7/27/2019 Solucion Fisica Un_22
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22.1: a) C.Nm75.160cos)m(0.250N/C)14( 22 === AE
b) As long as the sheet is flat, its shape does not matter.
ci) The maximum flux occurs at an angle = 0 between the normal and field.cii) The minimum flux occurs at an angle = 90 between the normal and field.
In part i), the paper is oriented to capture the most field lines whereas in ii) thearea is oriented so that it captures no field lines.
22.2: a) nAAE wherecos AEA ===
CmN329.36cosm)(0.1)CN104((back)
CmN329.36cosm)(0.1)CN104(front)(
090cosm)(0.1)CN104(bottom)(
CmN24)9.36(90cosm)(0.1)CN104(right)(
090cosm)(0.1)CN104(top)(
CmN24)93690(cosm)(0.1)CN104(left)(
223
223
23
223
23S
223
66
55
44
33
22
11
===
=+=+=
===
+=+=+=
==+=
===
SS
SS
SS
SS
S
SS .
in
in
kn
jn
kn
jn
b) The total flux through the cube must be zero; any flux entering the cube must alsoleave it.
22.3: a) Given that lengthedge,,DCB AEkjiE
=+= L, and
.BL
.BL
.DL
.CL
.DL
.CL
26
25
24
23
22
21
66
55
44
33
22
11
+===
==+=
+===
+==+=
==+=
===
SS
SS
SS
SS
SS
SS
A
A
A
A
A
A
nEin
nEin
nEkn
nEjn
nEkn
nEjn
b) Total flux = ==6
1 i0
i
22.4: C.Nm16.670cos)m(0.240)CN0.75( 22 === AE
7/27/2019 Solucion Fisica Un_22
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22.5: a) C.Nm1071.2)2( 25m)(0.400C/m)1000.6(2 06
00=====
lr
rlAE
b) We would get the same flux as in (a) if the cylinders radius was made largerthefield lines must still pass through the surface.
c) If the length was increased to m,800.0=l the flux would increase by a factor of
two: C.Nm105.42
25
=
22.6: a) C.Nm452C)1000.4( 209
011=== qS
b) C.Nm881C)1080.7( 209
022=== qS
c) C.Nm429C)10)80.700.4(()( 209
0213==+= qqS
d) C.Nm723C)10)40.200.4(()( 209
0214=+=+= qqS
e) C.Nm158C)10)40.280.700.4(()( 209
03215=+=++= qqqS
f) All that matters for Gausss law is the total amount of charge enclosed by the
surface, not its distribution within the surface.
22.7: a) C.Nm1007.4C)1060.3( 2506
0 === q
b) C.106.90)CNm780( 92000==== qq
c) No. All that matters is the total charge enclosed by the cube, not the details of
where the charge is located.
22.8: a) No charge enclosed so 0=
b) C.Nm678NmC108.85C1000.6 2
2212
9
0
2 =
==
q
c) C.Nm226NmC108.85
C10)00.600.4( 22212
9
0
21 =
=
+=
22.9: a) Since E
is uniform, the flux through a closed surface must be zero. That is:
===== .0000 1 dVdVd qAE
But because we can choose any volume we
want, must be zero if the integral equals zero.
b) If there is no charge in a region of space, that does NOT mean that the electric fieldis uniform. Consider a closed volume close to, but not including, a point charge. The fielddiverges there, but there is no charge in that region.
7/27/2019 Solucion Fisica Un_22
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22.10: a) If 0> and uniform, then q inside any closed surface is greater than zero.
>> 00 AE
d and so the electric field cannot be uniform, i.e., since an
arbitrary surface of our choice encloses a non-zero amount of charge, E must depend onposition.
b) However, inside a small bubble of zero density within the material with density ,
the field CAN be uniform. All that is important is that there be zero flux through thesurface of the bubble (since it encloses no charge). (See Exercise 22.61.)
