Download - Cap 21 Fisica Solucionario
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21.1:
C103.20chargeandg00.8 9lead
v!!m
a) .100.2C106.1
C1020.3 1019
9
e v!v
v!
n
b) 1058.8and1033.2207
g00.8 13
lead
e22lead
v!v!v!nnNn A
21.2: s10s100ands000,20current4!!! Qt
Q = It= 2.00 C
.1025.1C1060.1
19
19ev!
v!
Qn
21.3: The mass is primarily protons and neutrons of271067.1 v!m kg, so:
28
27nandp 1019.4kg101.67
kg70.0
v!v! n
About one-half are protons, so e28
p 1010.2 nn !v! and the charge on the electrons is
given by: .1035.3)1010.2()1060.1(92819 v!vvv! Q
21.4: Mass of gold = 17.7 g and the atomic weight of gold is 197 mol.g So the number
of atoms .1041.5)1002.6(mol 22molg197
g7.1723 v!vv!vA
N
a)2422
p 1027.41041.579 v!vv!n
C106.83C1060.1 519p v!vv!nq
b) .1027.424
pe v!! nn
21.5: .electrons101.08atomsH106.021.80mol80.12423 v!vv!
C.101.73C1060.1101.08charge 51924 v!vvv!
21.6: First find the total charge on the spheres:
1043.1)2.0)(1057.4(444
1 162210
2
02
2
0
v!v!!! Frqr
q
F
And therefore, the total number of electrons required is
890.C101.60C1043.1 1916 !vv!! eqn
21.7: a) Using Coulombs Law for equal charges, we find:
.C1042.7C105.5m)150.0(4
1N220.0
7213
2
2
0
v!v!!! qq
b) When one charge is four times the other, we have:
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1071.310375.1m)150.0(
4
4
1220.0
7213
2
2
0
v!v!!! qq
F
So one charge is71071.3 v C, and the other is C.10484.1 6v
21.8:a) The total number of electrons on each sphere equals the number of protons.
.1025.7molkg026982.0
kg0250.013 24pe v!vv!! ANnn
b) For a force of41000.1 v N to act between the spheres,
.1043.8m)08.0(N)10(44
1N10
424
02
2
0
4 v!!!! qr
q
F
15
e 1027.5 v!!d eqn
c)10
e 107.27isvdn of the total number.
21.9: The force of gravity must equal the electric force.
.m08.5m8.25)sm8.9(kg)1011.9(
C)1060.1(
4
1
4
1 231
219
0
2
2
2
0
!!v
v!!
r
rr
q
mg
21.10: a) Rubbing the glass rod removes electrons from it, since it becomes positive.
kg.1027.4)electronkg1011.9(electrons)104.69(
electrons1069.4)Celectrons1025.6(C)107.50(nC50.7
203110
10189
v!vv
v!vv!
The rods mass decreases by kg.1027.420v
b) The number of electrons transferred is the same, but they are addedto the mass of the
plastic rod, which increases by kg.1027.420v
21.11: positive.isanddirection-theinbemustsodirection,-theinis 112 qxx FFTT
n750.00400.00200.0
,
2
2
1
2
23
32
2
13
31
21
!!
!!
qq
r
qqk
r
qqkFF
21.12: a)2
2
6
0
2
21
0 m)30.0(
C)10550.0(
4
1N200.0
4
1 q
r
qq
v!!
C.1064.3 62 v! q b) 200.0!F N, and is attractive.
21.13: Since the charges are equal in sign the force is repulsive and of magnitude:
N172.0m)800.0(4
C)1050.3(2
0
26
2
2
!v
!!
r
kqF
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21.14: We only need the y-components, and each charge contributes equally.
0.6).sinsince(N173.0sinm)500.0(
C)104(C)100.2(
4
12
66
0
!!vv
!
F
Therefore, the total force is N35.02 !F , downward.
21.15: theinbothareand 32 FF TT x-direction.
N10124.1,N10749.64
2
13
31
3
5
2
12
21
2
v!!v!!r
qqkF
r
qqkF
v!! thein,108.1 432 FFF x-direction.
21.16:
N100.0m60.0
C)100.2(C)10.20()CmN109(2
66229
21 !vvv
!
F
1QF is equal and opposite to QF1 (Ex. 21.4), so N17.0
N23.0
1
1
!
!
yQ
xQ
F
F
Overall:
N27.0N17.0N100.0
N23.0
!!
!
y
x
F
F
The magnitude of the total force is N.35.0N27.0N23.0 22 ! The direction of theforce, as measured from the +y axis is
Q
4027.0
23.0
tan1
!!
21.17: 2FT
is in the direction. x
N10.37N37.3N00.7
N00.7and
N37.3so,N37.3
23
32
22
12
21
2
!!!
!!
!!!
xxx
xxxx
x
FFF
FFFF
Fr
qqkF
For xF3 to be negative, 3q must be on the x-axis.
m144.0som,144.0so,3
31
2
31
3 !!!! xF
qqkxx
qqkF
21.18: The charge 3q must be to the right of the origin; otherwise both 32 and qq would
exert forces in the + x direction. Calculating the magnitude of the two forces:
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direction.in the375.3
m)200.0(
C)1000.5C)(1000.3)(Cm109(
4
12
66229
2
12
21
0
21
x
r
qq
F
!
vvv!!
xr
rF
!
vvv
!
We need
! FF
!
r
r gfr g.
8...
.
!
