PHY 5346
HW Set 1 Solutions – Kimel
2
1.3 In general we take the charge density to be of the form ρ = fr⃗δ, where fr⃗ is determined by
physical constraints, such as ∫ ρd 3 x = Q.
a) variables: r, θ, φ. d 3 x = dφd cos θr 2 dr
ρ = fr⃗δr − R = frδr − R = fRδr − R
Q
∫ ρd 3 x = fR ∫ r 2 drdΩdr − R = 4πfRR 2 = Q → fR = 4πR
2
ρr⃗ =
Q
δr − R
4πR 2
b) variables: r, φ, z. d 3 x = dφdzrdr
ρr⃗ = fr⃗δr − b = fbδr − b
λ
∫ ρd 3 x = fb ∫ dzdφrdrδr − b = 2πfbbL = λL → fb = 2πb
λ δr − b
2πb
c) variables: r, φ, z. d 3 x = dφdzrdr Choose the center of the disk at the origin, and the z-axis
perpendicular to the plane of the disk
ρr⃗ = fr⃗δzθR − r = fδzθR − r
where θR − r is a step function.
2
∫ ρd 3 x = f ∫ δzθR − rdφdzrdr = 2π R2 f = Q → f = πRQ 2
Q
ρr⃗ =
δzθR − r
πR 2
d) variables: r, θ, φ. d 3 x = dφd cos θr 2 dr
ρr⃗ = fr⃗δcos θθR − r = frδcos θθR − r
ρr⃗ =
∫ ρd 3 x = ∫ frδcos θθR − rdφd cos θr 2 dr = ∫ 0 frrrdrdφ = 2πN ∫ 0 rdr = πR 2 N = Q
R
R
where I’ve used the fact that rdrdφ is an element of area and that the charge density is uniformly
distributed over area.
Q
ρr⃗ =
δcos θθR − r
πR 2 r
PHY 5346
HW Set 1 Solutions – Kimel
3. 1.4 Gauss’s Law:
∫ E⃗ ⋅ da⃗ = Q enclosed
0
Q
a) Conducting sphere: all of the charge is on the surface σ = 4πa
2
Q
E4πr 2 = 0, r < a
E4πr 2 =  0 , r ∑ a
Q ̂
⃗ =
E = 0, r < a
E
r, r ∑ a
4π 0 r 2
b) Uniform charge density: ρ =
Q
4 πa 3
3
, r < a, ρ = 0, r ∑ a.
Qr 3
Qr
→E=
, r<a
3
4π 0 a 3
0a
Q
Q
, r∑a
E4πr 2 =  0 → E =
4π 0 r 2
E4πr 2 =
c) ρ = A r n
Q = 4π ∫ r 2 drAr n = 4πAa n+3 /n + 3 → ρ =
n + 3Q n
r ,r<a
4πa n+3
ρ = 0, r ∑ a
E4πr 2 =
Q
n + 3Q
4πr n+3 /n + 3 → E =
n+3
4π 0 r 2
4π 0 a
Q
E=
, r∑a
4π 0 r 2
1. n = −2.
E=
Q
,
4π 0 ra
r<a
r n+3
a n+3
, r<a
E=
Q
, r∑a
4π 0 r 2
2. n = 2.
Qr 3
,
4π 0 a 5
Q
E=
,
4π 0 r 2
E=
r<a
r∑a
PHY 5346
HW Set 1 Solutions – Kimel
4. 1.5
qe −αr 1 + αr2
4π 0 r
ρ
∇ 2 φr⃗ = −  0 ,
∇ 2 = 12 ∂ r 2 ∂
∂r
r ∂r
−αr
q 1 ∂
∇2φ =
r 2 ∂ e r + α e −αr
2
4π 0 r 2 ∂r
∂r
φr⃗ =
Using ∇ 2 1r = −4πδr⃗
∇2φ =
=
q 1 ∂
−αre −αr + e −αr r 2 ∂
4π 0 r 2 ∂r
∂r
1
r
− α r 2 e −αr
2
2
2 −αr
−αr
2 −αr
3 −αr
q
− α2 e −αr + α er
+ αe 2 − 4πδr⃗ − α er
+ α e
2
4π 0
r
r
3
qα −αr
= − 10 qδr⃗ −
e
8π
3
qα −αr
ρr⃗ = qδr⃗ −
e
8π
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
⃗ ⋅ da
⃗ = Qenc
2. 1.8 We will be using Gauss’s law ∫ E
0
a) 1) Parallel plate capacitor
From Gauss’s law E = σ0 = AQ0 =
φ 12
d
→Q=
0
2

W = 0 ∫ E 2 d 3 x =  0 E Ad =
2
2
2) Spherical capacitor
Q
A 0
2
Ad
2
A 0 φ 12
d
2
Q
= 1
d= 1
2 0 A
2 0
A 0 φ 12
d
2
A
φ2
d = 1  0 A 12
d
2
From Gauss’s law, E = 4π1 0 rQ2 , a < r < b.
φ 12 = ∫ Edr =
b
a
W=
0
∫ E 2 d 3 x = 20
2
b
4π 0 baφ 12
Q
Q b − a
→Q=
r −2 dr = 1 π 0
∫
4π 0 a
ba
4
b − a
Q
4π 0
W= 1
8 0
3) Cylindrical conductor
2

= 0
∫ a 4π r rdr
4
2
b
2
4π 0 baφ 12
b−a
π
2
Q
4π 0
2
−4π −b + a
ba
φ 212
b − a
= 2π 0 ba
ba
b − a
=
2
1 Q b − a
8 0 π
ba
From Gauss’ s law, E2πrL = λL
= Q0
0
φ 12 = ∫ Edr =
b
a
W=
0
∫ E 2 d 3 x = 20
2
b
Q
Q
dr =
ln ba
∫
2π 0 L a r
2π 0 L
Q
2π 0 L
2π 0 Lφ 12
W=
0 2
E
2
b) w =
1) wr = 20
Q
A 0
2
2) wr = 20
Q
1
4π 0 r 2
3) wr = 20
Q
2π 0 Lr
ln
1
4π 0
b
a
L
2
2
b
Q
2πL ∫ rdr
= 1
ln ba
2
4π 0 L
a r
2
ln ba
= π 0 L
φ 212
ln ba
2
= 210 QA 2 0 < r < d, = 0 otherwise.
2
2
= 321 π 2 Qr 4 , a < r < b, = 0, otherwise
0
2
2
= 810 π 2QL 2 r 2 , = 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, Fδl is the work done by an
external force. If F is along δl (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 φ 12 .
Conservation of energy gives
Fδl = δW + δQφ 12
or
F = ∂W +
∂l
∂Q
φ 12
∂l
From problem 1.8,
a) Charge fixed.
1) Parallel plate capacitor
dQ
2
φ2
dQ
→ W = 1  0 A A 0
= d Q2
W = 1  0 A 12 , φ 12 =
A 0
d
2
2 0 A
2
d
2
Q
∂Q
= 0, F = ∂W =
(attractive)
2
∂l
∂l
0A
2) Parallel cylinder capacitor
φ 12 = λ0 ln da , a = a 1 a 2
λQ
W = 1 Qφ 12 → F = ∂W = 1 Q λ0 ∂ ln da  = 1
(attractive)
2
2
2 0d
∂l
∂d
b) Potential fixed
1) Parallel plate capacitor
Using Gauss’s law, Q =
φ 12 A 0
,
d
∂ Q = φ 12 A 0
∂l
d2
2
φ2
φ 212 A 0
1  0 A φ 12 = 1  0 A
→ F = − 1  0 A 12
+
=
2
2
2
d2
d2
d2
2) Parallel cylinder capacitor
 Lφ
W = 1 Qφ 12 , and Q = 0 d 12
2
ln a 
Qd
0A
2
d
2
=
1 Q2
2 0 A
so
 Lφ 2
φ 212
W = 1 0 d 12 , ∂W = − 1  0 L
2
2 ln a 
∂l
ln 2 da d
∂Q
∂l
= 0L
φ 12
ln 2 da d
φ 212
φ 212
φ 212
1 L
F = − 1 0L
+

L
=
0
0
2
2
ln 2 da d
ln 2 da d
ln 2 da d
PHY 5346
HW Set 2 Solutions – Kimel
4. 2.1 We will work in cylidrical coordinate, (ρ, z, φ, with the charge q located at the point
⃗
d = dẑ , and the conducting plane is in the z = 0 plane.
Then we know from class the potential is given by
Ez = −
q ∂
4π 0 ∂z
Ez =
q
4π 0
q
q
−
⃗
⃗
x−⃗
d
x+⃗
d
1
4π 0
φx⃗ =
1
z − d 2 + ρ 2
z−d
z − d 2 + ρ 2
z+d
z + d 2 + ρ 2
−
3/2
1
z + d 2 + ρ 2
−
1/2
1/2
3/2
a) σ =  0 E z z = 0
σ = 0
Plotting
q
4π 0
−1
1 2 +r 2
3
2
−d
−d + ρ 2
2
−
3/2
+d
+d + ρ 2
3/2
=−
r
4
5
2
q
2πd 2
1
1 +
2
ρ
d
2
3
2
gives
1
2
3
0
-0.2
-0.4
-0.6
-0.8
-1
b) Force of charge on plane
⃗=
F
q2 ̂
1 q−q −ẑ  =
1
z
2
4π 0 2d
4 × 4π 0 d 2
c)
2
F = w = 0 E2 = σ2 = 1
2 0
2 0
A
2
F = 2π
β
q2
2 4 ∫
8 0 π d 0
−
q
2πd 2
ρ2
3
ρ
1 + d2
1
12 +
dρ = 2π
ρ
d
2
q2
8 0 π 2 d 4
=
3
2
1 d2
4
1
8 0
=
q2
π 2 d 4 1 + ρd 2
q2
1
4 × 4π 0 d 2
2
3
d)
β
W = ∫ Fdz =
d
β
q2
q2
dz =
∫
4 × 4π 0 d z 2
4 × 4π 0 d
e)
qiqj
q2
W= 1 1 >
=−
2 4π 0 i,j,iΦj |x⃗i − ⃗
2 × 4π 0 d
xj |
Notice parts d) and e) are not equal in magnitude, because in d) the image moves when q moves.
f) 1 Angstrom = 10 −10 m, q = e = 1. 6 × 10 −19 C.
