Download PHYS 110B - HW #8

PHYS 110B - HW #8
Spring 2004, Solutions by David Pace
Any referenced equations are from Griffiths
Problem statements are paraphrased
[1.] Problem 11.3 from Griffiths
Calculate the radiation resistance of the wire connecting the charges that form the dipole
of section 11.1.2 in Griffiths. Radiation resistance is that value of resistance in the wire that
results energy dissipation (as heat) equal to the actual energy output of the system due to
radiation. Show,
2
d
Ω
(1)
R = 790
λ
where λ is the radiation wavelength.
For wires of length d = 5 cm (e.g. in a radio), is it necessary to worry about the contribution
of radiation to their resistance?
Solution
This problem is solved by comparing different models for energy dissipation. The power
of a circuit is given by,
P = I 2R
Eq. 7.7
(2)
where I have chosen the form that includes a resistance factor because that is what we seek
in this problem.
Solving for the equivalent current,
I =
=
dq
dt
(3)
d
qo cos(ωt)
dt
Eq. 11.3
= −qo ω sin(ωt)
(4)
(5)
This is an oscillating parameter so it makes sense to solve for the time-averaged power
dissipation in the system.
hP i = hI 2 Ri
(6)
= hqo2 ω 2 sin2 (ωt)Ri
(7)
qo2 ω 2
=
R
2
(8)
where the time averaging does not affect the constants and hsin2 (ωt)i = 1/2.
The time-averaged power radiated from the dipole is given.
hP i =
µo p2o ω 4
12πc
1
Eq. 11.22
(9)
where po ≡ qo d.
Equating (8) and (9),
µo p2o ω 4
qo2 ω 2
R =
2
12πc
(10)
µo p2o ω 2
R =
6πcqo2
(11)
=
µo qo2 d2 ω 2
6πcqo2
(12)
=
µo d2 ω 2
6πc
(13)
Radiation is simply electromagnetic waves. We know that the solution needs to be in terms
of the wavelength of these waves and only the angular frequency term in (13) can be rewritten in terms of wavelength.
ω
k
=
ω2
c
=
k
c2 k 2 =
=
2π
λ
(14)
4π 2 c2
λ2
(15)
The ω 2 term in (13) can now be replaced,
µo d2
R =
6πc
4π 2 c2
λ2
2
2µo πc d
=
3
λ
(16)
(17)
It remains to show that the prefactor in (17) equals 790.
2(4π × 10−7 )(3.0 × 108 )
2µo πc
=
3
3
(18)
≈ 789.6
(19)
this is close enough so we have shown that (1) is true.
We might be tempted to check whether the units make sense. This can be proven quickly.
In (17) the d/λ term is unitless because both variables are in units of length and therefore
cancel each other. Then,
[Ω] =
[V ]
[A]
[J]
[C][A]
=
(20)
2µo πc
[N ] [m]
=
3
[A]2 [s]
=
[J]
[C]2
[s]
[s]2
(21)
=
[J]
[C]
[C]
[s]
2
=
[J]
[A][C]
=
[Ω] X
(22)
Radio waves may be treated with λ ≈ 103 m. Combining this with the given value of
d = 0.05 m gives,
2
0.05
(23)
R = 790
103
= 1.975 × 10−6 Ω
(24)
this is an insignificant resistance and we are justified in ignoring this contribution when
designing radios.
[2.] Problem 11.10 from Griffiths
An electron is released from rest and allowed to fall under the influence of gravity. During
the first centimeter of its free fall what fraction of the lost potential energy is dissipated
through radiation?
Solution
Recall the kinematic equations in one dimension with constant acceleration.
1
v − vo = at
x − xo = vo t + at2
2
(25)
These equations underlie a physical discrepancy with this problem. The kinematic equations above rely on an ideal system in which all of the energy of the falling object is kinematic. There are corrections to these equations in the case where radiation is significant
compared to the kinetic energy. It turns out that the energy lost due to radiation in this
situation pales in comparison to the kinetic term and therefore these equations are valid.
If you considered this initially, then it is still best to assume the radiation term will be very
small and use the equations above. Should the radiation term turn out to be very large,
then you would have had to start over and think about the new physical issues.
