Classical_Electrodynamics Solved problems Jackson

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PHY 5346
HW Set 1 Solutions – Kimel
2
1.3 In general we take the charge density to be of the form _ = fÝr3ÞN, where fÝr3Þ is determined by
physical constraints, such as X_d3x = Q.
a) variables: r,S,d. d3x = ddd cosSr2dr
_ = fÝr3ÞNÝr ? RÞ = fÝrÞNÝr ? RÞ = fÝRÞNÝr ? RÞ
X _d3x = fÝRÞ X r2drdIdÝr ? RÞ = 4^fÝRÞR2 = Q ¸ fÝRÞ = Q4^R2
_Ýr3Þ = Q
4^R2
NÝr ? RÞ
b) variables: r,d, z. d3x = dddzrdr
_Ýr3Þ = fÝr3ÞNÝr ? bÞ = fÝbÞNÝr ? bÞ
X _d3x = fÝbÞ X dzddrdrNÝr ? bÞ = 2^fÝbÞbL = VL ¸ fÝbÞ = V2^b
_Ýr3Þ = V2^b NÝr ? bÞ
c) variables: r,d, z. d3x = dddzrdr Choose the center of the disk at the origin, and the z-axis
perpendicular to the plane of the disk
_Ýr3Þ = fÝr3ÞNÝzÞSÝR ? rÞ = fNÝzÞSÝR ? rÞ
where SÝR ? rÞ is a step function.
X _d3x = f X NÝzÞSÝR ? rÞdddzrdr = 2^ R22 f = Q ¸ f =
Q
^R2
_Ýr3Þ = Q
^R2
NÝzÞSÝR ? rÞ
d) variables: r,S,d. d3x = dddcosSr2dr
_Ýr3Þ = fÝr3ÞNÝcosSÞSÝR ? rÞ = fÝrÞNÝcosSÞSÝR ? rÞ
X _d3x = X fÝrÞNÝcosSÞSÝR ? rÞdddcosSr2dr = X
0
R
ßfÝrÞràrdrdd = 2^N X
0
R
rdr = ^R2N = Q
where I’ve used the fact that rdrdd is an element of area and that the charge density is uniformly
distributed over area.
_Ýr3Þ = Q
^R2r
NÝcosSÞSÝR ? rÞ
PHY 5346
HW Set 1 Solutions – Kimel
3. 1.4 Gauss’s Law:
X E3 6 da3 = QenclosedP0
a) Conducting sphere: all of the charge is on the surface a = Q
4^a2
E4^r2 = 0, r < a E4^r2 = QP0 , r > a
E = 0, r < a E3 = Q
4^P 0r2
r!, r > a
b) Uniform charge density: _ = Q4
3 ^a
3 , r < a, _ = 0, r > a.
E4^r2 = Qr
3
P0a
3 ¸ E =
Qr
4^P 0a3
, r < a
E4^r2 = QP0 ¸ E =
Q
4^P 0r2
, r > a
c) _ = A rn
Q = 4^ X r2drArn = 4^Aan+3/Ýn + 3Þ ¸ _ = Ýn + 3ÞQ4^an+3 r
n
, r < a
_ = 0, r > a
E4^r2 = Ýn + 3ÞQ
4^P 0an+3
4^rn+3/Ýn + 3Þ ¸ E = Q
4^P 0r2
rn+3
an+3
, r < a
E = Q
4^P 0r2
, r > a
1. n = ?2.
E = Q4^P 0ra , r < a
E = Q
4^P 0r2
, r > a
2. n = 2.
E = Qr
3
4^P 0a5
, r < a
E = Q
4^P 0r2
, r > a
PHY 5346
HW Set 1 Solutions – Kimel
4. 1.5
dÝr3Þ =
qe?Jr 1 + Jr2
4^P 0r
42dÝr3Þ = ? _P0 , 4
2 = 1
r2
/
/r
r2 /
/r
42d = q4^P 0
1
r2
/
/r
r2 /
/r
e?Jr
r +
J
2 e
?Jr
Using 42 1r = ?4^NÝr3Þ
42d = q4^P 0
1
r2
/
/r
?Jre?Jr + e?Jrr2 /
/r
1
r ?
J2
2 r
2e?Jr
=
q
4^P 0
? J
r2
e?Jr + J
2e?Jr
r +
Je?Jr
r2
? 4^NÝr3Þ ? J
2e?Jr
r +
J3e?Jr
2
= ? 1P0 qNÝr3Þ ?
qJ3
8^ e
?Jr
_Ýr3Þ = qNÝr3Þ ? qJ
3
8^ e
?Jr
That is, the charge distribution consists of a positive point charge at the origin, plus an
exponentially decreasing negatively charged cloud.
PHY 5346
HW Set 2 Solutions – Kimel
2. 1.8 We will be using Gauss’s law XE3 6 da3 = QencP0
a) 1) Parallel plate capacitor
From Gauss’s law E = aP0 =
Q
AP0
=
d12
d ¸ Q =
AP0d12
d
W = P02 X E
2d3x = P0E
2Ad
2 =
P0
Q
AP0
2
Ad
2 =
1
2P0
Q2
A d =
1
2P0
AP0d12
d
2
A d =
1
2 P0A
d122
d
2) Spherical capacitor
From Gauss’s law, E = 14^P0
Q
r2
, a < r < b.
d12 = X
a
b
Edr = Q4^P 0 Xa
b
r?2dr = 14
Q
^P 0
Ýb ? aÞ
ba ¸ Q =
4^P 0bad12
Ýb ? aÞ
W = P02 X E
2d3x = P02
Q
4^P 0
2
X
a
b
4^ r
2dr
r4
=
P0
2
Q
4^P 0
2
?4^ ?b + aba =
1
8P0
Q2
^
Ýb ? aÞ
ba
W = 18P0
4^P0bad12
Ýb?aÞ
2
^
Ýb ? aÞ
ba = 2^P 0ba
d122
Ýb ? aÞ
3) Cylindrical conductor
From Gauss’ s law, E2^rL = VL
P0
=
Q
P0
d12 = X
a
b
Edr = Q2^P 0L Xa
b dr
r =
Q
2^P 0L
ln ba
W = P02 X E
2d3x = P02
Q
2^P 0L
2
2^L X
a
b
rdr
r2
= 14^P 0
Q2
L ln
b
a
W = 14^P 0
2^P0Ld12
ln ba
2
L ln
b
a = ^P 0L
d122
ln ba
b) w = P02 E2
1) wÝrÞ = P02 QAP0
2
= 12P0
Q2
A2
0 < r < d, = 0 otherwise.
2) wÝrÞ = P02 14^P0
Q
r2
2
= 1
32P0^2
Q2
r4
, a < r < b, = 0, otherwise
3) wÝrÞ = P02 Q2^P0Lr
2
= 18P0
Q2
^2L2r2
, = 0 otherwise.
PHY 5346
HW Set 2 Solutions – Kimel
3. 1.9 I will be using the principle of virtual work. In the figure below, FNl is the work done by an
external force. If F is along Nl (ie. is positive), then the force between the plates is attractive. This
work goes into increasing the electrostatic energy carried by the electric field and into forcing charge
into the battery holding the plates at constant potential d12.
Conservation of energy gives
FNl = NW + NQd12
or
F = /W
/l +
/Q
/l d12
From problem 1.8,
a) Charge fixed.
1) Parallel plate capacitor
W = 12 P0A
d122
d , d12 =
dQ
AP 0
¸ W = 12 P0A
dQ
AP0
2
d =
d
2P0A
Q2
/Q
/l = 0, F =
/W
/l =
Q2
2P0A
(attractive)
2) Parallel cylinder capacitor
d12 = VP0 lnÝ
d
a Þ, a = a1a2
W = 12 Qd12 ¸ F =
/W
/l =
1
2 Q
V
P0
/
/d lnÝ
d
a Þ =
1
2
VQ
P0d
(attractive)
b) Potential fixed
1) Parallel plate capacitor
Using Gauss’s law, Q = d12AP 0d ,
/
/l Q =
d12AP 0
d2
¸ F = ? 12 P0A
d122
d2
+
d122 AP 0
d2
= 12 P0A
d122
d2
= 12 P0A
Qd
P0A
2
d2
= 12P0A
Q2
2) Parallel cylinder capacitor
W = 12 Qd12, and Q =
P0Ld12
lnÝ da Þ
so
W = 12
P0Ld122
lnÝ da Þ
,
/W
/l = ?
1
2 P0L
d122
ln2 da d
/Q
/l = P0L
d12
ln2 da d
F = ? 12 P0L
d122
ln2 da d
+ P0L
d122
ln2 da d
= 12 P0L
d122
ln2 da d
PHY 5346
HW Set 2 Solutions – Kimel
4. 2.1 We will work in cylidrical coordinate, (_, z,dÞ, with the charge q located at the point
d3 = dz!, and the conducting plane is in the z = 0 plane.
Then we know from class the potential is given by
dÝx3Þ = 14^P 0
q
x3 ? d3
? q
x3 + d3
Ez = ?
q
4^P 0
/
/z
1
Ýz ? dÞ2 + _2 1/2
? 1
Ýz + dÞ2 + _2 1/2
Ez =
q
4^P 0
z ? d
Ýz ? dÞ2 + _2 3/2
? z + d
Ýz + dÞ2 + _2 3/2
a) a = P0EzÝz = 0Þ
a = P0
q
4^P 0
?d
Ý?dÞ2 + _2 3/2
? +d
Ý+dÞ2 + _2 3/2
= ? q
2^d2
1
12 + _d
2 32
Plotting ?1
12+r2
3
2
gives
-1
-0.8
-0.6
-0.4
-0.2
0
1 2 3 4 5r
b) Force of charge on plane
F3 = 14^P 0
qÝ?qÞ
Ý2dÞ2
Ý?z!Þ = 14 × 4^P 0
q2
d2
z!
c)
F
A = w =
P0
2 E
2 = a
2
2P0
= 12P0
? q
2^d2
1
12 + _d
2 32
2
= 18P0
q2
^2d4 1 + _
2
d2
3
F = 2^ q
2
8P0^2d4
X
0
K _
1 + _
2
d2
3 d_ = 2^
q2
8P0^2d4
1
4 d
2 = 14 × 4^P 0
q2
d2
d)
W = X
d
K
Fdz = q
2
4 × 4^P 0 Xd
K dz
z2
=
q2
4 × 4^P 0d
e)
W = 12
1
4^P 0 >i,j,i®j
q iq j
|x3i ? x3j | = ?
q2
2 × 4^P 0d
Notice parts d) and e) are not equal in magnitude, because in d) the image moves when q moves.
f) 1 Angstrom = 10?10m, q = e = 1. 6 × 10?19C.
W = q
2
4 × 4^P 0d
= e e4 × 4^P 0d
= e 1.6 × 10
?19
4 × 10?10
9 × 109 V = 3.6 eV
PHY 5346
HW Set 2 Solutions – Kimel
5. 2.2 The system is described by
a) Using the method of images
dÝx3Þ = 14^P 0
q
|x3 ? y3| +
q v
|x3 ? y3v |
with y v = a2y , and q v = ?q ay
b) a = ?P0 //n d|x=a = +P0 //x d|x=a
a = P0 14^P 0
/
/x
q
Ýx2 + y2 ? 2xycosLÞ1/2
+ q
v
Ýx2 + y v2 ? 2xy v cosLÞ1/2
a = ?q 14^
a 1 ? y
2
a2
Ýy2 + a2 ? 2aycosLÞ3/2
Note
q induced = a2 X adI = ?q 14^ a
22^a 1 ? y
2
a2
X
?1
1 dx
Ýy2 + a2 ? 2ayxÞ3/2
, where x = cosL
q induced = ?
q
2 aÝa
2 ? y2 Þ 2
aÝa2 ? y2 Þ
= ?q
c)
|F| = qq v
4^P 0Ýy v ? yÞ2
= 14^P 0
q2ay
Ýa2 ? y2Þ
, the force is attractive, to the right.
d) If the conductor were fixed at a different potential, or equivalently if extra charge were put on
the conductor, then the potential would be
dÝx3Þ = 14^P 0
q
|x3 ? y3| +
q v
|x3 ? y3v | + V
and obviously the electric field in the sphere and induced charge on the inside of the sphere would
remain unchanged.
PHY 5346
HW Set 3 Solutions – Kimel
2. 2.3 The system is described by
a) Given thepotenial for a line charge in the problem, we write down the solution from the figure,
dT = V4^P 0
ln R
2
Ýx3 ? x3o Þ2
? ln R
2
Ýx3 ? x3o1 Þ2
? ln R
2
Ýx3 ? x3o2 Þ2
+ ln R
2
Ýx3 ? x3o3 Þ2
Looking at the figure when y = 0, Ýx3 ? x3o Þ2 = Ýx3 ? x3o1 Þ2, Ýx3 ? x3o2 Þ2 = Ýx3 ? x3o3 Þ2, so dT |y=0 = 0
Similarly, when x = 0, Ýx3 ? x3o Þ2 = Ýx3 ? x3o2 Þ2, Ýx3 ? x3o1 Þ2 = Ýx3 ? x3o3 Þ2, so dT |x=0 = 0
On the surface dT = 0, so NdT = 0, however,
NdT =
/dT
/x t
Nx t = 0 ¸ /dT/x t
= 0 ¸ E t = 0
b) We remember
a = ?P0
/dT
/y =
?V
^
y0
Ýx ? x0 Þ2 + y02
? y0
Ýx + x0 Þ2 + y02
where I’ve applied the symmetries derived in a). Let
a/V = ?1^
y0
Ýx ? x0 Þ2 + y02
? y0
Ýx + x0 Þ2 + y02
This is an easy function to plot for various combinations of the position of the original line charge
(x0,y0Þ.
c) If we integrate over a strip of width Az, we find, where we use the integral
X
0
K 1
Ýx @ x0 Þ2 + y02
dx = 12
^ ± 2arctan x0y0
y0
AQ = X
0
K
adxAz ¸ AQ
Az
= X
0
K
adx = ?2V^ tan?1
x0
y0
and the total charge induced on the plane is ?K, as expected.
d)
Expanding
ln R
2
Ýx ? x0 Þ2 + Ýy ? y0 Þ2
? ln R
2
Ýx ? x0 Þ2 + Ýy + y0 Þ2
? ln R
2
Ýx + x0 Þ2 + Ýy ? y0 Þ2
+ ln R
2
Ýx + x0 Þ2 + Ýy + y
to lowest non-vanishing order in x0, y0 gives
16 xy
Ýx2 + y2 Þ2
y0x0
so
d ¸ dasym = 4V^P 0
xy
Ýx2 + y2 Þ2
y0x0
This is the quadrupole contribution.