22.11: C.Nm1008.1C)1060.9( 2606
0sides6 === q But the box is
symmetrical, so for one side, the flux is: .CNm1080.1 25side1 =
b) No change. Charge enclosed is the same.
22.12: Since the cube is empty, there is no net charge enclosed in it. The net flux,according to Gausss law, must be zero.
22.13: 0encl QE =
The flux through the sphere depends only on the charge within the sphere.
nC3.19)CmN360( 200encl === Q E
22.14: a) .CN44.7m)(0.550
C)1050.2(
4
1
4
1m)0.1m450.0(
2
10
02
0
=
==+=
r
q
rE
b) 0=E
inside of a conductor or else free charges would move under the influence offorces, violating our electrostatic assumptions (i.e., that charges arent moving).
22.15: a) m.62.1CN614
C)10180.0(
4
1
4
1||
4
1||
6
002
0
=
===
E
q
r
r
q
E
b) As long as we are outside the sphere, the charge enclosed is constant and the sphereacts like a point charge.
22.16:a) C.1056.7)m(0.0610)CN1040.1(/
825
000
===== EAqqEA b) Double the surface area: C.1051.1)m(0.122)CN1040.1( 7250
== q
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22.17: C.1027.3m)(0.160)CN1150(44 9202
041
20
==== ErqEr
q
So the
number of electrons is: .1004.2 10C1060.1
C1027.3
e 19
9
==
n
22.18: Draw a cylindrical Gaussian surface with the line of charge as its axis. Thecylinder has radius 0.400 m and is 0.0200 m long. The electric field is then 840 N/C atevery point on the cylindrical surface and directed perpendicular to the surface. Thus
== )2)(())(( cylinder rLEAEdsE
/CmN42.2m)(0.0200m)(0.400)(2N/C)840( 2==
The field is parallel to the end caps of the cylinder, so for them 0= sE
d . From
Gausss law:
C1074.3
)C
mN2.42()
mN
C10854.8(
10
2
2
212
0
=
== Eq
22.19:
rE
2
1
0=
7/27/2019 Solucion Fisica Un_22
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22.20: a) For points outside a uniform spherical charge distribution, all the charge canbe considered to be concentrated at the center of the sphere. The field outside the sphereis thus inversely proportional to the square of the distance from the center. In this case:
CN53cm0.600
cm0.200)CN480(
2
=
=E
b) For points outside a long cylindrically symmetrical charge distribution, the field isidentical to that of a long line of charge:
,2
0rE=
that is, inversely proportional to the distance from the axis of the cylinder. In this case
CN160cm0.600
cm0.200)CN480( =
=E
c) The field of an infinite sheet of charge is ;2/ 0E= i.e., it is independent of the
distance from the sheet. Thus in this case .CN480=E
22.21: Outside each sphere the electric field is the same as if all the charge of the sphere
were at its center, and the point where we are to calculate E
is outside both spheres.
21 andEE
are both toward the sphere with negative charge.