!
!
r
r
21.19:
FFF !! FFFTTT
since they are acting in the same direction at
m!y so,
o
!
v!
v
vv!
"
21.20: #$
#
$
FFF !! FFFTTT
since they are acting in opposite directions at
x= 0 so,
righttheto0000
000
000
000000
%
&
'
&
'
'
0
v!
vvv!
(
21.21: a)
b) )0))1
))
1 )) xa
qQa
xa
qQ
22
yx
!!
c) c34
ya
q 5
6
7
8
x y !!
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d)
21.22: a)
b)))
9
/@99
A
99
A
!
!
! yx F
xa
qQx
xa
qQ
F
c) At x = 0, F= 0.
d)
21.23:
b) thebelow45ofangleanat24
1221
4
12
24
12
2
02
2
02
2
0
r!!L
q
L
q
L
q
F
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positive x-axis
21.24: a) particle.thetowarddown,CN432m)250.0(
C)1000.3(
4
1
4
12
9
0
2
0
!v
!
r
q
E
b) m.50.1)CN0.12(C)1000.3(
41
41CN00.12
9
0
2
0
!v!!!
r
rq
E
21.25: Let +x-direction be to the right. Find :xa
? A CN5.23C)10602.1(2N)10516.7(le t.thetoissopositive,ischarge,direction-le tthetois
N10516.7
sm101.132gives
?,s1065.2,sm101.50,sm1050.1
1918
18
29
0
633
0
!vv!!
v!!
v!!
!v!v!v!
qFB
x
C aF
atavv
atvv
xx
xxxx
xxx
EFTT
21.26: (a)2
21 atx !
CN69.5
C106.1
)sm101.00(kg)1011.9(
sm1000.1s)103.00(
m)50.4(22
19
21231
212
26-2
!
v
vv!!!
v!v
!!
q
ma
q
FE
t
xa
The force is up, so the electric field must be downwardsince the electron is negative.
(b) The electrons acceleration is ~
11
10 g, so gravity must be negligibly smallcompared to the electrical force.
21.27: a) negative.issign,C1019.2CN650
)sm8.9(kg)00145.0( 52
v!!! qmgEq
b) C,N1002.1C101.60
)sm9.8(kg)1067.1( 719
227
v!v
v!! EmgqE
21.28: a) .CN1004.1m)1000.6(
C)1060.126(
4
1
4
1 11210
19
0
2
0
v!v
vv!!
r
q
E
b) CN1015.5m)1029.5(
C)1060.1(
4
1
4
1 11211
19
0
2
0
proton v!vv
!!
r
q
E
21.29: a) Fq andC,100.556v! is downward with magnitude
C,N1013.1There ore,N.1020.6 49 v!!v qFD
upward.
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b) If a copper nucleus is placed at that point, it feels an upward force of magnitude
N.105.24CN101.13C106.129 22419 v!vv!! qEF
21.30: a) The electric field of the Earth points toward the ground, so a NEGATIVE charge
will hover above the surface.
C.92.3CN150
)sm8.9(kg)0.60( 2 !!! qqEmg
b) N1038.1m)00.100(
C)92.3(
4
1
4
1 72
2
0
2
2
0
v!!!r
q
F The magnitude of the charge is too
great for practical use.
21.31: a) Passing between the charged plates the electron feels a force upward, and justmisses the top plate. The distance it travels in the y-direction is 0.005 m. Time of flight
s1025.1 8sm1060.1
m0200.06
vv!!! t and initialy-velocity is zero. Now,
.sm1040.6s)1025.1(m005.0so21328
212
210 v!v!! aaattvy y But also.CN364
C1060.1
)sm1040.6kg)(1011.9(19
21331
!!!!
v
vvEa
emeE
mF
b) Since the proton is more massive, it will accelerate less, and NOT hit the plates. Tofind the vertical displacement when it exits the plates, we use the kinematic equations
again:
m.1073.2s)1025.1(2
1
2
1 628
p
2 v!v!!m
eEaty
c) As mention in b), the proton will not hit one of the plates because although the
electric force felt by the proton is the same as the electron felt, a smaller acceleration resultsfor the more massive proton.
d) The acceleration produced by the electric force is much greater thang; it isreasonable to ignore gravity.
21.32: a)
jjE )CN10813.2(
m0400.0
C)1000.5()CmN109(
4
4
2
9229
2
10
1
1v!
vv!!
r
qT
CN1008.1
m)0400.0(0.0300m
C)1000.3()CmN109( 4
22
9229
2
2
2
2v!
vv!!
r
qET
The angle of ,2ET
measured from the axis,-x is r! 9.126tan180cm3.00
cm00.41 Thus
ji
jiE
)CN10(8.64)CN106.485(
)9.126sin9.126cos()CN10080.1(
33
4
2
vv!
rrv!T
b) The resultant field is
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ji
jiEE
)CN1095.1()CN10485.6(
)CN1064.8CN10813.2()CN10485.6(
43
343
21
vv!
vvv!TT
21.33: Let x be to the right and y be downward.
Use the horizontal motion to find the time when the electron emerges from the field:
sm1079.1
sm1000.8gives2
,s1025.1,0m,0.0050
sm1060.1
s1025.1gives
,sm1060.1,0,m0200.0
622
50
0
8
0y0
6
82
21
00
6
00
v!!
v!
!
!v!!!
v!
v!!
