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Physics 104
21.8.
Assignment 21
IDENTIFY: Use the mass of a sphere and the atomic mass of aluminum to find the number of aluminum atoms
in one sphere. Each atom has 13 electrons. Apply Coulomb’s law and calculate the magnitude of charge q on
each sphere.
SET UP: NA  602  1023 atoms/mol. q  nee, where ne is the number of electrons removed from one
sphere and added to the other.
EXECUTE: (a) The total number of electrons on each sphere equals the number of protons.


0.0250 kg
24
ne  np  (13)( N A ) 
    10 electrons.
 0.026982 kg/mol 
1 q2
. This gives
4e0 r 2
(b) For a force of 100  104 N to act between the spheres, F  100  104 N 
q  4e0 (100  104 N)(0800 m)2  843  104 C. The number of electrons removed from one sphere and
added to the other is ne  q /e  527 1015 electrons
21.9.
(c) ne/ne  727 1010.
EVALUATE: When ordinary objects receive a net charge the fractional change in the total number of electrons
in the object is very small.
IDENTIFY: Apply Coulomb’s law.
SET UP: Consider the force on one of the spheres.
(a) EXECUTE: q1  q2  q
F
1 q1q2
q2
F
0220 N

so q  r
 0150 m
 742 107 C (on each)
2
9
2 2
4 e0 r
(1/4 e0 )
4 e0r 2
8988  10 N  m /C
(b) q2  4q1
F
1 q1q2
4q12
F
F

 1r
 1 (742  107 C)  371  107 C
so q1  r
4 e0 r 2
4(1/4 e0 ) 2 (1/4 e0 ) 2
4 e0r 2
And then q2  4q1  148  106 C
EVALUATE: The force on one sphere is the same magnitude as the force on the other sphere, whether the
spheres have equal charges or not.
21.15.IDENTIFY: Apply Coulomb’s law. The two forces on q3 must have equal magnitudes and opposite directions.
SET UP: Like charges repel and unlike charges attract.
q q
EXECUTE: The force F2 that q2 exerts on q3 has magnitude F2  k 2 2 3 and is in the x-direction.
r2
F1 must be in the x-direction, so q1 must be positive. F1  F2 gives k
q1 q3
r12
k
q2 q3
r22
.
2
2
r 
 200 cm 
q1  q2  1    300 nC  
  0750 nC.
 400 cm 
 r2 
EVALUATE: The result for the magnitude of q1 doesn’t depend on the magnitude of q2 .
21.17.IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on q1.
SET UP: Like charges repel and unlike charges attract, so F2 and F3 are both in the  x-direction.
EXECUTE: F2  k
q1q2
r122
 6749  102 5 N, F3  k
q1q3
r132
F  18  104 N and is in the  x-direction.
1
 1124  104 N. F  F2  F3  18  104 N.
EVALUATE: Comparing our results to those in Example 21.3, we see that F1 on 3   F3 on 1, as required by
Newton’s third law.
21.21.IDENTIFY: Apply Coulomb’s law to calculate the force each of the two charges exerts on the third charge. Add
these forces as vectors.
SET UP: The three charges are placed as shown in Figure 21.21a.
Figure 21.21a
EXECUTE: Like charges repel and unlike attract, so the free-body diagram for q3 is as shown in
Figure 21.21b.
1 q1q3
F1 
4 e0 r132
F2 
1 q2q3
2
4 e0 r23
Figure 21.21b
F1  8988  109 N  m 2 /C2 
F2  8988  109 N  m 2 /C2 
(150  109 C)(500  109 C)
(0200 m) 2
(320  109 C)(500  109 C)
(0400 m) 2
 1685  106 N
 8988  107 N
The resultant force is R  F1  F2 
Rx  0
Ry   F1  F2   1685  106 N  8988  107 N  258  106 N
The resultant force has magnitude 258  106 N and is in the  y-direction.
EVALUATE: The force between q1 and q3 is attractive and the force between q2 and q3 is replusive.
21.25.IDENTIFY: F  q E. Since the field is uniform, the force and acceleration are constant and we can use a constant
acceleration equation to find the final speed.
SET UP: A proton has charge e and mass 167  1027 kg.
EXECUTE: (a) F  160  1019 C275  103 N/C  440  1016 N
(b) a 
F 440  1016 N