W=
−19
q2
e
=e
= e 1. 6 × 10−10 9 × 10 9 V = 3.6 eV
4 × 4π 0 d
4 × 4π 0 d
4 × 10
PHY 5346
HW Set 2 Solutions – Kimel
5. 2.2 The system is described by
a) Using the method of images
φx⃗ =
1
4π 0
q
q′
+
y′ |
y|
|x⃗ − ⃗
|x⃗ − ⃗
with y ′ = ay , and q ′ = −q ay
b) σ = − 0 ∂n∂ φ|x=a = + 0 ∂x∂ φ|x=a
2
σ = 0
1 ∂
4π 0 ∂x
q
q′
+
1/2
1/2
x 2 + y 2 − 2xy cos γ
x 2 + y ′2 − 2xy ′ cos γ
2
a 1 − ay 2
σ = −q 1
4π y 2 + a 2 − 2ay cos γ 3/2
Note
y2
q induced = a 2 ∫ σdΩ = −q 1 a 2 2πa 1 − 2
4π
a
∫ −1
1
dx
, where x = cos γ
3/2
y 2 + a 2 − 2ayx
q
2
q induced = − aa 2 − y 2 
= −q
2
aa 2 − y 2 
c)
qq ′
q 2 ay
1
=
, the force is attractive, to the right.
4π 0 a 2 − y 2 
4π 0 y ′ − y 2
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
q′
q
φx⃗ = 1
+
+V
4π 0 |x⃗ − ⃗
y|
y′ |
|x⃗ − ⃗
and obviously the electric field in the sphere and induced charge on the inside of the sphere would
remain unchanged.
|F| =
PHY 5346
HW Set 3 Solutions – Kimel
2. 2.3 The system is described by
a) Given the potenial for a line charge in the problem, we write down the solution from the figure,
λ
4π 0
R2
R2
R2
R2
− ln
− ln
+ ln
2
2
2
xo 
x o1 
x o2 
x o3  2
x⃗ − ⃗
x⃗ − ⃗
x⃗ − ⃗
x⃗ − ⃗
x o3  2 , so φ T |y=0 = 0
x o2  2 = x⃗ − ⃗
x o1  2 , x⃗ − ⃗
Looking at the figure when y = 0, x⃗ − ⃗
x o  2 = x⃗ − ⃗
2
2
2
2
x o3  , so φ T |x=0 = 0
x o1  = x⃗ − ⃗
x o2  , x⃗ − ⃗
Similarly, when x = 0, x⃗ − ⃗
x o  = x⃗ − ⃗
On the surface φ T = 0, so δφ T = 0, however,
∂φ T
∂φ T
δφ T =
δx = 0 →
= 0 → Et = 0
∂x t t
∂x t
b) We remember
∂φ
y0
y0
−
σ = − 0 T = −λ
π
2
2
∂y
x + x 0  2 + y 20
x − x 0  + y 0
where I’ve applied the symmetries derived in a). Let
y0
y0
σ/λ = −1
π x − x  2 + y 2 − x + x  2 + y 2
0
0
0
0
This is an easy function to plot for various combinations of the position of the original line charge
(x 0 , y 0 .
φT =
ln
c) If we integrate over a strip of width Δz, we find, where we use the integral
x0
∞
y0
∫ 0 x ∓ x 1 2 + y 2 dx = 12 π ± 2 arctan
y0
0
0
∞
∞
ΔQ
−1 x 0
ΔQ = ∫ σdxΔz →
= ∫ σdx = −2λ
π tan
y0
Δz
0
0
and the total charge induced on the plane is −∞, as expected.
d)
Expanding
ln
R2
R2
R2
−
ln
−
ln
x − x 0  2 + y − y 0  2
x − x 0  2 + y + y 0  2
x + x 0  2 + y − y 0  2
to lowest non-vanishing order in x 0 , y 0 gives
xy
16 2
y x
2 0 0
x + y 2 
so
xy
4λ
y x
φ → φ asym = π
2 0 0
0
2
x + y 2 
This is the quadrupole contribution.
+ ln
R2
x + x 0  2 + y + y
PHY 5346
HW Set 3 Solutions – Kimel
3. 2.5
a)
∞
W = ∫ |F|dy =
r
q2a ∞
∫
4π 0 r
dy
2
y 3 1 − ay 2
=
2
q2a
8π 0 r 2 − a 2 
Let us compare this to disassemble the charges
aq 2
r
qq
− W′ = − 1 ∑ ⃗ i ⃗ j = 1
4π 0
8π 0 i≠j |x i −x j |
1
2
r 1 − ar 2
=
q2a
>W
4π 0 r 2 − a 2 
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 = ∫ |F|dy =
r
q
4π 0
∞
∫r
∞ 2y 2 − a 2 
Qdy
3
−
qa
∫ r 2 2 2 dy
y2
yy − a 
Using standard integrals, this gives
W=
1
4π 0
q2a
qQ
q2a
−
− r
2
2
2
2r
2r − a 
On the other hand
qQ + ar q
q2a
qQ
aq 2
aq 2
−
= 1
− 2 − r
2
2
2
2
r
4π 0 r − a 
r
r − a 
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.
− W′ =
1
4π 0
PHY 5346
HW Set 3 Solutions – Kimel
4. 2.6 We are considering two conducting spheres of radii r a and r b respectively. The charges on
the spheres are Q a and Q b .
a) The process is that you start with q a 1 and q b 1 at the centers of the spheres, and sphere a
then is an equipotential from charge q a 1 but not from q b 1 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) q a 1 and q b 1 are determined from the two requirements
β
β
j=1
j=1
> q a j = Q a and > q b j = Q b
As a program equation, we use a do-loop of the form
n
>q a j = −rdabqjb j− −11 
j=2
and similar equations for q b j, x a j, x b j, d a j, d b j. The potential outside the spheres is given
by
n
n
φx⃗ =
1
4π 0
> ⃗x −qxa jjk̂ + > ⃗x −qdb jjk̂
a
j=1
b
j=1
This potential is constant on the surface of the spheres by construction.
And the force between the spheres is
q a jq b k
F= 1 >
4π 0 j,k d − x a j − x b k 2
c) Now we take the special case Q a = Q b , r a = r b = R, d = 2R. Then we find, using the iteration
equations
x a j = x b j = xj
x1 = 0, x2 = R/2, x3 = 2R/3, or xj =
j − 1
R
j
q a j = q b j = qj
qj = q, q2 = −q/2, q3 = q/3, or qj =
−1 j+1
q
j
So, as n → β
β
β
Q
−1 j+1
= q ln 2 = Q → q =
ln
2
j
j=1
> qj = q >
j=1
The force between the spheres is
q2
−1 j+k
F= 1
>
4π 0 R 2 j,k jk 2 − j−1 − k−1
j
Evaluating the sum numerically
k
=
2
j+k
2
1 q > −1 jk
4π 0 R 2 j,k j + k 2
F=
2
2
1 0. 0739
1 q 0. 0739 = 1 Q
2
2
4π 0 R
4π 0 R ln 2 2
Comparing this to the force between the charges located at the centers of the spheres
Q2
Fp = 1
4π 0 R 2 4
Comparing the two results, we see
F = 4 1 2 0. 0739F p = 0. 615F p
ln 2
On the surface of the sphere
β
φ=
β
qj
q
1
=
>−1 j+1
4π 0 >
4π
R
−
xj
0R
j=1
j=1
β
Notice
So φ =
1 = >−1 j+1
1+1
j=1
Q
q
Q
1
C
= 1
=
→
= 2 ln 2 = 1. 386
4π 0 2R
C
4π 0 2 ln 2R
4π 0 R
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
Gx⃗, ⃗
x′ =
1
1
−
x′ |
x ′I |
|x⃗ − ⃗
|x⃗ − ⃗
where
⃗
x ′I = x ′ î + y ′ ̂ − z ′ k̂
x ′ = x ′ î + y ′ ̂ + z ′ k̂ , ⃗
b) There is no free charge distribution, so the potential everywhere is determined by the potential
on the surface. From Eq. (1.44)
∂Gx⃗, ⃗
x′ ′
φx⃗ = − 1 ∫ φx⃗′ 
da
′
4π S
∂n
Note that n̂ ′ is in the −z direction, so
2z
∂ Gx⃗, ⃗
x ′ |z ′ =0 = − ∂ Gx⃗, ⃗
x ′ |z ′ =0 = −
3/2
∂z
′ 2
∂n ′
x − x  + y − y ′  2 + z 2
So
a 2π
φx⃗ = z V ∫ ∫
2π
0 0
ρ ′ dρ ′ dφ ′
2
x − x ′  + y − y ′  2 + z 2
3/2
where x ′ =ρ ′ cos φ ′ , y ′ = ρ ′ sin φ ′ .