sphere.chargednegativelythetoward,CN1006.8
CN10471.5m)(0.250
C1080.3||
CN10591.2m)(0.250
C1080.1||
521
5
2
6
2
2
22
5
2
6
2
1
11
=+=
=
==
=
==
EEE
kr
qkE
kr
qkE
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22.22: For points outside the sphere, the field is identical to that of a point charge of thesame total magnitude located at the center of the sphere. The total charge is given bycharge density volume:
C1060.1m)150.0)(3
4)(mCn50.7( 1033 == q
a) The field just outside the sphere is
CN4.42m)(0.150
C)10(1.06)/CmN109(
4 2
10229
20
=
==
r
qE
b) At a distance of 0.300 m from the center (double the spheres radius) the field willbe 1/4 as strong: 10.6 CN
c) Inside the sphere, only the charge inside the radius in question affects the field. Inthis case, since the radius is half the spheres radius, 1/8 of the total charge contributes tothe field:
CN2.21m)(0.075
C)1006.1()8/1()C/mN109(2
10229
==
E
22.23: The point is inside the sphere, so 3/ RkQrE= (Example 22.9)
nC2.10)m100.0(
m)(0.220)CN950( 33===
kkr
ERQ
22.24: a) Positive charge is attracted to the inner surface of the conductor by the chargein the cavity. Its magnitude is the same as the cavity charge: nC,00.6inner +=q since
0=E inside a conductor.b) On the outer surface the charge is a combination of the net charge on the conductor
and the charge left behind when the nC00.6+ moved to the inner surface:
nC.1.00nC6.00nC00.5innertotouterouterinnertot ===+= qqqqqq
22.25: 32 SandS enclose no charge, so the flux is zero, and electric field outside the
plates is zero. For between the plates,1
S shows that: .000
EAqEA ===
7/27/2019 Solucion Fisica Un_22
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22.26: a) At a distance of 0.1 mm from the center, the sheet appears infinite, so:
=
====
.CN662m)800.0(2
C1050.7
22
20
9
00 A
qE
qAEdAE
b) At a distance of 100 m from the center, the sheet looks like a point, so:
.CN1075.6m)(100
C)1050.7(
4
1
4
132
9
02
0
=
= r
q
E
c) There would be no difference if the sheet was a conductor. The charge wouldautomatically spread out evenly over both faces, giving it half the charge density on anyas the insulator
00 2::).(
cE == near one face. Unlike a conductor, the insulatoris the
charge density in some sense. Thus one shouldnt think of the charge as spreading overeach face for an insulator. Far away, they both look like points with the same charge.
22.27: a) .2
2
RL
Q
RL
Q
A
Q ====
b) ==== .2
)2(000 r
RE
RL
QrLEdAE
c) But from (a), ,so,202
rER == same as an infinite line of charge.
7/27/2019 Solucion Fisica Un_22
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22.28: All the s' are absolute values.
(a) at0
1
0
4
0
3
0
2
2222:
AEA ++=
left.thetoCN1082.2
)mC6mC4mC2mC5(21
)(2
1
5
2222
0
1432
0
=
++=
++=
EA
(b)
left.thetoCN1095.3
)mC5mC4mC2mC6(2
1
)(2
1
2222
5
2222
0
2431
00
2
0
4
0
3
0
1
=
++=
++=++=
EB
(c)
leftthetoCN1069.1
)mC6mC4mC2mC5(2
1
)(2
1
2222
5
2222
0
1432
00
1
0
4
0
3
0
2
=
+=
+=+=
EC
22.29: a) Gausss law says +Q on inner surface, so 0=E inside metal.b) The outside surface of the sphere is grounded, so no excess charge.c) Consider a Gaussian sphere with the Q charge at its center and radius less than
the inner radius of the metal. This sphere encloses net charge Q so there is an electricfield flux through it; there is electric field in the cavity.
d) In an electrostatic situation 0=E inside a conductor. A Gaussian sphere withthe Q charge at its center and radius greater than the outer radius of the metal encloses
zero net charge (the Q charge and the Q+ on the inner surface of the metal) so there isno flux through it and 0=E outside the metal.
e) No, 0=E there. Yes, the charge has been shielded by the groundedconductor. There is nothing like positive and negative mass (the gravity force is alwaysattractive), so this cannot be done for gravity.
