Download Chapter 21 1. Use Coulomb`s law to calculate the magnitude of the

Chapter 21
1.
Use Coulomb’s law to calculate the magnitude of the force.
F k
2.
Q1Q2
r
2

 8.988  109 N  m 2 C 2
19

C 26 1.602 10 19 C

12
1.5  10 m

2
  2.7 10
3
N
Use the charge per electron to find the number of electrons.
1 electron
 38.0  10 C   1.602
 10
6
19

3.

1.602 10
  2.37  1014 electrons

C
Use Coulomb’s law to calculate the magnitude of the force.
F k
Q1Q2
r2

 8.988  10 N m C
9
2
2

 25  10 C  2.5  10 C   7200 N
6
3
 0.28 m 2
14.
(a)If the force is repulsive, both charges must be positive since the total charge is positive. Call
the total charge Q.
Q1  Q2  Q
F
Q  Q2  4
Q1 
2


kQ1Q2
kQ1  Q  Q1 
d2
 Q12  QQ1 
Fd 2
k
0
Fd 2
Fd 2
Q  Q2  4
k 
k
2

 12  90.0  106 C 

d2

 90.0  10 C 
6
2
4


8.988  109 N  m2 C2  
12.0N 1.16 m  2
 60.1  106 C , 29.9  106 C
(b)If the force is attractive, then the charges are of opposite sign. The value used for F must
then be negative. Other than that, the solution method is the same as for part (a).
Q1  Q2  Q
F
d2
Fd 2
Q Q 4
2
Q1 
kQ1Q2
k

kQ1  Q  Q1 
d2
Q Q 4
2

2

  90.0  106 C  

1
2
 Q  QQ1 
2
1
Fd 2
k
0
Fd 2
k
2
 90.0  10 C 
6
2
4
 12.0N 1.16 m  2 
8.988  10
9


N  m 2 C2 

 106.8  10 6 C ,  16.8  106 C
15.
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
Q1
Q4
Q3
Q2
F41  k
Q
d
2
 F41 x  k
Q2
F42  k
2d 2
Q2
F43  k
Q
2
, F41 y  0
d2
 F42 x  k
Q2
2d 2
cos45o  k
 F43 x  0 , F43 y  k
d2
2Q 2
4d 2
, F42 y  k
2Q 2
4d 2
Q2
d2
Add the x and y components together to find the total force, noting that
F4 x  F41x  F42 x  F43 x  k
F4  F42x  F42y  k

Q2 
2
1 
 8.988  10 N  m C
  tan 1
9
F4 y
F4 x
2
2

d
2
2
Q2 
4 
2
 2k
d 
Q2
2Q 2
4d
2
0 k
Q2 
2
1 
  F4 y
d 
4 
2
1
 2 
d 
2
 4.15  10 C 
3
 0.100 m 
k
2
2
 45o
above the x-direction.
 2  1   2.96  107 N


2

F41
d
represent the 0.100 m length of a side of the square, and let the variable
Q represent the 4.15 mC charge at each corner.
2
F42
F43
F4 x  F4 y
.
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.
17.
The spheres can be treated as point charges since they are spherical, and so Coulomb’s law may
be used to relate the amount of charge to the force of attraction. Each sphere will have a magnitude Q
of charge, since that amount was removed from one sphere and added to the other, being initially
uncharged.
F k
Q1Q2
r2
k
Q2
r2
F
 Qr
  0.12 m 
k
1.7  10 2 N
8.988  109 N  m 2 C 2
 1 electron 
12
  1.0  10 electrons
19
 1.602  10 C 
 1.650  10 7 C 
18.
The negative charges will repel each other, and so the third charge
Q
Q0
4Q0
must put an opposite force on each of the original charges. Consideration
x
l–x
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
l
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
4Q02
l2
4Q0Q
l  x
4Q02
l
2
2
right: k
 l  x 2
4Q02
l2

 Q  4Q0
x2
l
2
 Q0
4
9
Q0
4
 3
2
 94 Q0
, and a distance
1
3
l from  Q0 towards  4Q0
Use Eq. 21–3 to calculate the force. Take east to be the positive x direction.
E
22.
k
 x  13 l
Thus the charge should be of magnitude
21.
4Q0Q
F
q



 F  qE  1.602  1019 C 1920 N C ˆi  3.08  10 16 N ˆi  3.08  10 16 N west
Use Eq. 21–3 to calculate the electric field. Take north to be the positive y direction.
.
E
F
q
2.18  1014 N ˆj

 1.36  105 N C ˆj  1.36  105 N C south
19
1.602  10 C
Q1  0
28.
The electric field due to the negative charge will point
toward the negative charge, and the electric field due to the
positive charge will point away from the positive charge. Thus
both fields point in the same direction, towards the negative
charge, and so can be added.
Q1
E  E1  E2  k

k
2
1
r

Q2
r
4 8.988  109 N  m 2 C2
 0.080 m 
2
k
2
2

Q1
 l / 2
8.0  10
k
2
6
 l / 2
2

E2
l 2
4k
l2
Q
1
 Q2


C  5.8  106 C  7.8  107 N C
towards the negative charge
The direction is
Q2
Q2  0
E1
.
30.
Assuming the electric force is the only force on the electron, then Newton’s second law may be
used to find the electric field strength.
ma
Fnet  ma  qE  E 
33.
q
1.673  10