c) If ρ = 0, or equivalently x = y = 0,
a 2π
a
ρdρ
ρ ′ dρ ′ dφ ′
φz = z V ∫ ∫
=
zV
∫
3/2
3/2
2
′2
2
2π
0 0 ρ + z 
0 ρ + z 2 
φz = zV −
z − a 2 + z 2 
z a + z 
2
2
= V 1−
z
a + z 2 
2
d)
a 2π
φx⃗ = z V ∫ ∫
2π
0 0
ρ ′ dρ ′ dφ ′
⃗−ρ
⃗ ′ 2 + z2
ρ
3/2
⃗, then ρ
⃗⋅ρ
⃗ ′ = ρρ ′ cos φ ′
In the integration choose the x −axis parallel to ρ
a 2π
ρ ′ dρ ′ dφ ′
φx⃗ = z V ∫ ∫
2π
0 0 ρ 2 + ρ ′2 − 2ρ
⃗⋅ρ
⃗ ′ + z 2  3/2
Let r 2 = ρ 2 + z 2 , so
a 2π
ρ ′ dρ ′ dφ ′
φx⃗ = z V3 ∫ ∫
3/2
⃗⋅ρ
⃗′
2π r 0 0
ρ ′2 −2ρ
1+
r2
We expand the denominator up to factors of O1/r , ( and change notation
φ ′ → θ, ρ ′ → α, r 2 → β12 
4
a 2π
φx⃗ = z V3 ∫ ∫
2π r 0 0
αdαdθ
1+
⃗⋅α
⃗
α 2 −2ρ
r2
3/2
where the denominator in this notation is written
1
3/2
1 + β α − 2ρα cos θ
2
2
or, after expanding,
a
2π
φx⃗ = z V3 ∫ αdα ∫ 1 − 3 β 2 α 2 + 3β 2 ρα cos θ + 15 β 4 α 4 − 15 β 4 α 3 ρ cos θ + 15 β 4 ρ 2 α 2 cos 2 θ dθ
2π r 0
2
8
2
2
0
Integrating over θ gives
a
φx⃗ = z V3 ∫ α 2π + 15 β 4 α 4 π − 3β 2 α 2 π + 15 β 4 ρ 2 α 2 π dα
2π r 0
4
2
Integrating over α yields
φx⃗ = z V3 5 β 4 πa 6 − 3 a 4 β 2 π + 15 a 4 β 4 ρ 2 π + πa 2
4
8
2π r
8
or
φx⃗ =
Va 2
2
2ρ + z 2  3/2
a2
1− 3
+ 5
2
4 ρ + z 2 
8
a 4 + 3ρ 2 a 2
2
ρ 2 + z 2 
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
′
φr⃗ = 1 λ ln rr = V, a constant. Let us define V ′ = 4π 0 V
2π 0
Then the above equation can be written
r ′ 2 = e Vλ′ or r ′2 = r 2 e Vλ′
r
2
⃗ , the above can be written
Writing r ′2 = ⃗r − R
2
R
⃗r + ẑ
V′
e λ − 1
V′
R2e λ
V′
e λ − 1 2
=
The equation is that of a circle whose center is at −ẑ
R
V′
e λ −1
, and whose radius is a =
b) The geometry of the system is shown in the fugure.
Note that
d = R + d1 + d2
with
R
d1 =
e
V ′a
λ −1
R
, d2 =
e
−V ′b
λ −1
and
′
Va
−V ′b
2λ
2λ
a = Re
, b = Re
−V ′b
V ′a
e λ − 1
e λ − 1
Forming
V′
Re 2λ
V′
e λ −1
2
d −a −b =
2
2
2
R
R+
R
+
V ′a
e λ −1
e
−
−V ′b
λ −1
2
′
Va
Re 2λ
V ′a
e λ − 1
−
−V ′b
2
Re 2λ
−V ′b
e λ − 1
or
R2 e
d2 − a2 − b2 =
V ′a −V ′b
λ
+1
−V ′b
V ′a
e λ − 1e λ − 1
Thus we can write
e
d2 − a2 − b2 =
2ab
V ′a −V ′b
λ
+1
= e
−V ′b
V ′a
V ′a −V ′b
2λ
2e 2λ e 2λ
+e
2
− V ′a −V ′b
2λ
= cosh
V ′a − V ′b
2λ
or
V a − V b = 1 cosh −1 d 2 − a 2 − b 2
λ
2ab
2π 0
Q/L
2π 0
λ
Capacitance/unit length = C =
=
=
−1 d 2 −a 2 −b 2
Va − Vb
L
Va − Vb
cosh
2ab
′
c) Suppose a << d , and b << d , and a = ab , then
2π 0
2π 0
C =
=
2 1−a 2 +b 2 /d 2
d
−1 d 2 −a 2 −b 2
L
−1
cosh
cosh
′2
′2
2a
2
2
2
2
2a
cosh −1
d 2 1 − a 2 + b 2 /d 2 
2a ′2
d 2 1 − a 2 + b 2 /d 2 
2a ′2
= e
2π 0 L
C
2
= 2π 0 L
C
+ negligible terms if 2π 0 L >> 1
C
or
ln
d 2 1 − a 2 + b 2 /d 2 
a ′2
= 2π 0 L
C
or
C =
L
Let us defind α 2 = a 2 + b 2 /d 2 , then
2π 0
C =
2 1−α 2
d
L
ln
a ′2
ln
2π 0
d 2 1−a 2 +b 2 /d 2
a ′2
= 2π  0d 2 + 2π 2 0d 2 α 2 + Oα 4 
ln a ′2
ln a ′2
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 d 2 − a 2 − b 2 , since a 2 + b 2 > d 2 . Thus
2π 0
C =
−1
a 2 +b 2 −d 2
L
cosh
2a ′2
If we use the identiy, lnx + x 2 − 1  = cosh −1 x, G.&R., p. 50., then for d = 0
C =
L
ln
in agreement with problem 1.6.
2π 0
2 −b 2
+ a2ab
a 2 +b 2
2ab
=
2π 0
ln ab
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
σ = 3 0 E 0 cos θ
We also note the radial force/unit area outward from the surface is σ 2 /2 0 . Thus the force on the
right hand hemisphere is, using x = cos θ
1
⃗ = 1 3 0 E 0  2 2πR 2 ∫ x 3 dx = 1 3 0 E 0  2 2πR 2 /4 = 9 π 0 E 20 R 2
F z = 1 ∫ σ 2 ẑ ⋅ da
2 0
2 0
2 0
4
0
An equal force acting in the opposite direction would be required to keep the hemispheres from
sparating.
b) Now the charge density is
σ = 3 0 E 0 cos θ +
Fz =
Q
Q
Q
= 3 0 E 0 x +
= 3 0 E 0 x +
2
2
4πR
12π 0 E 0 R 2
4πR
1
Q
1 ∫ σ 2 ẑ ⋅ da
⃗ = 1 3 0 E 0  2 2πR 2 ∫ x x +
2 0
2 0
0
12π 0 E 0 R 2
2
dx
Thus
1
F z = 9 π 0 E 20 R 2 + 1 QE 0 +
Q2
2
4
32 0 πR 2
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 E 0 by two charges at ∞
σ = 3 0 E 0 cos θ
a) This charge distribution simulates the given system for cos θ > 0. We have treated this probem
in class. The potential is given by
3
φx⃗ = −E 0 1 − a3 r cos θ
r
Using
σ = − 0 ∂ φ|surface
∂n
We have the charge density on the plate to be
3
σ plate = − 0 ∂ φ|z=0 =  0 E 0 1 − a 3
∂z
ρ
σ
For purposes of plotting, consider  0plate
= 1 − x13
E0
1
0.8
0.6
0.4
0.2
0
2
4
x 6
8
σ boss = 3 0 E 0 cos θ =
For plotting, we use 3σ boss
0E0
= cos θ
10
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
b)
q = 3 0 E 0 2πa 2 ∫ xdx = 3π 0 E 0 a 2
1
0
c) Now we have
φr⃗ =
1
4π 0
q′
q
+
⃗r − ⃗
⃗r − ⃗
d
d′
−
q
q′
−
⃗r + ⃗
⃗r + ⃗
d
d′
where q ′ = −q ad , d ′ = ad .
2
−q
σ = − 0 ∂ φ|r=a =
4π
∂r
q ind =
d 2 − a 2 
d 2 − a 2 
−
3
3
a⃗
a−⃗
d
a+⃗
d
a⃗
d 2 − a 2 
d 2 − a 2 
−
3
3
a⃗
a−⃗
d
a+⃗
d
a⃗
q ind = 2πa 2
−q
4π
∫0
−qa 2 d 2 − a 2 
2a
1
da
1 −
d−a
2
2
q ind = −1 q d − a
2
d
1
2d −
d2 − a2
1
+ 1 −
d+a
2
a +d
2
2
2
a + d2
= −q 1 −
dx
1
a + d2
2
d 2 − a 2 
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)
φr⃗ = λ ln rr∞
2π 0
Thus for this system
φr⃗ =
τ ln
2π 0
r∞
⃗
⃗r − R
+
τ ′ ln
2π 0
To determine τ ′ and R ′ , we need two conditions:
I) As r → ∞, we want φ → 0, so τ ′ = −τ.
II) φr⃗ = b  = φr⃗ = −b  or
′
ln b − R
R−b
′
= ln b + R
b+R
or
b − R′
R−b
′
This is an equation for R with the solution
b + R′
b+R
=
2
R′ = b
R
The same condition we found for a sphere.
b)
r∞
⃗′
⃗r − R
4
2
r 2 + Rb 2 − 2r bR cos φ
φr⃗ = τ ln
4π 0
r 2 + R 2 − 2rR cos φ
as r → ∞
φr⃗ =
2
τ ln 1 − 2b cos φ/rR
4π 0
1 − 2R cos φ/r
=
τ ln 1 − 2 b 2 − R 2  cos φ
rR
4π 0
Using
ln1 + x = x − 1 x 2 + 1 x 3 + Ox 4 
3
2
1 b 2 − R 2  cos φ
φr⃗ = − τ
2π 0 rR
c)
4
σ = − 0 ∂ φ|r=b = − τ ∂ ln
4π ∂r
∂r
τ
2πb
σ=
where y = R/b. Plotting σ/
τ
2πb
gy =
r 2 + R 2 − 2rR cos φ
r=b
1−y
y + 1 − 2y cos φ
2
2
1−y 2
y 2 +1−2y cos φ
=
2
r 2 + Rb 2 − 2r bR cos φ
, for y = 2, 4, gives
1 − y2
y 2 + 1 − 2y cos φ
g2, g4 = − 5−43cos φ , − 17−815cos φ
0.5
1
1.5
2
2.5
3
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 τ from τ ′ is, using Gauss’s law
′
E= τ
2π 0 d
The force on τ is τLE, ie,
′
2
F = ττ L = − τ L , and the force is attractive.