7/27/2019 Solucion Fisica Un_22
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22.30: Given ,)m)CN(00.3()m)CN(00.5( kiE zx +=
edge length
,m300.0=L ,m300.0=L and .011 1
=== ASs nEjn
==+= )CN(00.3(21 2A
SS nEkn
== zz )m)CN(27.0()m300.0)(m 2
.m)CN(081.0)m300.0)(m)CN(27.0( 2=
.033 3
==+= ASS nEjn
).0(0)m)CN(27.0(44 4
===== zzASS nEkn
xxASS
)m)N/C(45.0()m300.0)(m)CN(00.5( 25 55 ===+= nEin
).m)CN(135.0()m300.0)(m)N/C(45.0( 2==
).0(0)m)CN(45.0(66 6
==+=== xxASS nEin
b) Total flux:
CNm054.0m)CN()135.0081.0( 2252 ==+=
C1078.4 130== q
22.31: a)
b) Imagine a charge q at the center of a cube of edge length 2L. Then: ./ 0q=
Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24of the flux. That is, .24 0q=
22.32: a) .CmN750)m0.6)(CN125( 22 === EA
b) Since the field is parallel to the surface, .0= c) Choose the Gaussian surface to equal the volumes surface. Then: 750
EA= ,CN577)750C1040.2( 08
m0.6
10 2 =+=
Eq in the positive x -direction.
Since 0
7/27/2019 Solucion Fisica Un_22
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22.33: To find the charge enclosed, we need the flux through the parallelepiped:
CmN5.3760cos)CN1050.2)(m0600.0)(m0500.0(60cos 2411 === AE
CmN10560cos)CN1000.7)(m0600.0)(m0500.0(120cos 2422 === AE
So the total flux is ,CmN5.67CmN)1055.37( 2221 ==+= and
.C1097.5)CmN5.67(10
0
2
0
=== q b) There must be a net charge (negative) in the parallelepiped since there is a netflux flowing into the surface. Also, there must be an external field or all lines would pointtoward the slab.
22.34:
The particle feels no force where the net electric field is zero. The fields cancancel only in regions A and B.
sheetline EE =
00 22
r=
cm16m16.0)C/m100(
C/m50
2====
r
The fields cancel 16 cm from the line in regions A andB.
7/27/2019 Solucion Fisica Un_22
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22.35:
The electric field 1E
of the sheet of charge is toward the sheet, so the electric
field 2E
of the sphere must be away from the sheet. This is true above the center of the
sphere. Let rbe the distance above the center of the sphere for the point where theelectric field is zero.
32
00
121
4
1
2
so
R
rQ
EE ==
m097.0C10900.0
)m120.0)(C/m1000.8(229
329
2
31 =
==
Q
Rr
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22.38: a) ,, 204
1
r
Q
Ear =< since the charge enclosed is Q.
,0, = , since the total enclosed charge is 2Q.
b) The surface charge density on inner surface: 24aQ = .
c) The surface charge density on the outer surface: .242
b
Q =
d)
e)
7/27/2019 Solucion Fisica Un_22
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22.39: a)(i) ,0, =< Ear since 0=Q
(ii) ,0, =
7/27/2019 Solucion Fisica Un_22
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22.41: a)(i) ,0, =< Ear since charge enclosed is zero.
(ii) ,0, =
7/27/2019 Solucion Fisica Un_22
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22.43: a) The sphere acts as a point charge on an external charge, so:
,204
1
r
qEF == radially inward.
(b) If the point charge was inside the sphere (where there is no electric field) itwould feel zero force.
22.44: a)
=
=
+=
333
343
34inner
1
4
3
ab
q
ab
q
VV
q
ab
=
=
=
333
343
34outer
1
4
3
cd
q
cd
q
VV
q
cd
b) (i) ==< .00 Edar AE
(ii) inner33
0
2inner
0
)(
3
44
1 ar
rEdV
dbra ==
7/27/2019 Solucion Fisica Un_22
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22.45: a) ,2
4
1,
0 rEbra
= radially outward, since again the charge enclosed is the
same as in part (a).
c)
d) The inner and outer surfaces of the outer cylinder must have the same amountof charge on them: .and, outerinnerinner === ll
22.46: a) (i) .2
)2(,000 r
E
l
qrlEar ===<
(ii) ,bra
b) (i) Inner charge per unit length is . (ii) Outer charge per length is .2+
7/27/2019 Solucion Fisica Un_22
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22.47: a) (i) ,ErlEarr
l
q
000 2)2(, ===< radially outward.