PHY 5346
HW Set 3 Solutions – Kimel
3. 2.5
a)
W = X
r
K|F|dy = q2a4^P 0 Xr
K dy
y3 1 ? a2
y2
2 =
q2a
8^P 0Ýr2 ? a2 Þ
Let us compare this to disassemble the charges
? Wv = ? 18^P 0 >i®j
q iq j
|x3i?x3j | =
1
4^P 0
aq2
r
1
r 1 ? a2
r2
= q
2a
4^P 0Ýr2 ? a2 Þ
> W
The reason for this difference is that in the first expression W, the image charge is moving and
changing size, whereas in the second, whereas in the second, they don’t.
b) In this case
W = X
r
K|F|dy = q4^P 0 Xr
K Qdy
y2
? qa3 X
r
K Ý2y2 ? a2 Þ
yÝy2 ? a2 Þ2
dy
Using standard integrals, this gives
W = 14^P 0
q2a
2Ýr2 ? a2 Þ
?
q2a
2r2
?
qQ
r
On the other hand
? Wv = 14^P 0
aq2
Ýr2 ? a2 Þ
?
qÝQ + ar qÞ
r =
1
4^P 0
aq2
Ýr2 ? a2 Þ
?
q2a
r2
?
qQ
r
The first two terms are larger than those found in W for the same reason as found in a), whereas the
last term is the same, because Q is fixed on the sphere.
PHY 5346
HW Set 3 Solutions – Kimel
4. 2.6 We are considering two conducting spheres of radii ra and rb respectively. The charges on
the spheres are Qa and Qb.
a) The process is that you start with qaÝ1Þ and qbÝ1Þ at the centers of the spheres, and sphere a
then is an equipotential from charge qaÝ1Þ but not from qbÝ1Þ and vice versa. To correct this we use
the method of images for spheres as discussed in class. This gives the iterative equations given in the
text.
b) qaÝ1Þ and qbÝ1Þ are determined from the two requirements
>
j=1
K
qaÝjÞ = Qa and >
j=1
K
qbÝjÞ = Qb
As a program equation, we use a do-loop of the form
>
j=2
n
ßqaÝjÞ = ?raqbÝj ? 1ÞdbÝj ? 1Þ à
and similar equations for qbÝjÞ, xaÝjÞ, xbÝjÞ, daÝjÞ,dbÝjÞ. The potential outside the spheres is given
by
dÝx3Þ = 14^P 0 >j=1
n
qaÝjÞ
x3 ? xaÝjÞk!
+>
j=1
n
qbÝjÞ
x3 ? dbÝjÞk!
This potential is constant on the surface of the spheres by construction.
And the force between the spheres is
F = 14^P 0 >j,k
qaÝjÞqbÝkÞ
ßd ? xaÝjÞ ? xbÝkÞà2
c) Now we take the special case Qa = Qb, ra = rb = R, d = 2R. Then we find, using the iteration
equations
xaÝjÞ = xbÝjÞ = xÝjÞ
xÝ1Þ = 0, xÝ2Þ = R/2, xÝ3Þ = 2R/3, or xÝjÞ = Ýj ? 1Þj R
qaÝjÞ = qbÝjÞ = qÝjÞ
qÝjÞ = q, qÝ2Þ = ?q/2, qÝ3Þ = q/3, or qÝjÞ = Ý?1Þ
j+1
j q
So, as n ¸ K
>
j=1
K
qÝjÞ = q>
j=1
K
Ý?1Þ j+1
j = q ln2 = Q ¸ q =
Q
ln2
The force between the spheres is
F = 14^P 0
q2
R2 >j,k
Ý?1Þ j+k
jk 2 ? Ýj?1Þj ? Ý
k?1Þ
k
2 =
1
4^P 0
q2
R2 >j,k
Ý?1Þ j+kjk
Ýj + kÞ2
Evaluating the sum numerically
F = 14^P 0
q2
R2
Ý0.0739Þ = 14^P 0
Q2
R2
1
Ýln2Þ2
Ý0.0739Þ
Comparing this to the force between the charges located at the centers of the spheres
Fp = 14^P 0
Q2
R24
Comparing the two results, we see
F = 4 1
Ýln2Þ2
Ý0.0739ÞFp = 0.615Fp
On the surface of the sphere
d = 14^P 0 >j=1
K
qÝjÞ
R ? xÝjÞ =
q
4^P 0R >j=1
K
Ý?1Þ j+1
Notice 11 + 1 = >j=1
K
Ý?1Þ j+1
So d = 14^P 0
q
2R =
1
4^P 0
Q
2 ln2R =
Q
C ¸
C
4^P 0R
= 2 ln2 = 1. 386
PHY 5346
HW Set 3 Solutions – Kimel
5. 2.7 The system is described by
a) The Green’s function, which vanishes on the surface is obviously
GÝx3,x3vÞ = 1|x3 ? x3v | ?
1
|x3 ? x3Iv |
where
x3v = x vî + y vc! + z vk!, x3Iv = x vî + y vc! ? z vk!
b) There is no free charge distribution, so the potential everywhere is determined by the potential
on the surface. From Eq. (1.44)
dÝx3Þ = ? 14^ XS dÝx3
vÞ /GÝx3,x3
vÞ
/n v
da v
Note that n! v is in the ?z direction, so
/
/n v
GÝx3, x3vÞ|zv=0 = ? //z GÝx3,x3vÞ|zv=0 = ? 2zÝx ? x v Þ2 + Ýy ? y vÞ2 + z2 3/2
So
dÝx3Þ = z2^ V X0
a
X
0
2^ _ vd_ vdd v
Ýx ? x v Þ2 + Ýy ? y vÞ2 + z2
3/2
where x v=_ v cosd v,y v = _ v sind v.
c) If _ = 0, or equivalently x = y = 0,
dÝzÞ = z2^ V X0
a
X
0
2^ _ vd_ vdd v
ß_ v2 + z2 à3/2
= zV X
0
a _d_
ß_2 + z2 à3/2
dÝzÞ = zV ?
z ? Ýa2 + z2 Þ
z Ýa2 + z2 Þ
= V 1 ? z
Ýa2 + z2 Þ
d)
dÝx3Þ = z2^ V X0
a
X
0
2^ _ vd_ vdd v
Ý_3 ? _3 v Þ2 + z2
3/2
In the integration choose the x ?axis parallel to _3, then _3 6 _3 v = __ v cosd v
dÝx3Þ = z2^ V X0
a
X
0
2^ _ vd_ vdd v
ß_2 + _ v2 ? 2_3 6 _3 v + z2 à3/2
Let r2 = _2 + z2, so
dÝx3Þ = z2^
V
r3
X
0
a
X
0
2^ _ vd_ vdd v
1 + _
v2?2_36_3 v
r2
3/2
We expand the denominator up to factors of OÝ1/r4Þ, ( and change notation
d v ¸ S,_ v ¸ J, r2 ¸ 1
K2
Þ
dÝx3Þ = z2^
V
r3
X
0
a
X
0
2^ JdJdS
1 + J
2?2_36J3
r2
3/2
where the denominator in this notation is written
1
ß1 + K2ÝJ2 ? 2_JcosSÞà3/2
or, after expanding,
dÝx3Þ = z2^
V
r3
X
0
a
JdJ X
0
2^
1 ? 32 K
2J2 + 3K2_JcosS + 158 K
4J4 ? 152 K
4J3_cosS + 152 K
4_2J2 cos2S dS
Integrating over S gives
dÝx3Þ = z2^
V
r3
X
0
a
J 2^ + 154 K
4J4^ ? 3K2J2^ + 152 K
4_2J2^ dJ
Integrating over J yields
dÝx3Þ = z2^
V
r3
5
8 K
4^a6 ? 34 a
4K2^ + 158 a
4K4_2^ + ^a2
or
dÝx3Þ = Va
2
2Ý_2 + z2 Þ3/2
1 ? 34
a2
Ý_2 + z2 Þ
+ 58
a4 + 3_2a2
Ý_2 + z2 Þ2
PHY 5346
HW Set 4 Solutions – Kimel
1. 2.8 The system is pictured below
a) Using the known potential for a line charge, the two line charges above give the potential
dÝr3Þ = 12^P 0
V ln r
v
r = V, a constant. Let us define V
v = 4^P 0V
Then the above equation can be written
r v
r
2
= e V
v
V or r v2 = r2e V
v
V
Writing r v2 = r3 ? R3
2
, the above can be written
r3 + z! R
Ýe
Vv
V ? 1Þ
2
= R
2e
Vv
V
Ýe
Vv
V ? 1Þ2
The equation is that of a circle whose center is at ?z! R
e
Vv
V ?1
, and whose radius is a = Re
Vv
2V
Ýe
Vv
V ?1Þ
b) The geometry of the system is shown in the fugure.
Note that
d = R + d1 + d2
with
d1 = R
e
Vav
V ?1
, d2 = R
e
?Vb
v
V ?1
and
a = Re
Vav
2V
Ýe
Vav
V ? 1Þ
, b = Re
?Vb
v
2V
Ýe
?Vb
v
V ? 1Þ
Forming
d2 ? a2 ? b2 = R + R
e
Vav
V ?1
+ R
e
?Vb
v
V ?1
2
? Re
Vav
2V
Ýe
Vav
V ? 1Þ
2
? Re
?Vb
v
2V
Ýe
?Vb
v
V ? 1Þ
2
or
d2 ? a2 ? b2 =
R2 e
Vav ?Vb
v
V + 1
Ýe
Vav
V ? 1ÞÝe
?Vb
v
V ? 1Þ
Thus we can write
d2 ? a2 ? b2
2ab =
e
Vav ?Vb
v
V + 1
2e
Vav
2V e
?Vb
v
2V
= e
Vav ?Vbv
2V + e
? Vav ?Vb
v
2V
2 = cosh
Vav ? Vbv
2V
or
Va ? Vb
V =
1
2^P 0
cosh?1 d
2 ? a2 ? b2
2ab
Capacitance/unit length = CL =
Q/L
Va ? Vb
= VVa ? Vb
= 2^P 0
cosh?1 d2?a2?b22ab
c) Suppose a2 << d2, and b2 << d2, and a v = ab , then
C
L =
2^P 0
cosh?1 d2?a2?b2
2a v2
= 2^P 0
cosh?1 d
2 1?Ýa2+b2Þ/d2
2a v2
cosh?1 d
2Ý1 ? Ýa2 + b2Þ/d2 Þ
2a v2
= 2^P 0LC
d2Ý1 ? Ýa2 + b2Þ/d2 Þ
2a v2
= e
2^P0L
C
2 + negligible terms if
2^P 0L
C >> 1
or
ln d
2Ý1 ? Ýa2 + b2Þ/d2 Þ
a v2
= 2^P 0LC
or
C
L =
2^P 0
ln d
2 1?Ýa2+b2Þ/d2
a v2
Let us defind J2 = Ýa2 + b2Þ/d2, then
C
L =
2^P 0
ln d
2 1?J2
a v2
= 2^ P0
ln d2
a v2
+ 2^ P0
ln2 d2
a v2
J2 + OÝJ4 Þ
The first term of this result agree with problem 1.7, and the second term gives the appropriate
correction asked for.
d) In this case, we must take the opposite sign for d2 ? a2 ? b2, since a2 + b2 > d2. Thus
C
L =
2^P 0
cosh?1 a2+b2?d2
2a v2
If we use the identiy, lnÝx + x2 ? 1 Þ = cosh?1ÝxÞ, G.&R., p. 50., then for d = 0
C
L =
2^P 0
ln a2+b22ab +
a2?b2
2ab
= 2^P 0
ln ab
in agreement with problem 1.6.
PHY 5346
HW Set 4 Solutions – Kimel
2. 2.9 The system is pictured below
a) We have treated this probem in class. We found the charge density induced was
a = 3P0E0 cosS
We also note the radial force/unit area outward from the surface is a2/2P0. Thus the force on the
right hand hemisphere is, using x = cosS
Fz = 12P0 X a
2z! 6 da3 = 12P0 Ý
3P0E0 Þ22^R2 X
0
1
x3dx = 12P0 Ý
3P0E0 Þ22^R2/4 = 94 ^P 0E0
2R2
An equal force acting in the opposite direction would be required to keep the hemispheres from
sparating.
b) Now the charge density is
a = 3P0E0 cosS + Q4^R2 = 3P0E0x +
Q
4^R2
= 3P0E0 x + Q12^P 0E0R2
Fz = 12P0 X a
2z! 6 da3 = 12P0 Ý
3P0E0 Þ22^R2 X
0
1
x x +
Q
12^P 0E0R2
2
dx
Thus
Fz = 94 ^P 0E0
2R2 + 12 QE0 +
1
32P0^R2
Q2
An equal force acting in the opposite direction would be required to keep the himispheres from
separating.
PHY 5346
HW Set 4 Solutions – Kimel
3. 2.10 As done in class we simulate the electric field E0 by two charges at K
a = 3P0E0 cosS
a) This charge distribution simulates the given system for cosS > 0.We have treated this probem
in class. The potential is given by
dÝx3Þ = ?E0 1 ? a
3
r3
r cosS
Using
a = ?P0 //n
d|surface
We have the charge density on the plate to be
aplate = ?P0 //z
d|z=0 = P0E0 1 ? a3
_3
For purposes of plotting, consider aplateP0E0 = 1 ?
1
x3
0
0.2
0.4
0.6
0.8
1
2 4 6 8 10x
aboss = 3P0E0 cosS =
For plotting, we use aboss3P0E0 = cosS
0
0.2
0.4
0.6
0.8
1
0.2 0.4 0.6 0.8 1 1.2 1.4
b)
q = 3P0E02^a2 X
0
1
xdx = 3^P 0E0a2
c) Now we have
dÝr3Þ = 14^P 0
q
r3 ? d3
+ q
v
r3 ? d3v
? q
r3 + d3
? q
v
r3 + d3v
where q v = ?q ad , d
v = a2d .
a = ?P0 //r
d|r=a = ?q4^
Ýd2 ? a2 Þ
a a3 ? d3
3 ?
