24 CHAPTER OUTLINE 24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions 24.4 Conductors in Electrostatic Equilibrium 24.5 Formal Derivation of Gauss‘s Law Gauss’s Law ANSWERS TO QUESTIONS Q24.1 The luminous flux on a given area is less when the sun is low in the sky, because the angle between the rays of the sun and the local area vector, dA, is greater than zero. The cosine of this angle is reduced. The decreased flux results, on the average, in colder weather. Q24.2 If the region is just a point, line, or plane, no. Consider two protons in otherwise empty space. The electric field is zero at the midpoint of the line joining the protons. If the field-free region is three-dimensional, then it can contain no charges, but it might be surrounded by electric charge. Consider the interior of a metal sphere carrying static charge. Q24.3 The surface must enclose a positive total charge. Q24.4 The net flux through any gaussian surface is zero. We can argue it two ways. Any surface contains zero charge so Gauss’s law says the total flux is zero. The field is uniform, so the field lines entering one side of the closed surface come out the other side and the net flux is zero. Q24.5 Gauss’s law cannot tell the different values of the electric field at different points on the surface. When E is an unknown number, then we can say E dA E dA cos cos θ θ z z = . When Ex y z , , b g is an unknown function, then there is no such simplification. Q24.6 The electric flux through a sphere around a point charge is independent of the size of the sphere. A sphere of larger radius has a larger area, but a smaller field at its surface, so that the product of field strength and area is independent of radius. If the surface is not spherical, some parts are closer to the charge than others. In this case as well, smaller projected areas go with stronger fields, so that the net flux is unaffected. Q24.7 Faraday’s visualization of electric field lines lends insight to this question. Consider a section of a vertical sheet carrying charge +1 coulomb. It has 1 0 ∈ field lines pointing out from it horizontally to the right and left, all uniformly spaced. The lines have the same uniform spacing close to the sheet and far away, showing that the field has the same value at all distances. 29
Transcript
24
CHAPTER OUTLINE24.1 Electric Flux24.2 Gauss’s Law24.3 Application of Gauss’s Law to Various Charge Distributions24.4 Conductors in Electrostatic Equilibrium24.5 Formal Derivation of Gauss‘s Law
Gauss’s Law
ANSWERS TO QUESTIONS
Q24.1 The luminous flux on a given area is less when the sun is low inthe sky, because the angle between the rays of the sun and thelocal area vector, dA, is greater than zero. The cosine of thisangle is reduced. The decreased flux results, on the average, incolder weather.
Q24.2 If the region is just a point, line, or plane, no. Consider twoprotons in otherwise empty space. The electric field is zero atthe midpoint of the line joining the protons. If the field-freeregion is three-dimensional, then it can contain no charges, butit might be surrounded by electric charge. Consider the interiorof a metal sphere carrying static charge.
Q24.3 The surface must enclose a positive total charge.
Q24.4 The net flux through any gaussian surface is zero. We can argue it two ways. Any surface containszero charge so Gauss’s law says the total flux is zero. The field is uniform, so the field lines enteringone side of the closed surface come out the other side and the net flux is zero.
Q24.5 Gauss’s law cannot tell the different values of the electric field at different points on the surface.When E is an unknown number, then we can say E dA E dAcos cosθ θz z= . When E x y z, ,b g is an
unknown function, then there is no such simplification.
Q24.6 The electric flux through a sphere around a point charge is independent of the size of the sphere. Asphere of larger radius has a larger area, but a smaller field at its surface, so that the product of fieldstrength and area is independent of radius. If the surface is not spherical, some parts are closer to thecharge than others. In this case as well, smaller projected areas go with stronger fields, so that thenet flux is unaffected.
Q24.7 Faraday’s visualization of electric field lines lends insight to this question. Consider a section of a
vertical sheet carrying charge +1 coulomb. It has 1
0∈ field lines pointing out from it horizontally to
the right and left, all uniformly spaced. The lines have the same uniform spacing close to the sheetand far away, showing that the field has the same value at all distances.
29
30 Gauss’s Law
Q24.8 Consider any point, zone, or object where electric field lines begin. Surround it with a close-fittinggaussian surface. The lines will go outward through the surface to constitute positive net flux. ThenGauss’s law asserts that positive net charge must be inside the surface: it is where the lines begin.Similarly, any place where electric field lines end must be just inside a gaussian surface passing netnegative flux, and must be a negative charge.
Q24.9 Inject some charge at arbitrary places within a conducting object. Every bit of the charge repelsevery other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surfaceof the conductor.
Q24.10 If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shellcarries no charge. The person is not harmed by touching this wall. If the person carries a (small)charge q, the electric field inside the sphere is no longer zero. Charge –q is induced on the inner wallof the sphere. The person will get a (small) shock when touching the sphere, as all the charge on hisbody jumps to the metal.
Q24.11 The electric fields outside are identical. The electric fields inside are very different. We have E = 0everywhere inside the conducting sphere while E decreases gradually as you go below the surface ofthe sphere with uniform volume charge density.
Q24.12 There is zero force. The huge charged sheet creates a uniform field. The field can polarize theneutral sheet, creating in effect a film of opposite charge on the near face and a film with an equalamount of like charge on the far face of the neutral sheet. Since the field is uniform, the films ofcharge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forces addto zero.
Q24.13 Gauss’s law predicts, as described in section 24.4, that excess charge on a conductor will reside onthe surface of the conductor. If a car is left charged by a lightning strike, then that charge will remainon the outside of the car, not harming the occupants. It turns out that during the lightning strike, thecurrent also remains on the outside of the conductor. Note that it is not necessarily safe to be in afiberglass car or a convertible during a thunderstorm.
