
In the diagram shown, the left face has a positive flux
and the right face has the opposite amount of negative
flux.


 left  EA  El 2  6.50  103 N C l 2


 right   6.50  103 N C l 2
47.
Equation 16-10 applies.
E
48.
l
 other  0
Q A
0


 Q   0 EA  8.85  1012 C 2 N  m 2 130 N C 1.0 m   1.15  10 9 C
2
The electric field can be calculated by Eq. 16-4a, and that can be solved for the charge.
Ek
Q
r
2
 Q
Er 2
k
 2.75 10

2

N C 3.50  10 2 m
8.988  10 N  m C
9
2
2

2
 3.75  10 11 C
This corresponds to about 2  10 electrons. Since the field points toward the ball, the
charge must be negative .
8
49.
See Example 16-11 for a detailed discussion related to this problem.
(a) Inside a solid metal sphere the electric field is 0 .
(b)
Inside a solid metal sphere the electric field is 0 .
(c)
Outside a solid metal sphere the electric field is the same as if all the charge were
concentrated
at the center as a point charge.
E k
(d)
50.
r
2

 8.988  109 N  m 2 C 2

 3.50 10 C   3.27 10

 3.50 10 C   8.74 10
Same reasoning as in part (c).
E k
(e)
Q
Q
r
2

 8.988  109 N  m 2 C 2
6
 3.10 m 
3
N C
2
N C
2
6
 6.00 m 
2
The answers would be no different for a thin metal shell.
See Figure 16-33 in the text for additional insight into this problem.
(a) Inside the shell, the field is that of the point charge, E  k
(b)
Q
r2
.
There is no field inside the conducting material: E  0 .
(c) Outside the shell, the field is that of the point charge, E  k
Q
.
r2
(d)
The shell does not affect the field due to Q alone, except in the shell material,
where the field is
0. The charge Q does affect the shell – it polarizes it. There will be an induced charge
of –Q uniformly distributed over the inside surface of the shell, and an induced charge
of +Q uniformly distributed over the outside surface of the shell.
51.
(a)
The net force between the thymine and adenine is due to the following forces.
 0.4e  0.2e  0.08ke 2
FOH  k

O – H attraction:
2
2
O – N repulsion:
FON  k
N – N repulsion:
FNN  k
H – N attraction:
FHN  k
 1.80 Ao 




 1.80 Ao 




 0.4e  0.2e 
0.08ke 2
 2.80 Ao 




2
 0.2e  0.2e 
 3.00 Ao 




2
 0.2e  0.2e 
 2.00 Ao 




2



 2.80 Ao 




2
0.04ke 2
 3.00 Ao 




2
0.04ke 2
 2.00 Ao 




2
2
2
1
 0.08 0.08 0.04 0.04 
 ke



 2
2
2
2
2 
10
 1.80 2.80 3.00 2.00  1.0  10 m  d
FA-T  FOH  FON  FNN  FHN  
8.988 10
 .02004 
9

N  m 2 C2 1.602 1019 C
1.0 10
10
m

2

2
 4.623 1010 N  4.6 1010 N
(b)
The net force between the cytosine and guanine is due to the following forces.
 0.4e  0.2e  0.08ke 2
FOH  k

O – H attraction:
(2 of
2
2
 1.90 Ao 




these)
O – N repulsion:
FON  k
H – N attraction:
FHN  k
N – N repulsion:
FNN  k
(2 of these)
 1.90 Ao 




 0.4e  0.2e 
 2.90 Ao 




2
 0.2e  0.2e 
 2.00 Ao 




2
 0.2e  0.2e 
 3.00 Ao 




2



0.08ke 2
 2.90 Ao 




2
0.04ke 2
 2.00 Ao 




2
0.04ke 2
 3.00 Ao 




2
2
FC-G
2
0.08 0.04 0.04 
1
 0.08
 ke
 2 FOH  2 FON  FNN  FHN   2

2




2
2.902 3.002 2.002  1.0  10 10 m  d 2
 1.90
8.988 10
 .03085 
9

N  m 2 C 2 1.602  10 19 C
1.0 10
10
m


2
2
 7.116  10 10 N  7.1  10 10 N
(c) For 10 5 pairs of molecules, we assume that half are A-T pairs and half are C-G pairs. We
average the above results and multiply by 10 5 .