Ýd2 ? a2 Þ
a a3 + d3
3
q ind = 2^a2 ?q4^ X0
1 Ýd2 ? a2 Þ
a a3 ? d3
3 ?
Ýd2 ? a2 Þ
a a3 + d3
3 dx
q ind =
?qa2Ýd2 ? a2 Þ
2a
1
da
1
d ? a ?
1
a2 + d2
+ 1d + a ?
1
a2 + d2
q ind = ?12 q
d2 ? a2
d
2d
d2 ? a2
? 2
a2 + d2
= ?q 1 ? Ýd
2 ? a2 Þ
d a2 + d2
PHY 5346
HW Set 4 Solutions – Kimel
4. 2.11 The system is pictured in the following figure:
a) The potential for a line charge is (see problem 2.3)
dÝr3Þ = V2^P 0
ln rKr
Thus for this system
dÝr3Þ = b2^P 0
ln rK
r3 ? R3
+ b
v
2^P 0
ln rK
r3 ? R3 v
To determine b v and R v, we need two conditions:
I) As r ¸ K, we want d ¸ 0, so b v = ?b.
II) dÝr3 = b� Þ = dÝr3 = ?b� Þ or
ln b ? R
v
R ? b = ln
b + R v
b + R
or
b ? R v
R ? b =
b + R v
b + R
This is an equation for R v with the solution
R v = b
2
R
The same condition we found for a sphere.
b)
dÝr3Þ = b4^P 0
ln
r2 + b4
R2
? 2r b2R cosd
r2 + R2 ? 2rRcosd
as r ¸ K
dÝr3Þ = b4^P 0
ln 1 ? 2b
2 cosd/rR
1 ? 2Rcosd/r =
b
4^P 0
ln 1 ? 2
rR Ýb
2 ? R2 Þcosd
Using
lnÝ1 + xÞ = x ? 12 x
2 + 13 x
3 + OÝx4 Þ
dÝr3Þ = ? b2^P 0
1
rR Ýb
2 ? R2 Þcosd
c)
a = ?P0 //r
d|r=b = ? b4^ //r ln
r2 + b4
R2
? 2r b2R cosd
r2 + R2 ? 2rRcosd
r=b
a = b2^b
1 ? y2
y2 + 1 ? 2ycosd
where y = R/b. Plotting a/ b2^b =
1?y2
y2+1?2ycosd
, for y = 2,4, gives
gÝyÞ = 1 ? y
2
y2 + 1 ? 2ycosd
gÝ2Þ,gÝ4Þ = ? 35?4 cosd , ?
15
17?8 cosd
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5 1 1.5 2 2.5 3
d) If the line charges are a distance d apart, then the electric field at b from b v is, using Gauss’s law
E = b
v
2^P 0d
The force on b is bLE, ie,
F = bb
vL
2^P 0d
= ? b
2L
2^P 0d
, and the force is attractive.
PHY 5346
HW Set 5 Solutions – Kimel
1. 2.13 The system is pictured in the following figure:
a) Notice from the figure, ®Ý_,?dÞ = ®Ý_,dÞ; thus from Eq. (2.71) in the text,
®Ý_,dÞ = a0 +>
n=1
K
an_n cosÝndÞ
X
?^/2
3^/2
®Ýb,dÞ = 2^a0 = ^V1 + ^V2 ¸ a0 = V1 + V22
Using
X
?^/2
3^/2
cosmd cosnddd = Nnm^
Applying this to ®, only odd terms m contribute in the sum and
am =
2ÝV1 ? V2 Þ
^mbm Ý?1Þ
m?1
2
Thus
®Ý_,dÞ = V1 + V22 +
2ÝV1 ? V2 Þ
^ Im >
m odd
im_me imd
mbm
Using
2 >
m odd
xm
m = ln
Ý1 + xÞ
Ý1 ? xÞ
and
Im lnÝA + iBÞ = tan?1ÝB/AÞ
we get
®Ý_,dÞ = V1 + V22 +
ÝV1 ? V2 Þ
^ tan
?1 2
_
b cosd
1 ? _
2
b2
as desired.
b)
a = ?P0 //_
®Ý_,dÞ|_=b
a = ?P0
ÝV1 ? V2 Þ
^
/
/_
tan?1
2 _b cosd
1 ? _
2
b2 _=b
a = ?2P0 V1 ? V2^ bÝcosdÞ
b2 + _2
b4 ? 2b2_2 + _4 + 4_2b2 cos2d
|_=b = ?P0 V1 ? V2^bcosd
:
PHY 5346
HW Set 5 Solutions – Kimel
2. 2.23 The system is pictured in the following figure:
a) As suggested in the text and in class, we will superpose solutions of the form (2.56) for the two
sides with VÝx,y, zÞ = V.
1) First consider the side VÝx,y, aÞ = z :
®1Ýx,y, zÞ = >
n,m=1
K
Anm sinÝJnxÞ sinÝKmyÞ sinhÝLnmzÞ
with Jn = n^a , Km = m^a , Lnm = ^a n2 + m2 . Projecting out Anm using the orthogonality of the
sine functions,
Anm = 16V
sinhÝLnmaÞnm^2
where both n, and m are odd. (Later we will use n = 2p + 1, m = 2q + 1Þ
2) In order to express ®2Ýx, y, zÞ in a form like the above, we make the coordinate transformation
x v = y, y v = x, z v = ?z + a
So
®2Ýx, y, zÞ = ®1Ýx v,y v, z vÞ = ®1Ýy,x,= z + aÞ
®Ýx, y, zÞ = ®1Ýx,y, zÞ + ®2Ýx,y, zÞ
b)
®Ý a2 ,
a
2 ,
a
2 Þ =
16 6 V
^2
>
p,q=0
K
Ý?1Þp+q
Ý2p + 1ÞÝ2q + 1Þcosh Lnm a2
where I have used the identity
sinhÝLnmaÞ = 2sinhÝ
Lnma
2 ÞcoshÝ
Lnma
2 Þ
Let fÝp,qÞ ¯ >p,q=0
K Ý?1Þp+q
Ý2p+1ÞÝ2q+1Þcosh Ý2p+1Þ2+Ý2q+1Þ2 ^2
(p,q) fÝp,qÞ Error Sum
0,0 0.213484 4.4% .214384
1,0 ?0.004641 2.13% 0.20974
0,1 ?0.004641 0.013% 0.20510
1,1 0.0002835 0.015% 0.20539
The first three terms give an accuracy of 3 significant figures.
®Ý a2 ,
a
2 ,
a
2 Þ =
16 6 0.20539
^2
V = 0.33296V
®avÝ a2 ,
a
2 ,
a
2 Þ =
2
6 V = 0.333. . . .V
c)
aÝx,y,aÞ = ?P0 //z ®|z=a
aÝx,y,aÞ = ? 16P0
^2
V
= ? 16P0
^2
V >
n,m odd
K
sinÝJnxÞ sinÝKmyÞ
ÝcoshÝLnmaÞ ? 1
sinhÝLnmaÞ
aÝx, y,aÞ = ? 16P0
^2
V >
n,m odd
K
sinÝJnxÞ sinÝKmyÞ tanh
Lnma
2
PHY 5346
HW Set 5 Solutions – Kimel
3. 3.1 The system is pictured in the following figure:
X
0
1 PlÝxÞdx
The problem is symmetric around the z axis so
dÝr,SÞ = >
l
ÝA lr l + B lr?l?1 ÞPlÝcosSÞ
The A l and B l are determined by the conditions
1)
X
?1
1
dÝa, xÞPlÝxÞdx = 22l + 1 ÝA la
l + B la?l?1 Þ
2)
X
?1
1
dÝb, xÞPlÝxÞdx = 22l + 1 ÝA lb
l + B lb?l?1 Þ
Solving these two equations gives
A l = 2l + 12Ýa2l+1 ? b2l+1 Þ
a l+1 X
?1
1
dÝa,xÞPlÝxÞdx ? b l+1X
?1
1
dÝb,xÞPlÝxÞdx
B l = a l+1 2l + 12 X?1
1
dÝa,xÞPlÝxÞdx ? A la2l+1
Using
X
?1
1
dÝa,xÞPlÝxÞdx = V X
0
1
PlÝxÞdx
X
?1
1
dÝb,xÞPlÝxÞdx = V X
?1
0
PlÝxÞdx = VÝ?1Þ l X
0
1
PlÝxÞdx
So
A l = 2l + 12Ýa2l+1 ? b2l+1 Þ
Vßa l+1 ? b l+1Ý?1Þ l à X
0
1
PlÝxÞdx
B l = a l+1 2l + 12 V X0
1
PlÝxÞdx ? A la2l+1
Note that
X
0
1
PlÝxÞdx = 12 X?1
1
PlÝxÞdx
for l even. For even l > 0, X
0
1 PlÝxÞdx = 0. Thus we have
X
0
1
P0ÝxÞdx = 1; X
0
1
P1ÝxÞdx = 12 , X0
1
P3ÝxÞdx = ? 18
and
A0 = V2 , A1 =
3
4Ýa3 ? b3 Þ
VÝa2 + b2 Þ, A3 = ? 716Ýa7 ? b7 Þ
VÝa4 + b4 Þ
B0 = 12 Va ?
1
2 Va = 0
B1 = 34 a
2V ? 3
4Ýa3 ? b3 Þ
VÝa2 + b2 Þa3 = 34 Va
2b2 b + a
?a3 + b3
B3 = ? 716 a
4V ? a7 ? 7
16Ýa7 ? b7 Þ
VÝa4 + b4 Þ = ? 716 Va
4b4 b
3 + a3
?a7 + b7
As b ¸ K, only the B l terms (and A0Þ survive. Thus using the general expression for X0
1 PlÝxÞdx
given by (3.26)
dÝr,SÞ = V2 P0ÝxÞ +
3
2
a3
r2
P1ÝxÞ ? 78
a4
r4
P3ÝxÞ +. . . . .
Let’s now solve the problem neglecting the outer sphere (since b ¸ KÞ using the Green’s function
result
(2.19) this integral give, for cosS = 1
dÝr,SÞ = V2 Ý1 ? _
2 Þ 1
Ý1 ? _Þ ?
1
Ý1 + _2 Þ
with _ = a/r. Expanding the above,
dÝr,SÞ = V2
a
r +
3
2
a2
r2
? 78
a4
r4
+. . . . . .
Comparing with our previous solution with x = 1, we see the Green’s function solution differs by
having a B0 term and by not having an A0 term. All the other higher power terms agree in the series.
This difference is due to having a potential at K in the original problem.
PHY 5346
HW Set 5 Solutions – Kimel
4. 3.4 Slice the sphere equally by n planes slicing through the z axis, subtending angle Ad about
this axis with the surface of each slice of the pie alternating as ±V.
dÝr,S,dÞ = >
l,m
A lmr lYlmÝS,dÞ
so
A lm = 1
a l
X dIÝYlmÝS,dÞÞDdÝa,S,dÞ
Symmetries:
A l?m = Ý?1ÞmÝA lm ÞD
dÝr,S,d + 2AdÞ = dÝr,S,dÞ
where
Ad = 2^2n
Thus
m = ±n, and integral multiples thereof
dÝ?r3Þ = ?dÝr3Þ, n = 1
dÝ?r3Þ = dÝr3Þ, n > 1
Since
PYlmÝS,dÞ = Ý?1Þ lYlmÝS,dÞ
Then
l is odd for n = 1; l is even for n > 1
Thus we only have contributions of l ³ n. Using
A lm = 1
a l
X dIÝYlmÝS,dÞÞDdÝa,S,dÞ
The integral over d can be done trivially, since the integrand is just e?imd leaving the desired answer
in terms of an integral over cosS.
n = 1 case: I am going to keep only the lowest novanishing terms, involving A11 and A1?1.
d = rÝA11Y11 + A1?1Y1?1 Þ = rÝA11Y11 + ÝA11Y11 Þ
D Þ = 2r ReÝA11Y11 Þ
Y11 = ? 38^ Ý1 ? x
2Þ1/2e id
A11 = ? 1a
3
8^ V X?1
1
Ý1 ? x2Þ1/2dx X
0
^
e?iddd ? X
^
2^
e?iddd
A11 = 2i^a
3
8^ V
d = 2r Re 2i^a
3
8^ V ?
3
8^ sinSe
id = 3r2a VsinS sind
From the figure
we see
sinS sind = cosS v
So
d = 3r2a VcosS
v = V 32
r
a P1ÝcosS
vÞ +. . . . . .
The other terms, for l = 2,3, can be obtained in the same way in agreement with the result of
(3.36)
PHY 5346
HW Set 6 Solutions – Kimel
2. 3.10 This problem is described by
a) From the class notes
®Ý_, z,dÞ = >
nX
ÝAnX sinXd + BnX cosXdÞIX
n^_
L sin
n^z
L
where
AnX = 2^L
1
IX n^bL
X
0
L X
0
2^
VÝd, zÞ sinÝ n^aL Þ sinÝXdÞdddz, X > 0
BnX = 2^L
1
IX n^bL
X
0
L X
0
2^
VÝd, zÞ sinÝ n^aL ÞcosÝXdÞdddz, X ® 0
BnX = 1^L
1
IX n^bL
X
0
L X
0
2^
VÝd, zÞ sinÝ n^aL Þdddz, X = 0
Noting
X
? ^2
^
2
sinXddd ? X
^
2
3^
2
sinXddd = 0
we conclude AnX = 0. Similarly, noting
X
? ^2
^
2
cosXddd ? X
^
2
3^
2
cosXddd = 4Ý?1Þ
m
2m + 1 , m = 0,1,2, . . . .
where I’ve recognized that X must be odd, ie, X = 2m + 1. Also
X
0
L
sinÝ n^zL Þdz =
2
Ý2l + 1Þ^ , l = 0,1,2, . . . . .
where again I’ve recognized that n must be odd, ie, n = 2l + 1. Thus
BnX =
16Ý?1ÞmV
^2I2m+1Ý n^bL ÞÝ2l + 1ÞÝ2m + 1Þ
b) Now z = L/2, L >> b, L >> _. Then from the class notes
I2m+1Ý
Ý2l + 1Þ^_
L Þ C
1
@Ý2m + 2Þ
Ý2l + 1Þ^_
2L
m+1
Also
sin Ý2l + 1Þ^L = Ý?1Þ
l
so
®Ý_, z,dÞ = >
l,m
16Ý?1Þ l+mV
^2Ý2l + 1ÞÝ2m + 1Þ
_
b
2m+1
cosßÝ2m + 1Þdà
Using
tan?1ÝxÞ = >
l=0
K
x2l+1
1l + 1 Ý?1Þ
l
^
4 = tan
?1Ý1Þ = >
l=0
K
Ý?1Þ l
2l + 1
so
®Ý_, z,dÞ = 4V^ >
m
Ý?1Þm
2m + 1
_
b
2m+1
cosßÝ2m + 1Þdà
Remembering from problem 2.13 that
>
m
Ý?1Þm
2m + 1
_
b
2m+1
cosßÝ2m + 1Þdà = 12 tan
?1 2
_
b cosd
1 ? _
2
b2
we find
®Ý_, z,dÞ = 2V^ tan
?1 2
_
b cosd
1 ? _
2
b2
which is the answer for problem 2.13.
