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Homework 2
1.
PHYS 212
Dr. Amir
What is the magnitude of the electric force of attraction between an iron nucleus
(q=+26e) and its innermost electron if the distance between them is 1.5 X 10-12 m.
Use Coulomb’s law to calculate the magnitude of the force.
F k
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
12. (II) Particles of charge +75 μC, +48 μC and -85μC are placed in a line (Fig. 21–52). The center
one is 0.35 m from each of the others. Calculate the net force on each charge due to the other
two.
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  ˆ  75C 85C  ˆ
ik
i  147.2 N ˆi  150 N ˆi
 0.35m 2
 0.70 m 2
 75C  48C  ˆ  48C 85C  ˆ
F48  k
ik
i  563.5 N ˆi  560 N ˆi
2
2
0.35m
0.35m




85C  75C  ˆ 85C  48C  ˆ
F85   k
ik
i  416.3 N ˆi  420 N ˆi
2
2
 0.70 m 
 0.35m 
F75   k
13. (II) Three charged particles are placed at the corners of an equilateral triangle of side 1.20
m (Fig. 21–53). The charges are +7.0μC,-8.0μC and -6.0μC. Calculate the magnitude and
direction of the net force on each due to the other two.
F13
d
F23

Q1Q2
F12  k
d

 8.988  109 N  m 2 C2
2
Q1
d
Q2
F21
 7.0  10 C 8.0  10 C 
6
F12
F32
Q3
d
F31
6
1.20 m 2
 0.3495 N

Q1Q3
F13  k
 8.988  10 N  m C
d2
9
2
2

 7.0  10 C  6.0  10 C 
6
6
1.20 m 2
 0.2622 N
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.
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  F  F  0.53 N
2
1x
2
1y
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 60o   0.2996 N   1.249  101 N
F2 y  F21 y  F23 y   0.3495 N  sin 60o  0  3.027  101 N
F2  F  F  0.33 N
2
2x
2
2y
 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
⃗ = (7.22 × 10−4 𝑁)𝒋. What is the electric field
25. (I) The electric force on a +4.20 μC charge is 𝑭
at the position of the charge?
Use the definition of the electric field, Eq. 21-3.
E
F
q
 7.22  10

4
N ˆj
6
4.20  10 C
  172 N C ˆj
31. (II) A long uniformly charged thread (linear charge density λ =2.5C/m) lies along the x axis in
Fig. 21–56. A small charged sphere (Q=-2.0C) is at the point x  0 cm, y=-5.0cm. What is the
electric field at the point x  7.0 cm, y  7.0 cm? Ethread and EQ represent fields due to the
long thread and the charge Q, respectively.
The field at the point in question is the vector sum of the two fields shown in Figure 21-56. Use the
results of Example 21-11 to find the field of the long line of charge.
Ethread 
1 ˆ
1 Q
j ; EQ 
cos ˆi  sin  ˆj
2 0 y
4 0 d 2



Q
1
E  
 4 0 d


 1 

1 Q

sin   ˆj
2
 2 0 y 4 0 d

cos   ˆi  
2

d 2   0.070 m    0.120 m   0.0193m 2 ; y  0.070 m ;   tan 1
2
Ex  
Ey 
Q
1
4 0 d
1 
2 0 y
2

2

cos    8.988  109 N  m 2 C 2
1
Q
4 0 d
2
sin  


 0.0193m
2.0 C
2
12.0 cm
7.0 cm
 59.7
cos 59.7  4.699  1011 N C
1  2 Q

 2 sin  

4 0  y d

 2  2.5C m 

 2.0 C 
 8.988  109 N  m 2 C 2 

sin 59.7   1.622  1011 N C
2
 0.070 cm 0.0193m




 

E  4.7  1011 N C ˆi  1.6  1011 N C ˆj
E
 4.699  10 N C    1.622  10
 1.622  10 N C   199
 4.699  10 N C 
E x2  E y2 
2
11
11
N C

2
 5.0  1011 N C
11
 E  tan
1
11
36. (II) Two point charges, Q1= -25 μC and Q2=+45 μC are separated by a distance of 12 cm. The
electric field at the point P (see Fig. 21–58) is zero. How far from Q1 is P?
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 l represent the 12 cm distance,
and note that Q1  12 Q2 .
E1  E2
 k
xl
Q1

Q2 
Q1
Q1
x
2

k
Q2
 x  l 2
 12 cm 


25C
45C  25C

 35cm