Homework 1
Ch16
1. Particles of charge 75,  48, and 85 C are placed in a line. 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 expressions,
k  8.988 109 N  m2 C2 .
 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
1
2. Three positive particles of equal charge, 11.0 C, are located at the corners of an equilateral
triangle of side 15.0 cm. 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
d2
Q2
d
2
 F12 x  k
F1 
d2
 F13 x   k
F1x  F12 x  F13 x  0
2
1y
Q2
d
Q2
d2
Q2
cos 60o , F12 y  k
2
d2
cos 60 , F13 y  k
o
F1 y  F12 y  F13 y  2k
F  F  3k
2
1x
Q2

Q
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 .
2
3. 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.
Solution
Method 1: 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
 F41 x  k
d2
Q
Q2
2
2d
2
Q2
d2
, F41 y  0
d2
 F42 x  k
Q
Q1
2
2d
2
cos45o  k
 F43 x  0 , F43 y  k
F42
F43
2Q
4d
2
2
, F42 y  k
2Q
Q4
F41
2
4d 2
Q2
d
Q2
d2
Q3
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  F  F  k
2
4x

2
4y
Q2
d
Q2 
F4 y
2Q 2
4d
2
0 k
2
Q2 
2
1 
  F4 y
d 
4 
2
Q2 
1
1 
 2 k 2  2 
d 
4 
d 
2
2
 8.988  109 N  m 2 C 2
  tan 1
2
k

 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.
Method 2: See notes of Lecture 1.
3
4. Repeat previous problem for the case when two of the positive charges, on opposite corners,
are replaced by negative charges of the same magnitude.
Solution
Method 1: 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.
F42
F41
Q2
Q2
Q1
F41  k 2  F41 x   k 2 , F41 y  0
Q4
d
d
F42  k
F43  k
Q2
2d 2
Q2
 F42 x  k
Q2
2d 2
2Q 2
cos45  k
o
 F43 x  0 , F43 y   k
4d 2
, F42 y  k
F43
2Q 2
d
4d 2
Q2
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  F4 x

Q2
d2
2k
2Q 2
k
Q2
d
2
4d 2
 8.988  10 N  m C
  tan 1
F4 y
F4 x
9
2
2

2

1


  F4 y
d2 
4 
Q2 
1
2k 2  2 
d 
2
0 k
2
4
1 
Q3
Q2 
 6.00 10 C 
3
 0.100 m 
2
2
1

7
 2    2.96  10 N
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.
Method 2: Is similar to the previous problem.
4
5. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m.
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
Method 1: 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  F21  k
Q1Q2
F13  F31  k
Q1Q3
F23  F32  k
Q2 Q3
d2
d2
d2

 8.988  10 N  m C
9

9

9
2
 8.988  10 N  m C
2
2
 8.988  10 N  m C
2
2
2

 4.0 10 C 8.0 10 C   0.1997 N
6
6
1.20 m 2

 4.0 10 C  6.0 10 C   0.1498 N

8.0 10 C  6.0 10 C   0.2996 N
6
6
1.20 m 2
6
6
1.20 m 
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
 tan 1
1.297  101 N
1
F3 x
2.247  10 N
Method 2: Use symmetry as in the previous problems.
 30o
5
6. 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 .
left: k
k
k
Q0Q
x
2
Q0Q
x
2
Q0Q
x2
k
k
k
3Q02
l2
3Q0Q
l  x
3Q02
l
2
2
right: k
l  x
l
 x
 Q  3Q0
3Q0Q
3 1
x2
l
2
2
k
3Q02
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 .
6