Let the distance fallen by the electron be equal to h. The actual value of this parameter
is given as 1 cm, but it is easier to use the variable throughout the algebra and then plug
in the numerical value for the final calculation. The goal of this problem is to determine
the fraction Urad /∆Upot where Urad is energy radiated during the 1 cm fall, and ∆Upot is the
total amount of gravitational potential energy that is lost during this fall. Treat ∆Upot as a
positive value because we are admitting that the energy is decreasing and we only wish to
know how this absolute amount relates to the radiated amount.
Power is a measure of energy per unit time. The total energy radiated by this falling electron is the power radiated multiplied by the time it takes for it to fall the specified distance.
Solving for this time requires the use of (25). Since the electron is dropped from rest we
have vo = 0. Setting the ground as my reference requires that the acceleration be a = −g,
where g is the acceleration due to gravity.
1
(26)
x − x0 = vo t + at2
|{z} 2
0
1
−h = − |g|t2
2
s
2h
t =
|g|
3
(27)
(28)
The radiated power is given by,
P =
µ o q 2 a2
6πc
Eq. 11.61
(29)
where this is known as the Larmor formula. Griffiths explains in section 11.2.1 that this
expression is valid for v c. This condition is met here because the final velocity of any
object falling (from rest) through a distance of 1 cm in the Earth’s gravitational field is not
going to be appreciable compared to the speed of light.
Multiplying (28) and (29) gives the total energy radiated by the electron in this measured
fall.
s
2h
µ o q 2 a2
·
(30)
Urad = P t =
6πc
|g|
s
µo q 2 g 2 2h
=
(31)
6πc
|g|
The total change in gravitational potential energy is (leaving this value positive for the
reason given above),
∆Upot = mgxf − mgxo
= mg(xf − xo )
(32)
=
(33)
mgh
The desired ratio is,
Urad
=
∆Upot
µo q 2 g 2
6πc
q
2h
|g|
(34)
mgh
r
µo q 2
2|g|
=
6πcm
h
(35)
(4π × 10−7 )(1.6 × 10−19 )2
=
6π(3.0 × 108 )(9.11 × 10−31 )
r
2(9.8)
0.01
≈ 2.8 × 10−22
(36)
(37)
The energy lost due to radiation is insignificant compared to the general kinetic loss.
[3.] Problem 11.13 from Griffiths
(a) Consider an electron decelerating (a = constant) from initial velocity vo to final velocity
v = 0. Assuming vo c so that you may apply the Larmor formula, what fraction of the
initial kinetic energy of this electron is radiated away?
(b) Apply the following values to the problem in part (a).
vo ≈ 105 m/s
4
d = 30 Å
(38)
where d is the total distance traveled. What does this result tell us about radiation losses
for electrons in ordinary conductors?
Solution
The initial kinetic energy of the electron is,
1
K.E.i = mvo2
2
(39)
where m is the electron mass.
The total power radiated away is found using (30), but with a different expression for the
time factor.
The time it takes for the electron to come to rest is found using (25). The a term is written
negative because this is a deceleration.
v − vo = −at
(40)
−vo
= t
−a
(41)
t =
vo
|a|
(42)
so the time is now written in terms of parameters given in the problem statement.
The total radiated power, Urad , is given from (29) multiplied by this time.
Urad =
=
µo q 2 a2 vo
·
6πc
|a|
(43)
µo q 2 vo |a|
6πc
(44)
Now to solve for the desired fraction,
µo q 2 vo |a|
6πc
1
2
mv
o
2
(45)
=
µo q 2 vo |a|
2
·
6πc
mvo2
(46)
=
µo q 2 |a|
3πcmvo
(47)
Urad
=
K.E.i
(b) In this part we want to apply numbers to (47). This requires solving for the absolute
value of the acceleration because it is not given.
Using (25), (42), and (38),
1
xf − xi = vo t + at2
2
vo
a vo2
d = vo
−
|a|
2 a2
5
(48)
(49)
vo2
1 vo2
d =
−
|a| 2 |a|
(50)
1 vo2
2 |a|
(51)
vo2
2d
(52)
d =
|a| =
where this derivation also assumes the initial location is the origin and the velocity is v ≥ 0.
Solving for the numerical fraction of kinetic energy lost to radiation,
µo q 2
v2
Urad
=
· o
K.E.i
3πcmvo 2d
(53)
=
µo q 2 v o
6πcmd
(54)
=
(4π × 10−7 )(1.6 × 10−19 )2 (105 )
6π(3.0 × 108 )(9.11 × 10−31 )(30 × 10−10 )
(55)
≈ 2.1 × 10−10
(56)
This is another instance in which the energy lost to radiative processes is negligible compared to other loss mechanisms.