 263  1011 m/s 2
m 167  1027 kg
(c) vx  v0 x  axt gives v  (263  1011 m/s2 )(100  106 s)  263  105 m/s
EVALUATE: The acceleration is very large and the gravity force on the proton can be ignored.
21.29.(a) IDENTIFY: Eq. (21.4) relates the electric field, charge of the particle, and the force on the particle. If the particle
is to remain stationary the net force on it must be zero.
SET UP: The free-body diagram for the particle is sketched in Figure 21.29. The weight is mg, downward. For the
net force to be zero the force exerted by the electric field must be upward. The electric field is downward. Since the
electric field and the electric force are in opposite directions the charge of the particle is negative.
2
mg  q E
Figure 21.29
mg (145  102 3 kg)(980 m/s 2 )

 219  105 C and q  219C
E
650 N/C
(b) SET UP: The electrical force has magnitude FE  q E  eE The weight of a proton is w  mg FE  w so
eE  mg
q 
EXECUTE:
mg (1673  1027 kg)(980 m/s 2 )

 102  107 N/C
e
1602  1019 C
This is a very small electric field.
EVALUATE: In both cases q E  mg and E  (m/ q ) g  In part (b) the m/ q ratio is much smaller ( 108 )
EXECUTE: E 
than in part (a) ( 102 ) so E is much smaller in (b). For subatomic particles gravity can usually be ignored
compared to electric forces.
q
21.31.IDENTIFY: For a point charge, E  k 2  The net field is the vector sum of the fields produced by each charge. A
r
charge q in an electric field E experiences a force F  qE 
SET UP: The electric field of a negative charge is directed toward the charge. Point A is 0.100 m from q2 and
0.150 m from q1. Point B is 0.100 m from q1 and 0.350 m from q2 .
EXECUTE: (a) The electric fields at point A due to the charges are shown in Figure 21.31a.
q
625  109 C
E1  k 21  899  109 N  m 2 /C2 
 250  103 N/C
rA1
(0150 m) 2
E2  k
q2
 (899  109 N  m 2 /C2 )
125  109 C
 1124  104 N/C
rA22
(0100 m) 2
Since the two fields are in opposite directions, we subtract their magnitudes to find the net field.
E  E2  E1  874  103 N/C, to the right.
(b) The electric fields at point B are shown in Figure 21.31b.
q
625  109 C
E1  k 21  899  109 N  m 2 /C2 
 5619  103 N/C
2
rB1
(0100 m)
E2  k
q2
 (899  109 N  m 2 /C2 )
125  109 C
 917  102 N/C
(0350 m) 2
Since the fields are in the same direction, we add their magnitudes to find the net field.
rB22
E  E1  E2  654  103 N/C, to the right.
(c) At A, E  874  103 N/C, to the right. The force on a proton placed at this point would be
F  qE  (160 1019 C)(874 103 N/C)  140 1015 N, to the right.
EVALUATE: A proton has positive charge so the force that an electric field exerts on it is in the same direction
as the field.
Figure 21.31
3
21.32.IDENTIFY: The electric force is F  qE .
SET UP: The gravity force (weight) has magnitude w  mg and is downward.
EXECUTE: (a) To balance the weight the electric force must be upward. The electric field is downward,
so for an upward force the charge q of the person must be negative. w  F gives mg  q E and
mg (60 kg)(980 m/s 2 )

 39 C.
E
150 N/C
qq
(39 C) 2
 14  107 N. The repulsive force is immense and this is not a
(b) F  k 2  899  109 N  m 2 /C2 
r
(100 m) 2
feasible means of flight.
EVALUATE: The net charge of charged objects is typically much less than 1 C.
IDENTIFY: Eq. (21.3) gives the force on the particle in terms of its charge and the electric field between the
plates. The force is constant and produces a constant acceleration. The motion is similar to projectile motion;
use constant acceleration equations for the horizontal and vertical components of the motion.
(a) SET UP: The motion is sketched in Figure 21.33a.
q 
21.33.
For an electron q  e
Figure 21.33a
F  qE and q negative gives that F and E are in opposite directions, so F is upward. The free-body
diagram for the electron is given in Figure 21.33b.
EXECUTE:  Fy  ma y
eE  ma
Figure 21.33b
Solve the kinematics to find the acceleration of the electron: Just misses upper plate says that x  x0  200 cm
when y  y0  0500 cm
x-component
v0 x  v0  160  106 m/s, ax  0,x  x0  00200 m, t  ?
x  x0  v0 xt  12 a xt 2
x  x0
00200 m