27

 9.80 m s   0.18 N C
C
kg 1.8  106
1.602  10
19
2
The field at the upper right corner of the square is the vector sum
of the fields due to the other three charges. Let the variable l represent
the 1.0 m length of a side of the square, and let the variable Q represent
the charge at each of the three occupied corners.
E1  k
E2  k
E3  k
Q
l
 E1 x  k
2
Q
2l
Q
l
2
2
Q
l2
 E2 x  k
, E1 y  0
Q
2l
2
cos45o  k
 E3 x  0 , E1 y  k
E3
Q1
E1
l
Q3
Q2
2Q
4l
2
, E2 y  k
E2
2Q
4l 2
Q
l2
Add the x and y components together to find the total electric field, noting that
Ex  E y
.
E x  E1 x  E2 x  E3 x  k
E  E x2  E y2  k
Ex
k
2Q
4l
2
0k
Q
2
1 
  Ey
l 
4 
2
2
Q
1
1


 2 k 2 2 
2
l 
4 
l 
2

Ey
l
2
Q
 8.988  109 N  m 2 C2
  tan 1
Q

 2.25  10 C  
6
1.22 m 
2
1
4
 2    2.60  10 N C
2

 45.0
from the x-direction.
37.
Make use of Example 21-11. From that, we see that the electric field due to the line charge
1 ˆ
E1 
i.
2 0 x
along the y axis is
In particular, the field due to that line of charge has no y dependence.
1 ˆ
E2 
j.
2 0 y
In a similar fashion, the electric field due to the line charge along the x axis is
Then the
total field at
 x, y  is the vector sum of the two fields.
E  E1  E2 
1 ˆ
1 ˆ
  1 ˆ 1 ˆ
i
j
 i  y j
2 0 x
2 0 y
2 0  x

1 
E

2 0
1
x
2

1
y
2

Ey

x
2 0 y
x 2  y 2 ;   tan 1
 tan 1
 tan 1
1

2 0 xy
Ex
y
2 0 x
45.
Because the distance from the wire is much smaller than the length of the wire, we can
approximate the electric field by the field of an infinite wire, which is derived in Example 21-11.
 4.75  106 C 
6
1 
1 2 
N m2   2.0 m  1.8  10 N C,
E

  8.988  109


2 0 x 4 0 x 
C2   2.4  102 m 
away from the wire
2
49.
Select a differential element of the arc which makes
an angle of  with the x axis. The length of this element is
Rd , and the charge on that element is dq   Rd . The
magnitude of the field produced by that element is
Rd
R
dEbottom
dEtop

0
x
dE 
1
 Rd
4 0
R2
.
From the diagram, considering pieces of the arc that are symmetric with respect to
the x axis, we see that the total field will only have an x component. The vertical components of the
field due to symmetric portions of the arc will cancel each other. So we have the following.
dEhorizontal 
1
 Rd
4 0
R2
0
Ehorizontal 
cos 
1
 4
 0
cos 
 Rd
0
R2
0



2 sin  0
cos  d 
sin 0  sin  0  

4 0 R 
4 0 R
4 0 R
0
E 
The field points in the negative x direction, so
2 sin  0 ˆ
i
4 0 R
Choose a differential element of the rod dx a distance x from the origin, as shown in the
Q
dq  dx .
l
diagram. The charge on that differential element is
The variable x is treated as positive, so
53.
dE 
that the field due to this differential element is
the rod to find the total field.
dx
dx 
1
dq
4 0  x  x
2

dx
Q
4 0 l  x  x
2
.
Integrate along
l
 1   Q  1  1 
E   dE  






2
2

4 0 l  x  x
4 0 l 0  x  x
4 0 l  x  x 0 4 0 l  x x  l 
0
l

Q
Q
l
Q
Q
4 0 x  x  l 
58.
(a)The electron will experience a force in the opposite direction to the electric field. Since the
electron is to be brought to rest, the electric field must be in the same direction as the initial velocity of
the electron, and so is to the right.
(b)Since the field is uniform, the electron will experience a constant force, and therefore have a constant
acceleration. Use constant acceleration relationships to find the field strength.
F  qE  ma  a 
qE
m
v 2  v02  2ax  v02  2
qE
m
x 
E
60.

m v 2  v02
2qx
  mv
 9.109  10 kg  7.5  10 m s 

2qx
2  1.602  10 C   0.040 m 
31
2
0
5
19
2
 40 N C
 2 sig. fig.
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
y0  0, vx 0  1.90 m s,
and
x0  0,
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
y0
1
2
2
y
y
11
2
31
2
 4.2 m