Fnet  12 105  FA-T  FC-G   105 4.623  1010 N  7.116  1010 N

 5.850  105 N  6  105 N
52.
Set the magnitude of the electric force equal to the magnitude of the force of gravity and
solve for
the distance.
e2
FE  FG  k 2  mg 
r
re
k
mg

19
 1.602  10 C

8.988 10
 9.1110
31
9
N  m 2 C2


kg 9.80 m s 2

 5.08 m
53.
Calculate the total charge on all electrons in 3.0 g of copper, and then compare the 38 C
to that
value.
 1 mole   6.02  1023 atoms   29 e   1.602  1019 C 
  atoms  


mole
1e

 63.5g 


Total electron charge  3.0 g 
 1.32  105 C
Fraction lost 
38  106 C
1.32  10 C
5
 2.9  1010
54.
Since the gravity force is downward, the electric force must be upward. Since the charge
is positive,
the electric field must also be upward. Equate the magnitudes of the two forces and solve
for the electric field.
FE  FG  qE  mg  E 
55.
mg
q
1.67 10 kg  9.80 m s   1.02 10

1.602 10 C 
27
2
7
19
N C , up
Use Eq. 16-4a to calculate the magnitude of the electric charge on the Earth.
Ek
150 N C   6.38 106 m 
Er 2
Q
2
 Q

 6.8  105 C
r2
k
8.988  109 N  m 2 C 2
Since the electric field is pointing towards the Earth’s center, the charge must be
negative .
56.
(a)
From problem 55, we know that the electric field is pointed towards the Earth’s
center. Thus an
electron in such a field would experience an upwards force of magnitude FE  eE .
The force of gravity on the electron will be negligible compared to the electric force.
FE  eE  ma 
a
eE
m
1.602 10 C  150 N C   2.638 10

 9.1110 kg 
19
13
31
m s 2  2.6  1013 m s 2 , up
(b)
A proton in the field would experience a downwards force of magnitude
FE  eE . The force of
gravity on the proton will be negligible compared to the electric force.
FE  eE  ma 
a
eE
m
1.602 10 C  150 N C   1.439 10

1.67 10 kg 
19
10
27
a
(c) For the electron:
For the proton:
g
a
g

2.638  1013 m s 2
9.80 m s
1.439  10 m s 2
2
 2.7  1012
10

9.80 m s
2
m s 2  1.4  1010 m s 2 , down
 1.5  109
57. For the droplet to remain stationary, the magnitude of the electric force on the droplet must
be the same as the weight of the droplet. The mass of the droplet is found from its volume
times the density of water. Let n be the number of excess electrons on the water droplet.
FE  q E  mg  neE  43  r 3  g 
n
4 r 3  g
3eE


 1.00 10 kg m 9.80 m s   9.96 10
3 1.602  10 C  150 N C 
4 1.8  105 m
3
3
3
2
6
19
 1.0  107 electrons
58.
There are four forces to calculate. Call the rightward direction the positive direction.
The value of k
is 8.988 109 N  m2 C2 and the value of e is 1.602 1019 C .
Fnet  FCH  FCN  FOH  FON 
k  0.40e  0.20e  
110 m 
9
2
1
1
1
1 




2
2
2
2 
  0.30   0.40   0.18   0.28  
 2.445  10 10 N  2.4  10 10 N
59.
orbit.
The electric force must be a radial force in order for the electron to move in a circular
FE  Fradial  k
rorbit  k
Q2
mv 2
Q2
2
rorbit

mv 2
rorbit

1.602 10 C 
C 
 9.1110 kg 1.110
19

 8.988  10 N  m
9
2
2
2
31
6
ms

2
 2.1  10 10 m
60.
Set the Coulomb electrical force equal to the Newtonian gravitational force on one of the
bodies (the
Moon).
FE  FG  k
Q
Q2
2
orbit
r
GM Moon M Earth
k
G

