Download Homework Set Solutions Chapter 20

P21 Homework Set #1 Solutions Chapter 20
P20.9. Prepare: We will model the charged masses as point charges. A visual overview of the forces and the
coordinate system is shown. The charge q1 exerts a force F1 on 2 on q2 to the right, and the charge q2 exerts a
force F2 on 1 on q1 to the left.
(a) Solve: Using Coulomb’s law,
K | q1|| q2| (9.0 10 N  m2 /C )(1.0 106 C) (1.0 10 6 C)
3

 9.0  10 N
2
2
r12
(1.0 m)
(b) Newton’s second law on either q1 or q2 is
9
2
F1 on 2  F2 on 1 
F1 on 2  m1a1 
a1 
9.0  103 N
3
 9.0  10 m/s2
1.0 kg
Assess:A relatively small force on a relatively large mass causes small acceleration.
P20.11. Prepare: We need to solve Coulomb’s law (Equation 20.1) for r :
FK
| q1|| q2 |
r2
where F  8  2  104 N and q1   50 nC, q2   12 nC, and K  9 0  109 N  m 2 /C2.
Solve:
| q || q |
r2  K 1 2
F
r K
| q1|| q2 |
|  0.5  109 C|| 12  10 9 C|
9
 (9 0  10 N  m 2 /C 2 )
 0 026 m  2 6 cm
F
8.2  104 N
Assess:Notice the N and C cancel out leaving units of m. Comparing with Problem 20.10, the answer of 2.6 cm
seems to be in the right ballpark.
P20.13. Prepare: We will model the glass bead and the ball bearing as point charges. A visual overview of
the forces and the coordinate system is shown. The ball bearing experiences a downward electric force F1on 2 . By
Newton’s third law, F2 on 1  F1 on 2 .
Solve:Using Coulomb’s law,
F1on 2  K
| q1|| q2|
(9.0 109 N  m2 /C2 )(20 109 C) | q2|
 0.018 N 
 | q2|  1.0  108 C
2
r12
(1.0  102 m)2
Because the force F1 on 2 is attractive and q 1 is a positive charge, the charge q 2 is a negative charge. Thus,
q2  1.0  108 C  10 nC.
P20.19. Prepare: The electric field is that of a positive charge on the glass bead. The charge is assumed to be
a point charge.
Solve:The electric field is


6.0  109 C
E   (9.0  109 N  m2 /C 2 )
, away from bead   (1.4  10 5 N/C, away from bead)
2
2
(2.0

10
m)


Assess:This is a typical electric field near objects that are charged by rubbing.
P20.21. Prepare: Protons and electrons are point charges and produce electric fields, according to Equation 20.6.
The charge on a proton is positive and on an electron is negative.
(a) Solve: The electric field of the proton is
19

 1 |q|
 
C
9
2   1.6  10
2
E 
,
away
from
q

(9.0

10
N

m
/C
)
, away from q 



3
2 
2


 (1.0  10 m) 
 4 0 r
 

 (1.4  103 N/C, away from proton)
(b) The electric field of the electron is
19

 1 |q|
 
C
9
2   1.6  10
2
E 
,
toward
q

(9.0

10
N

m
/C
)
, toward q 




3
2 
2


4

r
(1.0

10
m)
0



 

 (1.4  103 N/C, toward electron)
Assess:An electron charge is very small, so its electric field at a point 1 mm away was expected to be small.
P20.23. Model: The electric field is that of a negative charge located at the origin as shown below. We will
use Equation 20.6.
The positions (0 cm, 5 cm), (5 cm, 5 cm), and (5 cm, 5 cm) are denoted by A, B, and C, respectively.
(a) Solve: The electric field strength for a charge q is
EK
|q|
r2
Using K  9.0  10 9 N  m2/C2 and | q |  10.0  10 9 C,
90.0 N  m2 /C
r2
The electric field strengths at points A, B, and C are
90.0 N  m2 /C
4
EA 
 3.6  10 N/C
(5.0  10 2 m)2
E
EB 
90.0 N  m2 /C
 1.8  10 4 N/C
(5.0  10 2 m)2  (5.0  10 2 m)2
90.0 N  m2 /C
 1.8  10 4 N/C
(5.0  10 m)2  ( 5.0  10 2 m)2
(b) The three vectors are shown in the diagram.
EC 
2
Assess: Note that the vectors EA , EB, and EC are pointing toward the negative charge.
P20.25. Prepare: The electric field is that of the two charges placed on the y-axis. Please refer to Figure P20.25.
We denote the upper charge by q1 and the lower charge by q2. The electric field at the dot due to the positive
charge is directed away from the charge and making an angle of 45 below the +x axis, but the electric field due
to the negative charge is directed toward it making an angle of 45 below the –x axis.
Solve:The electric field strength of q1 is
E1  K
| q1| (9.0  109 N  m2/C2 )(1  109 C)

 1800 N/C
r 12
(0.050 m)2  (0.050 m) 2
Similarly, the electric field strength of q2 is
E2  K
| q2| (9.0  109 N  m2/C2 )(1109 C)