SOLUTIONS TO PROBLEMS
Section 24.1 Electric Flux
P24.1 (a) ΦE EA= = × × °= ⋅cos . . . cosθ 3 50 10 0 350 0 700 0 8583e ja f N m C2
(b) θ = °90 0. ΦE = 0
(c) ΦE = × × °= ⋅3 50 10 0 350 0 700 40 0 6573. . . cos .e ja f N m C2
P24.2 ΦE EA= = × °= ⋅cos . . cos .θ 2 00 10 18 0 10 0 3554 N C m kN m C2 2e je j
P24.3 ΦE EA= cosθ A r= = =π π2 20 200 0 126. .a f m2
5 20 10 0 126 05. . cos× = °Ea f E = × =4 14 10 4 146. . N C MN C
Chapter 24 31
P24.4 (a) ′ =A 10 0 30 0. . cm cma fa f′ = =
= ′
= × °
= − ⋅
′
′
′
AEAE A
E A
E A
300 0 030 0
7 80 10 0 030 0 180
2 34
4
cm m
kN m C
2 2
2
.cos
. . cos
.
,
,
,
Φ
Φ
Φ
θ
e jb g
(b) ΦE A EA A, cos . cos .= = × °θ 7 80 10 60 04e ja f
10.0 cm
30.0 cm
60.0Þ
FIG. P24.4
A w
E A
= =°
FHG
IKJ = =
= × °= + ⋅
30 0 30 010 0
60 0600 0 060 0
7 80 10 0 060 0 60 0 2 344
. ..
cos ..
. . cos . .,
cm cm cm
cm m
kN m C
2 2
2
a fa f a f
e jb gΦ
(c) The bottom and the two triangular sides all lie parallel to E, so ΦE = 0 for each of these. Thus,
ΦE, . .total2 2 kN m C kN m C= − ⋅ + ⋅ + + + =2 34 2 34 0 0 0 0 .
P24.5 (a) ΦE a b A aA= ⋅ = + ⋅ =E A i j ie j
(b) ΦE a b A bA= + ⋅ =i j je j
(c) ΦE a b A= + ⋅ =i j ke j 0
P24.6 Only the charge inside radius R contributes to the total flux.
ΦEq
=∈0
P24.7 ΦE EA= cosθ through the base
ΦE = °= − ⋅52 0 36 0 180 1 87. . cos .a fa f kN m C2 .
Note the same number of electric field lines go through the base as go through thepyramid’s surface (not counting the base).
For the slanting surfaces, ΦE = + ⋅1 87. kN m C2 .
FIG. P24.7
P24.8 The flux entering the closed surface equals the flux exiting the surface. The flux entering the left sideof the cone is ΦE d ERh= ⋅ =zE A . This is the same as the flux that exits the right side of the cone.
Note that for a uniform field only the cross sectional area matters, not shape.
32 Gauss’s Law
Section 24.2 Gauss’s Law
P24.9 (a) ΦEq
=∈
=+ − + −
× ⋅= − × ⋅−
in2 2
2 2 C C C C
C N m N m C
012
65 00 9 00 27 0 84 0
8 85 106 89 10
. . . .
..
µ µ µ µb g
ΦE = − ⋅6 89. MN m C2
(b) Since the net electric flux is negative, more lines enter than leave the surface.
P24.10 (a) Ek Qre= 2 : 8 90 10
8 99 10
0 7502
9
2..
.× =
×e ja f
Q
But Q is negative since E points inward. Q = − × = −−5 56 10 55 68. . C nC
(b) The negative charge has a spherically symmetric charge distribution.
P24.11 ΦEq
=∈
in
0
Through S1 ΦEQ Q Q
=− +
∈= −
∈2
0 0
Through S2 ΦEQ Q
=+ −∈
=0
0
Through S3 ΦEQ Q Q Q
=− + −
∈= −
∈2 2
0 0
Through S4 ΦE = 0
P24.12 (a) One-half of the total flux created by the charge q goes through the plane. Thus,
Φ ΦE Eq q
, , plane total= =∈FHGIKJ = ∈
12
12 20 0
.
(b) The square looks like an infinite plane to a charge very close to the surface. Hence,
Φ ΦE Eq
, , square plane≈ =∈2 0
.
(c) The plane and the square look the same to the charge.
P24.13 The flux through the curved surface is equal to the flux through the flat circle, E r02π .
P24.14 (a) ΦEq
,.
.. .shell
in 2 2 N m C MN m C=∈
=××
= × ⋅ = ⋅−
−0
6
12612 0 10
8 85 101 36 10 1 36
(b) ΦE, half shell2 2 2 N m C N m C kN m C= × ⋅ = × ⋅ = ⋅
12
1 36 10 6 78 10 6786 5. .e j
(c) No, the same number of field lines will pass through each surface, no matter how the
radius changes.
Chapter 24 33
P24.15 (a) With δ very small, all points on the hemisphere are nearly ata distance R from the charge, so the field everywhere on the
curved surface is k QRe
2 radially outward (normal to the
surface). Therefore, the flux is this field strength times thearea of half a sphere:
Φ
Φ
curved local hemisphere
curved
= ⋅ =
= FHGIKJFHG
IKJ = ∈
=+∈
z E Ad E A
kQR
R QQ
e 22
0 0
12
41
42
2π
ππa f
Qδ → 0
FIG. P24.15
(b) The closed surface encloses zero charge so Gauss’s law gives
Φ Φcurved flat+ = 0 or Φ Φflat curved= − =−∈Q
2 0.
