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 on position.. 22.20: a For points outsi
Trang 122.1: a) EA(14 N/C)(0.250 m2)cos601.75 Nm2 C.
b) As long as the sheet is flat, its shape does not matter
ci) The maximum flux occurs at an angle 0between 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) the area is oriented so that it “captures” no field lines
22.2: a) EAEAcosθ whereA A nˆ
CmN329
.36cosm)(0.1)CN104((back)
ˆˆ
CmN329.36cosm)(0.1)CN104(front)
(ˆ
090cosm)(0.1)CN104(bottom)
(ˆˆ
CmN24)9.36(90cosm)(0.1)CN104(right)
(ˆ
090cosm)(0.1)CN104(top)
(ˆˆ
CmN24)93690(cosm)(0.1)CN104(left)
(ˆ
2 2
3
2 2
3
2 3
2 2
3
2 3
S
2 2
3
6 6
5 5
4 4
3 3
2 2
1 1
S S
S S
S S
S
S
i n
i n
k n
j n
k n
j n
b) The total flux through the cube must be zero; any flux entering the cube must also leave it
22.3: a) Given thatE Biˆ Cˆj Dkˆ, E A,edgelength
ˆˆ
.BLˆ
ˆˆ
.DLˆ
ˆˆ
.CLˆ
ˆˆ
.DLˆ
ˆˆ
.CLˆ
ˆˆ
2 6
2 5
2 4
2 3
2 2
2 1
6 6
5 5
4 4
3 3
2 2
1 1
S S
S S
S S
S S
S S
A A A A A A
n E i
n
n E i
n
n E k
n
n E j
n
n E k
n
n E j
Trang 222.5: a) (2 ) ( 6 00 10 C/m) (0.400 m) 2.71 105 Nm2 C.
6 0
πε πrl
A
E b) We would get the same flux as in (a) if the cylinder’s radius was made larger—the field lines must still pass through the surface
c) If the length was increased to l0.800m, the flux would increase by a factor of two: 5.42105 Nm2 C
0
9 0
2 1
2 1
3 2 1
f) All that matters for Gauss’s law is the total amount of charge enclosed by the
surface, not its distribution within the surface
C1000
2 2 12 9
NmC108.85
C10)00.600.4
2 2 12
b) If there is no charge in a region of space, that does NOT mean that the electric field
is uniform Consider a closed volume close to, but not including, a point charge The field diverges there, but there is no charge in that region
Trang 322.10: a) If ρ0 and uniform, then q inside any closed surface is greater than zero
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 on
position
b) However, inside a small bubble of zero density within the material with density ρ, the field CAN be uniform All that is important is that there be zero flux through the surface of the bubble (since it encloses no charge) (See Exercise 22.61.)
22.11: (9.60 10 C) 1.08 106 Nm2 C
0
6 0
sides
symmetrical, so for one side, the flux is: 1.80 105 Nm2 C
side
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 Gauss’s law, must be zero
22.13: E Qencl ε0
The flux through the sphere depends only on the charge within the sphere
nC3.19)CmN360
0 0
encl ε ε
m)(0.550
C)1050.2(4
14
1m)0.1m450.0
0
2 0
q πε r
C)10180.0(4
14
1
|
|4
2 0
q πε
r r
q πε E
b) As long as we are outside the sphere, the charge enclosed is constant and the sphere acts like a point charge
22.16: a) / (1.40 105 N C)(0.0610m2) 7.56 10 8 C
0 0
Trang 422.17: 4 4 (1150N C)(0.160m)2 3.27 10 9 C.