PHY 5346
HW Set 6 Solutions – Kimel
3. 4.1
q lm = X r lYlmDÝS,dÞ_Ýx3Þd3x = >
i
q ir llYlmDÝS i,d iÞ
Using
YlmDÝS,dÞ =
Ý2l + 1ÞÝl ? mÞ!
4^Ýl + mÞ! Pl
mÝxÞe??md = N lmPlmÝxÞe??md
From the figure we get
q lm = a lN lmPlmÝ0ÞqßÝ1 ? Ý?1Þm ÞÝ1 ? im Þà = 0, for m even, so m = 2n + 1, n = 0,1,2, . . .
q lm = 2qa lN lmPlmÝ0ÞßÝ1 ? Ý?1ÞniÞà
b) The figure for this system is
Since the sum of the charges equals zero, l ³ 1.
q lm = qa lßYlmDÝx = 1,dÞ + YlmDÝx = ?1,dÞà = qa lN lmßPlmÝ1Þ + PlmÝ?1Þà
From the Rodrigues formula for PlmÝxÞ, we see Plmݱ1Þ = 0, for m ® 0. So
q lm = qa lN l0ß1 + Ý?1Þ l àPlÝ1Þ
Thus l is even, but l ® 0
q lm = 2qa lN l0
c) Using the fact that N l0 = 2l+14^ and Yl0 = 2l+14^ Pl
®Ýx3Þ = >
l=2
K
Ý2qa l Þ PlÝxÞ
r l+1
u
2qa2
r3
P2Ýx = 0 on x-y plane)
®Ýx3Þ = ? qa
2
r3
Let us plot ®Ýx3Þ/Ý?q/aÞ, ie, 1
Ý ra Þ
3 =
1
x3
0
1
2
3
4
5
6
7
8
0.5 1 1.5 2 2.5 3x
The exact answer on the x-y plane is
®Ýx3Þ = ?qa
2
x ?
2
x 1 + 1
x2
=
?q
a
1
x
3
? 34
1
x
5
+ 58
1
x
7
? 3564
1
x
9
+. . . .
So let’s plot 1
x3
,
2
x ?
2
x 1+ 1
x2
0
0.05
0.1
0.15
0.2
0.25
0.3
1 2 3 4 5x
where the smaller is the exact answer.
PHY 5346
HW Set 7 Solutions – Kimel
2. 4.2 We want to show that we can obtain the potential and potential energy of an elementary
diplole:
®Ýx3Þ = 14^P 0
p3 6 x3
r3
W = ?p3 6 E3Ý0Þ
from the general formulas
®Ýx3Þ = 14^P 0 X
_Ýx3vÞd3x v
|x3 ? x3v |
W = X _Ýx3Þ®Ýx3Þd3x
using the effective charge density
_eff = ?p3 6 43NÝx3Þ
where I’ve chosen the origin to be at x30.
®Ýx3Þ = 14^P 0 X
?p3 6 43 vNÝx3vÞd3x v
|x3 ? x3v | = ?
1
4^P 0
p3 6 X 43 1|x3 ? x3v | NÝx3
vÞd3x v
®Ýx3Þ = 14^P 0
p3 6 x3
r3
Similarly,
W = X _Ýx3Þ®Ýx3Þd3x = ?X p3 6 43NÝx3Þ®Ýx3Þd3x = p3 6 X NÝx3Þ43®Ýx3Þd3x = ?p3 6 E3Ý0Þ
PHY 5346
HW Set 7 Solutions – Kimel
3. 4.7 a) Since _ does not depend on d, we can write it in terms of spherical harmonics with
m = 0. First note
Y20 = 54^
3
2 1 ? sin
2S ? 12
or
sin2S = ? 23
4^
5 Y2
0 + 4^ 23 Y0
0
Thus only the m = 0, l = 0, 2 multipoles contribute.
q00 =
2 4^
3 X0
K
r2 164^ r
2e?r dr = 2 4^3
3
8^ =
1
2 ^
q20 = ? 23
4^
5 X0
K
r4 164^ r
2e?r dr = ? 23
4^
5
45
4^ = ?3
5
^
®Ýx3Þ = 14^P 0
4^q00
Y00
r + 4^q20
Y20
5r3
= 14^P 0
4^ q00 P0r +
4^
5 q20
P2
r3
®Ýx3Þ = 14^P 0
P0
r ? 6
P2
r3
b)
®Ýx3Þ = 14^P 0 X
_Ýx3vÞd3x v
|x3 ? x3v |
Using
1
|x3 ? x3v | = 4^>lm
1
Ý2l + 1Þr>
r<
r>
l
YlmDÝS v,d vÞYlmÝS,dÞ
, we see only the l = 0, 2 and m = 0 terms of the expansion contribute in the potential. Next take
r v > r.
®Ýx3Þ = 14^P 0
4^>
lm
1
Ý2l + 1Þ r
lYlmÝS,dÞ X YlmDÝS v,d vÞr v2dI v
_Ýx3vÞ
r vl+1
dr v
®Ýx3Þ = 14^P 0
4^ Y00 4^ 23 X0
K 1
64^ r
2e?r rdr + Y2
0
5 r
2 ? 23
4^
5 X0
K 1
64^ r
2e?r 1r dr
®Ýx3Þ = 14^P 0
4^ Y00 4^ 23
3
32^ +
Y20
5 r
2 ? 23
4^
5
1
64^
®Ýx3Þ = 14^P 0
4^ P0 23
3
32^ +
P2
5 r
2 ? 23
1
64^ =
1
4^P 0
P0
4 ?r2P2
120
PHY 5346
HW Set 7 Solutions – Kimel
4. 4.9 a) The system is described by
Since there is azimuthal symmetry, choosing the z-axis through q,
®out =
1
4^P 0 >l
B lr?l?1Pl +
q
|x3 ? x3v |
®out =
1
4^P 0 >l
B lr?l?1Pl +
q
r> >
l
r<
r>
l
Pl
® in =
1
4^P 0 >l
A lr lPl + qr> >
l
r<
r>
l
Pl
Boundary conditions: At the surface, r v = d = r>, r = a = r<.
1) ®out = ® in |r=a, or
B l = A la2l+1
2) P //r ® in = //r ®out |r=a, or letting k = PP0
k >
l
lAla l?1Pl + qd l
a l?1
d l
Pl = >
l
?Ýl + 1ÞB la?l?2Pl + qd l
a l?1
d l
Pl
= >
l
?Ýl + 1ÞA la l?1Pl + qd l
a l?1
d l
Pl
or
A l =
aÝ1 ? kÞl
ßÝ1 + kÞl + 1àd l+1
B l =
aÝ1 ? kÞla2l+1
ßÝ1 + kÞl + 1àd l+1
Remember that Pl = 4^2+1 Yl
0
, and substitute the above coefficients into the expansion to get the
answer requested by the problem.
PHY 5346
HW Set 6 Solutions – Kimel
3. 4.10 The system is described by
a) Since there is azimutal symmetry,
®Ýr,SÞ = >
l
ÝA lr l + B lr?l?1 ÞPlÝcosSÞ
Also
D3 = PÝx3ÞE3 = ?PÝx3Þ43®Ýr,SÞ
Dr = ?PÝx3Þ>
l
ÝlAlr l?1 ? Ýl + 1ÞB lr?l?2 ÞPlÝcosSÞ
between the spheres,
XDrdIr2 = Q, and is independent of r.
Thus
A l = 0, B l = 0, l ® 0 ¸ Dr = PÝx
3ÞB0
r2
XDrdIr2 = 2^B0 P0 X
?1
0
dcosS + P X
0
1
dcosS = 2^B0ÝP0 + PÞ = Q
B0 =
Q
2^P 0Ý1 + PP0 Þ
E3 = Q
2^P 0Ý1 + PP0 Þr2
r!
b)
XDrdA = DrA = a fA ¸ a f = Dr = PÝx3ÞEr
a f =
PQ
2^P 0Ý1 + PP0 Þr2
, cosS ³ 0
a f =
Q
2^Ý1 + PP0 Þr2
, cosS < 0
c)
X _poldV = apolA = X ?43 6 P3dV = ?PA ¸ apol = ?P = ?P0eeE
apol = ?ÝPÝx3Þ/P0 ? 1Þ Q2^Ý1 + PP0 Þr2
Notice
apol + a f = a tot =
Q
2^Ý1 + PP0 Þr2
= P0E, as expected.
PHY 5346
Homework Set 10 Solutions – Kimel
1. 5.1 The system is described by
We want to show
dm = ?
W0I
4^ I
Suppose the observation point is moved by a displacement Nx3, or equivlently that the loop is
displaced by ?Nx3.
If we are to have B3 = ?43dm, then
Ndm = ?Nx3 6 B3
Using the law of Biot and Savart,
Ndm = ?
W0I
4^ [ Nx3 6
dl3v × r3
r3
= ? W0I4^ [ r3 6
Nx3 × dl3v
r3
= ? W0I4^ [ r! 6
Nx3 × dl3v
r2
Ndm = ?
W0I
4^ [
r! 6 NÝdAÞ
r2
= ? W0I4^ NI
Or,
dm = ?
W0I
4^ I
PHY 5346
Homework Set 10 Solutions – Kimel
2. 5.2 a) The system is described by
First consider a point at the axis of the solenoid at point z0. Using the results of problem 5.1,
ddm = W04^ NIdzI
From the figure,
I = X r! 6 dA3
r2
= X dAcosS
r2
= 2^z X
0
R _d_
Ý_2 + z2 Þ3/2
= 2^ ? z
ÝR2 + z2 Þ
+ 1
dm =
W0
2 NI Xz0
K
z ? 1
ÝR2 + z2 Þ
+ 1z dz =
W0
2 NI ?z0 + ÝR
2 + z0
2 Þ
Br = ?
W0
2 NI
/
/z0
?z0 + ÝR2 + z02 Þ =
W0
2 NI
?z0 + ÝR2 + z02 Þ
ÝR2 + z02 Þ
In the limit z0 ¸ 0
Br =
W0
2 NI
By symmetry, thej loops to the left of z0 give the same contribution, so
B = B l + Br = W0NI
H = NI
By symmetry, B3 is directed along the z axis, so
Ndm = ?N_3 6 B3 = 0
if N_3 is directed ó to the z axis. Thus for a given z, dm is independent of _, and consequently
H = NI
everywhere within the solenoid.
If you are on the outside of the solenoid at position z0, by symmetry the magnetic field must be in
the z direction. Thus using the above argument, dm must not depend on _. Set us take _ far away
from the axis of the solenoid, so that we can replace the loops by elementary dipoles m3 directed along
the z axis. Thus for any point z0 we will have a contributions
dm 9
m3 6 r31
r1
3 +
m3 6 r32
r2
3
where m3 6 r31 = ?m3 6 r32 and r1 = r2. Thus
H = 0
PHY 5346
Homework Set 10 Solutions – Kimel
3. 5.8 Using the same arguments that lead to Eq. (5.35), we can write
Ad =
W0
4^ X
d3x v cosd vJdÝr v,S vÞ
|x3 ? x3v |
Choose x3 in the x ? z plane. Then we use the expansion
1
|x3 ? x3v | = >l,m
4^
2l + 1
r<
l
r>
l+1 Yl
mDÝS v,d vÞYlmÝS, 0Þ
The cosd v factor leads to only an m = 1 contribution in the expansion. Using
YlmÝS, 0Þ = 2l + 14^
Ýl ? mÞ!
Ýl + mÞ! Pl
mÝcosSÞ
and Ýl?1Þ!
Ýl+1Þ! =
1
lÝl+1Þ , we have on the inside
Ad =
W0
4^ >
l
1
lÝl + 1Þ r
lPl1ÝcosSÞ X d3x v
Pl1ÝcosS vÞJdÝr v,S vÞ
r vl+1
which can be written
Ad = ?
W0
4^ >
l
m lr
lPl1ÝcosSÞ
with
m l = ?
1
lÝl + 1Þ X d
3x v
Pl1ÝcosS vÞJdÝr v,S vÞ
r vl+1
A similar expression can be written on the outside by redefining r< and r>.
PHY 5346
Homework Set 9 Solutions – Kimel
1. 5.10
a) From Eq. (5.35)
AdÝr,SÞ =
W0
4^
I
a X
r v2dr vdI v sinS v cosj vNÝcosS vÞNÝr v ? aÞ
|x3 ? x3v |
Using the expansion of 1/|x3 ? x3v | given by Eq. (3.149),
1
|x3 ? x3v | =
4
^ X0
K
dkcosßkÝz ? z v Þà 12 I0Ýk_<ÞK0Ýk_>Þ +>
m=1
K
cosßmÝj ? j v ÞàImÝk_<ÞKmÝk_>Þ
We orient the coordinate system so j = 0, and because of the cosj v factor, m = 1. Thus,
AdÝr,SÞ =
W0
4^
I
a
4^
^ X0
K
dk X r v2dr vdcosS v sinS vNÝcosS vÞNÝr v ? aÞcosÝkzÞI1Ýk_<ÞK1Ýk_>Þ
AdÝr,SÞ =
W0
^ aI X0
K
dkcosÝkzÞI1Ýk_<ÞK1Ýk_>Þ
where _<Ý_>Þ is the smaller (larger) of a and _.
b) From problem 3.16 b),
1
|x3 ? x3v | = >
m=?K
K
X
0
K
dke imÝd?dvÞJmÝk_ÞJmÝk_ vÞe?k|z|
Note z v = 0, and d = 0, so
Ad =
W0Ia
2 X0
K
dke?k|z|J1Ýk_ÞJ1ÝkaÞ
PHY 5346
Homework Set 10 Solutions – Kimel
4. 5.16 a) The system is shown in the figure
I shall use the magnetic potential approach and will call inside the sphere region 1 and outside the
sphere region 2.
d1 = d loop +>
l
A lr lPl
d2 = d loop +>
l
B lr?l?1Pl
where H3 = ?43d, and we have the boundary conditions,
H1|| = H2|| ¸ d1Ýr = bÞ = d2Ýr = bÞ
W0 //r
d1Ýr = bÞ = W //r d2Ýr = bÞ
We are given that b >> a, so
d loop = 14^
m cosS
r2
with m = ^a2I. (From the form of d loop, only the l = 1 term contributes.) The boundary
conditions give
A1b1 = B1b?1?1
? 2W0m
4^b3
+ W0A1 = ?