!v!!!
yx
y
yy
y
x
xx
xx
vvv
vtvv
yy
vtvyy
v
ttatvxx
tvaxx
21.34: a) .CN8.17)14()11(so,CN14CN11 22 !!! EjiET
r!r!! 128so,8.51)1114(tan 1 counterclockwise from the x-axis
b) )repulsive(52at)iN,1045.4)C105.2()CN8.17(so89 rv!v!! FqEF
TT
).repulsive(128at)ii r
21.35: a) !!v!v!! eEFgmF e30231
eg .N1093.8)sm(9.8kg)1011.9(
.N1060.1)CN1000.1()C1060.1( 15419 v!vv Yes, ok to neglect gF because .ge FF ""
b) kg1063.1N106.1CN10
1615
e
4
v!!v!! mmgFF
.101.79 e
14mm v! c) No. The field is uniform.
21.36: a) .CN148)1050.1()C1060.1(
)kg1067.1()m0160.0(2
2
1
2
12619
2722 !
vv
v!!!
sEt
m
eEatx
p
b) .sm1013.24
0 v!!! tm
eEatvv
p
21.37: a) jr ,20
35.1tan 1 !!
T b) jir
22
22,
42.12tan 1 !!
c) jir 92.039.0,112.9radians97.110.1
6.2tan 1 !r!!
(Second quadrant).
21.38: a) .N1082.9,CN61417v!!! qEFE
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b) .N103.2)100.1(48210
0
2 v!v! eF c) Part (b) >> Part (a), so the electron hardly notices the electric field. A person in the
electric field should notice nothing if physiological effects are based solely on magnitude.
21.39: a) Let x be east.ET
is west and q is negative, so FT
is east and the electron speeds up.
sm1033.6gives)(2
?m,375.0,sm10638.2,sm1050.4
sm10638.2)kg10109.9()N10403.2(
N10403.2)m50.1()C10602.1(||
5
0
2
0
2
0
2115
0
2113119
1919
v!!
!!v!v!
v!vv!!
v!v!!
xxxx
xxx
xx
x
vxxavv
vxxav
mFa
EqF
b) FT
so0"q is west and the proton slows down.
sm1059.1gives)(2
?,m375.0,sm10436.1,sm1090.1
sm10436.1)kg10673.1()N10403.2(
N10403.2)m50.1()C10602.1(||
4
0022
0
284
0
282719
1919
v!!
!!v!v!
v!vv!!
v!v!!
xxxx
xxx
xx
x
vxxavv
vxxav
mFa
EqF
21.40: Point charges1
q (0.500 nC) and2
q (8.00 nC) are separated by .m20.1!x The
electric field is zero when !!!!
2
11
2
12)20.1(21)2.1(
21
2
21
1 rqrqEEr
kq
r
kq
072.02.17.5or0)2.1()2.1(2)(2.1)2.1(2 12
11
2
11
2
1121
2
11
2
11 !! rrqrqrqqqrqrq24.04.0,24.0
11!! rr is the point between.
21.41:Two positive charges, q , are on the x-axis a distance a from the origin.
a) Halfway between them, .0!E
b) At any position
"
!
axxa
q
xa
q
axxa
q
xa
q
axxa
q
xa
q
Ex
,)()(4
1
,)()(4
1
||,)()(4
1
,
22
0
22
0
22
0
For graph, see below.
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21.42: The point where the two fields cancel each other will have to be closer to the
negative charge, because it is smaller. Also, it cantt be between the two, since the twofields would then act in the same direction. We could use Coulombs 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 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:
m0.
0.
!
!
x
xx
21.43: a) Point chargeG
q (2.00 nC) is at the origin and nC00.5(2 q ) is at !x
i) At i t))
H
H
H
I !!!qkqk
EP
ii) At t))
Q
R
Q
Q
!!!qkqk
ES
iii) At t))
T
T
T
U !!!qkqkV
W
b) !v!v!!
FFeEF )iileft,N102.9N575106.1i) 1719
right.N1048.6405106.1iii)right,N103.4N269106.1 17191719 vXvXvXv F
21.44: A positive and negative charge, of equal magnitude q , are on the x-axis, a distanceY from the origin.
a) Halfway between them, ,a
b
q
E! to the left.
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!! E atQ6.12
4.128
8.28tan 1 !
! up from x axis.
21.47: a) At the origin,)
)
iiE !v
!
T
b) At !! yx
.C2133)m45.0(
1
)m15.0(
1)C100.6(
4
122
9
0
iiE !
v!
T
c) At !! y
v! jij
5.0
4.0
)m5.0(
15.0
3.0
)m5.0(
1)m4.0(
1)C100.6(
4
1222
9
0
T
!! EjiET
and r clockwise from x
axis.
d) ).
)..
.
.
iE !v!!!!
Eyx y
T
21.48: For a long straight wire,
!v
!!
r
rE
21.49: a) For a wire of length a2 centered at the origin and lying along the y-axis, theelectric field is given by Eq. (21.10).
iE
1
1
0
!
axx
T
b) For an infinite line of charge:
iE
2
0x
!T
Graphs of electric field versus position for both are shown below.
21.50: For a ring of charge, the electric field is given by Eq. (21.8).
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a) iE)4
12322
0 ax
Qx
!T
so with
iE !!!v!
TxaQ
b) ))
j
k
l
j m j
iiEFF vn
v
n
n
n
q
21.51: For a uniformly charged disk, the electric field is given by Eq. (21.11):
iE
1
11
2 220
o
xR
T
The x -component of the electric field is shown below.