0
2 0
C 10 27 3
22.18: Draw a cylindrical Gaussian surface with the line of charge as its axis The
cylinder has radius 0.400 m and is 0.0200 m long The electric field is then 840 N/C at every point on the cylindrical surface and directed perpendicular to the surface Thus
Ed s(E)(Acylinder)(E)(2πrL)
(840N/C)(2π)(0.400m)(0.0200m)42.2Nm2/CThe field is parallel to the end caps of the cylinder, so for them E s0
Gauss’s law:
C1074.3
)C
mN2.42()mN
C10854.8(10
2 2
2 12 0
Trang 522.20: a) For points outside a uniform spherical charge distribution, all the charge can
be considered to be concentrated at the center of the sphere The field outside the sphere
is thus inversely proportional to the square of the distance from the center In this case:
CN53cm
0.600
cm0.200)
CN480(
λ
0r πε
that is, inversely proportional to the distance from the axis of the cylinder In this case
CN160cm
0.600
cm0.200)
CN480
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 2
thetoward,
CN1006.8
CN10471.5m)(0.250
C1080.3
|
|
CN10591.2m)(0.250
C1080.1
|
|
5 2
1
5 2
6 2
2
2 2
5 2
6 2
1
1 1
E
k r
q k
E
k r
q k
E
Trang 622.22: For points outside the sphere, the field is identical to that of a point charge of the
same total magnitude located at the center of the sphere The total charge is given by charge density volume:
C1060.1m)150.0)(
3
4)(
mCn50.7
(0.150
C)10(1.06)/CmN109(
10 2
2 9
2 0
q E
b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will
be 1/4 as strong: 10.6 N C
c) Inside the sphere, only the charge inside the radius in question affects the field In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field:
CN2.21m)
(0.075
C)1006.1()8/1()C/mN109(
2
10 2
2 9
m100.0(
m)(0.220)CN950
ER Q
22.24: a) Positive charge is attracted to the inner surface of the conductor by the charge
in the cavity Its magnitude is the same as the cavity charge: qinner 6.00nC, since 0
6.00nC
00.5inner tot outer outer
Trang 722.26: a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so:
m)800.0(2
C1050.72
q A E
dA
b) At a distance of 100 m from the center, the sheet looks like a point, so:
.CN1075.6m)
(100
C)1050.7(4
14
2 9
0
2 0
q πε E
c) There would be no difference if the sheet was a conductor The charge would automatically spread out evenly over both faces, giving it half the charge density on any
as the insulator
0
0 2::)
ε σ c
E
σ near one face Unlike a conductor, the insulator is the
charge density in some sense Thus one shouldn’t think of the charge as “spreading over each face” for an insulator Far away, they both look like points with the same charge
22.27: a) 2
Q πRL
Q A
Q
b) (2 ) 2
0 0
σR E ε
πRL σ ε
Q πrL E
Trang 822.28: All the σ's are absolute values.
(a) at
0 1 0 4 0 3 0
2
2 2 2 2
ε
σ ε
σ ε
σ A
E
left
thetoCN1082.2
)mC6mC4mC2mC5(21
)(
21
5
2 2
2 2
0
1 4 3 2 0
μ μ
ε
σ σ σ σ ε
E A
(b)
left
thetoCN1095.3
)mC5mC4mC2mC6(21
)(
2
12
222
5
2 2
2 2
0
2 4 3 1 0 0
2 0
4 0
3 0 1
μ μ
ε
σ σ σ σ ε ε
σ ε
σ ε
σ ε
σ
E B
(c)
left thetoCN1069.1
)mC6mC4mC2mC5(21
)(
2
12
222
5
2 2
2 2
0
1 4 3 2 0 0
1 0
4 0
3 0 2
μ μ
ε
σ σ σ σ ε ε
σ ε
σ ε
σ ε
σ
E C
22.29: a) Gauss’s law says +Q on inner surface, so E0 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 electric
field flux through it; there is electric field in the cavity
d) In an electrostatic situation E0 inside a conductor A Gaussian sphere with the charge at its center and radius greater than the outer radius of the metal encloses Q
zero net charge (the charge and the Q Q on the inner surface of the metal) so there is
no flux through it and E 0outside the metal
e) No, E 0 there Yes, the charge has been shielded by the grounded
conductor There is nothing like positive and negative mass (the gravity force is always attractive), so this cannot be done for gravity
Trang 922.30: Given E(5.00(N C)m)x iˆ(3.00(N C)m)z kˆ,edge length
m300.0)(
.m)CN(081.0)m300.0)(
ˆˆ
ˆˆ
ˆˆ
b) Since the field is parallel to the surface, 0
c) Choose the Gaussian surface to equal the volume’s surface Then: 750 –
m 0 6
1
ε
Since q0 we must have some net flux flowing in so EA E A on second face
d) q0 but we have E pointing away from face I This is due to an external field that does not affect the flux but affects the value of E.