(ii) ,bra there is no net charge enclosed, so the electric field is zero.
b) (i) Inner charge per unit length is . (ii) Outer charge per length is ZERO.
22.48: a) ,)2(,00
2
0 2
r
lr
qErlERr ===< radially outward.
b) .)2(,and, 22
22
00
2
0
2
0 rk
rr
R
lR
q ErlERRr ======>
c) .Rr= the electric field for BOTH regions is ,02
RE= so they are consistent.
d)
22.49: a) The conductor has the surface charge density on BOTH sides, so it has twicethe enclosed charge and twice the electric field.
b) We have a conductor with surface charge density on both sides. Thus theelectric field outside the plate is .)2()2( 00 EAAE === To find the field
inside the conductor use a Gaussian surface that has one face inside the conductor, andone outside.Then:
.00but)( inin0out0inout ====+= EAEEAAEAE
7/27/2019 Solucion Fisica Un_22
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22.50: a) If the nucleus is a uniform positively charged sphere, it is only at its verycenter where forces on a charge would balance or cancel
b)3
03
3
0
2
0 44
R
erE
R
r
erE
qd =
=== AE
.4
13
2
0 R
re
qEF ==
So from the simple harmonic motion equation:
.4
1
2
1
4
1
4
13
2
03
2
03
2
0
2
mR
e
f
mR
e
R
re
rmF ====
c) If3
2
0
14
4
1
2
1Hz1057.4
mR
e
f
==
m.1013.3Hz)1057.4)(kg1011.9(4
C)1060.1(
4
1 103
214312
219
0
=
=
R
1Thompsonactual
rr
d) If Rr> then the electron would still oscillate but not undergo simple
harmonic motion, because for ,1, 2rFRr > and is not linear.
22.51: The electrons are separated by a distance ,2d and the amount of the positive
nucleuss charge that is within radius d is all that exerts a force on the electron. So:
.2/8/2)2(
332nucleus2
2
3 RdRdkeFd
keF
R
de =====
7/27/2019 Solucion Fisica Un_22
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22.52: ===
raxar
dxexa
QQdddrre
a
QQdVQrQ
0
/22
30
2/2
30
004
sin)(a)
].1)/(2)/(2[)222(4
)( 02
0/222
33
0
0 ++==
ararQerrea
QeQrQ
arrr
0.)(,ifNote rQr
b) The electric field is radially outward, and has magnitude:
).1)(2)(2( 02
02
/2 0
++=
ararr
kQeE
ar
22.53: a) At N.94,2 215219
02Fee
0 m)101.7(4
C)106.1)(82(4
144
1e =====
R
EqFRr
So: .m/s101.0kg1011.9N94 23231 === mFa
b) .m/s101.44,At 232(a) === aaRr
c) At8
1
8
1 ()82(,2 eQRr == because the charge enclosed goes like 3r ) so with the
radius decreasing by 2, the acceleration from the change in radius goes up by ,4)2( 2 =
but the charge decreased by 8, so .m/s101.2 232)b(84 == aa
d) At .0so,0,0 === FQr
7/27/2019 Solucion Fisica Un_22
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22.54: a) The electric field of the slab must be zero by symmetry. There is no preferreddirection in the y -z plane, so the electric field can only point in the x -direction. But at
the origin in the x -direction, neither the positive nor negative directions should besingled out as special, and so the field must be zero.
b) Use a Gaussian surface that has one face of area A on in the y -z plane at
,0=x and the other face at a general value .x Then:
,:000
xE
Ax
QEAdx encl ====
with direction given by .||i
x
x
Note that E is zero at .0=x Now outside the slab, the enclosed charge is constant with :x
,:000
encl
dE
Ad
QEAdx ====
again with direction given by .||i
x
x
22.55: a) Again, E is zero at 0=x , by symmetry arguments.
b) i||
in,33
'':2
0
30
20
30
0
2
20
0
0
encl
x
x
d
xE
d
Axdxx
d
A
QEAdx
x
===== direction.