*P24.16 Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.
E A⋅ =∈z dq
0: + + − =
∈120 100
100
0 N C N C
mb g b g a fA A
Aρ
ρ =× ⋅
= ×−
−20 8 85 10
1001 77 10
1212
N C C N m
m C m
2 23
b ge j..
The charge is positive , to produce the net outward flux of electric field.
P24.17 The total charge is Q q− 6 . The total outward flux from the cube is Q q−
∈
6
0, of which one-sixth goes
through each face:
ΦEQ qb gone face
=−
∈
6
6 0
ΦEQ qb g a f
one face
22 C N m
C kN m C=
−
∈=
− × ⋅ ⋅
× ×= − ⋅
−
−
6
65 00 6 00 10
6 8 85 1018 8
0
6
12 2
. .
.. .
P24.18 The total charge is Q q− 6 . The total outward flux from the cube is Q q−
∈
6
0, of which one-sixth goes
through each face:
ΦEQ qb gone face
=−
∈
6
6 0.
P24.19 If R d≤ , the sphere encloses no charge and ΦEq
=∈
=in
00 .
If R d> , the length of line falling within the sphere is 2 2 2R d−
so ΦER d
=−
∈2 2 2
0
λ.
34 Gauss’s Law
P24.20 ΦEek Q
Rr,
. .
.. hole hole
2 2 N m C C
m m= ⋅ = FHG
IKJ =
× ⋅ ×FHGG
IKJJ ×
−−E A 2
29 6
23 28 99 10 10 0 10
0 1001 00 10π πe j e je j
a f e j
ΦE, . hole2 N m C= ⋅28 2
P24.21 ΦEq
=∈
=×
× ⋅= × ⋅
−
−in
2 22 C
8.85 10 C N m N m C
0
6
127170 10
1 92 10.
(a) Φ ΦE Eb gone face
2 N m C= =
× ⋅16
1 92 106
7.ΦEb gone face
2 MN m C= ⋅3 20.
(b) ΦE = ⋅19 2. MN m C2
(c) The answer to (a) would change because the flux through each face of the cube wouldnot be equal with an asymmetric charge distribution. The sides of the cube nearer thecharge would have more flux and the ones further away would have less. The answerto (b) would remain the same, since the overall flux would remain the same.
P24.22 No charge is inside the cube. The net flux through the cube is zero. Positive fluxcomes out through the three faces meeting at g. These three faces together fillsolid angle equal to one-eighth of a sphere as seen from q, and together pass
flux 18 0
q∈FHGIKJ . Each face containing a intercepts equal flux going into the cube:
0 38
24
0
0
= = +∈
=−∈
Φ Φ
Φ
E E
E
q
q
, ,
,
net abcd
abcd
FIG. P24.22
Section 24.3 Application of Gauss’s Law to Various Charge Distributions
P24.23 The charge distributed through the nucleus creates a field at the surface equal to that of a point
charge at its center: Ek qr
e= 2
E =× × ×
×
−
−
8 99 10 82 1 60 10
208 1 20 10
9 19
1 3 15 2
. .
.
Nm C C
m
2 2e je ja f
E = ×2 33 1021. N C away from the nucleus
Chapter 24 35
P24.24 (a) Ek Qr
ae= =3 0
(b) Ek Qr
ae= =
× ×=
−
3
9 6
3
8 99 10 26 0 10 0 100
0 400365
. . .
.
e je ja fa f kN C
(c) Ek Qre= =
× ×=
−
2
9 6
2
8 99 10 26 0 10
0 4001 46
. .
..
e je ja f MN C
(d) Ek Qre= =
× ×=
−
2
9 6
2
8 99 10 26 0 10
0 600649
. .
.
e je ja f kN C
The direction for each electric field is radially outward .
*P24.25 mg qE q qQ A
= =∈FHGIKJ = ∈FHGIKJ
σ2 20 0
QA
mgq
=∈
=×
− ×= −
−
−
2 2 8 85 10 0 01 9 8
0 7 102 480
12
6
. . .
..
e ja fa f C m2µ
P24.26 (a) Ekre=
2 λ3 60 10
2 8 99 10 2 40
0 1904
9
.. .
.× =
×e jb gQ
Q = + × = +−9 13 10 9137. C nC
(b) E = 0
*P24.27 The volume of the spherical shell is
43
0 25 0 20 3 19 103 3 2π . . . m m m3a f a f− = × − .
Its charge is
ρV = − × × = − ×− − −1 33 10 3 19 10 4 25 106 2 8. . . C m m C3 3e je j .
The net charge inside a sphere containing the proton’s path as its equator is
− × − × = − ×− − −60 10 4 25 10 1 02 109 8 7 C C C. . .
The electric field is radially inward with magnitude
k q
r
q
re2
02
9 7
2 24
4
8 99 10 10
0 251 47 10=
∈=
× ×= ×
−
π
.
..
Nm 1.02 C
C m N C
2 e ja f .
For the proton
F ma∑ = eEmv
r=
2
veErm
= FHGIKJ =
× ×
×
FHGG
IKJJ = ×
−
−
1 2 19 4
27
1 2
51 60 10 1 47 10 0 25
105 94 10
. . ..
C N C m
1.67 kg m s
e j.