2π 0 d
2π 0 d
PHY 5346
HW Set 5 Solutions – Kimel
1. 2.13 The system is pictured in the following figure:
a) Notice from the figure, Φρ, −φ = Φρ, φ; thus from Eq. (2.71) in the text,
β
Φρ, φ = a 0 + ∑ a n ρ n cosnφ
n=1
∫ −π/2 Φb, φ = 2πa 0 = πV1 + πV2 → a 0 = V1 +2 V2
3π/2
Using
∫ −π/2 cos mφ cos nφdφ = δ nm π
3π/2
Applying this to Φ, only odd terms m contribute in the sum and
m−1
2V 1 − V 2 
am =
−1 2
πmb m
Thus
2V 1 − V 2 
i m ρ m e imφ
Φρ, φ = V 1 + V 2 +
Im ∑
π
2
mb m
m odd
Using
2 ∑ xm = ln
m
m odd
1 + x
1 − x
and
Im lnA + iB = tan −1 B/A
we get
V 1 − V 2 
V + V2
Φρ, φ = 1
+
tan −1
π
2
as desired.
b)
σ = − 0 ∂ Φρ, φ|ρ=b
∂ρ
ρ
2 b cos φ
1 − ρb 2
2
σ = − 0
σ = −2 0
:
ρ
2 b cos φ
V 1 − V 2  ∂
tan −1
π
∂ρ
ρ2
1 − b2
ρ=b
b +ρ
V1 − V2
V − V2
bcos φ 4
|ρ=b = − 0 1
π
2 2
4
2 2
2
πb cos φ
b − 2b ρ + ρ + 4ρ b cos φ
2
2
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 Vx, y, z = V.
1) First consider the side Vx, y, a = z :
∞
Φ 1 x, y, z = ∑ A nm sinα n x sinβ m y sinhγ nm z
n,m=1
with α n =
sine functions,
mπ
π
nπ
a , β m = a , γ nm = a
n 2 + m 2 . Projecting out A nm using the orthogonality of the
16V
sinhγ nm anmπ 2
where both n, and m are odd. (Later we will use n = 2p + 1, m = 2q + 1
2) In order to express Φ 2 x, y, z in a form like the above, we make the coordinate transformation
x ′ = y, y ′ = x, z ′ = −z + a
So
Φ 2 x, y, z = Φ 1 x ′ , y ′ , z ′  = Φ 1 y, x, = z + a
A nm =
Φx, y, z = Φ 1 x, y, z + Φ 2 x, y, z
b)
∞
−1 p+q
Φ a , a , a  = 16 ⋅2 V ∑
2 2 2
π
2p + 12q + 1 cosh γ nm a
p,q=0
where I have used the identity
sinhγ nm a = 2 sinh
∞
Let fp, q ≡ ∑ p,q=0
−1 p+q
2p+12q+1 cosh
2
γ nm a
γ a
 cosh nm 
2
2
2p+1 2 +2q+1 2 π2
(p,q)
fp, q
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.
Φ a , a , a  = 16 ⋅ 0. 20539
V = 0. 332 96V
2 2 2
π2
Φ av  a , a , a  = 2 V = 0. 333. . . . V
2 2 2
6
c)
σx, y, a = − 0 ∂ Φ|z=a
∂z
σx, y, a = − 162 0 V
π
∞
coshγ nm a − 1
= − 162 0 V ∑ sinα n x sinβ m y
sinhγ nm a
π
n,m odd
∞
σx, y, a = − 162 0 V ∑ sinα n x sinβ m y tanh
π
n,m odd
γ nm a
2
PHY 5346
HW Set 5 Solutions – Kimel
3. 3.1 The system is pictured in the following figure:
1
∫ 0 Pl xdx
The problem is symmetric around the z axis so
φr, θ = ∑A l r l + B l r −l−1 P l cos θ
l
The A l and B l are determined by the conditions
1)
∫ −1 φa, xPl xdx = 2l 2+ 1 A l a l + B l a −l−1 
1
2)
∫ −1 φb, xPl xdx = 2l 2+ 1 A l b l + B l b −l−1 
1
Solving these two equations gives
1
1
2l + 1
a l+1 ∫ φa, xP l xdx − b l+1 ∫ φb, xP l xdx
Al =
2l+1
2l+1
−1
−1
2a
−b 
B l = a l+1 2l + 1 ∫ φa, xP l xdx − A l a 2l+1
2
−1
1
Using
∫ −1 φa, xPl xdx = V ∫ 0 Pl xdx
1
1
∫ −1 φb, xPl xdx = V ∫ −1 Pl xdx = V−1 l ∫ 0 Pl xdx
1
0
1
So
Al =
1
2l + 1
l+1
l+1
l
Va
−
b
−1
P l xdx

∫
0
2a 2l+1 − b 2l+1 
B l = a l+1 2l + 1 V ∫ P l xdx − A l a 2l+1
2
0
1
Note that
∫ 0 Pl xdx = 12 ∫ −1 Pl xdx
1
1
1
for l even. For even l ∑ 0, ∫ P l xdx = 0. Thus we have
0
∫ 0 P0 xdx = 1; ∫ 0 P1 xdx = 12 , ∫ 0 P3 xdx = − 18
1
1
1
and
3
7
A0 = V , A1 =
Va 2 + b 2 , A 3 = −
Va 4 + b 4 
2
4a 3 − b 3 
16a 7 − b 7 
B 0 = 1 Va − 1 Va = 0
2
2
3
B1 = 3 a2V −
Va 2 + b 2 a 3 = 3 Va 2 b 2 b3 + a 3
4
4
4a 3 − b 3 
−a + b
7
B3 = − 7 a4V − a7 −
Va 4 + b 4 
16
16a 7 − b 7 
3
3
= − 7 Va 4 b 4 b 7 + a 7
16
−a + b
1
As b → ∞, only the B l terms (and A 0  survive. Thus using the general expression for ∫ P l xdx
0
given by (3.26)
3
4
φr, θ = V P 0 x + 3 a2 P 1 x − 7 a4 P 3 x +. . . . .
2 r
8 r
2
Let’s now solve the problem neglecting the outer sphere (since b → ∞ using the Green’s function
result
(2.19) this integral give, for cosθ = 1
φr, θ = V 1 − ρ 2 
2
1
−
1 − ρ
1
1 + ρ 2 
with ρ = a/r. Expanding the above,
2
4
φr, θ = V ar + 3 a2 − 7 a4 +. . . . . .
2 r
8 r
2
Comparing with our previous solution with x = 1, we see the Green’s function solution differs by
having a B 0 term and by not having an A 0 term. All the other higher power terms agree in the series.
This difference is due to having a potential at ∞ 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 Δφ about
this axis with the surface of each slice of the pie alternating as ±V.
φr, θ, φ = > A lm r l Y ml θ, φ
l,m
so
A lm = 1l ∫ dΩY ml θ, φ ∗ φa, θ, φ
a
Symmetries:
A l−m = −1 m A lm  ∗
φr, θ, φ + 2Δφ = φr, θ, φ
where
Δφ = 2π
2n
Thus
m = ±n, and integral multiples thereof
φ−r⃗ = −φr⃗, n = 1
φ−r⃗ = φr⃗, n > 1
Since
PY ml θ, φ = −1 l Y ml θ, φ
Then
l is odd for n = 1; l is even for n > 1
Thus we only have contributions of l ≥ n. Using
A lm = 1l ∫ dΩY ml θ, φ ∗ φa, θ, φ
a
The integral over φ can be done trivially, since the integrand is just e −imφ leaving the desired answer
in terms of an integral over cosθ.
n = 1 case: I am going to keep only the lowest novanishing terms, involving A 11 and A 1−1 .
1
1 ∗
1
φ = rA 11 Y 11 + A 1−1 Y −1
1  = rA 11 Y 1 + A 11 Y 1   = 2r ReA 11 Y 1 
Y 11 = −
A 11 = − 1a
3 1 − x 2  1/2 e iφ
8π
3 V ∫ 1 1 − x 2  1/2 dx
8π
−1
π
∫ 0 e −iφ dφ − ∫ π e −iφ dφ
2π
A 11 = 2iπ
a
φ = 2r Re
2iπ
a
3 V
8π
−
3 V
8π
3 sin θe iφ
8π
= 3r Vsin θ sin φ
2a
From the figure
we see
sin θ sin φ = cos θ ′
So
φ = 3r Vcos θ ′ = V 3 ar P 1 cos θ ′  +. . . . . .
2a
2
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
Φρ, z, φ = ∑A nν sin νφ + B nν cos νφI ν
nν
nπρ
L
sin nπz
L
where
1
A nν = 2
πL I ν nπb
 sinνφdφdz,
∫ 0 ∫ 0 Vφ, z sin nπa
L
ν∑0
1
B nν = 2
πL I ν nπb
 cosνφdφdz,
∫ 0 ∫ 0 Vφ, z sin nπa
L
νΦ0
L
L
2π
L
L
2π
dφdz,
∫ 0 ∫ 0 Vφ, z sin nπa
L
L
1
B nν = 1
πL I ν nπb
L
2π
ν=0
Noting
π
2
∫ − sin νφdφ − ∫
π
2
3π
2
π
2
sin νφdφ
=0
we conclude A nν = 0. Similarly, noting
π
2
∫ − cos νφdφ − ∫
π
2
3π
2
π
2
cos νφdφ
=
4−1 m
,
2m + 1
m = 0, 1, 2, . . . .
where I’ve recognized that ν must be odd, ie, ν = 2m + 1. Also
L
2
dz =
, l = 0, 1, 2, . . . . .
∫ 0 sin nπz
L
2l + 1π
where again I’ve recognized that n must be odd, ie, n = 2l + 1. Thus
16−1 m V
B nν = 2
nπb
π I 2m+1  L 2l + 12m + 1
b) Now z = L/2,
L ∑∑ b, L ∑∑ ρ. Then from the class notes
I 2m+1 
2l + 1πρ
2L
2l + 1πρ
1

L
Γ2m + 2
m+1
Also
2l + 1π
L
sin
= −1 l
so
Φρ, z, φ = ∑
l,m
ρ
b
16−1 l+m V
π 2l + 12m + 1
2
2m+1
cos2m + 1φ
Using
∞
tan
−1x
=∑ x
−1 l
1l
+
1
l=0
2l+1
∞
π = tan −1 1 = ∑ −1 l
4
2l + 1
l=0
so
Φρ, z, φ = 4V
π ∑
m
−1 m
2m + 1
ρ
b
2m+1
cos2m + 1φ
Remembering from problem 2.13 that
−1 m
∑ 2m
+1
m
ρ
b
2m+1
cos2m + 1φ = 1 tan −1
2
we find
−1
Φρ, z, φ = 2V
π tan
which is the answer for problem 2.13.
