Physics 102
Spring 2008
Homework 3 Solutions
Giancoli Chapter 16
Questions:
3.
Water is a polar molecule – it has a positive region and a negative region. Thus it is easily attracted
to some other charged object, like an ion or electron in the air.
4.
The positively charged rod slightly polarizes the molecules in the paper. The negative charges in the
paper are slightly attracted to the part of the paper closest to the rod, while the positive charges in the
paper are slightly repelled from the part of the paper closest to the rod. Since the opposite charges
are now closer together and the like charges are now farther apart, there is a net attraction between
the rod and the paper.
 

 
8.
The force of gravity pulling down on the leaves, tending to return them to the vertical position.
12. The charged plastic ruler has a negative charge residing on its surface. That charge polarizes the
charge in the neutral paper, producing a net attractive force. When the piece of paper then touches
the ruler, the paper can get charged by contact with the ruler, gaining a net negative charge. Then,
since like charges repel, the paper is repelled by the comb.
Problems:
6.
Since the magnitude of the force is inversely proportional to the square of the separation distance,
1
F  2 , if the distance is multiplied by a factor of 1/8, the force will be multiplied by a factor of 64.
r


F  64 F0  64 3.2  102 N  2.0 N
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
11
2
2
31
27
m
m
FG
Gm1m2
6.67  10 N  m kg 9.11  10 kg 1.67  10 kg
G 12 2
r
The electric force is about 2.3 1039 times stronger than the gravitational force for the given
scenario.






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.

F13
d
F12
Q1
d
Q2
Q3
d
F12  k
F13  k
Q2
d
2
Q
2
 F12 x  k
d2
F1 
F  F  3k
2
1y
Q2
2
d
 F13 x   k
F1x  F12 x  F13 x  0
2
1x
Q2
Q
Q2
cos 60 , F12 y  k
o
d
2
Q2
F1 y  F12 y  F13 y  2k

d2
 3 8.988  10 N  m C
d2
9
2
sin 60 o
Q2
cos 60o , F13 y  k
d2
2
d2
sin 60o
sin 60o  3k
2

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 .
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
k
Q2 Q
x
 18.5cm 
2

 xd

Q2
Q2

4.7  106 C  3.5  106 C

Q1 
3.5  106 C
 116 cm
29.
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.
Fnet  ma  qE  E 
m
q
 9.1110
a
31
 1.602 10
kg
19

C

115 m s
2

north  6.54  10 10 N C south
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.
Q
Q
EA  k 2  EAx  0 , EAx   k 2
l
l
EB  k
Q
l
Q
 EBx   k
2
2
EBy   k
Ex  EAx  EBx   k
E  Ex2  E y2 
  tan 1
Ey
Ex
l
Q
l2
cos 30o   k
sin 30o   k
3Q
2l
4l
4
2l
Q

9k 2 Q 2
4l
4
Q
l
B
l
EB
EA
2l 2
12k 2Q 2

l
2
3Q
E y  EAy  EBy   k
2
3k 2Q 2
3Q
Q
A
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
4l
4
k
 tan 1
k
3Q
2l
2
, EBy  k
E y  EAy  EBy   k
2
3k 2Q 2
l2
cos 30o  k
3Q
2l
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

E right
Eleft
A
d
Q

d

Q
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
EA  2
 5.0 cm   10.0 cm 
2
kQ
d
2
 0.1118 m
2

7.0 106 C

sin  2 8.988 109 N  m 2 C2
 0.1118 m 
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
E right
Q
2
d right



7.0  106 C 
Q
sin left  k
2
left
d right
d left
Q
 right
 left
Q
 0.1581m
cos left  k
2
left
d
Eleft
 0.0707 m
2
 A  90o
sin 26.6o  4.5 106 N C
2
cos  right
cos45o
  0.0707 m 
Q
2
d right
2

cos18.4o 
6
  6.51 10 N C
 0.1581m  
2
sin  right
 sin45o
sin18.4o 
  8.988  10 N  m C  7.0  10 C  

 9.69  106 N C
2
2 
  0.0707 m   0.1581m  
9
2
2
EB  Ex2  E y2  1.2  107 N C
6
 B  tan 1
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.
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 negative
Q
y-direction, and will be twice the y-component 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

3/ 2
sin 

1/ 2
in the negative y direction