[4.] Problem 11.14 from Griffiths
Consider the Bohr model of the hydrogen atom. In this model the ground state electron
travels around the proton in a circular orbit of radius ro = 5 × 10−11 m, due to the Coulomb
attraction between the particles. Classical electrodynamics requires that this accelerating
electron emit radiation and therefore continuously lose energy. As it loses energy this electron will move inward and eventually impact the proton. Show that the Larmor formula
is valid for most of the electron’s inward motion (i.e. v c). Calculate the typical lifetime
of such an atom if the electron’s orbits can always be considered circular.
Solution
To show that the Larmor formula is valid I will solve for the initial velocity of the electron
in functional form and then determine how this changes as the electron approaches the
proton. This is accomplished by looking at the forces involved,
F~
=
me~a
=
~
−q E
(57)
~ is the electric field due
where me is the electron’s mass, q is the electron’s charge, and E
only to the proton. Recall that the force experienced by the electron cannot include its own
fields.
The electric field of the proton is known, but it’s also equation 2.10 in the text in case you
wish to review.
~a = −
6
q2
r̂
4πo me r2
(58)
q2
(59)
4πo me r2
where the vector form is written in spherical coordinates. The magnitude of the acceleration is written because next we will solve for the electron velocity in its orbits.
a =
In the case of centripetal acceleration and circular orbits the tangential velocity of the particle is found using a = v 2 /r. The distance r changes as the electron moves closer to the
proton. As r decreases along this path the initial velocity must be the lowest velocity
throughout the trip. This initial velocity is found using,
q2
v2
=
r
4πo me r2
s
q2
v =
4πo me r
r
q
1
=
2 πo me r
(60)
(61)
(62)
All of the parameters in (62) are known numerically. The initial (and lowest) velocity of
this electron occurs when r = ro ,
s
1
1.6 × 10−19
(63)
vo =
−12
2
π(8.85 × 10 )(9.11 × 10−31 )(5 × 10−11 )
≈ 2.2 × 106
(64)
This is a large velocity compared to everyday experience, but vo /c ≈ 0.007, which satisfies
the small velocity
√ condition. As the electron approaches the proton this velocity increases
by a factor of 1/ r. This factor causes the velocity to slowly increase and we may therefore
claim that the electron’s velocity is less than the speed of light for most of the trip. Of
course, once the electron gets incredibly close to the proton the distance goes to zero and
the velocity goes to infinity. We ignore this situation here.
The lifetime of the electron in this configuration is given by the time necessary for the
electron to radiate away all of its energy. The total energy of the electron, Ee , is its kinetic
energy plus its potential energy. The kinetic energy of the electron is given by (39) replacing
the initial velocity there with the instantaneous velocity here, while its potential energy is
P E = −qV , where V is the scalar potential field due to the proton. This scalar potential is
known for a single charged particle.
q
1
2
(65)
Ee = me v − q
2
4πo r
This is simplified using (61),
me
q2
q2
−
2 4πo me r 4πo r
q2
me
=
−1
4πo r 2me
Ee =
7
(66)
(67)
q2
(68)
8πo r
This is the total amount of energy that must be lost to radiation before the electron impacts
the proton.
Ee = −
The energy radiated away is given by (29), and we have solved for the acceleration in (59).
2
q2
µo q 2
P =
(69)
6πc 4πo me r2
µo q 6
96π 3 2o cm2e r4
=
(70)
To be able to compare these energy terms we take advantage of what power represents.
Power is energy per unit time. This can be written mathematically as,
∆E
∆t
where E represents energy and not electric field.
P
=
(71)
As we reduce the ∆t to an infinitesimal time period we have,
dE
(72)
dt
where the negative sign is placed because we know the energy is decreasing in the system
but the radiated power must be positive.
P =−
Using our expression for the total energy that must be lost, (68), and noting that r = r(t),
we have,
d
q2
dEe
= −
−
(73)
−
dt
dt
8πo r
1 dr
q2
=
(74)
− 2
8πo
r
dt
= −
q 2 dr
8πo r2 dt
(75)
Now we equate (70) and (75) so that we can solve for the time it takes for the electron to
impact the proton.