2Wm
4^b3
? 2WB1b?3
So
A1 = ? 24^
m
b3
ÝW ? W0 Þ
Ý2W + W0 Þ
On the inside, at the center of the loop
H3 = ?43d loop ? 43A1r cosS
From Eq. (5.40), we are given ?43d loop at the center of the loop, which is directed in the z direction.
Hz = 1W0 Ý?BS Þ ? A1
If W >> W0
A1 ¸ ? 14^
m
b3
and from (5.40), at r = 0
Hz = I2a +
1
4^
m
b3
= I2a +
I
4
a2
b3
= I2a 1 +
a3
2b3
PHY 5346
Homework Set 9 Solutions – Kimel
2. 5.18
a) From the results of Problem 5.17, we can replace the problem stated by the system
where ID is equidistant from the interface and is equal to ID = Wr?1Wr+1 I. The radius of each current
loop is a. Now from Eq. (5.7)
F3Ýon IÞ = I X dl3× BÝr3Þ
dl3× B3 = dl3× B3 r + dl3× B3 S = dlBr ?S! + dlBSr!
By symmetry, only the z ? component survives, so, from the figure
dl3× B3 6 z! = dlBr a
4d2 + a2
+ dlBS 2d
4d2 + a2
So
Fz = 2^aI
4d2 + a2
ßaBr + 2dBS à
with Br and BS given by Eqs. (5.48) and (5.49) and cosS = 2d
4d2+a2
, r = 4d2 + a2 , and I ¸ ID.
c) To determine the limiting term, simply let r ¸ 2d and take the lowest non-vanishing term in the
expansion of the magnetic flux density.
Fz = ^aId ßaBr + 2dBS à
Fz = ^aId a
W0IDa
4d
a
Ý2dÞ2
+ 2d ? W0I
Da2
4
1
Ý2dÞ3
? a2d
Fz ¸ ?
3^W0
32
a4I × ID
d4
The minus sign shows the force is attractive if I and ID are in the same direction. This same result
can be gotten more directly, using
Fz = 4zÝmBz Þ
with m = ^a2I, and (from Eq. (5.64))
Bz =
W0
4^
2mD
z3
with mD = ^a2ID, and z = 2d
Fz =
W0
4^ 2^a
2ID^a2I ? 3
Ý2dÞ4
= ? 3^W032
a4I × ID
d4
with agrees with out previous result.
PHY 5346
Homework Set 9 Solutions – Kimel
3. 5.19 The system is described by
The effective volume magnetic charge density is zero, since M3 is constant within the cylinder. The
effective surface charge density (n!6 M3 from Eq. (5.99)) is M0, on the top surface and ?M0 on the
bottom surface. From the bottom surface the potential is (for z > 0Þ
®b = 14^ Ý?M0 Þ2^ X0
a _d_
Ý_2 + z2 Þ1/2
= ? M02 Ýa
2 + z2 Þ ? z
By symmetry, the potential from the top surface is (on the inside)
® t =
M0
2 a
2 + ÝL ? zÞ2 ? ÝL ? zÞ
The total magnetic potential is
® = ®b + ® t = ?
M0
2 Ýa
2 + z2 Þ ? z + M02 a
2 + ÝL ? zÞ2 ? ÝL ? zÞ
So, on the inside of the cylinder,
Hz = ? //z ?
M0
2 Ýa
2 + z2 Þ ? z + M02 a
2 + ÝL ? zÞ2 ? ÝL ? zÞ
Hz = ? M02 2 ?
z
Ýa2 + z2 Þ
? L ? z
a2 + ÝL ? zÞ2
while above the cylinder,
Hz = ? M02 ?
z
Ýa2 + z2 Þ
+ z ? L
a2 + ÝL ? zÞ2
with a similar expression below the cylinder.
B3 = W0 H3 + M3
Thus inside the cylinder,
Bz = W0 ? M02 2 ?
z
Ýa2 + z2 Þ
? L ? z
a2 + ÝL ? zÞ2
+ M0
Bz =
W0M0
2
z
Ýa2 + z2 Þ
+ L ? z
a2 + ÝL ? zÞ2
while above the cylinder,
Bz =
W0M0
2
z
Ýa2 + z2 Þ
? z ? L
a2 + ÝL ? zÞ2
First we plot Bz in units of a for L = 5a
gÝzÞ =
1
2
z
1+z2
+ 5?z
1+Ý5?zÞ2
if z < 5
1
2
z
1+z2
? z?5
1+Ý5?zÞ2
if 5 < z
gÝzÞ
0
0.2
0.4
0.6
0.8
1
2 4 6 8 10z
And similarly, Hz in units of a for L = 5a.
fÝzÞ =
? 12 2 ?
z
1+z2
? 5?z
1+Ý5?zÞ2
if z < 5
? 12 ?
z
1+z2
+ z?5
1+Ý5?zÞ2
if 5 < z
fÝzÞ
-0.4
-0.2
0
0.2
0.4
2 4 6 8 10z
PHY 5346
Homework Set 11 Solutions – Kimel
1. 5.26 The system is described by
Since the wires are nonpermeable, W = W0. The system is made of parts with cylindrical
symmetry, so we can determine B using Ampere’s law.
43 × B3 = W0J3, or X B3 6 dl3 = W0 X J3 6 da3
On the outside of each wire,
X B3 6 dl3 = B2^_ = W0I ¸ Bout = W0I2^_
On the inside of each wire
X B3 6 dl3 = B2^_ = W0I _
2
R2
, B in =
W0I
2^
_
R2
with R = a,b
From the right-hand rule, the B from each wire is in the d! direction. From the above figure, using
the general expression for the vector potential, we see A3 is in the ±z! direction. Since 43 × A3 = B3,
Bz = ? //_ Az ¸ Az = ?X Bzd_
Thus
Az =
? W0I2^ ln
_
R + C = ?
W0I
4^ ln
_2
R2
+ 1 on the outside
? W0I4^
_2
R2
, on the inside
where I’ve determined C = 1/2, from the requirement that Az be continuous at _ = R. Let l be the
length of the wire. Then we know the total potential energy is given by
W = 12 X J
3 6 A3d3x = l2 XßJaAdaa + JbAdab à
Consider the second term l2 X JbAdab. The system is pictured as
From the figure
_3a = d3 + _3b, _a2 = d2 + _b2 ? 2d_b cosd
so, since Jb = I^b2
l
2 X JbAdab =
l
2
I
^b2 XßAoutÝ_aÞ + A inÝ_bÞ à_bd_bdd
= l2
I
^b2
W0I
4^ X ln
_a2
a2
+ 1 ? _b
2
b2
_bd_bdd
p l2
I
^b2
W0I
4^ 2^ X0
b
ln d
2
a2
+ 1 ? _b
2
b2
_bd_b
= l2
I
^b2
W0I
4^ 2^
1
4 b
2 1 + 2 ln d
2
a2
= l2
W0
4^
1
2 + 2 ln
d
a I
2
The first term l2 X JaAdaa is equal to
l
2 X JaAdaa =
l
2
W0
4^
1
2 + 2 ln
d
b I
2
Thus
W = l2
W0
4^ 1 + 2 ln
d2
ab I
2 = l2
L
l I
2
or
L
l =
W0
4^ 1 + 2 ln
d2
ab
PHY 5346
Homework Set 11 Solutions – Kimel
2. 5.27 The system is described by
Using Ampere’s law in integral form
X B3 6 dl3 = W0Ienclosed
we get
B = W0I2^
_
b2
, _ < b
B = W0I2^
1
_ , b < _ < a
B = 0, _ > a
Now the energy in the magnetic field is given by ( l is the length of the wires)
W = 12 X B
3 6 H3d3x = 12W0 X B
2d3x
= 12W0
W0I
2^
2
l 2^ X
0
b _
b2
2
_d_ + 2^ X
b
a 1
_
2
_d_
= 12W0
W0I
2^
2
l^ 12 + 2 ln
a
b =
l
2
L
l I
2
¸ Ll =
W0
4^
1
2 + 2 ln
a
b
If the inner wire is hollow, B = 0, _ < b, so
L
l =
W0
2^ ln
a
b
PHY 5346
Homework Set 11 Solutions – Kimel
3. 5.29 The system is described by
This problem is very much like 5.26, except the wires are superconducting. We know from section
5.13 that the magnetic field within a superconductor is zero. We will be using
W = 12 X J3 6 A
3d3x = l2 XßJaAdaa + JbAdab à
Using the same arguments as applied in problem 5.26,
Az =
? WI2^ ln
_
R + C = ?
WI
4^ ln
_2
R2
+ 0 on the outside
0, on the inside
Thus if we consider the second term l2 X JbAdab,
l
2 X JbAdab =
l
2
I
^b2 XßAoutÝ_aÞ + A inÝ_bÞ à_bd_bdd
p l2
I
^b2
WI
4^ 2^ X0
b
ln d
2
a2
_bd_b = l2
W
4^ 2 ln
d
a I
2
The first term l2 X JaAdaa is equal to
l
2 X JaAdaa =
l
2
W
4^ 2 ln
d
b I
2
Thus
W = l2
W
4^ 2 ln
d2
ab I
2 = l2
L
l I
2
so
L
l =
W
4^ 2 ln
d2
ab
Now using the methods of problem 1.6, assuming the left wire has charge Q, and the right wire charge
?Q, we find
d12 = X
b
d?a
Edr =
Q
l
2^P Xb
d?a 1
r +
1
d ? r dr p
Q
l
2^P ln
d2
ab
C
l =
Q
l
d12 =
2^P
ln d2
ab
Thus
L
l ×
C
l =
W
4^ 2 ln
d2
ab ×
2^P
ln d2
ab
= WP
PHY 5346
Homework Set 12 Solutions – Kimel
1. 6.11
a) Consider the momentum contained in the volume
Ap = AtcAAg
F = ApAt = cAAg
P = F
AA = cg
where I’m using the time averaged quantities. In class we found
cg = 1c S =
1
2 P0 |E0 |
2 = u
Thus
P = u
b) We are given
S = 1.4 × 103W/m2
But we know P = u = Sc . From Newton’s second law
a = Fm =
F/A
m/A =
S/c
m/A =
1.4 × 103W/m2
3 × 108m/s × 1 × 10?3kg/m2
= 4.66 × 10?3m/s2
In the solar wind, there are approximately 10 × 104 protons/Ým2 ? secÞ, with average velocity
v = 4 × 105m/s.
Ap
AtA = P = 10 × 10
4 × 4 × 105 × 1.67 × 10?27 = 6. 68 × 10?17N/m2
:
a = Fm =
F/A
m/A =
P
m/A =
6. 68 × 10?17N/m2
1 × 10?3kg/m2
= 6. 68 × 10?14 m/s2
PHY 5346
Homework Set 12 Solutions – Kimel
2. 7.1 I shall apply Eqs.(26), (27), and (28)
s0 + s1
2 = a1
s0 ? s1
2 = a2 N l = N2 ? N1 = sin
?1 s3
2a1a2
s0 + s3
2 = a+
s0 ? s3
2 = a? Nc = N? ? N+ = sin
?1 s2
2a+a?
a) s0 = 3, s1 = ?1, s2 = 2, s3 = ?2
a1 = 1, a2 = 2
N l = sin?1 ?2
2 2
= ? 14 ^ rad
a+ =
1
2
, a? =
5
2
Nc = sin?1 2
2 1
2
5
2
= 1. 1071 rad
b) s0 = 25, s1 = 0, s2 = 24, s3 = 7
a1 =
25
2 , a2 =
25
2
N l = sin?1 s32a1a2
= sin?1 7
2 252
25
2
= 0.283 79 rad
a+ =
32
2 = 4, a? =
s0 ? s3
2 = 3
Nc = N? ? N+ = sin?1 242Ý4 × 3Þ =
1
2 ^ rad
To plot the two cases ReEx ¯ X = cosx, ReEy ¯ Y = rcoxÝx ? N lÞ, where r = a2/a1 and x = gt.
Case a) cosx, 2 cosÝx + ^4 Þ
-1
-0.5
0.5
1
-1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1
Case b) cosx, cosÝx ? 0.28379Þ
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1
PHY 5346
Homework Set 13 Solutions – Kimel
1. 7.2
a) The figure describes the muliple internal reflections which interfere to give the overall
reflection and refraction:
For the ij interface I shall use the notation
r ij =
E0v
E0
=
2n i
n i + n j
R ij =
E0vv
E0
=
n i ? n j
n i + n j
Thus from the figure
E0vv = E0R12 + r12E0R23r21e id + r12E0R23R21R23r21e i2d +. . . .
E0vv = E0R12 + r12E0R23r21e id>
n=0
K
ÝR21R23e id Þn
E0vv = E0 R12 +
r12r21R23
Ýe?id ? R21R23 Þ
Similarly
E0v = E0r12r23 + E0r12R23R21r23e id +. . . .