36 Gauss’s Law
P24.28 σ = × FHG
IKJ = ×− −8 60 10
1008 60 106
22. . C cm
cmm
C m2 2e j
E =∈
=×
×= ×
−
−
σ2
8 60 10
2 8 85 104 86 10
0
2
129.
..
e j N C away from the wall
The field is essentially uniform as long as the distance from the center of the wall to the field point ismuch less than the dimensions of the wall.
P24.29 If ρ is positive, the field must be radially outward. Choose as thegaussian surface a cylinder of length L and radius r, contained insidethe charged rod. Its volume is π r L2 and it encloses charge ρπ r L2 .Because the charge distribution is long, no electric flux passesthrough the circular end caps; E A⋅ = °=d EdA cos .90 0 0 . The curvedsurface has E A⋅ = °d EdA cos0 , and E must be the same strengtheverywhere over the curved surface. FIG. P24.29
Gauss’s law, E A⋅ =∈z dq
0, becomes E dA
r L
CurvedSurface
z =∈
ρπ 2
0.
Now the lateral surface area of the cylinder is 2π rL :
E r Lr L
22
0π
ρπb g =∈
. Thus, E =∈ρr
2 0 radially away from the cylinder axis .
*P24.30 Let ρ represent the charge density. For the field inside the sphere at r1 5= cm we have
E rq r
1 12
0
13
04
43
ππ ρ
=∈
=∈
inside Er
11
03=
∈ρ
ρ =∈
=× − ×
= − ×−
−3 3 8 85 10 86 10
0 054 57 100 1
1
12 35E
r C
.
..
C N
m Nm C m
2
23e je j
.
Now for the field outside at r3 15= cm
E rr
Ekr
e
3 32 2
3
0
332
3 5 9 74
3
443
43
0 10 4 57 10 8 99 10 1 91 107 64 10
76 4
ππ ρ
π
=∈
=− ×
=× − ×
= − ×
=
− −. . . ..
.
m C
m
Nm C
0.15 m C N C
kN C radially inward
3
2
2 2
a f e j e ja f
E
P24.31 (a) E = 0
(b) Ek Qre= =
× ×=
−
2
9 6
2
8 99 10 32 0 10
0 2007 19
. .
..
e je ja f MN C E = 7 19. MN C radially outward
P24.32 The distance between centers is 2 5 90 10 15× × −. m . Each produces a field as if it were a point chargeat its center, and each feels a force as if all its charge were a point at its center.
Fk q q
re= = × ⋅
×
× ×= × =
−
−
1 22
92 19 2
15 238 99 10
46 1 60 10
2 5 90 103 50 10 3 50.
.
.. . N m C
C
m N kN2 2e j
a f e je j
Chapter 24 37
P24.33 Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a0.05 m separation on the ends of strings making angles of 10° with the vertical.
(a) F T mg Tmg
y∑ = °− = ⇒ =°
coscos
10 010
F T F F Tx e e∑ = °− = ⇒ = °sin sin10 0 10 , so
Fmg
mg
F
e
e
=°
FHG
IKJ °= °= °
≈ × − −
cossin tan . . tan
1010 10 0 001 9 8 10
2 10 3 3
kg m s
N ~10 N or 1 mN
2b ge jFIG. P24.33
(b) Fk qree=
2
2
2 108 99 10
0 25
1 2 10 10
39 2
2
7 7
× ≈× ⋅
≈ ×
−
− −
N N m C
m
C ~ C or 100 nC
2 2.
.
.
e ja f
q
q
(c) Ek qr
e= ≈× ⋅ ×
≈ ×−
2
9 7
24
8 99 10 1 2 10
0 251 7 10 10
. .
.. ~
N m C C
m N C kN C
2 2e je ja f
(d) ΦEq
=∈
≈×
× ⋅= × ⋅ ⋅
−
−0
7
1241 2 10
1 4 10 10.
. ~ C
8.85 10 C N m N m C kN m C2 2
2 2
*P24.34 The charge density is determined by Q a=43
3π ρ ρπ
=3
4 3Qa
(a) The flux is that created by the enclosed charge within radius r:
ΦEq r r Q
aQr
a=∈
=∈
=∈
=∈
in
0
3
0
3
03
3
03
43
4 33 4
π ρ ππ
(b) ΦEQ
=∈0
. Note that the answers to parts (a) and (b) agree at r a= .
(c)
ar
ΦE
Q∈0
00
FIG. P24.34(c)
38 Gauss’s Law
P24.35 (a) Ekre= =
× ⋅ × −2 2 8 99 10 2 00 10 7 00
0 100
9 6λ . . .
.
N m C C m
m
2 2e j e j
E = 51 4. , kN C radially outward
(b) ΦE EA E r= = °cos cosθ π2 0b gΦE = × = ⋅5 14 10 2 0 100 0 020 0 1 00 6464. . . . N C m m N m C2e j a fb ga fπ
P24.36 (a) ρπ π
= =×
= ×−
−Qa4
33
6
43
325 70 10
0 040 02 13 10
.
..
b g C m3
q rin C nC= FHGIKJ = × F
HGIKJ = × =− −ρ π π
43
2 13 1043
0 020 0 7 13 10 7133 2 3 7. . .e j b g
(b) q rin C= FHGIKJ = × F
HGIKJ =−ρ π π µ
43
2 13 1043
0 040 0 5 703 2 3. . .e j b g
P24.37 E =∈
=×
× ⋅=
−
−
σ2
9 00 10
2 8 85 10508
0
6
12
.
.
C m
C N m kN C , upward
2
2 2e j
P24.38 Note that the electric field in each case is directed radially inward, toward the filament.