 1800 N/C
r 22
(0.050 m)2  (0.050 m) 2
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 in the
fourth quadrant and that due to q2 is toward q2 in the third quadrant. Their components are
E1 x  E1 cos 45 
E1 y  E1 sin 45 
E2 x  E2 cos 45 
E2 y  E2 sin 45 
The x and y components of the net electric field are:
( Enet ) x  E1x  E2 x  E1 cos 45   E2 cos 45   0 N/C
( Enet ) y  E1 y  E2 y   E1 sin 45   E2 sin 45   2500 N/C
 Enet at dot  (2500 N/C, along  y axis)
Thus, the strength of the electric field is 2500 N/C and its direction is vertically downward.
Assess:A quick visualization of the components of the two electric fields shows that the horizontal
components cancel.
P20.45. Prepare: The electric field is that of the two 1 nC charges located on the y-axis. Please refer to
Figure P20.45. We denote the top 1 nC charge by q 1 and the bottom 1 nC charge by q 2. The electric
fields (E1 and E2 ) of both the positive charges are directed away from their respective charges. With vector
addition, they yield the net electric field Enet at the point P indicated by the dot.
Solve:The electric fields from q1 and q2 are
 |q |
  (9.0  109 N  m2/C 2 )(1  10 9 C)

E1   K 21 , along  x -axis   
, along  x -axis 
2
(0.05 m)

 r1
 
 (3600 N/C, along  x -axis)
 1 | q2|

E2  
,  above  x -axis   (720 N/C,  above  x -axis)
2
4

r
0
2


Because tan   10 cm/5 cm,   tan 1 (2)  63.43 .
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 along
+ x and that due to q2 is away from q2 in the first quadrant. Their components are
E1 x  E1
E1 y  0
E2 x  E2 cos63.45 
E2 y  E2 sin 63.45 
The x and y components of the net electric field are:
( Enet ) x  E1x  E2 x  E1  E2 cos63.45   3922 N/C
( Enet ) y  E1 y  E2 y  0  E2 sin 63.45   644 N/C
Thus, the strength of the electric field at P is
Enet  (3922 N/C) 2  (644 N/C) 2  3975 N/C
which will be reported as 4000 N/C.
To find the angle this net vector makes with the x-axis, we calculate
tan  
644 N/C
3922 N/C    9.3 
Assess:Because of the inverse square dependence on distance, E2  E1. Additionally, because the point P has no
special symmetry relative to the charges, we expected the net field to be at an angle relative to the x-axis.
P20.47. Prepare: The charges are point charges. Please refer to Figure P20.47. The electric force on charge q1
is the vector sum of the forces F2 on 1 and F3 on 1, where q1 is the 1 nC charge, q2 is the left 2 nC charge, and q3 is the
right 2 nC charge.
Solve:We have
 K | q1|| q2|

F2 on 1  
, away from q2 
2
r


 (9.0  109 N  m2/C 2 )(1  10 9 C)(2  10 9 C)


, away from q2 
2
2
(1  10 m)


4
 (1.8  10 N, away from q2 )
( F2 on 1 ) x  (1.8  10 4 N)(cos60 )  (0.9  10 4 N)
( F2 on 1 ) y  (1.8  10 4 N)(sin 6 0 )  (1.56  104 N)
 K | q1|| q3|

F3 on 1  
, away from q3   (1.8  10 4 N, away from q3 )
r2


( F3 on 1 ) x  (1.8  10 4 N)(cos60 )   (0.9  10 4 N)
( F3 on 1 ) y  (1.8  104 N)(sin 60 )  (1.56  10 4 N)
( Fon 1 ) x  ( F2 on 1 ) x  ( F3 on 1 ) x  0
( Fon 1 ) y  ( F2 on 1 ) y  ( F3 on 1 ) y  3.12  10 4 N
So the force on the 1 nC charge is 3.1  10–4 N directed upward.
Assess:The magnitude and symmetry of q2 and q3 ensure that their x-component of the net force is zero.
P20.51. Prepare: The charges are point charges. Please refer to Figure P20.51. Placing the 1 nC charge at the origin
and calling it q1, the 6 nC is q3, the q2 charge is in the first quadrant, and the q4 charge is in the second quadrant. The net
electric force on q1 is the vector sum of the electric forces from the other three charges q2, q3, and q4.
Solve: We have
 K | q1|| q2|

F2 on 1  
, away from q2 
2
 r

 (9.0  109 N  m2 /C 2 )(1  10 9 C)(2 10 9 C)


, away from q2 
2
2
(5.0

10
m)


 (0.72  10 5 N, away from q2 )
 K | q1|| q3|

F3 on 1  
, toward q3 
2
 r

9
2
2
 (9.0  10 N  m /C )(1  10 9 C)(6 10 9 C)


, toward q3 
2
2
(5.0

10
m)


 (2.16  10 5 N, away from q3 )
 K | q1|| q4|

F4 on 1  
, away from q4   (0.72  10 5 N, away from q4 )
2
 r

( F2 on 1 ) x   (0.72  105 N)(cos 45 )  (0.509 10 5 N)
( F2 on 1 ) y   (0.72  105 N)(sin 45 )  (0.509 10 5 N)
( F3 on 1 ) x  0 N
( F3 on 1 ) y  (2.16  105 N)
( F4 on 1 ) x  (0.72  105 N)(cos 45 )  (0.509 10 5 N)
( F4 on 1 ) y   (0.72  1 05 N)(sin 45 )   (0.509  10 5 N)
( Fon 1 ) x  ( F2 on 1 ) x  ( F3 on 1 ) x  ( F4 on 1 ) x  0 N
( Fon 1 ) y  ( F2 on 1 ) y  ( F3 on 1 ) y  ( F4 on 1 ) y  1.14  105 N
So the force on the 1 nC charge is 1.14  10–5 N directed vertically up.