21.52: The earths electric field is 150 , directly downward. So,
,10.00150
0
0
v!!!!
E and is negative.
21.53: For an infinite plane sheet, E is constant and is given by
E! directed
perpendicular to the surface.
c
e.
c.
z!
z
!e
so{
|
}
~
!
!
directed toward the surface.
21.54: By superposition we can add the electric fields from two parallel sheets of charge.
a) .!E
b) !E c)
E !! directed downward.
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21.55:
21.56: The field appears like that of a point charge a long way from the disk and an infiniteplane close to the disks center. The field is symmetrical on the right and left (not shown).
21.57: An infinite line of charge has a radial field in the plane through the wire, and
constant in the plane of the wire, mirror-imaged about the wire:
Cross section through the wire: Plane of the wire:
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Length of vector does not depend on angle. Length of vector gets shorter at pointsfurther away from wire.
21.58: a) Since field lines pass from positive charges and toward negative charges, we can
deduce that the top charge is positive, middle is negative, and bottom is positive.
b) The electric field is the smallest on the horizontal line through the middle charge, at
two positions on either side where the field lines are least dense. Here the y-components ofthe field are cancelled between the positive charges and the negative charge cancels the x-component of the field from the two positive charges.
21.59: a) ))
!v!qdp , in the direction from and
towards q .
b) IfET
is at r and the torque sinJpE! then:
5.85636.9sin104.1(
mN102.7
sin 11
9
!rv
v!!
Jp
21.60: a) 105.56)10(1.6m)109.(1119
0 v!vv!! qpd
b) ))
v!vv!!
pE
Maximum torque:
21.61: a) Changing the orientation of a dipole from parallel to perpendicular yields:
!vv!rr!!( )CN10m)(1.6C100.5()0cos90cos(
630pEpEUUU if
J
v
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b) K.384.0)K103(1.38
)108(2108
2
323
2424 !
vv
!v!
TkT
21.62: 11.4
m)100.3(2
mC106.17m)1000.3(
2
)(39
0
309
dipole3
0
dipole !
vTI
v!v!
E
x
pxE
.CN106v The electric force !vv!! )CN10C)(4.111060.1( 619qEF N106.58 13v and is toward the water molecule (negativex-direction).
21.63: a)222222
22
22)4(
2
)4(
)2()2(
)2(
1
)2(
1
dy
yd
dy
dydy
dydy !
!
}
!
!
3
0
222
0
222
0 2)4(2)4(
2
4 y
p
dy
y
qd
dy
yd
qEy
b) This also gives the correct expression for yE since y appears in the full expressions
denominator squared, so the signs carry through correctly.
21.64: a) The magnitude of the field the due to each charge is
,))2(
1
44
122
0
2
0
!!
xd
q
r
q
E
where dis the distance between the two charges. The x-components of the forces due to the
two charges are equal and oppositely directed and so cancel each other. The two fields have
equaly-components, so:
xd
qEE
ysin
)2(
1
4
22
22
0
!!
where is the angle below the x-axis for both fields.
;)2(
2sin
22 xd
d
!
thus
2322
02222
0
dipole))2((4)2(
2
)2(
1
4
2
xd
qd
xd
d
xd
qE
!
!
The field is the y directions.
b) At large ,)2(,22
dxx "" so the relationship reduces to the approximations
3
0
dipole4 x
qdE }
21.65:
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The dipoles attract.
!! xxx FFF yyyy FFFF !!
b)
Opposite charges are closest so the dipoles attract.
21.66: a)
The torque is zero when pT
is aligned either in the samedirection as ET
or in the
opposite directions
b) The stable orientation is when p
T
is aligned in the same direction as E
T
c)
21.67:
r!! 48.6so00.250.1si
Opposite charges attract and like charges repel.
!! xxx FFF
!!!
!!
v!vv
!d
!
yyyy
y
FFFF
FF
k
r
qqkF
(in the direction from the C-00.5 charge toward the -.5 charge).
b)
-
8/4/2019 Cap 21 Fisica Solucionario
18/32
The y components have zero moment arm and therefore zero torque.
xF and xF2 both produce clockwise torques.
.cos
!! FFx
N
xF clockwise
21.68: a) jiF sin41cos
41 2
13
31
0
2
13
31
0
13 r
r
!
T
j
i
5
3
m)1016.0)((9.00
nC)nC)(6.0000.5(
4
15
4
m)1016.0)((9.00
nC)nC)(6.0000.5(
4
14
0
4
0
13 v
v
! FT
ji
vv!
T
Similarly for the force from the other charge:
jjjF )1020.1(m)(0.0300
nC)nC)(6.0000.2(
4
14
1 42
0
2
23
32
0
23 Nr
v!
!
!T
Therefore the two force components are:
48
v
x
v!vv!yF
b) Thus, ))
v!vv!! yx FFF and
the angle is ,.)a tan( !! x FF below the xaxis
21.69: a)
!
!
222
0
2
0
2
0 )1(
1
)1(
1
4
1
)(4
1
)(4
1
aaa
qQ
xa
qQ
xa
qQ
q
.44
1.))..21(...21(
4
13
0
2
0
2
0
xa
qQ
a
x
a
qQ
a
x
a
x
a
qQ
q
!