µo q 6
q 2 dr
=
−
96π 3 2o cm2e r4
8πo r2 dt
dt = −
= −
Z
Z
q2
96π 3 2o cm2e r4
·
dr
8πo r2
µo q 6
(77)
12π 2 o cm2e r2
dr
µo q 4
(78)
0
−
dt =
ro
8
(76)
12π 2 o cm2e r2
dr
µo q 4
(79)
In (79) the limits of the r integral reflect the fact that the electron begins at the ground state
radius and moves into the proton.
12π 2 o cm2e
t = −
µo q 4
Z
0
r2 dr
(80)
ro
To be more mathematically rigorous we may require that the left side of (80) be t − to , but
since we can always adjust the initial time to be to = 0 it will not matter.
0
12π 2 o cm2e r3
t = −
µo q 4
3 ro
= −
=
4π 2 o cm2e
−ro3
4
µo q
(81)
(82)
4π 2 o cm2e ro3
µo q 4
(83)
All of these values are known,
t =
4π 2 (8.85 × 10−12 )(3 × 108 )(9.11 × 10−31 )2 (5 × 10−11 )3
(4π × 10−7 )(1.6 × 10−19 )4
≈ 1.3 × 10−11 s
(84)
(85)
If this truly explained hydrogen, then it would be an incredibly short lived atom.
[5.] Problem 11.15 from Griffiths
Regarding example 11.3 (page 463), find the angle, θmax , at which the maximum radiation
emission occurs. Show that,
r
1−β
for v ≈ c
(86)
θmax ∼
=
2
Calculate the intensity of the radiation at θmax for the ultrarelativistic case (i.e. where v ≈ c),
but write this value in proportion to the same quantity for an instantaneous velocity v = 0.
State the ratio in terms of γ. Reference figure 11.14.
Solution
The parameters in the problem statement are,
β≡
v
c
1
γ≡p
1 − v 2 /c2
Griffiths, page 463
(87)
Example 11.3 provides an expression for the power radiated as a function of angle θ.
dP
µ o q 2 a2
sin2 θ
=
dΩ
16π 2 c (1 − β cos θ)5
9
Eq. 11.74
(88)
To find the angle at which the maximum radiation occurs we take the relevant derivative
of (88) and set it equal to zero. We already know that the minimum in radiation (equal to
zero) occurs at θ = 0. So although this method can only show us where extrema occur,
unless the solution is θ = 0 we can safely assume we have found a maximum. If you are
still unsure, then you can take the second derivative of (88) at the value of θ found from
setting the first derivative to zero and if this result is negative (positive) then the point
represents a maximum (minimum).
Finding θmax ,
0
=
sin2 θ
µo q 2 a2 d
d µ o q 2 a2
=
dθ 16π 2 c (1 − β cos θ)5
16π 2 c dθ
sin2 θ
(1 − β cos θ)5
µo q 2 a2 (1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ))
=
16π 2 c
(1 − β cos θ)10
(89)
(90)
Since we are setting this derivative equal to zero the prefactor in (90) does not matter.
(1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ))
(1 − β cos θ)10
(91)
= (1 − β cos θ)5 (2 sin θ cos θ) − sin2 θ(5(1 − β cos θ)4 (−β)(− sin θ))
(92)
= (1 − β cos θ)4 (1 − β cos θ)(2 sin θ cos θ) − 5β sin3 θ
(93)
= sin θ (1 − β cos θ)(2 cos θ) − 5β sin2 θ
(94)
= 2 cos θ − 2β cos2 θ − 5β(1 − cos2 θ)
(95)
= 2 cos θ − 2β cos2 θ − 5β + 5β cos2 θ
(96)
= 3β cos2 θ + 2 cos θ − 5β
(97)
0 =
This is the form of a common quadratic equation and may be solved as such. I have not
labeled this angle as θmax , but by the method taken we are finding that value.
p
−2 ± 4 − 4(3β)(−5β)
cos θ =
(98)
6β
p
−2 ± 2 1 + 15β 2
=
(99)
6β
p
−1 ± 1 + 15β 2
=
(100)
3β
We choose the sign in (100) based on information about θmax provided in example 11.3.
