PHY 5346
Homework Set 13 Solutions Kimel
2. 7.2
a) The gure describes the muliple internal reections which interfere
to give the overall reection and refraction:
For the ij interface I shall use the notation
rij =
0
E0
2ni
=
E0
ni + nj
Rij =
00
E0
ni nj
=
PHY 5346
HW Set 2 Solutions Kimel
2. The system is described by
a) The capacitance is dened by
Q
12
is the magnitude of the potential difference between the inner and outer spheres. From Gausss
Z
~ a Qenc
E d~ =
0
C=
law
where 12
or
Q
0
when r is between
PHY 5346
HW Set 4 Solutions Kimel
1. The system is described by
a) The Greens function, which vanishes on the surface is obviously
G(~ , ~ 0 ) =
xx
1
1
0|
|~ ~
xx
|~ ~ 0 |
x xI
where
x
~ 0 = x0 + y 0 + z 0 k, ~ 0 = x0 + y 0 z 0 k
x
I
b) There is no free c
PHY 5346
HW Set 5 Solutions Kimel
1. 3.2 The charge distribution is shown by
a) We see the charge distribution is given by
(~) = N (cos cos ) (r a)
r
R
r
where N is determined by the requirement d3 r (~) = Q, or
(~) =
r
Q
(cos cos ) (r R)
4 R2
Expanding
PHY 5346
HW Set 5 Solutions Kimel
2. 3.3 The system is described by
a) We will calculate the potential when the eld point is along the z axis,
then generalize to any ~ .
x
(~ = z z ) =
x
1
2
40
Z
0
R
C
0 d0
p
=
20
z 2 + 02
=
Z
R
0
C
R
tan1 ( )
20
z
0 d0
p
PHY 5346
HW Set 6 Solutions Kimel
2. a) We can expand the potential in terms of spherical harmonics as
X
(~ ) =
x
Alm rl Ylm (, )
lm
We are given
(R) = V0 cos sin
Projecting out the coecients Alm
Z
V0
dYlm (, ) cos sin
Alm = l
R
The above integral vani
PHY 5346
HW Set 7 Soutions Kimel
1. The problem can be visualized as
The potential energy is
W = p E (~ = x)
~
r
where, for an elementary q20 multipole
1 4
Y10
~
~ =
E
q20 3
40 5
r
If we write Y20 in terms of cartesian coordinates
r
5 2z 2 x2 y 2
0
Y2 =
PHY 5346
HW Set 8 Solutions - D. Kimel
1. a) Let us apply the method of images. The images will be located as
shown in the gure.
where the vector ~ a + a, with magnitude b = 2a, and ~ 0 = a a
b=
b
with magnitude b0 = 2a. The potential is just the linear
PHY 5346
Homework Set 9 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,
dm = 0 N Idz
4
From the gure,
!
Z
ZR
Z
z
d
dA cos
r dA
~
+1
=
= 2 z
= 2 p
=
3
PHY 5346
HW Set 8 Solutions Kimel
3. 5.13 We may choose the coordinate system so the ~ lies in the x z plane:
r
The vector potential is given by
~
A= 0
4
Z
~
Jd3 x0
|~ ~0 |
rr
~
Noting Jd3 x0 Id~0 , where
l
I =
Since
Q
a2 d0
=
2 /
0
d~0 = a sin 0 d0
l
0
PHY 5346
Homework Set 10 Solutions Kimel
~
4. 5.14 This problem corresponds to J = 0, so we have the equations
~
B = 0,
~
~
and H = 0
from which it follow that
~
~
H =
~
~
where B = H.
From the rst two equations we have the boundary
conditions at an int
PHY 5346
Homework Set 11 Solutions Kimel
2. 5.19 The system is described by
~
The eective volume magnetic charge density is zero, since M is constant
~ from Eq. (5.99)
within the cylinder. The eective surface charge density (n M
is M0 , on the top surface
PHY 5346
Homework Set 11 Solutions Kimel
3. 5.26 The system is described by
Since the wires are nonpermeable, = 0 . The system is made of parts
with cylindrical symmetry, so we can determine B using Amperes law.
Z
Z
~a
~
~ d~ = 0 J d~
~ B = 0 J, or
~
Bl
O
PHY 5346
Homework Set 12 Solutions Kimel
1. 6.8
The physical system is shown as
~
We know from Maxwells equations that M plays the role of the eective
magnetic charge density. Using the fact that
1
~
M=
~ J
x~
2
~
and the fact that J = pol~ , where pol =
PHY 5346
Homework Set 12 Solutions Kimel
1. 6.21
a) Im going to represent the dipole as a charge q at ~0 and a charge
r
l.
q at ~0 + ~ We take the limit
r
q~ p
l
~
Thus
h
i
= q (~ ~0 ~) (~ ~0 )
xr
l
xr
Expanding around ~ = 0 give
l
~xr
l
~ x r
(~ ) = q (
PHY 5346
Homework Set 13 Solutions Kimel
1. 6.11
a) Consider the momentum contained in the volume
p = tcAg
p
= cAg
t
F
P=
= cg
A
where Im using the time averaged quantities.
F=
cg =
In class we found
1
1
2
S = 0 |E0 | = u
c
2
Thus
P =u
b) We are given
But
PHY 5346
HW Set 2 Solutions Kimel
1. By symmetry the electric eld is in the z direction, and each of the line segments of charge gives equal
contribution to the potential. Let us work out the potential from each segment at a position r away from the cente