 125  108 s
v0 x
160  106 m/s
In this same time t the electron travels 0.0050 m vertically:
y-component
t
t  125  108 s, v0 y  0, y  y0  00050 m, a y  ?
y  y0  v0 yt  12 a yt 2
2(00050 m)

 640  1013 m/s2
t2
(125  108 s)2
(This analysis is very similar to that used in Chapter 3 for projectile motion, except that here the acceleration is
upward rather than downward.) This acceleration must be produced by the electric-field force: eE  ma
ay 
2( y  y0 )
E
ma (9109  1031 kg)(640  1013 m/s 2 )

 364 N/C
e
1602  1019 C
4
Note that the acceleration produced by the electric field is much larger than g, the acceleration produced by
gravity, so it is perfectly ok to neglect the gravity force on the elctron in this problem.
(b) a 
eE (1602 1019 C)(364 N/C)

 349 1010 m/s 2
27
mp
1673 10
kg
This is much less than the acceleration of the electron in part (a) so the vertical deflection is less and the
proton won’t hit the plates. The proton has the same initial speed, so the proton takes the same time
t  125  108 s to travel horizontally the length of the plates. The force on the proton is downward (in the
same direction as E , since q is positive), so the acceleration is downward and a y  349  1010 m/s2 
y  y0  v0 yt  12 a yt 2  12 (349  1010 m/s 2 )(125  108 s)2  273  106 m The displacement is
273  106 m, downward.
(c) EVALUATE: The displacements are in opposite directions because the electron has negative charge and the
proton has positive charge. The electron and proton have the same magnitude of charge, so the force the electric
field exerts has the same magnitude for each charge. But the proton has a mass larger by a factor of 1836 so its
acceleration and its vertical displacement are smaller by this factor.
(d) In each case a g and it is reasonable to ignore the effects of gravity.
21.39.IDENTIFY: Find the angle  that r̂ makes with the x-axis. Then rˆ  (cos ) iˆ  sin   ˆj. .
SET UP: tan  y/x

 135 
ˆ
EXECUTE: (a) tan 1 
   rad. rˆ   j.
2
 0 
2ˆ
2 ˆ
 12  
i
j.
(b) tan 1    rad. rˆ 
2
2
12
4
 
 26 
ˆ
ˆ
(c) tan 1 
  197 rad  1129 rˆ  039i  092 j (Second quadrant).
 110 
EVALUATE: In each case we can verify that r̂ is a unit vector, because rˆ  rˆ  1.
21.45.IDENTIFY: Eq. (21.7) gives the electric field of each point charge. Use the principle of superposition and add the
electric field vectors. In part (b) use Eq. (21.3) to calculate the force, using the electric field calculated in part
(a).
(a) SET UP: The placement of charges is sketched in Figure 21.45a.
\
Figure 21.45a
The electric field of a point charge is directed away from the point charge if the charge is positive and toward
1 q
the point charge if the charge is negative. The magnitude of the electric field is E 
, where r is the
4e0 r 2
distance between the point where the field is calculated and the point charge.
(i) At point a the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45b.
Figure 21.45b
EXECUTE: E1 
9
1 q1
C
9
2 2 200  10


8

988

10
N

m
/C

 4494 N/C
2
4 e0 r12
(0200 m)
5
E2 
9
1 q2
C
9
2 2 500  10

(8

988

10
N

m
/C
)
 1248 N/C
2
2
4 e0 r2
(0600 m)
E1x  4494 N/C, E1 y  0
E2 x  1248 N/C, E2 y  0
Ex  E1x  E2 x  4494 N/C  1248 N/C  5742 N/C
E y  E1 y  E2 y  0
The resultant field at point a has magnitude 574 N/C and is in the x-direction.
(ii) SET UP: At point b the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45c.
Figure 21.45c
EXECUTE: E1 
E2 
1 q1
200  109 C
 8988  109 N  m 2/C 2 
 125 N/C
2
4 e0 r1
(120 m) 2
1 q2
500  109 C
 8988  109 N  m 2 /C2 
 2809 N/C
2
4 e0 r2
(0400 m) 2
E1x  125 N/C, E1 y  0
E2 x  2809 N/C, E2 y  0
Ex  E1x  E2 x  125 N/C  2809 N/C  2684 N/C
E y  E1 y  E2 y  0
The resultant field at point b has magnitude 268 N/C and is in the  x-direction.
(iii) SET UP: At point c the fields E1 of q1 and E2 of q2 are directed as shown in Figure 21.45d.
Figure 21.45d
EXECUTE: E1 
E2 
9
1 q1
C
9
2 2 200  10

