PHYSICS 172
WQ 2010
Solutions to Homework
#1
1. Giancoli Chapter 21, Problem 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. Since the triangle is equilateral, each angle is
60o. First calculate the magnitude of each individual force.
Q1Q2
F12  k
d2

 8.988  10 N  m C
9
2

2
 7.0  10 C 8.0  10 C 
6
6
1.20 m 2
 0.3495 N
F13  k
Q1Q3
d
2

 8.988  109 N  m 2 C2

F13
 7.0  10 C  6.0  10 C 
6
6
1.20 m 
d
2
 0.2622 N
F23  k
Q2Q3
d2

F23
 8.988  10 N  m C
9
2
2

F12
Q1
d
Q2
F21
Q3
d
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.
F1 x  F12 x  F13 x    0.3495 N  cos 60o   0.2622 N  cos 60o  4.365  10 2 N
F1 y  F12 y  F13 y    0.3495 N  sin 60o   0.2622 N  sin 60o  5.297  10 1 N
F1 
F12x  F12y  0.53 N
1  tan 1
F1 y
F1 x
 tan 1
5.297  10 1 N
 265
4.365  10 2 N
F2 x  F21 x  F23 x   0.3495 N  cos 60 o   0.2996 N   1.249  10 1 N
F2 y  F21 y  F23 y   0.3495 N  sin 60 o  0  3.027  10 1 N
F2 
F22x  F22y  0.33 N
 2  tan 1
F2 y
F2 x
 tan 1
3.027  10 1 N
1.249  10 1 N
 112
F3 x  F31 x  F32 x    0.2622 N  cos 60o   0.2996 N   1.685  10 1 N
F3 y  F31 y  F32 y   0.2622 N  sin 60o  0  2.271  10 1 N
F3 
F32x  F32y  0.26 N
 3  tan 1
F3 y
F3 x
 tan 1
2.271  101 N
1.685  10 1 N
 53
2. Giancoli Chapter 21, Problem 18
The negative charges will repel each other, and so the third charge must put an opposite
force on each of the original charges. Consideration of the various possible configurations
F32
F31
leads to the conclusion that the third charge must be positive and must be between 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 .
QQ
4Q 2
4Q0Q
4Q02
left: k 02  k 2 0
right: k

k

x
l
l2
 l  x 2
k
k
Q0Q
x
2
Q0Q
x2
k
k
4Q0Q
l  x
4Q02
l2
 x  13 l
2
 Q  4Q0
x2
l2
4
9
Thus the charge should be of magnitude
1
3
 Q0
4
 3 2
Q0
 Q0
4Q0
Q
4
9
l–x
x
l
Q0 , and a distance
l from  Q0 towards  4Q0 .
3. Giancoli Chapter 21, Problem 38
(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
E A  k 2  E Ax  0 , E A x   k 2
l
l
EB  k
Q
 E Bx   k
E x2  E y2 
  tan 1
Ey
Ex
3Q
2l
 tan 1
4l
4
k
k

9 k 2Q 2
4l
4
12k 2Q 2

4l
4
l
Q
l
B
l
EA
E y  EAy  E By   k
2
3k 2Q 2
o

3Q
2l 2
3kQ
l2
3Q
2 l 2  tan 1 3  tan 1 3  240o
3Q
 3
2l 2
(b) Now reverse the direction of E A
EA  k
Q
EB  k
Q
l
l
2
2
 EAx  0 , EAx   k
 E Bx  k
E x  EAx  EBx  k
3Q
2l
2
Q
l
2
Q
l2
cos 30o  k
3Q
2l
2
Q
EB
3Q
cos 30   k
,
l2
2l 2
Q
Q
EBy   k 2 sin 30 o   k 2
l
2l
l2
E x  EAx  E Bx   k
E
Q
A
, E By  k
E y  EAy  E By   k
Q
2l 2
Q
l
2
sin 30 o  k
Q
2l 2
E
E x2  E y2 
  tan
1
Ey
Ex
 tan
3k 2Q 2
4l
1
4
k
k