(a) Ekre= =
× ⋅ ×=
−2 2 8 99 10 90 0 10
0 10016 2
9 6λ . .
..
N m C C m
m MN C
2 2e je j
(b) Ekre= =
× ⋅ ×=
−2 2 8 99 10 90 0 10
0 2008 09
9 6λ . .
..
N m C C m
m MN C
2 2e je j
(c) Ekre= =
× ⋅ ×=
−2 2 8 99 10 90 0 10
1 001 62
9 6λ . .
..
N m C C m
m MN C
2 2e je j
Section 24.4 Conductors in Electrostatic Equilibrium
P24.39 EdA E rlqz = =∈
20
πb g in Eq l
r r=
∈=
∈in
2 20 0πλ
π
(a) r = 3 00. cm E = 0
(b) r = 10 0. cm E =×
×=
−
−
30 0 10
2 8 85 10 0 1005 400
9
12
.
. .π e ja f N C , outward
(c) r = 100 cm E =×
×=
−
−
30 0 10
2 8 85 10 1 00540
9
12
.
. .π e ja f N C , outward
Chapter 24 39
P24.40 From Gauss’s Law, EAQ
=∈0
σ = =∈ = × − = − × = −− −QA
E012 98 85 10 130 1 15 10 1 15. . .e ja f C m nC m2 2
P24.41 The fields are equal. The Equation 24.9 E =∈
σ conductor
0 for the field outside the aluminum looks
different from Equation 24.8 E =∈
σ insulator
2 0 for the field around glass. But its charge will spread out to
cover both sides of the aluminum plate, so the density is σ conductor =QA2
. The glass carries charge
only on area A, with σ insulator =QA
. The two fields are Q
A2 0∈ the same in magnitude, and both are
perpendicular to the plates, vertically upward if Q is positive.
*P24.42 (a) All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside iszero .
(b) The charged sphere creates field at exterior points as if it were a point charge at the center:
σ = 708 nC m2 , positive on one face and negative on the other.
(b) σ =QA
Q A= = × −σ 7 08 10 0 5007 2. .e ja f C
Q = × =−1 77 10 1777. C nC , positive on one face and negative on the other.
P24.44 (a) E = 0
(b) Ek Qre= =
× ×= ×
−
2
9 6
27
8 99 10 8 00 10
0 030 07 99 10
. .
..
e je jb g
N C E = 79 9. MN C radially outward
(c) E = 0
(d) Ek Qre= =
× ×= ×
−
2
9 6
26
8 99 10 4 00 10
0 070 07 34 10
. .
..
e je jb g
N C E = 7 34. MN C radially outward
40 Gauss’s Law
P24.45 The charge divides equally between the identical spheres, with charge Q2
on each. Then they repel
like point charges at their centers:
Fk Q Q
L R R
k Q
L Re e=+ +
=+
=× ⋅ ×
=−2 2
4 2
8 99 10 60 0 10
4 2 012 002
2
2
9 6 2
2
b gb ga f a f
e ja f
. .
..
N m C
C m N
2
2.
P24.46 The electric field on the surface of a conductor varies inversely with the radius of curvature of thesurface. Thus, the field is most intense where the radius of curvature is smallest and vice-versa. Thelocal charge density and the electric field intensity are related by
E =∈σ
0or σ =∈0 E .
(a) Where the radius of curvature is the greatest,
σ =∈ = × ⋅ × =−0
12 48 85 10 2 80 10 248Emin2 2 2 C N m N C nC m. .e je j .
(b) Where the radius of curvature is the smallest,
σ =∈ = × ⋅ × =−0
12 48 85 10 5 60 10 496Emax . . C N m N C nC m2 2 2e je j .
P24.47 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside the conductingshell is zero, the total charge inside the gaussian surface must be zero, so the insidecharge/length = −λ .
0 = +λ qin soqin = −λ
Outside surface: The total charge on the metal cylinder is 2λ = +q qin out
qout = +2λ λ so the outside charge/length is 3λ .
(b) Ek
rkr r
e e= = =∈
2 3 6 32 0
λ λ λπ
b g radially outward
P24.48 (a) Ek Qre= =
× ×=
−
2
9 6
2
8 99 10 6 40 10
0 1502 56
. .
..
e je ja f MN C , radially inward
(b) E = 0
P24.49 (a) The charge density on each of the surfaces (upper and lower) of the plate is:
σ = FHGIKJ =
×= × =
−−1
212
4 00 10
0 5008 00 10 80 0
8
28q
A
.
.. .
C
m C m nC m2 2e j
a f .
(b) E k k k=∈FHGIKJ =
×
× ⋅
FHG
IKJ =
−
−σ
0
8
12
8 00 108 85 10
9 04.
..
C m C N m
kN C2
2 2 b g
(c) E k= −9 04. kN Cb g
Chapter 24 41
P24.50 (a) The charge +q at the center induces charge −q on the inner surface of the conductor,where its surface density is:
σπa
qa
=−
4 2 .
(b) The outer surface carries charge Q q+ with density
σπb
Q qb
=+
4 2 .
P24.51 Use Gauss’s Law to evaluate the electric field in each region, recalling that the electric field is zeroeverywhere within conducting materials. The results are:
E = 0 inside the sphere and within the material of the shell
E kQre= 2 between the sphere and shell, directed radially inward
E kQ
re=2
2 outside the shell, directed radially outward .
Charge −Q is on the outer surface of the sphere .
Charge +Q is on the inner surface of the shell ,
and +2Q is on the outer surface of the shell.