=====d
o
x
xin
dE
Addxx
d
A
QEAdx
0 00
02
20
0
0
encl ||
,33
'': i direction.
22.56: a) We could place two charges Q+ on either side of the charge :q+
b) In order for the charge to be stable, the electric field in a neighborhood around itmust always point back to the equilibrium position.
c) Ifq is moved to infinity and we require there to be an electric field always
pointing in to the region where q had been, we could draw a small Gaussian surface
there. We would find that we need a negative flux into the surface. That is, there has to be
a negative charge in that region. However, there is none, and so we cannot get such astable equilibrium.
d) For a negative charge to be in stable equilibrium, we need the electric field toalways point away from the charge position. The argument in (c) carries through again,this time inferring that a positive charge must be in the space where the negative chargewas if stable equilibrium is to be attained.
7/27/2019 Solucion Fisica Un_22
22/27
22.57: a) The total charge: ]/[4)/1(40
3
0 0
220 RdrrdrrdrrRrq
RR R
==
.3
12
4
12
4]43[4
3
30
333
0 Q
R
QRRRRq ====
b) ,Rr all the charge Q is enclosed, and: ,/)4( 204
10
2
r
Q
EQrE ===
the same as a point charge.
c) ).then, 33 Rrq(Q(r)Rr =
( ).4)1(4)(Also, 4302043
RrrdrrRrrQ ==
=
=
=
R
r
R
rkQ
R
r
R
r
r
kQ
R
r
R
r
r
kQrE 34
34
4
1
3
112)(
34
4
3
3
24
4
3
3
2
d)
e)22maxmax43 3
4)24(
3
2So.
3
20
64)(0
R
kQ
R
kQERrr
R
kQ
R
kQRr
r
E=====
7/27/2019 Solucion Fisica Un_22
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22.58: a)
=
==
RR
R
drrR
drrdrrR
rdrrrQ
00
32
0
02
02
03
44
3
414)(4
= 043
43
443
0 =
QR
RR
b) 00,0
encl === Eq
dRr AE
c)
==r r r
rdrR
rdr
rErdrr
dRr
0 0 0
32
0
022
0 3
444)(
4, AE
=
=
R
rr
R
rr
r
E 1
333
1
0
043
2
0
0
d)
e)2
03
2
30 max
0
max0
0
0 RrR
r
r
E===
0
0
0
0
122
11
232
RR
RrE =
=
22.59: a) === .4sin
2
2
Gmr
dddrrGmdg
Ag
b) For any closed surface, mass OUTSIDE the surface contributes zero to the fluxpassing through the surface. Thus the formula above holds for any situation where m isthe mass enclosed by the Gaussian surface.
That is: == .encl4GMdg Ag
7/27/2019 Solucion Fisica Un_22
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22.60: a) .masspointaforassametheiswhich,442
2
r
GMgGMrgg ===
b) Inside a hollow shell, the .0so,0encl == gM
c) Inside a uniform spherical mass:
,444 33
3
encl2R
GMrgR
rMGGMrgg =
===
which is linear in .r
22.61: a) For a sphere NOT at the coordinate origin:
,33
44
0
3
00
encl2
rE
r
QEr
=
==== brr
in the direction.-r
.3
)(
0
brE
=
b) The electric field inside a hole in a charged insulating sphere is:
.33
)(
3000
(a)spherehole
bbrrEEE
=
==
Note that E
is uniform.
22.62: Using the technique of22.61, we first find the field of a cylinder off-axis, then theelectric field in a hole in a cylinder is the difference between two electric fieldsthat of asolid cylinder on-axis, and one off-axis.
.2
)(
22
00
2
00
encl
rErl
QElr
brEbrr
=
=====
uniform.isthatNote.