!} But this is the
equation of a simple harmonic oscillator, so:
am
kqQ
am
qQ
f
am
qQf !!!!
b) If the charge was placed on they-axis there would be no restoring force ifq and Q
had the same sign. It would move straight out from the origin along the y-axis, since the x-
components of force would cancel.
-
8/4/2019 Cap 21 Fisica Solucionario
19/32
21.70: Examining the forces: cosansin e !!!! mgTFFTF x
So
o
d
kq
mgF !! But .t 31
2
23
2 0
22
mg
Lq
mg
Lkq
Ld dd !!}
21.71: a)
b) Using the same force analysis as in problem 21.70, we find:
gdq tan4
2
0
2
!and
!
rr!
! qqd )sm80.9)(kg015.0(25tan)25sin)2.1(2(4(1.2)sin25222
0
.1079.2 6v
c) From Problem 21.70, .ta 22
d
kqmg !
Lmg
kq
L
d
222
2-629
22
2
sin)smkg)(9.8015.0(m)6.0(4
C)10)(2.79CNm1099.8(
sin)2(tan
2sin
vv!!!
Therefore
2sin331.0tan ! . Numerical solution of this transcendental equation leads to
..
!
21.72: a) Free body diagram as in 21.71. Each charge still feels equal and opposite electric
forces.
b) .sinso.cos
r
q
qT
mgT e !!r!!r! (Note:
)342.0m)sin20500.0(21 !r!r
-
8/4/2019 Cap 21 Fisica Solucionario
20/32
c) From part (b), .C1071.3213
21
v!qq d) The charges on the spheres are made equal by connecting them with a wire, but we
still have 22
2
041
e N0453.0tan rQ
mgF !!! where .
221 qqQ
! But the separation2r is
known: m)500.0(22!r m.500.030sin !r Hence: 2202 4
21 rFQ eqq !! .C1012.1 6v!
This equation, along with that from part (b), gives us two equations in1
q and
C1024.2. 6212v! qqq and .C1070.3 21321
v!qq By elimination, substitution and
after solving the resulting quadratic equation, we find: C1006.26
1
v!q and
C1080.1 72v!q .
21.73: a) g2.30)molgmol)(22.99100.0(mNaClmol100.0 Na !!
g3.55)molgmol)(35.45100.0(mCl !!
Also the number of ions is22
A 1002.6mol)100.0( v!N so the charge is:
C.9630C)1060.1)(1002.6(1922
!vv!
q The force between two such charges is:
N.1009.2m)0200.0(
)9630(
4
1
4
1 212
2
0
2
2
0
v!!!r
q
F
b) .sm1089.5)kg10(3.55N)1009.2(223321 v!vv!! mFa
c) With such a large force between them, it does not seem reasonable to think the
sodium and chlorine ions could be separated in this way.
21.74: a)
v
v
!!v!
2
9
2
9
32
23
32
2
13
316
3m)0.2(
C105.4
m)0.3(
C)105.2(N100.4 kq
r
qkq
r
qkqF
nC.2.3)CN(1262
N100.4 63 !v!
q
b) The force acts on the middle charge to the right.
c) The force equals zero if the two forces from the other charges cancel. Because of the
magnitude and size of the charges, this can only occur to the left of the negative charge .2
q
Then:2
2
2
12313
)200.0()300.0( x
kq
x
kqFF
!
! where x is the distance from the origin.
Solving forx we find: m.76.1!x The other value ofx was between the two chargesand is not allowed.
21.75: a) ,6
4
1
)2(
)3(
4
12
2
02
0 L
q
L
qq
F !! toward the lower the left charge. The other two
forces are equal and opposite.
-
8/4/2019 Cap 21 Fisica Solucionario
21/32
b) The upper left charge and lower right charge have equal magnitude forces at rightangles to each other, resulting in a total force of twice the force of one, directed toward thelower left charge. So, all the forces sum to:
.2
323
4)2(
)3(2)3(
4
12
0
2
22
0
!
!
L
q
L
qq
L
qq
F
21.76: a) .2
)()(4
1)(
222
0
!
y
q
ay
q
ay
q
pE
b) ).2)1()1((
4
1)( 22
2
0
! yaya
y
q
p
Using the binomial expansion:
.4
12
321
321
4
1)(
4
2
0
2
2
2
2
2
0 y
qa
y
a
y
a
y
a
y
a
y
q
p
!
}
Note that a point charge drops off like2
1
yand a dipole like .
13y
21.77: a) The field is all in the x -direction (the rest cancels). From the charges:
.)(4
1
4
1
4
12322
02222
0
22
0 xa
qx
xa
x
xa
q
E
xa
q
E x
!
!
!
(Each q contributes this). From the q
!
!! xa
x
q
x
q
xa
qx
E
x
q
E
x
b) ..for,3
4
11
2
31
2
4
14
2
0
2
2
2
0
total axx
qa
x
a
x
q
E ""!
}
-
8/4/2019 Cap 21 Fisica Solucionario
22/32
Note that a point charge drops off like x
and a dipole like x
.
21.78: a) 20.0 g carbon .og0.2
0.20!
gmol carbon 0.10)67.1(6 ! mol electrons
.100. 6)1060.1()0.10( 61
A
!
!