E0v = E0
r12r23
1 ? R21R23e id
where the phase shift for the internally reflected wave is given by
d = 2^Ý2dÞV2 =
gn2Ý2dÞ
c
Now for a plane wave
S i = 12v i
|E0i |2
Thus
R = S
vv
S =
|E0vv |2
|E0 |2
T = v1v3
S v
S =
n3
n1
S v
S
From the above
R = R122 +
2r12r21R23R12Ýcosd ? R21R23 Þ + ÝR12r21R23 Þ2
1 + ÝR21R23 Þ2 ? 2R21R23 cosd
T = n3n1
Ýr12r23 Þ2
1 + ÝR21R23 Þ2 ? 2R21R23 cosd
Since these two equations are simple functions of d, which is linearly proportional to the
frequency,they are simple functions of frequency which you should plot.
b) Since in part a) we used the convention that the incident wave is from the left, I will rephrase
this question so that n1 is are, n2 is the coating, and n3 is glass. In this case, we will have
n1 < n2 < n3, and R21R23 < 0. Thus for T to be a maximum, from the above equation cosd = ?1, or
d = ^.
d = 2^Ý2dÞV2 = ^ ¸ d =
V2
4
where V2 is the wavelength in the medium = V1n2
PHY 5346
Homework Set 13 Solutions – Kimel
2. 7.4 We have a nonpermeable conducting material, so W = W0, and we have J = aE, where a is
the conductivity. The following figure describes the system:
The two boundary conditions that we must satisfy for plane waves are
E0 + E0vv ? E0v = 0
kÝE0 ? E0vv Þ ? k vE0v = 0
Or
E0vv
E0
= k ? k
v
k + k v
We must take into account the fact that J3 = aE3. Adding in this term in Maxwell’s equations for a
plane wave, we get
k = gc
k v2 = PWg2 1 + i agP
Thus we can write
k v = PWgÝJ + iKÞ
with
J =
1 + Ý agP Þ2 + 1
2
1/2
K =
1 + Ý agP Þ2 ? 1
2
1/2
Thus
E0vv
E0
=
1 ? PW0 cJ ? i PW0 cK
1 + PW0 cJ + i PW0 cK
1) For a very poor conductor a is very small, so keeping only first order in a
J =
1 + Ý agP Þ2 + 1
2
1/2
u 1
K =
1 + Ý agP Þ2 ? 1
2
1/2
u a2gP
2) For the case of a very good conductor, agP >> 1, so
J u a2gP =
2
W0gN2
2gP =
1
gN W0P
K u a2gP =
1
gN W0P
where I have used (5.165) to relate the conductivity to the skin depth.
a = 2
W0gN2
E0vv
E0
=
1 ? cgN ? i
c
gN
1 + cgN + i
c
gN
=
N ? cg ? i cg
N + cg + i cg
u ?1 + gc
2
1 + i N = ?1 +
g
c Ý1 ? iÞN
R = E0
vv
E0
2
= Ý?1 + Ng/cÞ2 + gNc
2
u 1 ? 2Ng/c
PHY 5347
Homework Set 1 Solutions – Kimel
1. 8.3
x
y
z
a)
Ý4t2 + L2 Þf = 0, f = EzÝTMÞ or f = HzÝTEÞ
As in class, we will use cylindrical coordinates, and assume
fÝ_,dÞ = RÝ_ÞQÝdÞ
We get the two equations
/2
/d2
QÝdÞ = ?m2QÝdÞ with solns QÝdÞ = e±imd, m = 0,1,2, . . .
d2
dx2
RÝxÞ + 1x
dRÝxÞ
dx + 1 ?
m2
x2
RÝxÞ Bessel eqn.
with regular solutions JmÝxÞ, and singular solution (which we reject as nonphysical) NmÝxÞ. Here
x = L_.
Solutions:
TM: BC: JmÝxmnÞ = 0, and
EzÝ_,dÞ = E0JmÝLmn_Þe±imd, m = 0,1,2, . . . . ; n = 1,2,3, . . . ; Lmn = xmn/R
Lowest cutoff frequencies:
gmn =
Lmn
PW
= xmnR PW
Using the results of Jackson, p. 114,
x0n = 2.405,5.52,8.654, . . .
x1n = 3.832,7.016,10.173, . . . .
x2n = 5.136,8.417,11.620, . . . .
TE: BC: Jmv Ýxmnv Þ = 0, and
EzÝ_,dÞ = E0JmÝLmnv _Þe±imd, m = 0,1,2, . . . . ; n = 1,2,3, . . . ; Lmnv = xmnv /R
Lowest cutoff frequencies:
gmn =
Lmnv
PW
= xmn
v
R PW
Using the results of Jackson, p. 370,
x0n
v = 3.832,7.016,10.173, . . .
x1n
v = 1.841,5.331,8.536, . . . .
x2n
v = 3.054,6.706,9.970, . . . .
From the above we see the lowest cutoff frequency is the TE mode
g11v = 1.841K, with K = 1/ R PW
The next four lowest cutoff frequencies are:
g01 = 2.405K = 1.31g11v
g21v = 3.054K = 1.66g11v
g01v = 3.832K = 2.08g11v
g11 = 3.832K = 2.08g11v
b) From Eq. (8.63) in the text
KV 9 g
1 ? gV
2
g2
1/2
YV + RV
gV
g
2
For TM modes, RV = 0, and for TE mode, YV + RV is of order unity. So for comparison purposes,
I’ll take
K11ÝTEÞ = f1ÝxÞ = x
1 ? 1.8412
x2
1/2
1 + 1.841
2
x2
K01ÝTMÞ = f2ÝxÞ = x
1 ? 2.4052
x2
1/2
where I’ve expressed the functions in terms of x = g/K.
1 + 1.8412
x2
x
1? 1.8412
x2
1/2
0
2
4
6
8
10
12
14
2 4 6 8 10x
x
1? 2.4052
x2
1/2
0
2
4
6
8
10
12
14
3 4 5 6 7 8 9 10x
PHY 5347
Homework Set 1 Solutions – Kimel
1. 8.4
a) TM:
Ý4t2 + L2 Þf = 0; f|B = 0; Ez = fÝx,yÞe±ikz?igt; Bz = 0
Since we have a node along y = x, then we just take the antisymmetrized version for the square
waveguide, developed in class, ie,
fÝx, yÞ = E0 sinÝ m^xa Þ sinÝ
n^y
a Þ ? sinÝ
n^x
a Þ sinÝ
m^y
a Þ
Again
Lmn2 = c^
2
a2
Ým2 + n2Þ, m,n = 1,2,3. . . . , but m ® n
TE:
Ý4t2 + L2 Þf = 0;
/f
/n
|B = 0; Hz = fÝx,yÞe±ikz?igt; Ez = 0
Now the BC require /f
/n
|B = 0, but using a 450 rotation of coordinates, we see
/
/n
= 1
2
? /
/x
+ /
/y
Thus the combination
fÝx,yÞ = H0 cosÝ m^xa ÞcosÝ
n^y
a Þ + cosÝ
n^x
a ÞcosÝ
m^y
a Þ
satisfies the above BC on the diagonal, as you can see by direct substitution.
Lmn2 = c^
2
a2
Ým2 + n2Þ, m,n = 0,1,2,3. . . . , but m ® n = 0
b) The lowest cutoff freq. are: TM: g12 or g21. TE: g01 or g10. From Eq. (8.63) in the text
K12ÝTMÞ 9 g1 ? g122 /g2
1/2
K01ÝTEÞ 9 g1 ? g122 /g2
1/2
1 + g01
2
g2
For the square wave guide, we don’t have the antisymmetrization, but the formulas for the cutoff
frequencies are the same without the present restrictions on m and n. So for the square guide, the cut
off frequencies are
TM: g11
TE: g01 (as before)
PHY 5347
Homework Set 2 Solutions – Kimel
1. 8.5 a)
R
L
For the TM modes, we saw in class the resonance frequencies are
TM:
gmnp =
1
PW
xmn
2
R2
+
p2^2
L2
p = 0,1,2, . . . .
m = 0,1,2, . . . .
n = 1,2,3, . . . .
TE:
gmnp
v = 1
PW
xmn
v2
R2
+
p2^2
L2
p = 1,2,3, . . . .
m = 0,1,2, . . . .
n = 1,2,3, . . . .
Thus
gmnp
1
PW
=
xmn
2
R2
+
p2^2
L2
gmnp
v
1
PW
=
xmn
v2
R2
+
p2^2
L2
The lowest four frequencies are (in these units)
g010 = 2.405
g110 = 3.832
g111
v = 1.8412 + ^2 RL
2
g211
v = 3.0542 + ^2 RL
2
2.405,3. 832, 1. 8412 + ^2x2 , 3. 0542 + ^2x2
0
2
4
6
8
10
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2x
where x = R/L.
The answer is ”No.” g111v and g010 cross when
1.8412 + ^2x2 = 2.405
or x = 0. 49258. For frequencies smaller than this cross over frequency, g111v is lowest, whereas for
larger frequencies, g010 is lowest.
PHY 5347
Homework Set 3 Solutions – Kimel
3. 9.1 a)
_Ýx3, tÞ = qNÝzÞNÝy ? sing0tÞNÝx ? dcosgtÞ
To illustrate the equivalence of the two methods, I’ll consider the lowest two moments.
n = 0 : QÝtÞ = X _Ýx3, tÞd3x = q = ReÝqe?i06gtÞ
n = 1 : p3ÝtÞ = X _Ýx3, tÞx3d3x = qdÝîcosgt + � singtÞ = ReßqdÝî + i� Þe?i16gt à
So we identify p3 = qdÝî + i� Þ as the quantity to be used in Jackson’s formulas.
Arbitrary n: The n’th multipoles will contribute with maximum frequencies of gn = ng.
b) The proof that we can write
_Ýx3, tÞ = _0Ýx3Þ +>
n=1
K
Reß2_nÝx3Þe?ingt à
with
_nÝx3Þ = 1b X0
b
_Ýx3, tÞe ingtdt
was presented in lecture and will not be repeated here.
c) We have already calculated the n = 0,1 moments by the method of part a). Now we
compute these moments by the method of part b).
n = 0 :
_0Ýx3Þ = 1b X0
b
ßqNÝzÞNÝy ? sing0tÞNÝx ? dcosgtÞàdt
Q = X _0Ýx3Þd3x = qb X0
b
dt X d3xßNÝzÞNÝy ? sing0tÞNÝx ? dcosgtÞà = q
n = 1 :
_1Ýx3Þ = 1b X0
b
ßqNÝzÞNÝy ? sing0tÞNÝx ? dcosgtÞàe igtdt
p3Ýx3Þ = X d3xx3Ý2_1Ýx3ÞÞ = 2qb X0
b
dt X d3xx3ßNÝzÞNÝy ? sing0tÞNÝx ? dcosgtÞà
= 2qdb X0
b
dte igtÝîcosgtÞ + � singtÞ = qdÝî + i� Þ
as before.
PHY 5347
Homework Set 3 Solutions – Kimel
3. 9.2 First consider a rotating charge which is at an angle J at time t = 0.
Compared to the lecture notes for this problem, where we assumed J = 0, we should let
gt ¸ gt + J. Thus using the result developed in class, we can write for this problem
QJÝtÞ = Re 32 qd
2
1 i 0
i ?1 0
0 0 0
e?i2ae?i2gt
From the figure
Q totÝtÞ = QJ1ÝtÞ + QJ2ÝtÞ + QJ3ÝtÞ + QJ4ÝtÞ
= Re 32 qd
2 ?e?i
^
2 + e?i
3^
2 ? e?i
5^
2 + e?i
7^
2
1 i 0
i ?1 0
0 0 0
e?i2gt
Q tot = 32 qd
2Ý4iÞ
1 i 0
i ?1 0
0 0 0
Thus from the class notes
dP
dI =
c2Z0k6
1152^2
qd2 32
2
16Ý1 ? cos4SÞ = 132^2
c2Z0k6q2d4Ý1 ? cos4SÞ
P = c
2Z0k6360^ qd
2 3
2
2
16 = 110^ c
2Z0k6q2d4
And, of course, the frequency of the radiation is 2g.
PHY 5347
Homework Set 3 Solutions – Kimel
3. 9.3
Since the problem has azimuthal symmetry, we can expand VÝr3, tÞ (in the radiation zone) in terms
of Legendre polynomials:
VÝr3, tÞ = >
l
b lÝtÞr?l?1PlÝcosSÞ
Using the orthogonality of the Legendre polynomials, the leading term of the expansion in the
radiation zone will be the l = 1 term.
b1ÝtÞ = 32 R
2 X
?1
1
xVÝr3, tÞdx = 32 VR
2 cosgt
So,
VÝr3, tÞ = 32 VR
2 cosgt /r2 = p3 6 r!
r2
cosgt = Re p3 6 r!
r2
e?igt
with p3 = 32 VR
2z!, which should be used in the radiation formulas developed in lecture.
dP
dI =
c2Z0k4
32^2 |p3|
2
sin2S
P = c
2Z0k4
32^2
8^
3 =
1
12^ c
2Z0k4 |p3|2
with p3 = 32 VR
2z!.
PHY 5347
Homework Set 4 Solutions – Kimel
1. 9.11 We are working with small sources in the radiation zone.
From the notes and Eqs (9.170) and (9.172),
Q lm = X r lYlmD_d3x
M lm = ? 1l + 1 X r
lYlmD43 6 r3 × J3 d3x
a) Electric Dipole Radiation:
Q1m = X rY1mD_d3x
= X rNÝxÞNÝyÞß?qNÝz ? acosg0tÞ ? qNÝz + acosg0tÞ + 2qNÝzÞàY1mDdxdydz
= ?qa cosg0tßY1mDÝ0,dÞ + Y1mDÝ^,dÞà = ?qaNm0 cosg0tßY10Ý0Þ + Y10Ý^Þà = 0
b) Magnetic Dipole Radiation: Since the particles move in an orbit with no area,
r3 × J3 = 0 ¸ M1m = 0
c) Electric Quadrupole Radiation:
Q2m = X r2Y2mD_d3x
= ?qNm0ßa2 cos2g0tY20ÝS = 0Þ + a2 cos2g0tY20ÝS = ^Þà
= ?qNm0a2Y20Ý0ÞÝcos2g0t + 1Þ
where I have used cos2g0t = Ýcos2g0t + 1Þ/2. Thus the Fourier Series decomposition of this
moment yields terms with frequency 0, and 2g0. The first term does not contribute to radiation, and
the second can be written
Q20ÝtÞ = Reß?2qa2Y20Ý0Þe?2ig0t à
so Q20 = ?2qa2Y20Ý0Þ is the quantity that is used in the radiation formulas of Jackson. Using Eq.