(8

988

10
N

m
/C
)
 4494 N/C
2
2
4 e0 r1
(0200 m)
9
1 q2
C
9
2 2 500  10

(8

988

10
N

m
/C
)
 449 N/C
2
4 e0 r22
(100 m)
E1x  4494 N/C, E1 y  0
E2 x  449 N/C, E2 y  0
Ex  E1x  E2 x  4494 N/C  449 N/C  4045 N/C
E y  E1 y  E2 y  0
The resultant field at point b has magnitude 404 N/C and is in the  x-direction.
(b) SET UP: Since we have calculated E at each point the simplest way to get the force is to use F  eE
EXECUTE: (i) F  (1602 1019 C)(5742 N/C)  920 1017 N,  x-direction
(ii) F  (1602 1019 C)(2684 N/C)  430 1017 N,  x-direction
(iii) F  (1602 1019 C)(4045 N/C)  648 1017 N,  x -direction
6
EVALUATE: The general rule for electric field direction is away from positive charge and toward negative
charge. Whether the field is in the x- or x-direction depends on where the field point is relative to the charge
that produces the field. In part (a), for (i) the field magnitudes were added because the fields were in the same
direction and in (ii) and (iii) the field magnitudes were subtracted because the two fields were in opposite
directions. In part (b) we could have used Coulomb’s law to find the forces on the electron due to the two
charges and then added these force vectors, but using the resultant electric field is much easier.
q
21.47.IDENTIFY: E  k 2 . The net field is the vector sum of the fields due to each charge.
r
SET UP: The electric field of a negative charge is directed toward the charge. Label the charges q1, q2 and q3 ,
as shown in Figure 21.47a. This figure also shows additional distances and angles. The electric fields at point P
are shown in Figure 21.47b. This figure also shows the xy coordinates we will use and the x and y components
of the fields E1 , E 2 and E3.
EXECUTE: E1  E3  (899  109 N  m 2 /C2 )
E2  (899  109 N  m 2 /C2 )
200  106 C
(00600 m)
2
500  106 C
(0100 m) 2
 449  106 N/C
 499  106 N/C
E y  E1 y  E2 y  E3 y  0 and Ex  E1x  E2 x  E3x  E2  2E1 cos531  104  107 N/C
E  104  107 N/C, toward the 200C charge.
EVALUATE: The x-components of the fields of all three charges are in the same direction.
Figure 21.47
21.52.
IDENTIFY: For a long straight wire, E 
SET UP:
.
1
 180  1010 N  m2 /C2 .
2 e0
EXECUTE: r 
21.54.

2 e0 r
15  1010 C/m
 108 m
2e0 (250 N/C)
EVALUATE: For a point charge, E is proportional to 1/r 2  For a long straight line of charge, E is proportional
to 1/r
(a) IDENTIFY: The field is caused by a finite uniformly charged wire.
SET UP: The field for such a wire a distance x from its midpoint is
E
 1 
1


 2
.

2
2 e0 x ( x/a)  1
 4 e0  x ( x/a) 2  1
7
EXECUTE: E 
(180  109 N  m2 /C2 )(175  109 C/m)
2
 3.03  104 N/C, directed upward.
 600 cm 
(00600 m) 
 1
 425 cm 
(b) IDENTIFY: The field is caused by a uniformly charged circular wire.
SET UP: The field for such a wire a distance x from its midpoint is E 
1
Qx
. We first find the
4e0 ( x2  a2 )3/2
radius of the circle using 2 r  l.
EXECUTE: Solving for r gives r l/2  (8.50 cm)/2  1.353 cm.
The charge on this circle is Q  l  (175 nC/m)(0.0850 m)  14.88 nC.
The electric field is
E
1
Qx
(900 109 N  m2/C2 )(1488 102 9 C/m)(00600 m)

3/2
4 e0 ( x 2  a 2 )3/2
(00600 m)2  (001353 m)2 


E = 3.45  104 N/C, upward.
EVALUATE: In both cases, the fields are of the same order of magnitude, but the values are different because
the charge has been bent into different shapes.
8