N This much charge is placed at the earths
poles (negative at north, positive at south), leading to a force:
.1013.)m10276.1(
C)10963.0(
4
1
)2(4
1 727
26
0
2
earth
2
0
vv
v
R
F
b) A positive charge at the equator of the same magnitude as above will feel a forcein the south-to-north direction, perpendicular to the earths surface:
m)
)
sin)
earth
v
v
v
F
R
q
F Q
21.79: a) With the mass of the book about 1.0 kg, most of which is protons and neutrons,we find: #protons = .100.3kg)1067.1(kg)0.1(
2627
21
v!v
Thus the charge difference
present if the electrons charge was %999.99 of the protons is
.0)106.1)(00001.0)(100.( 1926 !vv!( q
b) raThr pu s v38))48) 3
v
( krqkF
sm103.8)kg1()N103.8(21313
v!v!! mF
c) Thus even the slightest charge imbalance in matter would lead to explosive
repulsion!
21.80:(a)
-
8/4/2019 Cap 21 Fisica Solucionario
23/32
22
0
2
22
22
0
2
22
0
2
2
2
0
2
2
0
net
)()(
4
4)()(
)()(
4
)(
1
)(
1
4
)(4
1
)(4
1charge)central(on
xbxb
bx
q
xbxb
xbxb
q
xbxb
q
xb
q
xb
q
F
!
!
!
!
For ,bx this expression becomes
.tooppositeisDirection3
0
2
22
0
2
net xxb
q
bb
bx
qF !
(b) 22
30
2
:dt
xd
b
qmxmaF !!
TI
3
0
2
3
0
2
3
0
2
2
2
2
12bm
q
ff
bm
q
xbm
q
dt
xd
!p!!
!
(c) kg)1066.1(12amu12m,100.4,2710 v!!v!! mbeq
z1028.4)m100.4)(NmC1085.8()kg1066.1(12
)C106.1(
2
1 12310221227
219
v!vvv
v!
f
21.81: a) !vv!!! 3334333
34 )m1000.1)()(mkg109.8()( rVm
kg10728.3 5v
mol10867.5)molkg10546.63)(kg10728.3( 435 v!vv!! Mmn
atoms105.3 20A v!! nNN (b) protonsandelectrons10015.1)105.3)(29(
2220
e v!v!N
C6.1)10015.1)(C10602.1)(10100.0()99900.0( 22192eenet !vvv!!
eNeNq
N103.2)m00.1(
)C6.1( 102
2
2
2
v!!! kr
qkF
21.82: First, the mass of the drop:
kg.1041.13
)m100.15(4)mkg1000( 11
363
v!
v!!
m
Next, the time of flight: :onacceleratitheands00100.02002.0 !!! vDt
.sm600)s001.0(
)m1000.3(22
2
1 22
4
2
2 !v
!!!
t
daatd
So:
C.1006.1CN1000.8
)sm600)(kg1041.1( 134
211
v!v
v!!!! EmaqmqEmFa
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8/4/2019 Cap 21 Fisica Solucionario
24/32
21.83: eEFy ! 0!xF
pp m
eE
m
Fa
y
y!! 0!
xa
a) ymeEvyavvp
yyy (!(!2
sin2 220202 0when ax !!( yvhy
eE
vh
p
2
sin 22omax !
b) 202
1tatvy yy !(
ig ig
ig
i
whe
tta
tt
y
!
!(!
ya
vt sin2,so
orig !
or
eE
mvtvd
eE
mvt
x sinos2
sin2
p
2
0
orig0
p0
orig
!!
!
c)
d) m42.0)N500)(106.1(2
30sin)kg1067.1()sm104(19
22725
m x !v
rvv!
h
m89.2)CN500)C106.1
30s n30cos)kg1067.1)sm104219
2725
!v
rrvv
! d
21.84: a) !
!!! 22
2
2
2
1
1
0
2
2
2
0
2
1
1
0 4
1
4
1
4
1CN50 q
r
q
r
q
r
q
r
q
-
8/4/2019 Cap 21 Fisica Solucionario
25/32
C.108)m6.0(
)C1000.4(CN0.504)m2.1(4
9
2
9
0
2
22
1
10
2
2
v!
v!
q
r
qEr
b) !
!!! 22
2
2
2
1
1
0
2
2
2
0
2
1
1
04
1
4
1
4
1CN50 q
r
q
r
q
r
q
r
q
E
C.100.24)m6.0(
)C1000.4()0.50(4)m2.1(
50 92
9
0
2
22
1
12
2
v!
v!
q
r
q
kr
21.85:222
)m00.5()m00.8(
nC)0.12(
)m00.3(
)nC0.16(CN0.12
kqkkE
!!
nC.1.73C1031.7m00.64
C1020.1
m0.9
C1060.112m0.25 8
2
8
2
82 !v!
v
v!
kq
21.86: a) On the x-axis: !!!a
xxxraa
Qdx
Era
dq
dE0
2
0
2
0)(4
1)(4
1
0nd.11
4
1
0
!
yE
rara
Q
.
b) .11
4
111
4
1then,I
00
iEF
!!
!!
xaxa
qQ
q
xaxa
Q
Exra
TT
c) }}!!"" 2
1 )11()1)1((,Forx
kqQxa
ax
kqQxa
ax
kqQFax
pointalikelooksondistributiCharge.),forthatNote(.4
12
0 xaxraxr
qQ
}!""
from far away.
21.87: a) !
!
!
! yx dEyxa
dykQxdE
yxa
dykQ
yx
dqkdE and
(with
)()(23222222
.)(
2322yxa
dyKQy
Thus:
TI!TI!!a
xaxx
Q
x
a
axa
Qx
yx
dy
a
Qx
E
0
2122
0
22122
0
2322
0)(4
1
)(
1
4
1
)(4
1
!