From figure 11.14 and a statement on the bottom of page 463 we know that the radiation
10
is concentrated through a narrow cone near the forward direction. Figure 11.14 is turned
sideways, but if we look at the location of θmax it is evident that concentrating the radiation
in a narrow region near the z axis means that θmax < 45◦ . This requires that cos θ > 1.
Meeting this condition requires that we take the plus sign in (100).
Finally,
θmax
p
1 + 15β 2
3β
"
#
p
2
1
+
15β
−1
+
= cos−1
3β
cos θ =
−1 +
(101)
(102)
For the next part of the problem it is more direct to use the cosine form of θmax as written in
(101). Since we are in the ultrarelativistic case β ≡ v/c ≈ 1, but slightly less than one since
the velocity is still less than the speed of light. When the velocity is very close to the speed
of light we know that β is so close to unity that it may be written as,
β ≈1−
(103)
where is an expansion parameter and is very small. The rest of this problem is rewriting
terms and solving for this expansion parameter.
To begin, replace β with its expansion in (101). When writing out the terms we should
neglect anything that is of order 2 or higher. Since 1, any term containing an 2 is
so small it should be ignored. This process is known as expanding to order one in , a
common procedure in asymptotic analysis.
p
−1 + 1 + 15β 2
(104)
cos θmax =
3β
p
−1 + 1 + 15(1 − )2
=
(105)
3(1 − )
hp
i
1
2
=
1 + 15(1 − 2 + ) − 1
(106)
3(1 − )
√
1
1 + 15 − 30 − 1
3(1 − )
" r
#
1
30
=
4 1+ −1
3(1 − )
16
" #
1/2
1
15
=
4 1+ −1
3(1 − )
8
=
(107)
(108)
(109)
Recall the binomial expansion, which will be applied multiple times in the steps to follow,
(1 + x)n ≈ 1 + nx
for small x.
11
(110)
Applying a binomial expansion to the square root term in (109) provides,
1 15
1
4 1+
−1
cos θmax =
3(1 − )
2 8
15
1
4+ −1
=
3(1 − )
4
1
5
=
1− 1−
4
(111)
(112)
(113)
We can solve further by using the binomial expansion on the 1/(1 − ) term.
1
= (1 − )−1
1−
(114)
≈ 1+
cos θmax
(115)
5
≈ (1 + ) 1 − 4
(116)
5
5
≈ 1 − + − 2
4
4
(117)
1
≈ 1− 4
(118)
We invoke another expansion of the cosine function for comparison with (118).
cos x =
∞
X
(−1)n x2n
n=0
cos θmax ≈ 1 −
(2n)!
=
1−
2
θmax
2
x2 x4
+
− ...
2!
4!
(119)
(120)
4
since keeping the θmax
term seems too much.
Equate (118) and (120),
1
θ2
1 − = 1 − max
4
2
(121)
2
= θmax
2
(122)
= θmax
2
(123)
r
Using (103) to rewrite the above,
r
θmax ≈
12
1−β
2
(124)
In the next part of the problem we want to write the radiated power in the direction of θmax
as a fraction of the radiated power in the equivalent maximum direction for a particle of
instantaneous velocity equal to zero. Notice that the particle is still accelerating, as it must
in order to radiate at all, but that we are concerned with the instant at which its velocity
reaches zero.
Begin by writing (88) for v = 0, which is the same as setting β = 0.
µ o q 2 a2
sin2 θ
dP
=
dΩ
16π 2 c (1 − (0) cos θ)5
(125)
µ o q 2 a2
sin2 θ
16π 2 c
(126)
=
The direction at which the maximum power is radiated occurs at θ = 90◦ . This is known
because the maximum must occur when the sine term is equal to unity. The maximum
power radiated in this zero velocity instant is,
max
µ o q 2 a2
dP
(127)
=
dΩ v=0
16π 2 c
The other radiated power term is (88) evaluated at the already determined θmax .
max
dP
sin2 θmax
µ o q 2 a2
=
dΩ θ=θmax
16π 2 c (1 − β cos θmax )5
(128)
The desired ratio is,
dP max
dΩ θ=θmax
dP max
dΩ v=0
=
µo q 2 a 2
sin2 θmax
16π 2 c (1−β cos θmax )5
µo q 2 a 2
16π 2 c
(129)
=
sin2 θmax
(1 − β cos θmax )5
(130)
Since this is evaluated at θmax we know that β can be expanded as in (103). The plan is
to write (130) in terms of and then relate this result to γ as requested. The numerator of
(130) can be rewritten using yet another trigonometric property; sin x ≈ x for small x. The
denominator is rewritten using (118).