2
ρ
b
cos φ
1 − ρb 2
2
2
ρ
b
cos φ
1−
ρ2
b2
PHY 5346
HW Set 6 Solutions – Kimel
3. 4.1
⃗d 3 x = > q i r ll Y m∗
q lm = ∫ r l Y m∗
l θ, φρx
l θ i , φ i 
i
Using
Y m∗
l θ, φ =
2l + 1l − m! m
P l xe −−mφ = N ml P ml xe −−mφ
4πl + m!
From the figure we get
q lm = a l N ml P ml 0q1 − −1 m 1 − i m  = 0, for m even, so m = 2n + 1, n = 0, 1, 2, . . .
q lm = 2qa l N ml P ml 01 − −1 n i
b) The figure for this system is
Since the sum of the charges equals zero, l ≥ 1.
m
m
m∗
l m
q lm = qa l Y m∗
l x = 1, φ + Y l x = −1, φ = qa N l P l 1 + P l −1
From the Rodrigues formula for P ml x, we see P ml ±1 = 0, for m ≠ 0. So
q lm = qa l N 0l 1 + −1 l P l 1
Thus l is even, but l ≠ 0
q lm = 2qa l N 0l
c) Using the fact that N 0l =
2l+1
4π
and Y 0l =
β
2l+1
Pl
4π
2qa
P x
P 2 x = 0 on x-y plane)
Φx⃗ = >2qa l  ll+1 ≈
r3
r
l=2
2
2
Let us plot Φx⃗/−q/a, ie,
qa
Φx⃗ = − 3
r
1
 ar  3
= x13
8
7
6
5
4
3
2
1
0 0.5
1
1.5
x
2
2.5
− 3
4
1
x
3
The exact answer on the x-y plane is
−q
Φx⃗ = a
2 −
x
2
x 1 + x12
So let’s plot x13 , 2x −
−q
= a
3
1
x
5
+ 5
8
2
x 1+ 12
x
0.3
0.25
0.2
0.15
0.1
0.05
0 1
2
where the smaller is the exact answer.
3
x
4
5
1
x
7
− 35
64
1
x
9
+. . . .
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:
⃗
p⋅⃗
x
Φx⃗ = 1
4π 0 r 3
⃗ 0
⃗⋅E
W = −p
from the general formulas
Φx⃗ =
1 ∫ ρx⃗ d x
4π 0
x′ |
|x⃗ − ⃗
′
3 ′
W = ∫ ρx⃗Φx⃗d 3 x
using the effective charge density
⃗⋅⃗
ρ eff = −p
∇δx⃗
where I’ve chosen the origin to be at ⃗
x0.
′
⃗
⃗ ⋅ ∇ δx⃗′ d 3 x ′
−p
1
Φx⃗ = 1 ∫
=− 1 ⃗
p ⋅ ∫⃗
∇
4π 0
4π 0
x′ |
x′ |
|x⃗ − ⃗
|x⃗ − ⃗
Φx⃗ =
δx⃗′ d 3 x ′
p⋅⃗
x
1 ⃗
3
4π 0 r
Similarly,
⃗ Φx⃗d 3 x = −p
⃗ 0
⃗⋅E
W = ∫ ρx⃗Φx⃗d 3 x = − ∫ ⃗
p⋅⃗
∇δx⃗Φx⃗d 3 x = ⃗
p ⋅ ∫ δx⃗∇
PHY 5346
HW Set 7 Solutions – Kimel
3. 4.7 a) Since ρ does not depend on φ, we can write it in terms of spherical harmonics with
m = 0. First note
Y 02 =
5
4π
3 1 − sin 2 θ − 1
2
2
or
sin 2 θ = − 2 4π Y 02 + 4π 2 Y 00
3 5
3
Thus only the m = 0, l = 0, 2 multipoles contribute.
2 4π β 2 1 2 −r
3 = 1
q 00 =
∫ 0 r 64π r e dr = 2 34π 8π
3
2 π
q 20 = − 2
3
Φx⃗ =
1
4π 0
4π ∫ β r 4 1 r 2 e −r dr = − 2
5
3
64π
0
Y0
Y0
4πq 00 r0 + 4πq 20 23
5r
=
4π 45 = −3 5
5 4π
π
P
4π q 00 r0 +
1
4π 0
4π q P 2
5 20 r 3
P0 − 6 P2
r
r3
Φx⃗ =
1
4π 0
Φx⃗ =
1 ∫ ρx⃗ d x
4π 0
x′ |
|x⃗ − ⃗
b)
′
3 ′
Using
1
1
= 4π >
+
1r >
2l
x′ |
|x⃗ − ⃗
lm
r<
r>
l
m
′
′
Y m∗
l θ , φ Y l θ, φ
, we see only the l = 0, 2 and m = 0 terms of the expansion contribute in the potential. Next take
r ′ > r.
⃗′  ′
1
′
′ ′2
′ ρx
Φx⃗ = 1 4π >
r l Y ml θ, φ ∫ Y m∗
dr
l θ , φ r dΩ
′l+1
4π 0
2l + 1
r
lm
Φx⃗ =
1 4π Y 0 4π 2 ∫ β 1 r 2 e −r rdr + Y 2 r 2 − 2
0
5
4π 0
3 0 64π
3
0
Φx⃗ =
Φx⃗ =
0
1 4π Y 0 4π 2 3 + Y 2 r 2 − 2
0
5
4π 0
3 32π
3
1 4π P 2 3 + P 2 r 2 − 2
0
5
4π 0
3
3 32π
1
64π
=
4π ∫ β 1 r 2 e −r 1 dr
r
5
64π
0
4π
5
1
64π
1
4π 0
P0 − r 2 P2
120
4
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,
Φ out =
Φ out =
Φ in =
∑ B l r −l−1 Pl + |x⃗ −q⃗x ′ |
1
4π 0
l
1
4π 0
∑ B l r −l−1 Pl + rq> ∑ rr <>
1
4π 0
∑ A l r l Pl + rq> ∑ rr <>
l
l
Pl
l
l
l
Pl
l
Boundary conditions: At the surface, r ′ = d = r > , r = a = r < .
1) Φ out = Φ in |r=a , or
B l = A l a 2l+1
2)  ∂r∂ Φ in = ∂r∂ Φ out |r=a , or letting k = 0
k
∑ lA l a l−1 Pl + qd l ad l
l−1
Pl
l
=
=
∑ −l + 1B l a −l−2 Pl + qd l ad l
l−1
Pl
l
∑ −l + 1A l a l−1 Pl + qd l ad l
l−1
Pl
l
or
Al =
a1 − kl
1 + kl + 1d l+1
Bl =
a1 − kla 2l+1
1 + kl + 1d l+1
4π
Remember that P l = 2+1
Y 0l , 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,
Φr, θ = ∑A l r l + B l r −l−1 P l cos θ
l
Also
⃗ Φr, θ
⃗ = −x⃗∇
⃗ = x⃗E
D
D r = −x⃗ ∑lA l r l−1 − l + 1B l r −l−2 P l cos θ
l
between the spheres,
∫ D r dΩr 2 = Q, and is independent of r.
Thus
A l = 0, B l = 0, l Φ 0 → D r =
∫ D r dΩr 2 = 2πB 0  0 ∫ −1 d cos θ +  ∫ 0 d cos θ
0
1
B0 =
⃗ =
E
x⃗B 0
r2
= 2πB 0  0 +  = Q
Q
2π 0 1 + 0 
Q
r̂
2π 0 1 + 0 r 2
b)
∫ D r dA = D r A = σ f A → σ f = D r = x⃗E r
σf =
Q
,
2π 0 1 + 0 r 2
cos θ ≥ 0
σf =
Q
,
2π1 + 0 r 2
cos θ < 0
c)
∫ ρ pol dV = σ pol A = ∫ −∇⃗ ⋅ P⃗dV = −PA → σ pol = −P = − 0 χ e E
σ pol = −x⃗/ 0 − 1
Q
2π1 + 0 r 2
Notice
σ pol + σ f = σ tot =
Q
=  0 E, as expected.
2π1 + 0 r 2
PHY 5346
Homework Set 10 Solutions – Kimel
1. 5.1 The system is described by
We want to show
φm = −
μ0I
Ω
4π
Suppose the observation point is moved by a displacement δx⃗, or equivlently that the loop is
displaced by −δx⃗.
⃗ φ m , then
⃗ = −∇
If we are to have B
⃗
δφ m = −δx⃗ ⋅ B
Using the law of Biot and Savart,
δφ m = −
μ0I
∮ δx⃗ ⋅
4π
dl⃗′ × ⃗r
r3
δφ m = −
=−
μ0I ⃗
∮r ⋅
4π
δx⃗ × dl⃗′
r3
=−
μ 0 I r̂ ⋅ δdA
∮ r 2 = − μ4π0 I δΩ
4π
Or,
φm = −
μ0I
Ω
4π
μ0I
∮ r̂ ⋅
4π
δx⃗ × dl⃗′
r2
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 z 0 . Using the results of problem 5.1,
μ0
NIdzΩ
4π
dφ m =
From the figure,
⃗
ρdρ
z
θ = 2πz ∫ R
Ω = ∫ r̂ ⋅ 2dA = ∫ dA cos
= 2π −
+1
2
3/2
2
2
2
r
0 ρ + z 
r
R + z 2 
φm =
β
μ0
1
NI ∫ z −
+ 1z
2
2
2
z0
R + z 
Br = −
μ0
NI −z 0 + R 2 + z 20 
2
dz =
μ0
NI ∂ −z 0 + R 2 + z 20 
2
∂z 0
=
−z 0 + R 2 + z 20 
μ0
NI
2
R 2 + z 2 
0
In the limit z 0 → 0
Br =
μ0
NI
2
By symmetry, thej loops to the left of z 0 give the same contribution, so
B = B l + B r = μ 0 NI
H = NI
⃗ is directed along the z axis, so
By symmetry, B
⃗ =0
⃗⋅B
δφ m = −δρ
⃗ is directed ⊥ to the z axis. Thus for a given z, φ m is independent of ρ, and consequently
if δρ
H = NI
everywhere within the solenoid.