(9.151)
dP
dI Ý2,0Þ =
Z0
2k2 |aÝ2,0Þ|
2 X3 20
2
and from Eq. (9.169)
aÝ2, 0Þ = ck
4
iÝ5 × 3Þ
3
2 Q20 =
ck4
iÝ5 × 3Þ
3
2 Ý?2qa
2 Þ 54^
where I have used Y20Ý0Þ = 54^ . Thus
|aÝ2,0Þ|2 = 130^ c2k8q2a4
dP
dI Ý2, 0Þ =
Z0
2k2
1
30^ c
2k8q2a4 158^ sin
2Scos2S = 1
32^2
Z0k6c2q2a4 sin2Scos2S
PÝ2,0Þ = 2^
32^2
Z0k6c2q2a4 X
?1
1
Ý1 ? x2 Þx2dx = 2^32^2
Z0k6c2q2a4 × 415
PÝ2,0Þ = 160^ Z0k
6c2q2a4
PHY 5347
Homework Set 5 Solutions – Kimel
1. The system is described by
and is azimuthally symmetric
RÝSÞ = R0ß1 + KÝtÞP2ÝcosSÞà; KÝtÞ = K0 cosgt; kR << 1
Q = X _r2drdddcosS = 2^ X
?1
1
dcosS_ X
0
RÝSÞ
r2dr
= 2^3 _ X?1
1
R03Ý1 + 3KP1P2ÞdcosS + OÝK2Þ = 4^3 _R0
3 ¸ _ = 3
4^R03
Q
where I’ve used the fact that 1 = P0. Since the system is azimuthally symmetric, Q lm = Nm0Q l0.
Q lm = 2^_Nm0 X
?1
1
dxYl0 X
0
RÝSÞ
r l+2dr = 2^_Nm0l + 3 X?1
1
dxR0l+3ß1 + Ýl + 3ÞKP2àYl0
Using Yl0 = 2l+14^ Pl and 1 = P0,
Q lm = 2^_Nm0l + 3
2l + 1
4^ R0
l+3 2N l0 + Ýl + 3ÞK 22l + 1 N l2
Notice that the l = 0 term is time independent and thus does not contribute to the radiation.
Next consider the l = 2 term.
Q20ÝtÞ = 25 ^ _ 5 R0
5K = _ = 3
4^R03
Q 25 ^ _ 5 R0
5K = 3
20^
R02QKÝtÞ
Q20ÝtÞ = Re 3
20^
R02QK0e?igt
Q20 = 3
20^
R02QK0
dPÝ2, 0Þ
dI =
Z0
2k2 |aÝ2,0Þ|
2 X3 20
2
aEÝ2,0Þ = ck
4
iÝ5 × 3Þ
3
2 Q20
X3 20
2
= 158^ sin
2Scos2S
dPÝ2,0Þ
dI =
Z0
2k2
ck4
Ý5 × 3Þ
3
2 Q20
2
× 158^ sin
2Scos2S
= 1160
Z0
k2
|ck4Q20 |2
^ sin
2Scos2S = 1160
Z0
k2
Ýck4 Þ2 3
20^
R02QK0
2
^ sin
2Scos2S
= 9
3200^2
Z0k6c2R04Q2K02 sin2Scos2S
P = 9
3200^2
Z0k6c2R04Q2K02 × 2^ X
?1
1
Ý1 ? x2 Þx2dx = 93200^2
Z0k6c2R04Q2K02 × 2^ × 415
P = 32000^ Z0k
6c2R04Q2K02
PHY 5347
Homework Set 5 Solutions – Kimel
2. The system is described by
IÝtÞ = I0 cosgt = ReßI0e?igt à
J3ÝtÞ = 1a IÝtÞNÝr ? aÞNÝcosSÞd!
where I determined the normalization constant 1a by the condition X J3 6 da3 = I
J3ÝtÞ = Re I0a NÝr ? aÞNÝcosSÞd!e
?igt ¸ J3 = I0a NÝr ? aÞNÝcosSÞd!
We use the general expression for H3 and E3 in the radiation zone given by Eq.(9.149). Since this
system has no net charge density and there is no intrisic magnetization, the the expansion coefficients in
these equations are given by
aEÝl,mÞ = k
2
i lÝl + 1Þ
X YlmDik r3 6 J3 jlÝkrÞd3x
aMÝl,mÞ = k
2
i lÝl + 1Þ
X YlmD43 6 r3 × J3 jlÝkrÞd3x
a) r3 6 J3 = 0 in the first equation, so there is no electric multipole radiation. In spherical coordinates
r3 × J3 = ?aJS!
Using the formulas for 43 6 A3 in spherical coordinates given in the back of the book,
43 6 r3 × J3 = ? 1
sinS
/
/S Ý
J sinSJÞ = ? cosS
sinS J ?
/
/S
J
The first term does not contribute, because cosS = 0, while the second term can be written, using
the chain rule,
43 6 r3 × J3 = sinS /
/cosS
J
The problem has azimuthal symmetry, so m = 0. Realizing derivatives of N ?functions are defined
by integration by parts,
aMÝl, mÞ = Nm0k
2
i lÝl + 1Þ
X Yl0D sinS //cosS J jlÝkrÞd
3x = ik
2
lÝl + 1Þ
X //cosS ÝsinSYl
0D Þ Jj lÝkrÞd3x
aMÝl, 0Þ = i2^k
2
lÝl + 1Þ
I0
a a
2jlÝkaÞ //cosS ÝsinSYl
0 Þ|cosS=0
aMÝl, 0Þ = i2^k
2
lÝl + 1Þ
I0
a a
2jlÝkaÞÝ1 ? x2 Þ1/2 ddx Yl
0ÝxÞ|x=0
Since Yl0ÝxÞ is either an even or odd polynomial in x, then only odd l contribute to aMÝl, 0Þ. This
determines the expansion coefficients, and thus H3 and E3 in the radiation zone are known through
Eq.(9.149). The power distribution is given by Eq. (9.151)
b) From our previous answers, we see aEÝl,mÞ = 0, and that the lowest magnetic multipole
contribution is aMÝ1,0Þ.
aMÝ1, 0Þ = i2^k
2
2
I0aj1ÝkaÞÝ1 ? x2 Þ1/2 ddx Y1
0ÝxÞ|x=0
Using
j1ÝkaÞ ¸ ka3 ;
d
dx Y1
0ÝxÞ|x=0 = 34^
aMÝ1, 0Þ = i2^k3I0a2 124^ =
ik3
3 2 M l0
M l0 =
i2^k3I0a2 124^
ik3
3 2
= 34^ I0^a
2
Note that you would get the same answer, if you used Eq. (9.172) directly.
From Eq. (9.151)
dP
dI =
Z0
2k2
2^k3Fa2 124^
2
3
8^ sin
2S = 1
32^2
Z0k4ÝI0^a2 Þ2 sin2S
If we compare this result with the one that we get for an elementary magnetic dipole, which is
given by Eq. (9.23)
with the substitution p3 ¸ m3 /c,
dP
dI =
1
32^2
Z0k4 |m3 |2 sin2S
Thus we may identify
|m3 | = I0^a2
as would be expected.
PHY 5347
Homework Set 6 Solutions – Kimel
1. 10.1
a) Let us first simplify the expression we want to get for the cross section. Using n! 0 = z!,
da
dI ÝP! 0, n! 0,n!Þ = k
4a6 54 ? |P! 0 6 n! |2 ? 14 |n! 6 Ýz! × P! 0 Þ|
2 ? z! 6 n!
Orienting the system as
and using
P! 0 = J0x! + K0� , with |J0 |2 + |K0 |2 = 1
n! = cosSz! + sinSx!
then
da
dI ÝP! 0, n! 0,n!Þ = k
4a6 54 ? |J0 |2 sin2S ? 14 |K0 |
2 sin2S ? cosS
Using the result for the perfectly conducting sphere Eq. (10.14)
da
dI Ýn! ,P!,P! 0,n! 0Þ = k
4a6 P!D 6 P! 0 ? 12 Ýz! × P! 0 Þ 6 Ýn! × P!
D Þ
2
Using P!ó = �
da
dI Ýn! ,P!ó,P! 0, n! 0Þ = k
4a6 K0 ? 12 K0 cosS
2
= k4a6 |K0 |2 1 ? 12 cosS
2
Similarly P! || = x! v
da
dI Ýn! ,P! ||,P! 0, n! 0Þ = k
4a6 J0 cosS ? 12 J0
2
= k4a6 |J0 |2 cosS ? 12
2
By definition
da
dI ÝP! 0,n! 0,n!Þ =
da
dI Ýn! ,P!ó,P! 0,n! 0Þ +
da
dI Ýn! ,P! ||,P! 0,n! 0Þ
= k4a6 |K0 |2 1 ? 12 cosS
2
+ |J0 |2 cosS ? 12
2
which simplifies to
da
dI ÝP! 0, n! 0,n!Þ = k
4a6 54 ? |J0 |
2 sin2S ? 14 |K0 |
2 sin2S ? cosS
using |J0 |2 + |K0 |2 = 1, and cos2S = 1 ? sin2S.
b) If P! 0 is linearly polarized making an angle d with respect to the x axis, then
P! 0 = J0x! + K0� = cosdx! + sind� , so J0 = cosd, K0 = sind
Then from part a)
da
dI ÝP! 0, n! 0,n!Þ = k
4a6 54 ? |J0 |2 sin2S ? 14 |K0 |
2 sin2S ? cosS
= k4a6 54 ? cos
2d sin2S ? 14 sin
2d sin2S ? cosS
Using cos2d = cos2d ? sin2d, this expression simplifies to
da
dI ÝP! 0,n! 0, n!Þ = k
4a6 58 Ý1 + cos
2SÞ ? 38 sin
2 cos2d ? cosS
as desired.PHY 5347
Homework Set 7 Solutions – Kimel
1. 11.3 Let us just focus on the 0, 1 component transformation, since the 2, 3 components
remain unchanged, if we take the relative velocities between Lorentz frames to be along the x -
direction. We want to relate a single Lorentz transformation to two sequential transformations as
described by
Thus we require
A = A2A1
where A is a Lorentz transformation. Rewritten explicitly, the above equation reads
L ?KL
?KL L
=
L2 ?K2L2
?K2L2 L2
L1 ?K1L1
?K1L1 L1
=
L2L1 + K2L2K1L1 ?L2K1L1 ? K2L2L1
?L2K1L1 ? K2L2L1 L2L1 + K2L2K1L1
So
L = L2L1 + K2L2K1L1
KL = L2K1L1 + K2L2L1
K = KLL =
L2K1L1 + K2L2L1
L2L1 + K2L2K1L1
=
K1 + K2
1 + K2K1
Or
v =
v1 + v2
1 + v1v2
c2
as required.
PHY 5347
Homework Set 7 Solutions – Kimel
2. 11.5
From the class notes, and Eq. (11.31), we know
u || =
u ||v + v
1 + vu ||
v
c2
; uó =
uó
v
LÝ1 + vu ||
v
c2
Þ
and
dt = dt vLÝ1 +
vu ||v
c2
Þ
Thus taking the differential of the first equation above and using the second equation for dt,
du ||
dt ¯ a || =
Ý1 + vu ||
v
c2
Þa ||v ? Ýu ||v + vÞ vc2 a ||
v
LÝ1 + vu ||
v
c2
Þ3
=
1 ? v2
c2
3/2
Ý1 + vu ||
v
c2
Þ3
a ||v
Similarly,
duó
dt ¯ aó =
1 + vu ||
v
c2
aó
v ? uóv v
c2
a ||v
L2Ý1 + vu ||
v
c2
Þ3
This is equal to the expression we want to prove,
duó
dt ¯ aó =
1 ? v2
c2
Ý1 + vu ||
v
c2
Þ3
aó
v + v3
c2
× Ýa3v × u3v Þ
since the BAC - CAB theorem shows that
aó
v + v3
c2
× Ýa3v × u3v Þ = 1 +
vu ||v
c2
aó
v ? uóv v
c2
a ||v
PHY 5347
Homework Set 7 Solutions – Kimel
3. 11.15
From Eq. (11.149), it is clear that we should take K3 || z!, so K3 6 E3 = K3 6 B3 = 0. Then
E v = LÝE3 + K3 × B3Þ
B3 v = LÝB3 ? K3 × E3Þ
The vectors in parentheses should make the same angle wrt the x axis S v if they are to be parallel.
This can best be seen from the figure,
From the figure
E3 v = LÝE0î ? KÝ2E0Þ sinSî + K2E0 cosSc!Þ
B3 v = LÝcosS2E0î + sinS2E0c! ? KE0c!Þ
Thus
tanS v = 2KcosS1 ? 2K sinS =
2sinS ? K
2cosS
Ý2cosSÞ 6 Ý2KcosSÞ ? Ý1 ? 2K sinSÞ 6 Ý2sinS ? KÞ = 0
or
2K2 sinS ? 5K + 2sinS = 0
This quadratic equation has the solution
K = 14sinS 5 ? 25 ? 16sin
2S
where I’ve chosen the solution which give K = 0 if S = 0.
If S << 1, then K ¸ 0, and the original fields are parallel.
If S ¸ ^/2 then K = 14 Ý5 ? 3Þ = 1/2. L =
2
3
E3 v = 0 + OÝS ? ^2 Þc!
B3 v = B3 v = LÝ2E0c! ? 12 E0c!Þ = LE0
3
2 c!
So in these two limits, the fields are parallel to the x and y axes, respectively.
PHY 5347
Homework Set 8 Solutions – Kimel
2. 12.1 (a)
L = ? 12 muJu
J ? qc uJA
J (invariant Lagrangian)
Show this Lagrangian gives the correct eqn. of motion, ie, Eq. (12.2)
duJ
db =
e
mc F
JKuK
The Action is A = X
b1
b2 Ldb.
NA = 0 yields the Lagrange equations of motion
d
db
/L
/uJ
? /L
/xJ
= 0
d
db
/L
/uJ
= ?m ddb u
J ? qc
/AJ
/xW
dxW
db
/L
/xJ
= ? qc uK/
JAK
So ddb
/L
/uJ
? /L/xJ = 0 yields
m ddb u
J =
q
c uK/
JAK ? qc /WA
JuW =
q
c uKÝ/
JAK ? /KAJÞ
or
d
db u
J =
q
mc F
JKuK
PHY 5347
Homework Set 8 Solutions – Kimel
3. 12.2 (a)
L v = L + ddt VÝt,x3Þ
N X
t1
t2
ÝL v ? LÞdt = ßNÝVÝt2, x3Þ ? VÝt1,x3Þà = 0 ¸ L and L v yield the same Euler-Lagrange Eqs. of Mot.
where the last equality follows from the fact that variation at the end points is zero since the end
points are held fixed.
