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Physics 51
Chapter 21 (parts A and B), 13th ed, Freedman
Homework Solutions
21.9.
IDENTIFY: Apply Coulomb’s law.
SET UP: Consider the force on one of the spheres.
(a) EXECUTE: q1  q2  q
F
1 q1q2
q2

so
4 e0 r 2
4 e0 r 2
qr
F
0220 N
 0150 m
 742 107 C (on each)
9
2 2
(1/4 e0 )
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
2
4 e0 r
4(1/4 e0 ) 2 (1/4 e0 ) 2
4 e0 r 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
21.16.
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 .
IDENTIFY: Apply Coulomb’s law and find the vector sum of the two forces on Q.
SET UP: The force that q1 exerts on Q is repulsive, as in Example 21.4, but now the force that
q2 exerts is attractive.
EXECUTE: The x-components cancel. We only need the y-components, and each charge contributes
1 (20  106 C)(40  106 C)
sin   0173 N (since sin  0600)
equally. F1 y  F2 y  
4 e0
(0500 m) 2
Therefore, the total force is 2F  035 N, in the  y-direction.
EVALUATE: If q1 is 20C and q2 is 20C, then the net force is in the  y-direction.
21.19.
IDENTIFY and SET UP: Apply Coulomb’s law to calculate the force exerted by q2 and q3 on q1 Add
these forces as vectors to get the net force. The target variable is the x-coordinate of q3 
EXECUTE: F2 is in the x-direction.
F2  k
q1q2
r122
 337 N, so F2 x  337 N
Fx  F2 x  F3x and Fx  700 N
F3x  Fx  F2 x   700 N  337 N   1037 N
For F3x to be negative, q3 must be on the  x-axis.
F3  k
q1q3
x
2
, so x 
k q1q3
F3
 0144 m, so x  0144 m
EVALUATE: q2 attracts q1 in the x-direction so q3 must attract q1 in the x-direction, and q3 is at
negative x.
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.
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
EXECUTE:
q 
FE  w so eE  mg
mg (1673  1027 kg)(980 m/s 2 )

 102  107 N/C
19
e
1602  10
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
EXECUTE: E 
( 108 ) than in part (a) ( 102 ) so E is much smaller in (b). For subatomic particles gravity can
usually be ignored compared to electric forces.
21.35.
IDENTIFY: Apply constant acceleration equations to the motion of the electron.
SET UP: Let x be to the right and let  y be downward. The electron moves 2.00 cm to the right and
0.50 cm downward.
EXECUTE: Use the horizontal motion to find the time when the electron emerges from the field.
x  x0  00200 m, ax  0, v0 x  160  106 m/s. x  x0  v0 xt  12 axt 2 gives t  125  108 s. Since
 v0 y  v y 
ax  0, vx  160  106 m/s. y  y0  00050 m, v0y  0, t  125  108 s. y  y0  
 t gives
2


v y  800  105 m/s. Then v  vx2  v 2y  179  106 m/s.
EVALUATE: v y  v0 y  a yt gives a y  64  1013 m/s2. The electric field between the plates is
E
21.46.
ma y
e

(911  1031 kg)(64  1013 m/s2 )
160  1019 C
 364 V/m. This is not a very large field.
IDENTIFY: Apply Eq. (21.7) to calculate the field due to each charge and then require that the vector
sum of the two fields to be zero.
SET UP: The field of each charge is directed toward the charge if it is negative and away from the
charge if it is positive.
EXECUTE: The point where the two fields cancel each other will have to be closer to the negative
charge, because it is smaller. Also, it can’t be between the two charges, since the two fields would then
act in the same direction. We could use Coulomb’s law to calculate the actual values, but a simpler
way is to note that the 8.00 nC charge is twice as large as the 400 nC charge. The zero point will
therefore have to be a factor of
21.47.
2 farther from the 8.00 nC charge for the two fields to have equal
magnitude. Calling x the distance from the –4.00 nC charge: 120  x  2 x and x  290 m.
EVALUATE: This point is 4.10 m from the 8.00 nC charge. The two fields at this point are in opposite
directions and have equal magnitudes.
q
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 /C 2 )
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:

2 e0 r
.
1
 180  1010 N  m2 /C2 .
2e0
EXECUTE: 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
21.54.
(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
EXECUTE: E 
 1 
1


 2
.

2 e0 x ( x/a)2  1
 4 e0  x ( x/a) 2  1
(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
2
4e0 ( x  a2 )3/2
find the 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)

2
2
3/2
3/2
4 e0 ( x  a )
(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.
21.73.
IDENTIFY: The electric field exerts a horizontal force away from the wall on the ball. When the ball
hangs at rest, the forces on it (gravity, the tension in the string, and the electric force due to the field)
add to zero.
SET UP: The ball is in equilibrium, so for it  Fx  0 and  Fy  0. The force diagram for the ball is
given in Figure 21.73. FE is the force exerted by the electric field. F  qE . Since the electric field is
horizontal, FE is horizontal. Use the coordinates shown in the figure. The tension in the string has
been replaced by its x- and y-components.
Figure 21.73
EXECUTE:  Fy  0 gives Ty  mg  0. T cos  mg  0 and T 
mg
.  Fx  0 gives
cos 
FE  Tx  0. FE  T sin   0. Combing the equations and solving for FE gives
 mg 
3
2
2
FE  
 sin   mg tan   (12.3  10 kg)(9.80 m/s )(tan17.4)  3.78  10 N. FE  q E so
 cos 
FE 3.78  102 N

 3.41  104 N/C. Since q is negative and FE is to the right, E is to the left in the
q
1.11  106 C
figure.
E
EVALUATE: The larger the electric field E the greater the angle the string makes with the wall.
21.89.
IDENTIFY: Divide the charge distribution into infinitesimal segments of length dx. Calculate E x and
E y due to a segment and integrate to find the total field.
SET UP: The charge dQ of a segment of length dx is dQ  (Q/a)dx. The distance between a
segment at x and the charge q is a  r  x. (1  y)1  1  y when y
EXECUTE: (a) dEx 
a  r  x, so Ex 
(b) F  qE 
1.
1 a
Qdx
1 Q1
1 
1
dQ

so Ex 
 
.

2
2
0
4 e0 a(a  r  x)
4 e0 a  r a  r 
4 e0 (a  r  x)
1 Q 1
1
  . E y  0.

4 e0 a  x  a x 
1 qQ  1
1 ˆ
 
i .
4 e0 a  r a  r 
kqQ
kqQ
kqQ
1 qQ
((1  a/x)1  1) 
(1  a/x    1)  2 
. (Note
ax
ax
4 e0 r 2
x
that for x a, r  x  a  x.) The charge distribution looks like a point charge from far away, so the
force takes the form of the force between a pair of point charges.
EVALUATE: (c) For x
21.103.
a, F 
IDENTIFY and SET UP: Example 21.11 gives the electric field due to one infinite sheet. Add the two
fields as vectors.
EXECUTE: The electric field due to the first sheet, which is in the xy-plane, is E  ( /2e )kˆ for z  0
1
0
and E1  ( /2e0 )kˆ for z  0 We can write this as E1   /2e0  z/ z kˆ, since z/ z  1 for z  0 and
z/ z   z/z  1 for z  0 Similarly, we can write the electric field due to the second sheet as
E2  ( /2e0 )( x/ x )iˆ, since its charge density is   The net field is
E  E  E  ( /2e )( x/ x iˆ  ( z/ z )kˆ)
1
2
0
EVALUATE: The electric field is independent of the y-component of the field point since displacement in
the  y-direction is parallel to both planes. The field depends on which side of each plane the field is
located.