Trang 1022.33: To find the charge enclosed, we need the flux through the parallelepiped:
CmN5.3760cos)CN1050.2)(
m0600.0)(
m0500.0(60
cos)CN1000.7)(
m0600.0)(
m0500.0(120
CmN5.67
0
2 0
22.34:
The particle feels no force where the net electric field is zero The fields can cancel only in regions A and B
sheet line E
0
0 22
λ
ε
σ r
cm16m16.0)C/m100(
C/m50
Trang 11The electric field E1
of the sheet of charge is toward the sheet, so the electric field E2
of the sphere must be away from the sheet This is true above the center of the
sphere Let r be the distance above the center of the sphere for the point where the
electric field is zero
3 2 0 0
1 2
12
so
R
r Q πε ε
σ E
m097.0C
10900.0
)m120.0)(
C/m1000.8(22
9
3 2
r
Trang 1222.36: a) For ra,E0,since no charge is enclosed.
0
41 r
q πε
E b r
a since there is +q inside a radius r.
For brc,E 0, since now the –q cancels the inner +q.
0
41 r
q πε
E
c
r since again the total charge enclosed is +q.
b)
c) Charge on inner shell surface is –q.
d) Charge on outer shell surface is +q.
E R r
0
2
41 r
Q πε
E R r
charge enclosed is 2Q.
Trang 1322.38: a) , 2,
0
41 r
Q πε
E a
r since the charge enclosed is Q.
,0,
r b E
a since the –Q on the inner surface of the shell cancels the +Q at the
center of the sphere
2 0
r , since the total enclosed charge is –2Q.
b) The surface charge density on inner surface: 4πa2
d)
e)
Trang 14E c r
E d
r since Q6q
b)(i) small shell inner: Q0
(ii) small shell outer: Q2q
(iii) large shell inner: Q2q
(iv) large shell outer: Q6q
22.40: a)(i) r a,E 0, since the charge enclosed is zero
(ii) arb,E0, since the charge enclosed is zero
E c r
b since charge enclosed is 2q.(iv) cr d,E0, since the net charge enclosed is zero.(v) r d,E0, since the net charge enclosed is zero
b)(i) small shell inner: Q0
(ii) small shell outer: Q2q
(iii) large shell inner: Q2q
(iv) large shell outer: Q0
Trang 1522.41: a)(i) r a,E 0, since charge enclosed is zero.
(ii) arb,E0, since charge enclosed is zero
E c r
b since charge enclosed is 2q.(iv) crd,E0, since charge enclosed is zero
E d
r since charge enclosed is 2q
b)(i) small shell inner: Q0
(ii) small shell outer: Q2q
(iii) large shell inner: Q2q
(iv) large shell outer: Q2q
22.42: a) We need:
.28
33
28)
)2((
3
3 3
ρR π Q
R R
)(
3
4)
4(,
2 0
2 0
3 3 0 0
r ε
ρ r
πε
Q E
R r ε
ρ π ε
Q πr E R
r
Substituting ρ from (a) 3
0 2
0 28
72 πε R
Qr r
Q πε
c) We see a discontinuity in going from the conducting sphere to the insulator due to the thin surface charge of the conducting sphere—but we see a smooth transition from the uniform insulator to the outside
Trang 1622.43: a) The sphere acts as a point charge on an external charge, so:
,2 0
41 r
qQ πε
14
3
a b π
q πa πb
q V
V
q ρ
a b
14
3
c d π
q πc
πd
q V
V
q
c d
0
)(
3
44
ε r E dV ε
d b r
)(
)(
4
)(
3
1
3 3
3 3
0 2
3 3 inner
a r ε
q r
q r ε
2
q E ε
q πr E ε
q d c r
ε ε
q d d r
0 0
1
A
)(
4
)(
4so)
(3
4
0
3 3 2
0 outer,
3 3 0 0
2
c d r πε
c r q r
πε
q E ρ
c r ε
π ε
q r E
q d d
Trang 1722.45: a) λ,
4
1,
0 r E
b r a
r 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 amount
of charge on them: λl λinnerlλinner λ,andλouter λ
2)
2(,
0 0
α E ε
αl ε
q πrl E a
(ii) arb, there is no net charge enclosed, so the electric field is zero
0 0
α E ε
αl ε
q πrl E b