P24.52 An approximate sketch is given at the right. Note that the electric field linesshould be perpendicular to the conductor both inside and outside.
FIG. P24.52
Section 24.5 Formal Derivation of Gauss‘s Law
P24.53 (a) Uniform E, pointing radially outward, so ΦE EA= . The arc length is ds Rd= θ ,and the circumference is 2 2π π θr R= sin
A rds R Rd R d R R= = = = − = −z zz 2 2 2 2 2 10
2
0
20
2π π θ θ π θ θ π θ π θθ θ
θsin sin cos cosb g a f b g
ΦEQR
RQ
=∈
⋅ − =∈
−1
42 1
21
02
2
0ππ θ θcos cosa f a f [independent of R!]
FIG. P24.53
(b) For θ = °90 0. (hemisphere): ΦEQ Q
=∈
− ° =∈2
1 9020 0
cosa f .
(c) For θ = °180 (entire sphere): ΦEQ Q
=∈
− ° =∈2
1 1800 0
cosa f [Gauss’s Law].
42 Gauss’s Law
Additional Problems
P24.54 In general, E i j k= + +ay bz cx
In the xy plane, z = 0 and E i k= +ay cx
Φ
Φ
E
Ex
w
x
w
d ay cx dA
ch xdx chx chw
= ⋅ = + ⋅
= = =
z zz= =
E A i k ke j
0
2
0
2
2 2
�������������������������������������
x
y
z
x = 0
x = w
y = 0 y = h
dA = hdx
FIG. P24.54
P24.55 (a) q Q Q Qin = + − = +3 2
(b) The charge distribution is spherically symmetric and qin > 0 . Thus, the field is directedradially outward .
(c) Ek q
rk Qr
e e= =in2 2
2 for r c≥ .
(d) Since all points within this region are located inside conducting material, E = 0 for
b r c< < .
(e) Φ ΦE Ed q= ⋅ = ⇒ =∈ =z E A 0 00in
(f) q Qin = +3
(g) Ek q
rk Qr
e e= =in2 2
3 (radially outward) for a r b≤ < .
(h) q VQa
r Qrain = =
+FHGIKJFHGIKJ = +ρ
ππ
3 43
343
33
3
3
(i) Ek q
rkr
Qra
k Qr
ae e
e= = +FHG
IKJ =
in2 2
3
3 33 3 (radially outward) for 0 ≤ ≤r a .
(j) From part (d), E = 0 for b r c< < . Thus, for aspherical gaussian surface with b r c< < ,q Q qin inner= + + =3 0 where qinner is thecharge on the inner surface of theconducting shell. This yields q Qinner = −3 .
(k) Since the total charge on the conductingshell is q q q Qnet outer inner= + = − , we have
q Q q Q Q Qouter inner= − − = − − − = +3 2b g .
(l) This is shown in the figure to the right.
E
ra b c
FIG. P24.55(l)
Chapter 24 43
P24.56 The sphere with large charge creates a strong field to polarize the other sphere. That means itpushes the excess charge over to the far side, leaving charge of the opposite sign on the near side.This patch of opposite charge is smaller in amount but located in a stronger external field, so it canfeel a force of attraction that is larger than the repelling force felt by the larger charge in the weakerfield on the other side.
P24.57 (a) E A⋅ = =∈z d E rq
4 2
0πe j in
For r a< , q rin = FHGIKJρ π
43
3
so Er
=∈ρ
3 0.
For a r b< < and c r< , q Qin = .
So EQr
=∈4 2
0π.
FIG. P24.57
For b r c≤ ≤ , E = 0 , since E = 0 inside a conductor.
(b) Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside theconductor, the total charge enclosed by a spherical surface of radius b r c≤ ≤ must be zero.
Therefore, q Q1 0+ = and σπ π1
12 24 4
= =−q
bQb
.
Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollowsphere is uncharged, we require
q q1 2 0+ = and σπ π2
12 24 4
= =q
cQ
c.
P24.58 E A⋅ = =∈z d E rq
4 2
0πe j in
(a) − × =× ⋅−3 60 10 4 0 100
8 85 103 2
12. ..
N C m C N m2 2e j a fπ
Qa r b< <a f
Q = − × = −−4 00 10 4 009. . C nC
(b) We take ′Q to be the net charge on the hollow sphere. Outside c,
+ × =+ ′
× ⋅−2 00 10 4 0 5008 85 10
2 212. .
. N C m
C N m2 2e j a fπQ Q
r c>a f
Q Q+ ′ = + × −5 56 10 9. C , so ′ = + × = +−Q 9 56 10 9 569. . C nC
(c) For b r c< < : E = 0 and q Q Qin = + =1 0 where Q1 is the total charge on the inner surface of
the hollow sphere. Thus, Q Q1 4 00= − = + . nC .
Then, if Q2 is the total charge on the outer surface of the hollow sphere,
*P24.59 The vertical velocity component of the moving chargeincreases according to
mdv
dtFy
y= mdv
dxdxdt
qEyy= .
Now dxdt
vx= has the nearly constant value v. So
dvq
mvE dxy y= v dv
qmv
E dxy y
v
y
y
= =z z−∞
∞
0
.
v x v y
x
y
d q
v 0
θ
Q
FIG. P24.59
The radially outward compnent of the electric field varies along the x axis, but is described by
E dA E d dxQ
y y−∞
∞
−∞
∞
z z= =∈
20
πb g .