(sin θmax )2
sin2 θmax
≈
(1 − β cos θmax )5
(1 − β(1 − 4 ))5
=
=
13
(131)
2
θmax
(1 − (1 − )(1 − 4 ))5
p 2
2
4
(1 − (1 − − +
2 5
))
4
(132)
(133)
Once again drop the 2 term.
p 2
2
(1 − (1 − 4 − +
2
=
( 54 )5
5
1 4
=
−4
2 5
2 5
))
4
= 0.16384−4
(134)
(135)
(136)
where inverse powers of small numbers are large and therefore kept.
It is necessary to express in terms of γ.
γ
=
1
p
= p
1 − β2
= p
1
1 − (1 − )2
1
1 − (1 − 2 + 2 )
1
≈ √
2
≈
(137)
(138)
(139)
1
2γ 2
(140)
Our final answer is given in (136), and we can now write this in terms of γ,
−4
dP max
1
dΩ θ=θmax
= 0.16384
dP max
2γ 2
dΩ v=0
(141)
= 0.16384(24 )γ 8
(142)
≈ 2.6γ 8
(143)
There are probably more equations in this solution where it is appropriate to have an “approximately equal” sign instead of an “equal to” sign, but in all the steps above the approximations are so good that there is little error. You can use the equal signs in this case,
ultrarelativistic, without large error.
[6.] Problem 11.25 from Griffiths
A charged particle (mass m and charge q) is a distance z from a conducting surface. As
this particle moves toward (or away) from the conducting surface it emits radiation. This
process may be treated as an interaction between the particle and its image due to the
conducting surface. This pair forms an effective dipole, the moment of which changes
in time. Find the total radiated power as a function of the particle’s height z above the
conducting surface.
Griffiths Hint: The solution is,
P (z)
=
1
6m2 z 4
14
µo cq 2
4π
3
(144)
Solution
The problem statement gives the method by which to solve it. Treat this situation as an
electric dipole where the separation distance is changing in time. The general expression
for the power radiated by an electric dipole is,
µo p̈2
P ∼
=
6πc
11.60
(145)
where p is the dipole moment and the double dot represents a double time derivative.
This may be treated as a one dimensional problem along the line that connects the charged
particle to its image. This particle will accelerate as it approaches the conducting surface
due to the greater amount of induced surface charge that will exert a force on it. The dipole
moment of this particle/image pair is,
p~ = q d~
(146)
(147)
= q(2~z)
(148)
p(t) = 2qz(t)
where in the final step we keep the magnitude of the dipole moment and I indicated the
time dependence contained therein. The factor of 2 comes from the separation between the
particle and its image.
For the term in (145),
(149)
p̈ = 2qz̈
It remains to solve for z̈. This is done by examining the force on the particle due to its
image. Considering magnitudes only,
F
=
ma
=
m
d2
z
dt2
=
mz̈
(150)
This force is due to the electric influence of the image. The primed field is that due to the
image particle, which must have charge −q (review ch. 3 in the text if you do not believe
this).
F = qE 0
(151)
= q
−q
4πo r2
= −
q2
4πo (2z)2
(153)
= −
q2
16πo z 2
(154)
15
(152)
Equating (150) and (154),
mz̈ = −
z̈ = −
q2
16πo z 2
(155)
q2
16mπo z 2
(156)
Solving for the second time derivative of the dipole moment provides,
q2
p̈ = 2q −
16mπo z 2
= −
q3
8mπo z 2
(157)
(158)
Back to (145),
µo
P =
6πc
µo
=
6πc
=
q3
−
8mπo z 2
2
q6
64m2 π 2 2o z 4
(159)
µo q 6
384cm2 π 3 2o z 4
(160)
(161)
This does not look like the solution Griffiths provided. His solution, (144), has no o
dependence so it is probably best to remove this in our solution noting the relationship
µo o = 1/c2 .
1
= c4 µ2o
2o
P =
µo q 6
(c4 µ2o )
384cm2 π 3 z 4
µ3o q 6 c3
384m2 π 3 z 4
3
µo cq 2
1
=
6m2 z 4
4π
=
Our solution is equivalent to that provided by Griffiths.
16
(162)
(163)
(164)
(165)