b) (i) Inner charge per unit length is α. (ii) Outer charge per length is 2
Trang 1822.47: a) (i) r a E πrl E πε α r ,
ε
αl ε q
0 0
)2(
(ii) ar b,there is not net charge enclosed, so the electric field is zero (iii) r there is no net charge enclosed, so the electric field is zero.b,
b) (i) Inner charge per unit length is
(ii) Outer charge per length is ZERO
0 0
2
ρr ε
l ρπr ε
πrl E R
2λ2
2
0 0 2 0
2
ρR ε
l ρπR ε
πrl E ρπR R
c) rR.the electric field for BOTH regions is ,
0
2ε ρR
E so they are consistent.d)
22.49: a) The conductor has the surface charge density on BOTH sides, so it has twice
the enclosed charge and twice the electric field
b) We have a conductor with surface charge density σ on both sides Thus the
electric field outside the plate is E(2A)(2σA) ε0 Eσ ε0 To find the field inside the conductor use a Gaussian surface that has one face inside the conductor, and one outside
Then:
.00
but)
Trang 1922.50: a) If the nucleus is a uniform positively charged sphere, it is only at its very
center where forces on a charge would balance or cancel
0 3
R
r ε
e πr E ε
13 2
0 R
r e πε qE
So from the simple harmonic motion equation:
.4
12
14
14
1
3 2
0 3
2
0 3
f mR
e πε
ω R
r e πε r
1Hz1057.4
mR
e πε
kg1011.9(4
C)1060.1(4
3
2 14 31
2
2 19
1Thompson actual r
r
d) If r then the electron would still oscillate but not undergo simple R
harmonic motion, because for rR,F1 r2, and is not linear
22.51: The electrons are separated by a distance 2d and the amount of the positive ,
nucleus’s charge that is within radius d is all that exerts a force on the electron So:
.2/8
/2
)
2
(
3 3 2
nucleus 2
2
ke F
d
ke
Trang 2022.52: r a r x e x a dx
a
Q Q d dθ dr
r e πa
Q Q ρdV Q
r
Q
0
/ 2 2 3 0
2 / 2 3
0
0
sin)
(
].1)/(2)/(2[)
222
(
4)
0 /
2 2
2 3
3 0
Note r Q r
b) The electric field is radially outward, and has magnitude:
)
1)(2)(2
0 2
a r
2 19 0
2 Fe e
0 4 ( 7 1 10 m)
C) 10 6 1 )(
82 (
414
q πε
E q F R r
R
r because the charge enclosed goes like r ) so with the 3radius decreasing by 2, the acceleration from the change in radius goes up by (2)2 4,but the charge decreased by 8, so 2.1 1032m/s2
) b ( 8
a
a
d) At r0,Q0,soF 0
Trang 2122.54: a) The electric field of the slab must be zero by symmetry There is no preferred
direction 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 be
singled 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 0
ρx E ε
ρAx ε
Q EA d
with direction given by x x|i.ˆ
Note that E is zero at x0
Now outside the slab, the enclosed charge is constant with :x
,:
0 0
0
encl
ε
ρd E ε
ρAd ε
Q EA d
again with direction given by x x|i.ˆ
22.55: a) Again, E is zero at x0, by symmetry arguments
|
|in,33
''
0
3 0 2
0
3 0 0
2 2 0
0 0
encl
x
x d
ε
x ρ E d
ε
Ax ρ dx x d ε
A ρ ε
Q EA d
d ρ E ε
Ad ρ dx x d ε
A ρ ε
Q EA d
x
0 2
2 0
0 0
|
|
,33
''
22.56: a) We could place two charges on either side of the charge Q q:
b) In order for the charge to be stable, the electric field in a neighborhood around it must always point back to the equilibrium position
c) If q 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 a stable equilibrium
d) For a negative charge to be in stable equilibrium, we need the electric field to always 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 charge was if stable equilibrium is to be attained