M Moon M Earth
2
rorbit
 6.67 10
11

N m 2 kg 2
 7.35 10
8.988 10
9
Nm
2

C 
22
kg 5.98 10 24 kg
2
  5.7110
13
C
61.
(a)
The electron will experience a force in the opposite direction to the electric field.
Thus the
acceleration is in the opposite direction to the initial velocity. The force is constant, and
so constant acceleration equations apply. To find the stopping distance, set the final
velocity to 0.
F  eE  ma  a 
x 
(b)
to return.
v 2  v02
2a
v 2  v02  2ax 
m
 9.1110 kg  21.5 10 m s   0.115 m


2eE
2  1.602  10 C 11.4  10 N C 
31
mv02
2
6
19
3
To return to the starting point, the velocity will reverse. Use that to find the time
v  v0  at 
t
eE
v  v0
a
v0  v0

a

2mv0
qE



C 11.4  10
  2.14  10
N C
2 9.11  10 31 kg 21.5  106 m s
 1.602 10
19
3
8
s
62.
Because of the inverse square nature of the electric
Q1
Q2
field,
any location where the field is zero must be closer to the
d
l
weaker charge  Q2  . Also, in between the two charges,
the fields due to the two charges are parallel to each other and cannot cancel. Thus the only
places where the field can be zero are closer to the weaker charge, but not between them. In
the diagram, this means that x must be positive.
Q
Q1
2
E   k 22  k
 0  Q2  l  d   Q1l 2 
2
l
l  d 
l
Q2
Q1 
d
Q2
5.0  106 C
2.5  105 C  5.0  106 C
 2.0 m  
1.6 m from Q2 ,
2.6 m from Q1
63.
The sphere will oscillate sinusoidally about the equilibrium point, with an amplitude of
5.0 cm. The
angular frequency of the sphere is given by   k m  126 N m 0.800 kg  12.5rad s .
The distance of the sphere from the table is given by r  0.150  0.050cos 12.5t  m . Use
this distance and the charge to give the electric field value at the tabletop. That electric field
will point upwards at all times, towards the negative sphere.
8.988  109 N  m 2 C 2 3.00  106 C
Q
2.70  104
Ek 2 

N C
2
2
r
0.150  0.050 cos 12.5t  m2
0.150  0.050 cos 12.5t 


64.

1.08  107
3.00  cos 12.5t 
2

N C , upwards
The wires form two sides of an equilateral triangle, and so the two charges are
separated by a distance d  78 cm and are directly horizontal from each other. Thus
the electric force on each charge is horizontal. From the free-body diagram for one of
the spheres, write the net force in both the horizontal and vertical directions and solve
for the electric force. Then write the electric force by Coulomb’s law, and equate the
two expressions for the electric force to find the charge.
FT
FE

mg
mg
F
 FT cos   mg  0  FT 
F
 FT sin   FE  0  FE  FT sin  
y
x
FE  k
 Q 2 2
d
2
cos 
 mg tan   Q  2d

 2 7.8  10
cos 
k
 24 10 kg  9.80 m s  tan 30
m
8.988 10 N  m C 
2
9
sin   mg tan 
mg tan 
3
1
mg
2
2
o
 6.064  106 C  6.1  10 6 C
65.
The electric field at the surface of the pea is given by Equation (16-4a). Solve that
equation for the
charge.
Ek
Q
 Q
 3 10

Er 2
6

N C 3.75  10 3 m
r
k
8.988  10 N  m C
This corresponds to about 3 billion electrons.
2
9
2
2

2
 5  10 9 C
66.
There will be a rightward force on Q1 due to Q2 , given by Coulomb’s law. There will be
a leftward
force on Q1 due to the electric field created by the parallel plates. Let right be the positive
direction.
QQ
 F  k x1 2 2  Q1 E
6.7  106 C 1.8  106 C
9
2
2
 8.988  10 N  m C
 6.7  106 C 7.3  104 N C
2
 0.34 m 








 0.45 N, right
67.
the
Since the electric field exerts a force on the charge in
same direction as the electric field, the charge is positive.
Use the free-body diagram to write the equilibrium
equations for both the horizontal and vertical directions,
and use those equations to find the magnitude of the
charge.
43
  cos 1  38.6o
55
 Fx  FE  FT sin   0  FE  FT sin   QE
F
y
Q
 FT cos   mg  0  FT 
mg tan 
E
1.0 10