So E dxQdy
−∞
∞z =∈2 0π
and vqQ
mv dy = ∈2 0π. The angle of deflection is described by
tanθπ
= =∈
v
vqQ
dmvy
2 02 θ
π=
∈−tan 1
022
qQdmv
.
P24.60 First, consider the field at distance r R< from the center of a uniform sphere of positive chargeQ e= +b g with radius R.
4 2
0 043
3
43
3
0π
ρπ
πr E
q V eR
re j =∈
=∈
=+FHG
IKJ ∈
in so Ee
R=
∈
FHG
IKJ4 0
3π r directed outward
(a) The force exerted on a point charge q e= − located at distance r from the center is then
F qE ee
Rr
eR
r Kr= = −∈
FHG
IKJ = −
∈
FHG
IKJ = −
4 403
2
03π π
.
(b) Ke
Rk eRe=
∈=
2
03
2
34π
(c) F m ak eR
rr e re= = −FHGIKJ
2
3 , so ak e
m Rr rr
e
e
= −FHGIKJ = −
2
32ω
Thus, the motion is simple harmonic with frequency fk e
m Re
e
= =ωπ π2
12
2
3 .
(d) fR
= × =× ⋅ ×
×
−
−2 47 10
12
8 99 10 1 60 10
9 11 1015
9 19 2
31 3.
. .
. Hz
N m C C
kg
2 2
πe je j
e jwhich yields R3 301 05 10= × −. m3 , or R = × =−1 02 10 10210. m pm .
Chapter 24 45
P24.61 The field direction is radially outward perpendicular to the axis. The field strength depends on r butnot on the other cylindrical coordinates θ or z. Choose a Gaussian cylinder of radius r and length L.If r a< ,
ΦEq
=∈
in
0and E rL
L2
0π
λb g =∈
Er
=∈
λπ2 0
or E r=∈
<λ
π2 0rr aa f .
If a r b< < , E rLL r a L
22 2
0π
λ ρπb g e j=
+ −
∈
E r=+ −
∈< <
λ ρπ
π
r a
ra r b
2 2
02e j a f .
If r b> , E rLL b a L
22 2
0π
λ ρπb g e j=
+ −
∈
E r=+ −
∈>
λ ρπ
π
b a
rr b
2 2
02e j a f .
P24.62 Consider the field due to a single sheet and let E+ and E−
represent the fields due to the positive and negative sheets. Thefield at any distance from each sheet has a magnitude given byEquation 24.8:
E E+ −= =∈σ
2 0.
(a) To the left of the positive sheet, E+ is directed toward theleft and E− toward the right and the net field over this
region is E = 0 .
(b) In the region between the sheets, E+ and E− are bothdirected toward the right and the net field is
E =∈σ
0 to the right .
FIG. P24.62
(c) To the right of the negative sheet, E+ and E− are again oppositely directed and E = 0 .
46 Gauss’s Law
P24.63 The magnitude of the field due to the each sheet given byEquation 24.8 is
E =∈σ
2 0 directed perpendicular to the sheet.
(a) In the region to the left of the pair of sheets, both fields aredirected toward the left and the net field is
FIG. P24.63
E =∈σ
0 to the left .
(b) In the region between the sheets, the fields due to the individual sheets are oppositelydirected and the net field is
E = 0 .
(c) In the region to the right of the pair of sheets, both are fields are directed toward the rightand the net field is
E =∈σ
0 to the right .
P24.64 The resultant field within the cavity is the superposition of twofields, one E+ due to a uniform sphere of positive charge of radius2a, and the other E− due to a sphere of negative charge of radius acentered within the cavity.
43
43
0
2π ρπ
rr E
∈
FHGIKJ = + so E r
r+ = ∈
=∈
ρ ρr3 30 0
– −∈
FHGIKJ = −
43
413
012π ρ
πr
r E so E r r− = ∈− =
−∈
ρ ρr1
01
013 3
b g .
Since r a r= + 1 , Er a
− =− −
∈ρa f3 0
FIG. P24.64
E E Er r a a
i j= + =∈
−∈
+∈
=∈
= +∈+ −
ρ ρ ρ ρ ρ3 3 3 3
030 0 0 0 0
a.
Thus, Ex = 0
and Ea
y = ∈ρ
3 0 at all points within the cavity.
*P24.65 Consider the charge distribution to be an unbroken charged spherical shell with uniform chargedensity σ and a circular disk with charge per area −σ . The total field is that due to the whole sphere,
QR
RR4
440
2
2
02
0ππ σπε
σ∈
= =∈
outward plus the field of the disk −∈
=∈
σ σ2 20 0
radially inward. The total
field is σ σ σ∈
−∈
=∈0 0 02 2
outward .
Chapter 24 47
P24.66 The electric field throughout the region is directed along x; therefore, E will beperpendicular to dA over the four faces of the surface which are perpendicularto the yz plane, and E will be parallel to dA over the two faces which are parallelto the yz plane. Therefore,
ΦE x x a x x a cE A E A a ab a c ab abc a c= − + = − + + + + = +
= = +e j e j e j a fe j a f3 2 3 2 2 22 2 .
Substituting the given values for a, b, and c, we find ΦE = ⋅0 269. N m C2 .
Q E=∈ = × =−0
122 38 10 2 38Φ . . C pC
FIG. P24.66
P24.67 E A⋅ = =∈z d E rq
4 2
0πe j in
(a) For r R> , q Ar r drARR
in = =z 2 2
0
5
4 45
π πe j
and EAR
r=
∈
5
025
.
