Homework 1
Ch16: P 12, 13, 15, 17, 19
12. (II) Particles of charge 75,  48, and 85 C are placed in a line (Fig. 16–49). The
center one is 0.35 m from each of the others. Calculate the net force on each charge
due to the other two.
Solution
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. (II) Three positive particles of equal charge, 11.0 C, are located at the corners of
an equilateral triangle of side 15.0 cm (Fig. 16–50). Calculate the magnitude and
direction of the net force on each particle.
Solution
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.
F12  k
F13  k
Q2
d
2
Q2
d
2
 F12 x  k
F1 
d
 F13 x   k
F1x  F12 x  F13 x  0
2
1y
2
Q2
d
Q2
d2
Q2
cos 60o , F12 y  k
2
d
cos 60 , F13 y  k
o
F1 y  F12 y  F13 y  2k
F  F  3k
2
1x
Q2

Q
2
sin 60 o
Q2
d
2
sin 60
d
o
2
d2
sin 60  3k
o
 3 8.988  10 N  m C
9
F13
2
2

F12
Q1
d
Q2
Q2
Q3
d
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. (II) A charge of 6.00 mC is placed at each corner of a square 0.100 m on a side.
Determine the magnitude and direction of the force on each charge.
15. (II) Repeat Problem 14 for the case when two of the positive charges, on opposite
corners, are replaced by negative charges of the same magnitude (Fig. 16–51).
Solution
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.
F41
Q1
F42
Q4
F43
d
F41  k
F42  k
F43  k
Q
2
d
2
Q
 F41 x   k
2
2d
2
Q2
d2
 F42 x  k
Q
2
, F41 y  0
d2
Q
Q2
2
2d
2
2Q
cos45o  k
 F43 x  0 , F43 y   k
4d
2
2
, F42 y  k
2Q
Q3
2
4d 2
Q2
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
Q2
d2
d2
k
2Q 2
4d 2
0 k
 0.64645 2  k
Q2
d2

9
F4 y
F4 x
2
Q2 
2
Q2
d2 
4 
d2
 1 
  0.64645k
 F4 y
 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.
17. (II) Three charged particles are placed at the corners of an equilateral triangle of side
1.20 m (Fig. 16–53). The charges are 4.0 C,  8.0 C, and 6.0 C. Calculate the
magnitude and direction of the net force on each due to the other two.
F13
d
F23
F12
Q1
d
Q2
F21
Q3
d
F32
F31
Solution
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
 8.988  10 N  m C
d2
9
2

2
 0.1997 N  F21
F13  k

Q1Q3
 8.988  10 N  m C
d2
9
2
2

 0.1498 N  F31
F23  k
Q2Q3
d2

 8.988  109 N  m 2 C 2

 4.0 10 C 8.0 10 C 
6
6
1.20 m 
2
 4.0  10 C  6.0 10 C 
6
6
1.20 m 
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
F1  F12x  F12y  0.30 N
1  tan 1
F1 y
F1x
 tan 1
3.027  101 N
 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
F2  F  F  0.26 N
2
2x
2
2y
 2  tan
1
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
F3  F32x  F32y  0.26 N
3  tan 1
F3 y
F3 x
 tan 1
1.297  101 N
2.247  101 N
 30o
19. (III) Two charges, Q0 and 3Q0 , are a distance l apart. These two charges are free
to move but do not because there is a third charge nearby. What must be the charge
and placement of the third charge for the first two to be in equilibrium?
Solution
Q0
3Q0
Q
l–x
x
l
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
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 .
Q0Q
3Q02
3Q0Q
3Q02
left: k 2  k 2
right: k
k 2

2
x
l
l
l  x
k
k
Q0Q
x
2
Q0Q
x
2
k
k
3Q0Q
l  x
3Q02
l
2
2
l
 x
 Q  3Q0
3 1
x2
l
2
 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 .