3

mg
cos 

N C
1.2 10
FT
43cm
mg
FE
 QE  mg tan 
kg 9.80 m s 2 tan 38.6 o
4

 6.5  10 7 C

L  55cm
68.
than
The weight of the mass is only about 2 N. Since the tension in the string is more
that, there must be a downward electric force on the positive charge, which means that
the electric field must be pointed down . Use the free-body diagram to write an
expression for the magnitude of the electric field.
 F  FT  mg  FE  0  FE  QE  FT  mg 
E
FT  mg
Q


5.67 N   0.210 kg  9.80 m s 2
7
3.40  10 C
  1.06 10
7
FT
mg
N C
69.
To find the number of electrons, convert the mass to moles, the moles to atoms, and then
multiply by
the number of electrons in an atom to find the total electrons. Then convert to charge.
 1 mole Al   6.02  1023 atoms   13 electrons   1.602  10 19 C 
15 kg Al  15 kg Al  
  1 molecule  


2
1 mole
electron

 2.7  10 kg 


 7.0  108 C
The net charge of the bar is 0 C , since there are equal numbers of protons and electrons.
70.
(a)
The force of sphere B on sphere A is given by Coulomb’s law.
FAB 
kQ 2
, away from B
R2
The result of touching sphere B to uncharged sphere C is that the charge on B is
(b)
shared between
the two spheres, and so the charge on B is reduced to Q 2 . Again use Coulomb’s law.
FAB  k
QQ 2
kQ 2

, away from B
R2
2R2
The result of touching sphere A to sphere C is that the charge on the two spheres
(c)
is shared, and
so the charge on A is reduced to 3Q 4 . Again use Coulomb’s law.
FAB  k
 3 Q 4  Q 2 
R2

3kQ 2
8R 2
, away from B
71.
On the x-axis, the electric field can only be zero at a location closer to the smaller
magnitude charge.
Thus the field will never be zero to the left of the midpoint between the two charges. Also,
in between the two charges, the field due to both charges will point to the left, and so the
total field cannot be zero. Thus the only place on the x-axis where the field can be zero is to
the right of the negative charge, and so x must be positive. Calculate the field at point P and
set it equal to zero.
Ek
  Q 2
x
2
k
Q
x  d
2
 0  2x2   x  d 
2
 x
d
2 1
 2.41 d
FE
The field cannot be zero at any points off the x-axis. For any point off the x-axis, the electric
fields due to the two charges will not be along the same line, and so they can never combine
to give 0.
72.
73.
The electric field will put a force of magnitude FE  QE on
each charge. The distance of each charge from the pivot point is
QEL
.
L 2 , and so the torque caused by each force is   FE r 
2
Both torques will tend to make the rod rotate counterclockwise
 QEL 
in the diagram, and so the net torque is  net  2 
  QEL .
 2 
Q
FE
A negative charge must be placed at the center of the square. Let
Q  8.0  C be the charge at each corner, let -q be the magnitude of
negative charge in the center, and let d  9.2 cm be the side length of
the square. By the symmetry of the problem, if we make the net force
on one of the corner charges be zero, the net force on each other
corner charge will also be zero.
Q2
Q2
F41  k 2  F41 x  k 2 , F41 y  0
d
d
F42  k
F43  k
F4 q  k
Q2
 F42 x  k
2d 2
Q2
Q2
2d 2
cos45o  k
 F43 x  0 , F43 y  k
d2
qQ
 F4 qx   k
2
2qQ
2
2Q 2
4d 2
Q2
d
2
k
2Q 2
4d
2
0k
E
Q1
d
2qQ
2
2qQ
d2
F42
F43
Q4
F41
F4q
q
Q2
2Q 2
4d 2
d2
cos 45o   k
d
Q
Q2
d 2
d
The net force in each direction should be zero.
 Fx  k
, F42 y  k
FE
 F4 qy
 1 1
   7.66  106 C
 2 4
 0  q  Q
So the charge to be placed is q  7.66  106 C .
This is an unstable equilibrium . If the center charge were slightly displaced, say towards
the right, then it would be closer to the right charges than the left, and would be attracted
more to the right. Likewise the positive charges on the right side of the square would be
closer to it and would be attracted more to it, moving from their corner positions. The
system would not have a tendency to return to the symmetric shape, but rather would have a
tendency to move away from it if disturbed.
Q3