(b) For r R< , q Ar r drArr
in = =z 2 2
0
5
44
5π
πe j
and EAr
=∈
3
05.
P24.68 The total flux through a surface enclosing the charge Q is Q∈0
. The flux through the
disk is
Φdisk = ⋅z E Ad
where the integration covers the area of the disk. We must evaluate this integral
and set it equal to 14
0
Q∈
to find how b and R are related. In the figure, take dA to be
the area of an annular ring of radius s and width ds. The flux through dA isFIG. P24.68
E A⋅ = =d EdA E sdscos cosθ π θ2b g .
The magnitude of the electric field has the same value at all points within the annular ring,
EQr
Qs b
=∈
=∈ +
14
140
20
2 2π πand cosθ = =
+
br
b
s b2 2 1 2e j.
Integrate from s = 0 to s R= to get the flux through the entire disk.
ΦE
R RQb sds
s b
Qbs b
Q b
R b, disk =
∈ +=
∈− +LNM
OQP =
∈−
+
L
NMMM
O
QPPP
z2 2 21
0 2 2 3 20 0
2 2 1 2
0 0 2 2 1 2e je j
e jThe flux through the disk equals
Q4 0∈
provided that b
R b2 2 1 212+
=e j
.
This is satisfied if R b= 3 .
48 Gauss’s Law
P24.69 E A⋅ =∈
=∈z zd
q ar
r drr
in
0 0
2
0
14π
E ra
rdra r
Ea
r
44 4
2
2
2
0 0 0
2
0
ππ π
=∈
=∈
=∈
=
zconstant magnitude
(The direction is radially outward from center for positive a; radially inward for negative a.)
P24.70 In this case the charge density is not uniform, and Gauss’s law is written as E A⋅ =∈z zd dV1
0ρ . We
use a gaussian surface which is a cylinder of radius r, length , and is coaxial with the chargedistribution.
(a) When r R< , this becomes E r arb
dVr
2 0
0 0
πρb g =∈
−FHGIKJz . The element of volume is a cylindrical
shell of radius r, length , and thickness dr so that dV r dr= 2π .
E rr a r
b2
22 3
20
0π
π ρb g =∈
FHG
IKJ −FHG
IKJ so inside the cylinder, E
ra
rb
=∈
−FHGIKJ
ρ0
0223
.
(b) When r R> , Gauss’s law becomes
E r arb
r drR
2 20
0 0
πρ
πb g b g=∈
−FHGIKJz or outside the cylinder, E
Rr
aRb
=∈
−FHGIKJ
ρ02
0223
.
P24.71 (a) Consider a cylindrical shaped gaussian surface perpendicularto the yz plane with one end in the yz plane and the other endcontaining the point x :
Use Gauss’s law: E A⋅ =∈z dqin
0
By symmetry, the electric field is zero in the yz plane and isperpendicular to dA over the wall of the gaussian cylinder.Therefore, the only contribution to the integral is over the endcap containing the point x :
E A⋅ =∈z dqin
0 or EA
Ax=
∈ρa f
0
so that at distance x from the mid-line of the slab, Ex
The acceleration of the electron is of the form a x= −ω 2 with ωρ
=∈e
me 0.
Thus, the motion is simple harmonic with frequency fe
me= =
∈ωπ π
ρ2
12 0
.
Chapter 24 49
P24.72 Consider the gaussian surface described in the solution to problem 71.
(a) For xd
>2
, dq dV Adx CAx dx= = =ρ ρ 2
E A⋅ =∈
=∈
=∈FHGIKJFHGIKJ
z zz
d dq
EACA
x dxCA dd
1
13 8
0
0
2
0
2
0
3
ECd
=∈
3
024or E i E i=
∈> = −
∈< −
Cdx
d Cdx
d3
0
3
024 2 24 2; for for
(b) For − < <d
xd
2 2E A⋅ =
∈=∈
=∈z z zd dq
CAx dx
CAxx130 0
2
0
3
0
E i E i=∈
> = −∈
<Cx
xCx
x3
0
3
030
30; for for
P24.73 (a) A point mass m creates a gravitational acceleration g r= −Gmr 2 at a distance r.
The flux of this field through a sphere is g A⋅ = − = −z dGmr
r Gm224 4π πe j .
Since the r has divided out, we can visualize the field as unbroken field lines. The same fluxwould go through any other closed surface around the mass. If there are several or nomasses inside a closed surface, each creates field to make its own contribution to the net fluxaccording to
g A⋅ = −z d Gm4π in .
(b) Take a spherical gaussian surface of radius r. The field is inward sog A⋅ = °= −z d g r g r4 180 42 2π πcos
and − = −4 443
3π π π ρGm G rin .
Then, − = −g r G r4 443
2 3π π π ρ and g r G=43π ρ .
Or, since ρπ
=M
RE
E43
3 , gM Gr
RE
E
= 3 or g =M Gr
RE
E3 inward .
ANSWERS TO EVEN PROBLEMS
P24.2 355 kN m C2⋅ P24.10 (a) −55 6. nC ; (b) The negative charge has aspherically symmetric distribution.
P24.4 (a) − ⋅2 34. kN m C2 ; (b) + ⋅2 34. kN m C2 ;P24.12 (a)
q2 0∈
; (b) q
2 0∈; (c) Plane and square
both subtend a solid angle of a hemisphereat the charge.
(c) 0
P24.6q∈0
P24.14 (a) 1 36. MN m C2⋅ ; (b) 678 kN m C2⋅ ;P24.8 ERh(c) No; see the solution.