(b) For simplicity of notation in this part, I’m going to set c = 1.
L = ?m 1 ? u2 + eu3 6 A3 ? ed with AW = Ýd,A3Þ
If AJ ¸ AJ + /JC, then
d ¸ d + /0C
A3 ¸ a3 ? 43C
where the minus sign in the second equation should be noticed. Thus
L ¸ ?m 1 ? u2 + eu3 6 A3 ? ed ? eu3 6 43C ? e/0C
Now
d
dt CÝt, x3Þ =
/
/xW
CÝt,x3Þ /x
W
/t
= /
/t
C + 43C 6 u3
Or
L ¸ ?m 1 ? u2 + eu3 6 A3 ? ed ? e ddt CÝt,x3Þ
By the argument of part (a), this Lagrangian gives the same equations of motion as the original
Lagrangian.
PHY 5347
Homework Set 9 Solutions – Kimel
1. 12.3
a) Take E3 0 along z, v30 along y
Generally
dp3
dt = e E
3 + v3c × B
3 ; dEdt = ev3 6 E
3
Since B3 = 0, and E3 = E0z!
dpz
dt = eE0
dpy
dt = 0
The initial condition p3Ý0Þ = mv0� and the above equations show the subsequent motion is in the
y ? z plane. Consistent with this initial condition, we have
pyÝtÞ = mv0; pzÝtÞ = eE0t
EÝtÞ = p32ÝtÞc2 + m2c4 = m2v02c2 + m2c4 + ÝceE0tÞ2 = g02 + ÝceE0tÞ2
Using
p3 = mLv3 = E
c2
v3
vyÝtÞ =
pyÝtÞ
EÝtÞ/c2
= mv0c
2
g02 + ÝceE0tÞ2
yÝtÞ = mv0c
2
ceE0 X0
t dt
_2 + t2
= mv0c
eE0
sinh?1Ýt/_Þ
where _ ¯ g0
ceE0
. So
yÝtÞ = mv0c
eE0
sinh?1Ý tceE0g0 Þ Eq. (1)
Similarly
vzÝtÞ =
pzÝtÞ
EÝtÞ/c2
= eE0tc
2
g02 + ÝceE0tÞ2
Thus
zÝtÞ = eE0c
2
ceE0 X0
t tdt
_2 + t2
= c Ý_2 + t2 Þ ? _ Eq.(2)
b) From Eq. (1)
t =
g0
ceE0
sinhÝ eE0ymv0c Þ = _ sinhÝkyÞ, with k =
eE0
mv0c
Then from Eq.(2)
z = c_ sinh2ÝkyÞ + 1 ? 1 Eq.(3)
Let us plot sinh2ÝxÞ + 1 ? 1
0
2
4
6
8
10
12
14
16
18
20
1 2 3 4 5x
For small t : t/_ << 1, and ky << 1. Thus we can taylor expand Eq.(3) and get
z = c_k2y2/2
which is quadratic in y giving a parabolic shape.
For large t : t/_ >> 1, and we see the sinh term dominates in Eq.(3) and we get
z C
c_eky
2
which is an exponential shape.
PHY 5347
Homework Set 9 Solutions – Kimel
2. 12.5
a) The system is described by
Background: particle having m, e. Choose u3 ó to B3 and E3. We want E3óv = 0 = LÝE + u3c × B3Þ.
Thus u3c × B3 = ?E3; now B3 × u3 × B3 = cE3 × B3. Using BAC - CAB on the lhs of the equation gives
u3 = c E
3×B3
B2
. Thus from the figure,
u3 = c E
3 × B3
B2
= c EB z!
Then, using Eq. Ý11. 149Þ
E3 || = 0; E3óv = 0; B3 ||v = 0; B3óv = 1L B3 = 1 ?
u
c
2 B3
So
B3óv = 1 ? EB
2
B3 = B
2 ? E2
B2
BP! 2
Now from the class notes,
du3v
dt v = u
3v × g3 B v , where g3 B v = eB
v
E v
, where in this case E v is the energy of the particle.
I’ll choose the same boundary conditions as in class, decribed in the figure.
So
u3ó
v = gB vaßcosÝgB v t vÞP! 3 ? sinÝgB v t vÞP! 1 à
where uóv Ýt = 0Þ = gB va (ie, the BC determine aÞ.
x3vÝt vÞ = u ||v t vP! 2 + aÝP! 3 singB v t v + P! 1 cosgB v t v Þ
Consider the inverse Lorentz transformation between the frames,
ct
z
x
y
=
L KL 0 0
KL L 0 0
0 0 1 0
0 0 0 1
ct v
z v
x v
y v
or
t = ÝLt v + KLz v Þ/c = ÝLt v + KLa singB v t v Þ/c ¯ fÝt vÞ ¸ t v = f?1ÝtÞ
So
zÝtÞ = KLcf?1ÝtÞ + La singB v f?1ÝtÞ
xÝtÞ = acosgB v f?1ÝtÞ
yÝtÞ = u ||0v f?1ÝtÞ
b) If |E| > |B|, one can transform to a frame where the field is a static E3 field alone. Then the
solution is as we found in section 12.3 of the text, with the above transformation taking you to the
unprimed frame.
PHY 5347
Homework Set 9 Solutions – Kimel
3. 12.14
a) We are given
L = ? 18^ /JAK/
JAK ? 1c JJA
J
which can be rewritten
L = ? 18^ /KA
J/KAJ ? 1c JJA
J
Using the Euler-Lagrange equations of motion,
/K /L
/Ý/KAJ Þ
? /L
/AJ
= 0
Noting
/L
/Ý/KAJ Þ
= ? 14^ /KA
J
/L
/AJ
= ? 1c J
J
The Euler-Lagrange equations of motion are
/KÝ/KAJ Þ = /KÝ/KAJ Þ = 4^c J
J
or
/KÝ/KAJ ? /JAK + /JAK Þ = /KFKJ + /K/JAK = 4^c J
J
If we assume the Lorentz gauge, /KAK = 0, then the above reduces to
/KFKJ = 4^c J
J
Maxwell’s equations, given by Eq. (11.141).
b) Eq. Ý12. 85Þ gives
? 116^ ÝFJKF
JK Þ ? 1c JJA
J
The term in parentheses can be written
FJKFJK = 2/JAK/JAK ? 2/JÝAK/KAJ Þ + 2AK/K/JAJ
The last term vanishes if we choose the Lorentz gauge, and the second term is of the form of a
4-divergence.Thus the Lagrangian of this problem differs from the usual one, of Eq. (12.85) by a
4-divergence /JÝAK/KAJ Þ.
The 4-divergence does not change the euations of motion since the fields vanish at the limits of
integration given by the action. Using the generalized Gauss’s theorem or by integrating by parts, we
see the 4-divergence gives zero contribution to the action.
PHY 5347
Homework Set 10 Solutions – Kimel
2. 14.2 Background. In the nonrelativistic approximation the Lienard-Wiechert potenials are
dÝx3, tÞ = eR |ret, A3Ýx3, tÞ =
eK3
R |ret
Let us assume that we observe the radiation close enough to source so R/c << 1 and t v r t. Then
in the radiation zone
B3 = 43 × A3 = n! /
/R × A
3 = ?n! /
c/t v
× A3 = eK
% × n!
cR
E3 = B3 × n!
dPÝtÞ
dI =
c
4^ RB
3 2 = e
2
4^c K
% × n!
2
= e
2
4^c3
|v% |2 sin2S
where S is the angle between n! and K% (assuming here the particle is moving linearly)
PÝtÞ = 2e
2
3c3 |v% |
2
Let the time-average be defined by
ÖfÝtÞ× ¯ 1b X0
b fÝtÞdt
Then
dPÝtÞ
dI =
e2
4^c3
sin2S |v% |2
ÖPÝtÞ× = 2e
2
3c3 |v% |
2
a) Suppose x3ÝtÞ = z!acosg0t. Then v% = d2zdt2 = ?ag02 cosg0t
|v% |2 = Ýag02 Þ2 1b X0
b
cos2g0tdt = Ýag02 Þ
2/2
So
dPÝtÞ
dI =
e2
8^c3 Ý
ag02 Þ
2
sin2S
ÖPÝtÞ× = e
2
3c3 Ý
ag02 Þ
2
b) Suppose x3ÝtÞ = RÝîcosg0t + � sing0tÞ. Then
v% ÝtÞ = ?Rg02Ýîcosg0t + � sing0tÞ
n! × v% =
î � k!
sinScosd sinS sind cosS
?Rg02 cosg0t ?Rg sing0t 0
dPÝtÞ
dI =
e2
4^c3 Ý
Rg02 Þ
2
cos2S sin2g0t + cos2g0t + sin2S sin2Ýg0t + d Þ
dPÝtÞ
dI =
e2
4^c3 Ý
Rg02 Þ
2 Ý1 + cos2SÞ
2
ÖPÝtÞ× = e
2
4^c3 Ý
Rg02 Þ
22^ X
?1
1 Ý1 + x2 Þ
2 dx =
2e2
3c3 Ý
Rg02 Þ
2
PHY 5347
Homework Set 11 Solutions – Kimel
2. 14.12
a) From Jackson, Eq (14.38)
dP
dI =
e2
4^c
n! × Ýn! ? K3Þ × K%
2
Ý1 ? n! 6 K3Þ5
Using azimuthal symmetry, we can choose n! in the x-z plane.
From the figure
n! = cosSz! + sinSx!
K3Ýt vÞ = ? ac g0 sing0t
vz! = ?K sing0t vz!
K% Ýt vÞ = ? ag0
2
c cosg0t
vz! = ?g0Kcosg0t vz!
Using K3 × K& = 0
n! × Ýn! ? K3Þ × K%
2
= n! × K% 2
and
n! × K% = � g0K sinScosg0t v
So
dP
dI Ýt
vÞ = e
2cK4
4^a2
sin2Scos2g0t v
Ý1 + KcosS sing0t v Þ5
Defining d = g0t v,
Ö dPdI Ýt
vÞ× = e
2cK4
4^a2
1
2^ X0
2^ sin2Scos2d
Ý1 + KcosS sindÞ5
dd
Or, doing the integral,
Ö dPdI × =
e2cK4
32^a2
4 + K2 cos2S
Ý1 ? K2 cos2SÞ7/2
sin2S
c) 4+.052 cos2S
1?.052 cos2S 7/2
sin2S
0
1
2
3
4
5
0.5 1 1.5 2 2.5 3
4+.952 cos2S
1?.952 cos2S 7/2
sin2S
0
50
100
150
200
250
300
0.5 1 1.5 2 2.5 3
PHY 5347
Homework Set 12 Solutions – Kimel
1. 16.1 It’s useful to apply in this case the Virial Theorem, familiar from classical mechanics:
ÖT× = 12 Ö
dV
dr ×r
If V = arn, then
ÖT× = n2 ÖV×
In our case V = 12 kr
2
, with k = mg02, so n = 2 and
ÖT× = ÖV×
Or,
Ö dVdr × =
E
r
We are given
dE
dt = ?
b
m Ö
dV
dr
2
×
This can be rewritten
dE
dt = ?
b
m kE
So
E = E0e?
b
m kt = E0e?bg0
2t = E0e?@t
Similarly,
dL3
dt = ?
b
m Ö
1
r
dV
dr ×L
3
But 1r dVdr = mg0
2
, so
L = L0e?bg0
2t = L0e?@t
PHY 5347
Homework Set 12 Solutions – Kimel
2. 16.2 V = ?, q = ?e
dE
dt = ?
b
m Ö
dV
dr
2
×
If V = arn, then the Virial theorem tells us
ÖT× = n2 ÖV×
In the present case, n = ?1, so
E = ÖT× + ÖV× = 12 ÖV× = ?
Ze2
2r
dV
dr =
Ze2
r2
Now
dE
dt = ?
b
m Ö
dV
dr
2
×
gives
d
dr
1
rÝtÞ
= 2Ze
2b
mr4ÝtÞ
or
r2dr = ?2Ze2bdt/m
But b = 23
e2
c3m
, so
r2dr = ?3ZÝcbÞ3 tb
Integrating both sides gives
r3ÝtÞ = r03 ? 9ZÝcbÞ3 tb
b) At this point, for simplicity of notation, I’m going to take c = i = 1. Then from problem
14.21,
1
T =
2
3 e
2ÝZe2 Þ4 m
n5
We are given
r = n
2a0
Z
Where a0 = 1
me2
= Bohr radius, and b = 23
e2
m
? dndt = ?
Z
2a0n
dr
dt =
Z
2a0n
3
r2
Zb2 = Z2a0n
3 Z
a0n
2
2
Z 23
e2
m
2
= 23
Z4
m
e6
n5
in agreement with the result of problem 14.21.
c) From part bÞ
t =
r0
3 ? r3ÝtÞ
9Zb2
But rÝtÞ = nf
2a0
Z , r0 =
ni
2a0
Z
t =
ni
2a0
Z
3
?
nf2a0
Z
3
9Zb2
= 19 a0
3 n i
6 ? n f6
Z4b2
In our present case Z = 1, so
t = 19
1
me2
3 n i
6 ? n f6
2
3
e2
m
2 =
1
4m e
?10Ýn i
6 ? n f6 Þ
In these units, (from the particle data book) MeV?1 = 6.6 × 10?22s. e2 = J = 1/137, and
m = 207 ×. 511MeV.
t = 1 × 6.6 × 10
?22s
4 × 207 ×. 511Ý1/137Þ5
Ýn i
6 ? n f6 Þ = 7. 53 × 10?14Ýn i6 ? n f6 Þs
For the cases desired,
t1 = 7. 53 × 10?14Ý106 ? 46 Þs = 7. 5 × 10?8s
t2 = 7. 53 × 10?14Ý106 ? 16 Þs = 7. 5300 × 10?8s
Just as a check on working with these units, notice
b = 23
e2
m =
2
3
1
. 511 × 137 6.6 × 10
?22s = 6. 29 × 10?24s
in agreement with what we found before.
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