If you are on the outside of the solenoid at position z 0 , by symmetry the magnetic field must be in
the z direction. Thus using the above argument, φ m must not depend on ρ. Set us take ρ far away
⃗ directed along
from the axis of the solenoid, so that we can replace the loops by elementary dipoles m
the z axis. Thus for any point z 0 we will have a contributions
φm 
⃗ ⋅ ⃗r 2
⃗ ⋅ ⃗r 1
m
m
+
3
r 32
r1
⃗ ⋅ ⃗r 1 = −m
⃗ ⋅ ⃗r 2 and r 1 = r 2 . Thus
where m
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
′
Aφ =
d 3 x cos φ ′ J φ r ′ , θ ′ 
μ0
∫
4π
x′ |
|x⃗ − ⃗
Choose ⃗
x in the x − z plane. Then we use the expansion
r l< m∗ ′ ′ m
1
= > 4π
Y l θ , φ Y l θ, 0
′
2l + 1 r l+1
x|
|x⃗ − ⃗
>
l,m
The cos φ ′ factor leads to only an m = 1 contribution in the expansion. Using
Y ml θ, 0 =
2l + 1 l − m! P m cos θ
4π l + m! l
1
and l−1!
= ll+1
, we have on the inside
l+1!
Aφ =
1
′
′ ′
μ0
1 r l P 1 cos θ ∫ d 3 x ′ P l cos θ J φ r , θ 
>
l
4π l ll + 1
r ′l+1
which can be written
Aφ = −
μ0
m l r l P 1l cos θ
4π >
l
with
ml = −
′
′
′
P cos θ J φ r , θ 
1
d3x′ l
∫
ll + 1
r ′l+1
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)
A φ r, θ =
μ 0 I r ′2 dr ′ dΩ ′ sin θ ′ cos ϕ ′ δcos θ ′ δr ′ − a
∫
4π a
x′ |
|x⃗ − ⃗
Using the expansion of 1/|x⃗ − ⃗
x ′ | given by Eq. (3.149),
β
4 ∫ β dk coskz − z ′ 
1
= π
′
x|
0
|x⃗ − ⃗
1 I 0 kρ < K 0 kρ >  + > cosmϕ − ϕ ′ I m kρ < K m kρ > 
2
m=1
We orient the coordinate system so ϕ = 0, and because of the cosϕ ′ factor, m = 1. Thus,
A φ r, θ =
μ 0 I 4π β
∫ dk ∫ r ′2 dr ′ d cos θ ′ sin θ ′ δcos θ ′ δr ′ − a coskzI 1 kρ < K 1 kρ > 
4π a π 0
β
μ
A φ r, θ = π0 aI ∫ dk coskzI 1 kρ < K 1 kρ > 
0
where ρ < ρ >  is the smaller (larger) of a and ρ.
b) From problem 3.16 b),
β
β
′
1
= > ∫ dke imφ−φ  J m kρJ m kρ ′ e −k|z|
′
x|
|x⃗ − ⃗
0
m=−β
Note z ′ = 0, and φ = 0, so
Aφ =
μ 0 Ia β
∫ 0 dke −k|z|J 1 kρJ 1 ka
2
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.
φ 1 = φ loop + > A l r l P l
l
φ 2 = φ loop + > B l r −l−1 P l
l
⃗ φ, and we have the boundary conditions,
⃗ = −∇
where H
H 1|| = H 2|| → φ 1 r = b = φ 2 r = b
μ 0 ∂ φ 1 r = b = μ ∂ φ 2 r = b
∂r
∂r
We are given that b >> a, so
θ
φ loop = 1 m cos
4π r 2
with m = πa 2 I. (From the form of φ loop, only the l = 1 term contributes.) The boundary
conditions give
A 1 b 1 = B 1 b −1−1
−
2μ 0 m
2μm
+ μ0A1 = −
− 2μB 1 b −3
3
4πb
4πb 3
So
μ − μ 0 
A 1 = − 2 m3
4π b 2μ + μ 0 
On the inside, at the center of the loop
⃗ φ loop − ⃗
⃗ = −∇
H
∇A 1 r cos θ
⃗ φ loop at the center of the loop, which is directed in the z direction.
From Eq. (5.40), we are given −∇
H z = μ10 −B θ  − A 1
If μ >> μ 0
A 1 → − 1 m3
4π b
and from (5.40), at r = 0
2
3
H z = I + 1 m3 = I + I a 3 = I 1 + a 3
2a
4π b
2a
4 b
2a
2b
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 I ∗ is equidistant from the interface and is equal to I ∗ = μμ rr −1
I. The radius of each current
+1
loop is a. Now from Eq. (5.7)
⃗on I = I ∫ dl⃗ × Br⃗
F
⃗ θ = dlB r −θ̂ + dlB θ r̂
⃗ = dl⃗ × B
⃗ r + dl⃗ × B
dl⃗ × B
By symmetry, only the z − component survives, so, from the figure
⃗ ⋅ ẑ = dlB r
dl⃗ × B
a
4d + a 2
2
2d
4d 2 + a 2
+ dlB θ
So
Fz =
2πaI aB + 2dB 
r
θ
4d 2 + a 2
with B r and B θ given by Eqs. (5.48) and (5.49) and cos θ =
2d
4d 2 +a 2
,r=
4d 2 + a 2 , and I → I ∗ .
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.
F z = πaI aB r + 2dB θ 
d
F z = πaI a
d
μ0I∗ a a
4d 2d 2
Fz → −
+ 2d −
μ0I∗ a2
4
1
2d 3
− a
2d
3πμ 0 a 4 I × I ∗
32
d4
The minus sign shows the force is attractive if I and I ∗ are in the same direction. This same result
can be gotten more directly, using
F z = ∇ z mB z 
with m = πa 2 I, and (from Eq. (5.64))
Bz =
μ0
4π
2m ∗
z3
with m ∗ = πa 2 I ∗ , and z = 2d
Fz =
μ0
2πa 2 I ∗ πa 2 I − 3 4
4π
2d
with agrees with out previous result.
=−
3πμ 0 a 4 I × I ∗
32
d4
PHY 5346
Homework Set 9 Solutions – Kimel
3. 5.19 The system is described by
⃗ is constant within the cylinder. The
The effective volume magnetic charge density is zero, since M
⃗ from Eq. (5.99)) is M 0 , on the top surface and −M 0 on the
effective surface charge density (n̂ ⋅ M
bottom surface. From the bottom surface the potential is (for z > 0
a
ρdρ
Φ b = 1 −M 0 2π ∫
= − M0
1/2
2
2
4π
2
0 ρ + z 
a 2 + z 2  − z
By symmetry, the potential from the top surface is (on the inside)
Φt = M0
2
a 2 + L − z 2
− L − z
The total magnetic potential is
Φ = Φb + Φt = −
M0
2
M
a 2 + z 2  − z + 0
2
a 2 + L − z 2
− L − z
So, on the inside of the cylinder,
M
Hz = − ∂ − 0
2
∂z
Hz = −
while above the cylinder,
M0
2
a 2 + L − z 2
z
−
a + z 2 
L−z
a + L − z 2
a 2 + z 2  − z +
M0
2
2−
2
2
− L − z
Hz = − M0
2
−
z−L
a + L − z 2
z
+
a + z 2 
2
2
with a similar expression below the cylinder.
⃗ = μ0 H
⃗ +M
⃗
B
Thus inside the cylinder,
Bz = μ0 −
z
−
a + z 2 
L−z
a + L − z 2
μ0M0
2
z
+
a + z 2 
L−z
a + L − z 2
μ0M0
2
z
−
a + z 2 
z−L
a + L − z 2
M0
2
Bz =
2−
2
2
2
2
while above the cylinder,
Bz =
2
2
First we plot B z in units of a for L = 5a
1
2
1+z 2
1
2
1+z 2
gz =
+
z
5−z
1+5−z 2
−
z
z−5
1+5−z 2
if z < 5
if 5 < z
gz
1
0.8
0.6
0.4
0.2
0
2
And similarly, H z in units of a for L = 5a.
4
z
6
8
10
+ M0
− 12
2−
− 12
−
fz =
−
z
1+z 2
5−z
1+5−z 2
+
z
if 5 < z
z−5
1+z 2
if z < 5
1+5−z 2
fz
0.4
0.2
0
-0.2
-0.4
2
4
z
6
8
10
PHY 5346
Homework Set 11 Solutions – Kimel
1. 5.26 The system is described by
Since the wires are nonpermeable, μ = μ 0 . The system is made of parts with cylindrical
symmetry, so we can determine B using Ampere’s law.