22
)(
2 000abovecylinderhole E
bbrrEEE
=
==
7/27/2019 Solucion Fisica Un_22
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22.63: a) :0=x no field contribution from the sphere centered at the origin, and theother sphere produces a point-like field:
iiE 44
1)2(4
1)0(
20
20 R
Q
R
Q
x ===
.
b) :2Rx = the sphere at the origin provides the field of a point charge of charge
8Qq = since only one-eighth of the charges volume is included. So:
.184
1)9/42/1(4
1)23()2(
)8(
4
1)2(
20
20
220
iiiER
Q
R
Q
R
Q
R
Q
Rx ==
==
c) :Rx = the two electric fields cancel, so .0=E
d) :3Rx = now both spheres contribute fields pointing to the right:
.9
10
4
1)3(4
1)3(
20
220
iiER
Q
R
Q
R
Q
Rx =
+==
22.64: (See Problem 22.63 with QQ for terms associated with right sphere)
a) iE 44
1)0(
20 R
Q
x +==
b) iiiE 18
17
4
19
4
24
1)23()2(
)8(
4
1
2 2022
022
0 R
Q
R
Q
R
Q
R
Q
R
Q
Rx =
+=
+=
=
c) iiE 2
4
1)(
20
220 R
Q
R
Q
R
Q
Rx =
+==
d) iiiE 9
8
4
194
1)3(4
1)3(
2
0
22
0
22
0
R
Q
R
Q
R
Q
R
Q
R
Q
Rx
=
=
==
7/27/2019 Solucion Fisica Un_22
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22.65: a) The charge enclosed:
.5
8
24
1524
11
4
)16(
3
)8(8
)/()2(4and,63
)2(4where,
3
3
34433
2/
320
33
0
R
Q
RQ
R
R
RRRR
drRrrQRR
QQQQR
Rii
==
=
=
===+=
b) .15
8
33
44:2
3000
32
R
Qr
rE
rrERr ====
enclosed.ischargeallsince,4
:
).1)(48)(64(15
)1)(48)(64()4(24
4
)16(
3
)8(8
14:2
20
43
2
43
20
3
4433
00
2
r
QERr
RrRrr
kQRrRr
r
RE
R
RrRr
QrERrR i
=
==
+==
c) .267.015
4)15/4(===
Q
Q
Q
Qi
d) ,:2/ 3015
8reEFRr
R
eQ== so the restoring force depends upon displacement to
the first power, and we have simple harmonic motion.
e) .8
152
2,
15
8,
15
8,
30
30
30 eQ
mR
T
mR
eQ
m
k
R
eQkkrF e
ee
======
f) If the amplitude of oscillation is greater than ,2/R the force is no longer linear in,r and is thus no longer simple harmonic.
7/27/2019 Solucion Fisica Un_22
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22.66: a) Charge enclosed:
.233
480
480
233
120
47
32
3,Therefore
.120
47
160
31
24
74))(1(4and
.32
3
164
16
2
34where
3
33
2/
33220
343
2/
00
R
QRRQ
RRdrrRrQ
RR
R
dr
R
rQQQQ
R
R
R
ii
==
+=
=
==
===+=
b) ===
==r
R
Qr
R
rE
R
rrd
R
r
rERr
0 40
2
0
2
0
43
0
2 .233
180
16
6
2
3
2
344:2
enclosed.ischargeallsince,4
:
.1920
23
5
1
3
1
233
480
480
17
5
1
3
14
4
128
3
1605243
4
))/(1(4
4:2
20
53
20
53
0
3
0
33
2
533
00
2/
22
00
2
r
QERr
R
r
R
r
r
QE
R
r
R
r
R
RR
R
rRr
Q
rdrRr
QrERrR
i
r
R
i
=
=
+
=
++=
+==
c) The fraction ofQ between :2 RrR
.807.0233
480
120
47==
Q
Qo
d) ,)2/( 204233
180
R
QRrE == using either of the electric field expressions above,
evaluated at .2/Rr= e) The force an electron would feel never is proportional to r which is necessary for
simple harmonic oscillations. It is oscillatory since the force is always attractive, but ithas the wrong power of r to be simple harmonic.