! 212200
2322
0)(
11
4
1
)(4
1
axxa
Q
yx
ydy
a
Q
E
a
y
b) (a).ingivenareandhereand yxyyxx EEqEFqEF !!
c) .24
1
24
1))1(1(
4
1,or
3
0
2
2
0
2122
0 x
qQa
x
a
ax
qQ
xa
ax
qQ
Fax y !}!""
-
8/4/2019 Cap 21 Fisica Solucionario
26/32
Looks dipole-like iny-direction .4
14
1
2
0
21
2
2
2
0 x
qQ
x
a
x
qQ
Fx }
!
Looks point-like along x-direction
21.88: a) iE 4
1),9.22(.From22
0 axxQ
Eq
!T
.)CN1029.1(
)m025.0()m105.2()m105.2(
)1000.9(
4
1 62233
9
0
iiE v!vv
v!
C
T
(b) The electric field is less than that at the same distance from an infinite line of
charge ).CN1030.12
2
4
12
4
1( 6
00
v!
!!gpax
Q
xEa This is because in the
approximation, the terms left off were negative. !
}
22
0
21
02
1212
12
2
ax
xxa
x
gLine
E (Higher order terms).
c) For a 1 difference, we need the next highest term in the expansion that was left off
to be less than 0.01:
cm.0.352(0.01)m025.0)01.0(201.02 2
2
! xaxa
x
21.89: (a) From Eq. (22.9), iE 4
122
0 axx
Q
!T
.C)N7858()m025.0(m)(0.100m)(0.100
C)100.9(
4
122
9
0
iE !
v!
T
b) The electric field is less than that at the same distance from a point charge (8100
point2
2
2
02
14
1ESinceC).N E
x
a
x
Q
x
!
!gp (Higher order terms).
c) For a 1 difference, we need the next highest term in the expansion that was left
off to be less than 0.01:
.m177.002.01025.0))01.0(2(101.02 2
2
}!}} xaxx
a
-
8/4/2019 Cap 21 Fisica Solucionario
27/32
21.90: (a) On the axis,
!
!
21
2
2
0
221
2
2
0
1)m0020.0(
)m025.0(1
2
)m025.0(Cp00.411
2
x
R
E
! thein,CN106E x-direction.
b) The electric field is less than that of an infinite sheet .CN1152 0
!!g
E
c) Finite disk electric field can be expanded using the binomial theorem since the
expansion terms are small:
}3
3
02
12 R
x
R
x
E So the difference between the
infinite sheet and finite disk goes like .R
xThus:
.1616.05.24.0
cm)40.0(and80.082.50.2cm)20.0(
!!}
!(!!}!( xExE
21.91: (a) As in 22.72:
!
21
2
2
0
112 x
E
E
!
21
2
2
0
2
1)m200.0(
m)025.0(1
2
m)(0.025pC00.4
! in theCN89.0 x-direction.
b) )]8321(1[2
, 4422
0
!"" xRxR
ERx
.4422
2
0
2
0
2
2
2
0 x
Q
x
R
x
R
!!}
c) The electric field of (a) is less than that of the point charge (0.90 )CN since the
correction term that was omitted was negative.
d) From above m1.0For.101.00.89
0.89)(0.9m2.0 !!!
! xx
CN6.3
CN43.3
point
disk
!
!
E
E
-
8/4/2019 Cap 21 Fisica Solucionario
28/32
so .5047.06.3
)43.36.3(}!
21.92: a)
!!!
a aa
a axdxfdxxfdxxfdxxfxfxf
0 0
0
)()()()()(:)()(
!!
a
a a
a
a a
dxxf
dxxfydyfdxxfyxdxxf
0
0 0 0
.)(2
)()()()(:ithreplaceNo.)(
b)
!!!a
a a
a a
xdxgdxxgdxxgdxxgxgxg0
0 0))()(()()()(:)()(
!!a
a
a aa
dxxgydygdxxgyxdxxg0 00
.0)()()()(:withreplaceNow.)(
c) The integrand in yE for Example 21.11 is odd, so yE =0.
21.93: a) The y-components of the electric field cancel, and the x-components from bothcharges, as given in problem 21.87 is:
.)(
112
4
1
)(
112
4
12122
0
2122
0
iF
!
!ayya
Qq
ayya
Q
Ex
T
If .4
1))21(1(2
4
1,
3
0
22
0 y
Qqa
ya
ay
Qq
ay !
}"" iFT
b) If the point charge is now on thex-axis the two charged parts of the rods provide
different forces, though still along the x-axis (see problem 21.86).
iEFiEF 11
4
1and
11
4
1
00
!!
!! axxa
Qq
q
xaxa
Qq
q
TTTT
So,
iFFF 121
4
1
0
!! axxaxa
Qq
TTT
For .2
4
1...12...14
1,
3
0
2
2
2
2
0
iiFx
Qqa
x
a
x
a
x
a
x
a
ax
Qq
ax !
}""
T
21.94: The electric field in the x-direction cancels the left and right halves of the
semicircle. The remaining y-component points in the negative y-direction. The charge per
unit length of the semicircle is:
-
8/4/2019 Cap 21 Fisica Solucionario
29/32
.sin
sinbutand2
a
dkdEdE
a
dk
a
dlkdE
a
Qy
!!!!!