∫ B⃗ ⋅ dl⃗ = μ 0 ∫ ⃗J ⋅ da⃗
⃗
⃗ = μ 0⃗
∇×B
J, or
On the outside of each wire,
μ0I
∫ B⃗ ⋅ dl⃗ = B2πρ = μ 0 I → B out = 2πρ
On the inside of each wire
∫ B⃗ ⋅ dl⃗ = B2πρ = μ 0 I ρR 2 , B in = μ2π0 I Rρ2 with R = a, b
2
From the right-hand rule, the B from each wire is in the φ̂ direction. From the above figure, using
⃗ is in the ±ẑ direction. Since ⃗
⃗ =B
⃗,
the general expression for the vector potential, we see A
∇×A
B z = − ∂ A z → A z = − ∫ B z dρ
∂ρ
Thus
μ I
Az =
ρ
μ I
− 2π0 ln R + C
ρ2
= − 4π0 ln R 2 + 1 on the outside
− μ4π0 I ρR 2 ,
2
on the inside
where I’ve determined C = 1/2, from the requirement that A z be continuous at ρ = R. Let l be the
length of the wire. Then we know the total potential energy is given by
⃗ d 3 x = l ∫J a Ada a + J b Ada b 
W = 1 ∫⃗
J⋅A
2
2
Consider the second term 2l ∫ J b Ada b . The system is pictured as
From the figure
⃗ b,
⃗a = ⃗
d+ρ
ρ
ρ 2a = d 2 + ρ 2b − 2dρ b cos φ
so, since J b = πbI 2
l ∫ J b Ada b = l I ∫A out ρ a  + A in ρ ρ b dρ b dφ
b
2 πb 2
2
ρ2
μ I
ρ2
= l I 2 0 ∫ ln 2a + 1 − 2b
2 πb 4π
b
a
ρ b dρ b dφ
b
2
ρ2
μ I
≃ l I 2 0 2π ∫ ln d 2 + 1 − 2b
2 πb 4π
b
a
0
2
μ I
= l I 2 0 2π 1 b 2 1 + 2 ln d 2
2 πb 4π
4
a
= l
2
μ0
4π
ρ b dρ b
1 + 2 ln d I 2
a
2
The first term 2l ∫ J a Ada a is equal to
l ∫ J a Ada a = l
2
2
μ0
4π
1 + 2 ln d I 2
b
2
Thus
W= l
2
μ0
4π
2
1 + 2 ln d I 2 = l L I 2
ab
2 l
or
L = μ 0 1 + 2 ln d 2
l
ab
4π
PHY 5346
Homework Set 11 Solutions – Kimel
2. 5.27 The system is described by
Using Ampere’s law in integral form
∫ B⃗ ⋅ dl⃗ = μ 0 I enclosed
we get
μ0I ρ
,ρ<b
2π b 2
B=
B=
μ0I 1
,b<ρ<a
2π ρ
B = 0, ρ > a
Now the energy in the magnetic field is given by ( l is the length of the wires)
⃗⋅H
⃗ d3x = 1 ∫ B2d3x
W = 1 ∫B
2μ 0
2
=
μ0I
2π
1
2μ 0
=
1
2μ 0
2
l 2π ∫
μ0I
2π
b
0
2
ρ
b2
2
ρdρ + 2π ∫
lπ 1 + 2 ln a
2
b
μ
→ L = 0
l
4π
1 + 2 ln a
b
2
If the inner wire is hollow, B = 0, ρ < b, so
L = μ 0 ln a
l
2π b
a
b
1
ρ
= l L I2
2 l
2
ρdρ
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
⃗ d 3 x = l ∫J a Ada a + J b Ada b 
W = 1 ∫⃗
J⋅A
2
2
Using the same arguments as applied in problem 5.26,
μI
Az =
ρ
− 2π ln R + C
μI
ρ2
= − 4π ln R 2 + 0 on the outside
0,
on the inside
Thus if we consider the second term 2l ∫ J b Ada b ,
l ∫ J b Ada b = l I ∫A out ρ a  + A in ρ ρ b dρ b dφ
b
2 πb 2
2
b
2
μI
≃ l I2
2π ∫ ln d 2 ρ b dρ b = l
2 πb 4π
2
0
a
μ
4π
2 ln da I 2
The first term 2l ∫ J a Ada a is equal to
l ∫ J Ada = l
a
a
2
2
μ
4π
2 ln d I 2
b
Thus
W= l
2
μ
4π
2
2 ln d I 2 = l L I 2
ab
2 l
so
L =
l
μ
4π
2
2 ln d
ab
Now using the methods of problem 1.6, assuming the left wire has charge Q, and the right wire charge
−Q, we find
φ 12 = ∫
d−a
b
Edr =
Q
l
2π
∫b
d−a
Q
2
1 + 1
dr ≃ l ln d
r
d−r
2π
ab
Q
C = l = 2π
2
φ 12
l
ln d
ab
Thus
L × C =
l
l
μ
4π
2
2 ln d
ab
× 2π
= μ
2
ln dab
PHY 5346
Homework Set 12 Solutions – Kimel
1. 6.11
a) Consider the momentum contained in the volume
Δp = ΔtcΔAg
F=
Δp
= cΔAg
Δt
P = F = cg
ΔA
where I’m using the time averaged quantities. In class we found
cg = 1c S = 1  0 |E 0 |2 = u
2
Thus
P=u
b) We are given
S = 1. 4 × 10 3 W/m 2
But we know P = u = Sc . From Newton’s second law
1. 4 × 10 3 W/m 2
F = F/A = S/c =
a= m
= 4. 66 × 10 −3 m/s 2
2
8
−3
m/A
m/A
3 × 10 m/s × 1 × 10 kg/m
In the solar wind, there are approximately 10 × 10 4 protons/m 2 − sec, with average velocity
v = 4 × 10 5 m/s.
Δp
= P = 10 × 10 4 × 4 × 10 5 × 1. 67 × 10 −27 = 6. 68 × 10 −17 N/m 2
ΔtA
:
2
−17
F = F/A = P = 6. 68 × 10 N/m = 6. 68 × 10 −14 m/s 2
a= m
m/A
m/A
1 × 10 −3 kg/m 2
PHY 5346
Homework Set 12 Solutions – Kimel
2. 7.1 I shall apply Eqs.(26), (27), and (28)
s0 + s1 = a
1
2
s0 − s1
= a2
2
δ l = δ 2 − δ 1 = sin −1
s3
2a 1 a 2
s0 + s3 = a
+
2
s0 − s3
= a−
2
δ c = δ − − δ + = sin −1
s2
2a + a −
a) s 0 = 3, s 1 = −1, s 2 = 2, s 3 = −2
a 1 = 1, a 2 =
−2
2 2
δ l = sin −1
a+ =
1 ,
2
2
= − 1 π rad
4
5
2
a− =
2
δ c = sin −1
2
= 1. 107 1 rad
1
2
5
2
b) s 0 = 25, s 1 = 0, s 2 = 24, s 3 = 7
a1 =
δ l = sin −1
25 , a =
2
2
s3
2a 1 a 2
= sin −1
a+ =
32 = 4,
2
7
2
δ c = δ − − δ + = sin −1
25
2
25
2
a− =
25
2
= 0. 283 79 rad
s0 − s3 = 3
2
24
24 × 3
= 1 π rad
2
To plot the two cases Re E x ≡ X = cos x, ReE y ≡ Y = rcoxx − δ l , where r = a 2 /a 1 and x = ωt.
Case a) cos x, 2 cosx + π4 
1
0.5
-1
-0.8 -0.6 -0.4 -0.2
0.2
0.4
0.6
0.8
1
0.2
0.4
0.6
0.8
1
-0.5
-1
Case b) cos x, cosx − 0. 283 79
1
0.8
0.6
0.4
0.2
-1
0
-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 =
E ′0
i
= n 2n
i + nj
E0
R ij =
n −n
E ′′0
= ni + nj
i
j
E0
Thus from the figure
E ′′0 = E 0 R 12 + r 12 E 0 R 23 r 21 e iφ + r 12 E 0 R 23 R 21 R 23 r 21 e i2φ +. . . .
β
E ′′0 = E 0 R 12 + r 12 E 0 R 23 r 21 e iφ >R 21 R 23 e iφ 
n=0
E ′′0 = E 0 R 12 +
r 12 r 21 R 23
−iφ
e − R 21 R 23 
Similarly
E ′0 = E 0 r 12 r 23 + E 0 r 12 R 23 R 21 r 23 e iφ +. . . .
n
E ′0 = E 0
r 12 r 23
1 − R 21 R 23 e iφ
where the phase shift for the internally reflected wave is given by
φ=
ωn 2 2d
2π2d
=
c
λ2
Now for a plane wave
S i = 1 |E 0i |2
2v i
Thus
′′
|E ′′ |
R= S = 02
S
|E 0 |
2
′
′
T = vv 13 S = nn 3 S
1
S
S
From the above
R=
R 212 +
2r 12 r 21 R 23 R 12 cos φ − R 21 R 23  + R 12 r 21 R 23  2
1 + R 21 R 23  2 − 2R 21 R 23 cos φ
T = nn 3
1
r 12 r 23  2
1 + R 21 R 23  2 − 2R 21 R 23 cos φ
Since these two equations are simple functions of φ, 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 n 1 is are, n 2 is the coating, and n 3 is glass. In this case, we will have
n 1 < n 2 < n 3 , and R 21 R 23 < 0. Thus for T to be a maximum, from the above equation cosφ = −1, or
φ = π.
φ=
2π2d
λ
=π→d= 2
λ2
4
where λ 2 is the wavelength in the medium = λn 21
PHY 5346
Homework Set 13 Solutions – Kimel
2. 7.4 We have a nonpermeable conducting material, so μ = μ 0 , and we have J = σE, where σ is
the conductivity. The following figure describes the system:
The two boundary conditions that we must satisfy for plane waves are
E 0 + E ′′0 − E ′0 = 0
kE 0 − E ′′0  − k ′ E ′0 = 0
Or
′
E ′′0
= k − k′
E0
k+k
⃗ . Adding in this term in Maxwell’s equations for a
We must take into account the fact that ⃗
J = σE
plane wave, we get
k= ω
c
σ
k ′2 = μω 2 1 + i ω
Thus we can write
k′ =
μ ωα + iβ
with
α=
σ 2
1 +  ω
 +1
2
1/2
1/2
σ 2
1 +  ω
 −1
2
β=
Thus
1 − μ 0 cα − i μ 0 cβ
E ′′0
=
E0
1 + μ 0 cα + i μ 0 cβ
1) For a very poor conductor σ is very small, so keeping only first order in σ
α=
β=
1/2
σ 2
1 +  ω
 +1
2
≈1
1/2
σ 2
1 +  ω
 −1
2
≈
σ
2ω
σ
2) For the case of a very good conductor, ω
>> 1, so
α≈
σ =
2ω
β≈
2
μ 0 ωδ 2
2ω
=
1
ωδ μ 0 
σ =
1
2ω
ωδ μ 0 
where I have used (5.165) to relate the conductivity to the skin depth.
σ=
2
μ 0 ωδ 2
c
c
− i ωδ
1 − ωδ
E ′′0
δ − ωc − i ωc
2
ω
=
=
≈ −1 + ω
c
c
c 1 + i δ = −1 + c 1 − iδ
E0
1 + ωδ
+ i ωδ
δ + ωc + i ωc
R=
E ′′0
E0
2
= −1 + δω/c 2 +
ωδ
c
2
≈ 1 − 2δω/c
PHY 5347
Homework Set 1 Solutions – Kimel
1. 8.3
x
z
y
a)
∇ 2t + γ 2 ψ = 0,
ψ = E z TM or ψ = H z TE
As in class, we will use cylindrical coordinates, and assume
ψρ, φ = RρQφ
We get the two equations
∂ 2 Qφ = −m 2 Qφ with solns Qφ = e ±imφ , m = 0, 1, 2, . . .