k 2Q 2
4l
4
4k 2Q 2

4l

4
kQ
l2
Q
2 l 2  tan 1 1  330o
3Q
 3
2l 2
4. Giancoli Chapter 21, Problem 42


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

 5.0cm 2  10.0cm 2
d
EA  2
kQ
d
2

 5.7  10
6
C
 0.1118 m 
2
 5.0 cm 
d right 
 5.0 cm 2  15.0 cm 2
d
d


Q
sin 26.6  3.7  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
 45
 right  tan 1
 18.4
5.0
15.0
d left 
Eleft
Q
 0.1118 m
sin  2 8.988  109 N  m 2 C 2
2
A
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
 26.6
10.0
  5.0 cm   0.0707 m
2
 0.1581m
 A  90
Eleft
E right
d left
Q
 left
d right
 right
Q
E x   Eleft  x   Eright  x  k
Q
d

 8.988  109 N  m 2 C 2
E y   Eleft  y   Eright  y  k

EB 
Q
2
d right

cos  right
5.7  10 C  
cos45
Q
sin  right
6
  0.0707 m 
d
 8.988  109 N  m 2 C 2
cos left  k
2
left
2
left
sin left  k
Q
2
d right

5.7  10 C  
6

2

sin45
  0.0707 m 
 B  tan 1
E x2  E y2  9.5  106 N C
2
cos18.4 
6
  5.30  10 N C
 0.1581m  
2
sin18.4 
6
  7.89  10 N C
 0.1581m  
2
Ey
 56
Ex
The results are consistent with Figure 21-34b. 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.
5. Giancoli Chapter 21, Problem 47
If we consider just one wire, then from the answer to problem 46, we would have the
following. Note that the distance from the wire to the point in question is x  z 2   l 2  .
2
Ewire 

2 0
l
 4  z

  l 2    l 2
But the total field is not simply four times the above expression, because the fields due to the
four wires are not parallel to each other. Consider a side view of the problem. The two dots
represent two parallel wires, on opposite sides of the square. Note that only the vertical
component of the field due to each wire will actually contribute to the total field. The
horizontal components will cancel.
z
Ewire  4  Ewire  cos   4  Ewire 
2
z 2   l 2
Ewire


 4
 2 0


z 2   l 2
2
z2   l 2
2

2
2
1/ 2

z

1/ 2
2
 z 2   l 2 2
4  z 2   l 2    l 2

l
Eright 
Eleft
wire
wire


z2   l 2
z
8 l z
 0  4 z 2  l 2  4 z 2  2 l 2 
1/ 2
The direction is vertical, perpendicular to the loop.
l 2
l 2
2
6. Giancoli Chapter 21, Problem 60
Since the field is constant, the force on the electron is constant, and so the acceleration is
constant. Thus constant acceleration relationships can be used. The initial conditions are
x0  0, y0  0, v x 0  1.90 m s, and v y 0  0.
F  ma  qE  a 
q
m
E
x  x0  v x 0t  12 a x t 2  v x 0t 
e
m
eE x
2m
E ; ax  
e
m
Ex , a y  
e
m
Ey
t2
1.60  10 C  2.00  10 N C   2.0s   3.2 m
 1.90 m s  2.0s  
2  9.11  10 kg 
eE
1.60  10 C  1.20  10 N C   2.0s 
y  y v t a t 
t 
2m
2  9.11  10 kg 
19
11
2
31
19
0
1
2
y0
2
y
11
2
31
y
2
 4.2 m
7. Giancoli Chapter 21, Problem 62
(a) The dipole moment is given by the product of the positive charge and the separation
distance.



p  Ql  1.60  1019 C 0.68  109 m  1.088  1028 C m  1.1  1028 C m
(b) The torque on the dipole is given by Eq. 21-9a.
  pE sin   1.088  1028 C m  2.2  104 N C  sin 90   2.4  10 24 C m



(c)   pE sin   1.088  1028 C m 2.2  104 N C  sin 45   1.7  1024 N m
(d) The work done by an external force is the change in potential energy. Use Eq. 21-10.
W  U    pE cos final     pE cos  initial   pE  cos  initial  cos final 



 1.088  1028 C m 2.2  104 N C 1   1  4.8  10 24 J