So, !!U!!
o,o
ya
kQ
a
k
a
kd
a
kE
21.95: By symmetry, yx EE ! For yE compared to problem 21.94, the integral over the
angle is halved but the charge density doublesgiving the same result. Thus,
.22
2a
kQ
a
kEE yx !!!
21.96:
mgTmgTF
x
cos
cos0
!
!!
!
!
!
mg
q
mg
q
q
mg
q
T
q
TFy
0
00
00
2arctan
2tan
sin2cos
sin22sin0
21.97: a) 29)sm8.9(0
CN104.1
g1010
2
5
!v
!!!E
q
mmgqE
b) 1429 .e e
rb11.1
e e
1.1.
l
rb12..
112
l1. 1
192
v
vv
v ee
21.98: a) yx EE ! and
ere,4
122
2
2
20
c
ar
e, ireoflen
t
aQax
xx
Q
EE
!!
.,
a
QE
a
QE
ax yx
-
8/4/2019 Cap 21 Fisica Solucionario
30/32
b) If all edges of the square had equal charge, the electric fields would cancel by
symmetry at the center of the square.
21.99: a)0
2
0
2
0
2
0
3
0
2
0
1
2
mC0200.0
2
mC0100.0
2
mC0200.0
2
2
2
)(
PE !!
v!! in theC,N1065.52
mC0100.0)(
8
0
2
PE x-direction.
b)0
2
0
2
0
2
0
3
0
2
0
1
2
mC0200.0
2
mC0100.0
2
mC0200.0
2
2
2
)(
RE !!
v!! in theC,N1069.12
mC0300.0)(
9
0
2
RE x-direction.
c)0
2
0
2
0
2
0
3
0
2
0
1
2
mC0200.0
2
mC0100.0
2
mC0200.0
2
2
2
)(
SE !!
v!! in theC,N1082.22
mC0500.0)( 9
0
2
SE x-direction.
d)0
2
0
2
0
2
0
3
0
2
0
1
2
mC0200.0
2
mC0100.0
2
mC0200.0
2
2
2
)(
TE !!
v!! in theC,N1065.52
mC0100.0)(
8
0
2
SE x-direction.
21.100: m.N1013.12
mC1000.2
2
7
0
24
0
32
1IatIon v!
v!
!!
A
qE
A
F
mN1026.22
mC1000.4
2
7
0
24
0
312IIatIIon v!v
!
!!
A
qE
A
F
mN1039.32
mC1000.6
2
7
0
24
0
21
3IIIatIIIon v!
v!
!!
A
qE
A
F
(Note that + means toward the right, and is toward the left.)
21.101: By inspection the fields in the different regions are as shown below:
).(2
)(2
),(2
)(2
),(2
0
00
00
kiE
kiki
kiki
z
z
x
x
E
E
E
E
IVIII
III
!@
!
!
!
!
T
-
8/4/2019 Cap 21 Fisica Solucionario
31/32
21.102: a) RRAQ )(2
2
2
2
b) Recall the electric field of a disk, Eq. (21.11): ? A o,.1)(112
2
0
xR
E
? A ? A
.)()(
)()()()(
i
iE
x
xxRxR
xE
x
xxRxR
x
v!
!T
c) Note that
}! ...
2
)1))11)1
2
1
1
212
1
1
2
1
Rx
R
xRx
R
xxR
,11
2
2)(
2210210
ii
x
x
x
x
x
x
xE
!
!
Xand sufficiently close means that
Rx
d) )(
!!!
!!
RRm
q
fmx
RR
qxq
F
21.103: a) The four possible force diagrams are:
Only the last picture can result in an electric field in the x-direction.b) .and.4. 3 "!! qqq
c)22
2
0
12
1
0
sinm)0300.0(4
1sin
)m0400.0(4
10
q
q
Ey !!
C.43.064
27
54
53
16
9
sin
sin
16
911
2
112 qq
qq !!!!
-
8/4/2019 Cap 21 Fisica Solucionario
32/32
d) .565
3
9.5
4
16.4
1 1
333 !
!!qq
qEqF
x
21.104: (a) The four possible diagrams are:
The first diagram is the only one in which the electric field must point in the negativey-direction.
b) .0nd00.3 21 ! qq
c)12
5
)050.0()120.0(13
12
)120.0(13
5
)050.0(0
2
1
2
2
2
2
2
1 kqkqkqkqEx
!!!
!!!13
5
12
5
13
12
m)05.0(13
5
m)120.0(13
12
m)050.0( 21
2
2
2
1 kqkqkqEEy
C.N1017.17
v!! yEE
21.105: a) For a rod in general of length
!rLrL
kQEL
11, and here .
2
axr !
So, .ro!" # $ %
!
!axLaxL
kQ
axLaxL
kQE
b)
v!!!&!2
2
2
22
22aL
a
aL
aL
k'dx
L
E'dqE
(
Edqd(
dxaxLax
22
1
2
1
)
)
)
)
)
)
)][l ()](l[ aLaaL
a axLxaL
k0F !1
.)(
)(2
2
2
2
2
3
3
4
3
3
33
4
5
Laa
La
L
kQ
aL
aL
a
aLa
L
kQF
c) For ))21ln()2(1n12()21(
)1(n1:
2
2
2
22
2
2
aLaLL
kQ
aLa
aLa
L
kQFLa !
!""
6
6
6
6
6
6
6
6
a
kQF
a
L
a
L
a
L
a
L
L
kQF }
}