∂φ 2
d 2 Rx + 1 dRx + 1 − m 2
x dx
x2
dx 2
Rx
Bessel eqn.
with regular solutions J m x, and singular solution (which we reject as nonphysical) N m x. Here
x = γρ.
Solutions:
TM: BC: J m x mn  = 0, and
E z ρ, φ = E 0 J m γ mn ρe ±imφ ,
m = 0, 1, 2, . . . . ; n = 1, 2, 3, . . . ; γ mn = x mn /R
Lowest cutoff frequencies:
ω mn =
Using the results of Jackson, p. 114,
γ mn
= x mn
R μ
μ
x 0n = 2. 405, 5. 52, 8. 654, . . .
x 1n = 3. 832, 7. 016, 10. 173, . . . .
x 2n = 5. 136, 8. 417, 11. 620, . . . .
TE: BC: J ′m x ′mn  = 0, and
E z ρ, φ = E 0 J m γ ′mn ρe ±imφ ,
m = 0, 1, 2, . . . . ; n = 1, 2, 3, . . . ; γ ′mn = x ′mn /R
Lowest cutoff frequencies:
γ ′mn
x ′mn
=
R μ
μ
ω mn =
Using the results of Jackson, p. 370,
x ′0n = 3. 832, 7. 016, 10. 173, . . .
x ′1n = 1. 841, 5. 331, 8. 536, . . . .
x ′2n = 3. 054, 6. 706, 9. 970, . . . .
From the above we see the lowest cutoff frequency is the TE mode
ω ′11 = 1. 841K, with K = 1/ R μ
The next four lowest cutoff frequencies are:
ω 01 = 2. 405K = 1. 31ω ′11
′
= 3. 054K = 1. 66ω ′11
ω 21
ω ′01 = 3. 832K = 2. 08ω ′11
ω 11 = 3. 832K = 2. 08ω ′11
b) From Eq. (8.63) in the text
1/2
βλ 
ξ λ + η λ ωωλ
ω
ω2
1 − ω λ2
2
For TM modes, η λ = 0, and for TE mode, ξ λ + η λ is of order unity. So for comparison purposes,
I’ll take
1/2
β 11 TE = f 1 x =
x
1−
1.841 2
x2
1 + 1. 841
x2
1/2
β 01 TM = f 2 x =
x
2
1 − 2.405
x2
where I’ve expressed the functions in terms of x = ω/K.
2
1/2
1+
1.841 2
x2
x
1− 1.841
2
2
x
14
12
10
8
6
4
2
0 2
4
x6
8
10
1/2
x
1− 2.405
2
2
x
14
12
10
8
6
4
2
0
3
4
5
6x
7
8
9
10
PHY 5347
Homework Set 1 Solutions – Kimel
1. 8.4
a) TM:
∇ 2t + γ 2 ψ = 0;
ψ|B = 0; E z = ψx, ye ±ikz−iωt ; B z = 0
Since we have a node along y = x, then we just take the antisymmetrized version for the square
waveguide, developed in class, ie,
nπy
mπy
nπx
ψx, y = E 0 sin mπx
a  sin a  − sin a  sin a 
Again
2
γ 2mn = cπ2 m 2 + n 2 , m, n = 1, 2, 3. . . . , but m ≠ n
a
TE:
∇ 2t + γ 2 ψ = 0;
∂ψ
| = 0; H z = ψx, ye ±ikz−iωt ; E z = 0
∂n B
Now the BC require ∂ψ
| = 0, but using a 45 0 rotation of coordinates, we see
∂n B
∂ =
∂n
1
2
− ∂ + ∂
∂x
∂y
Thus the combination
nπy
mπy
nπx
ψx, y = H 0 cos mπx
a  cos a  + cos a  cos a 
satisfies the above BC on the diagonal, as you can see by direct substitution.
2
γ 2mn = cπ2 m 2 + n 2 , m, n = 0, 1, 2, 3. . . . , but m ≠ n = 0
a
b) The lowest cutoff freq. are: TM: ω 12 or ω 21 .
TE: ω 01 or ω 10 . From Eq. (8.63) in the text
β 12 TM 
β 01 TE 
ω
1 − ω 212 /ω 2
ω
1 − ω 212 /ω 2
1/2
1/2
1+
ω 201
ω2
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: ω 11
TE: ω 01 (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:
ω mnp =
1
μ
x 2mn
p2π2
+
R2
L2
p = 0, 1, 2, . . . .
m = 0, 1, 2, . . . .
n = 1, 2, 3, . . . .
TE:
ω ′mnp =
1
μ
p2π2
x ′2
mn
+
L2
R2
p = 1, 2, 3, . . . .
m = 0, 1, 2, . . . .
n = 1, 2, 3, . . . .
Thus
ω mnp
1
μ
ω ′mnp
1
μ
=
x 2mn
p2π2
+
2
R
L2
=
x ′2
p2π2
mn
+
R2
L2
The lowest four frequencies are (in these units)
ω 010 = 2. 405
ω 110 = 3. 832
ω ′111 =
1. 841 2 + π 2
R
L
2
ω ′211 =
3. 054 2 + π 2
R
L
2
2. 405, 3. 832, 1. 841 2 + π 2 x 2 , 3. 054 2 + π 2 x 2
10
8
6
4
2
0
0.2
0.4
0.6
0.8
1
x
1.2
1.4
1.6
1.8
2
where x = R/L.
The answer is ”No.” ω ′111 and ω 010 cross when
1. 841 2 + π 2 x 2 = 2. 405
or x = 0. 492 58. For frequencies smaller than this cross over frequency, ω ′111 is lowest, whereas for
larger frequencies, ω 010 is lowest.
PHY 5347
Homework Set 3 Solutions – Kimel
3. 9.1 a)
ρx⃗, t = qδzδy − sin ω 0 tδx − d cos ωt
To illustrate the equivalence of the two methods, I’ll consider the lowest two moments.
n = 0 : Qt = ∫ ρx⃗, td 3 x = q = Reqe −i0⋅ωt 
n=1:⃗
pt = ∫ ρx⃗, tx⃗d 3 x = qdî cos ωt +
sin ωt = Reqdî + i e −i1⋅ωt 
So we identify ⃗
p = qdî + i  as the quantity to be used in Jackson’s formulas.
Arbitrary n: The n’th multipoles will contribute with maximum frequencies of ω n = nω.
b) The proof that we can write
β
ρx⃗, t = ρ 0 x⃗ + ∑ Re2ρ n x⃗e −inωt 
n=1
with
τ
ρ n x⃗ = 1τ ∫ ρx⃗, te inωt dt
0
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:
τ
ρ 0 x⃗ = 1τ ∫ qδzδy − sin ω 0 tδx − d cos ωtdt
0
q τ
Q = ∫ ρ 0 x⃗d 3 x = τ ∫ dt ∫ d 3 xδzδy − sin ω 0 tδx − d cos ωt = q
0
n=1:
τ
ρ 1 x⃗ = 1τ ∫ qδzδy − sin ω 0 tδx − d cos ωte iωt dt
0
τ
2q
⃗
px⃗ = ∫ d 3 xx⃗2ρ 1 x⃗ = τ ∫ dt ∫ d 3 xx⃗δzδy − sin ω 0 tδx − d cos ωt
0
=
as before.
2qd τ
iωt
τ ∫ 0 dte î cos ωt +
sin ωt = qdî + i 
PHY 5347
Homework Set 3 Solutions – Kimel
3. 9.2 First consider a rotating charge which is at an angle α at time t = 0.
Compared to the lecture notes for this problem, where we assumed α = 0, we should let
ωt → ωt + α. Thus using the result developed in class, we can write for this problem
1
Q α t = Re
3 qd 2
2
i
0
i −1 0
0
0
e −i2a e −i2ωt
0
From the figure
Q tot t = Q α1 t + Q α2 t + Q α3 t + Q α4 t
1
= Re
3 qd 2 −e −i π2 + e −i 3π2 − e −i 5π2 + e −i 7π2
2
1
0
i −1 0
0
Thus from the class notes
i
0
0
i −1 0
0
Q tot = 3 qd 2 4i
2
i
0
0
0
e −i2ωt
dP = c 2 Z 0 k 6 qd 2 3
2
dΩ
1152π 2
P=
2
161 − cos 4 θ =
c2Z0k6
qd 2 3
360π
2
2
16 =
And, of course, the frequency of the radiation is 2ω.
1 c 2 Z k 6 q 2 d 4 1 − cos 4 θ
0
32π 2
1 c2Z0k6q2d4
10π
PHY 5347
Homework Set 3 Solutions – Kimel
3. 9.3
Since the problem has azimuthal symmetry, we can expand Vr⃗, t (in the radiation zone) in terms
of Legendre polynomials:
Vr⃗, t = ∑ b l tr −l−1 P l cos θ
l
Using the orthogonality of the Legendre polynomials, the leading term of the expansion in the
radiation zone will be the l = 1 term.
1
b 1 t = 3 R 2 ∫ xVr⃗, tdx = 3 VR 2 cos ωt
2
2
−1
So,
Vr⃗, t =
⃗
p ⋅ r̂
p ⋅ r̂ −iωt
3 VR 2 cos ωt /r 2 = ⃗
cos ωt = Re
e
2
2
r
r2
with ⃗
p = 32 VR 2 ẑ , which should be used in the radiation formulas developed in lecture.
dP = c 2 Z 0 k 4 |p
⃗ |2 sin 2 θ
dΩ
32π 2
2
4
⃗ |2
P = c Z 0 k2 8π = 1 c 2 Z 0 k 4 |p
3
12π
32π
with ⃗
p = 32 VR 2 ẑ .
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),
3
Q lm = ∫ r l Y m∗
l ρd x
⃗ ⃗ ⃗ 3
M lm = − 1 ∫ r l Y m∗
l ∇⋅ r×J d x
l+1
a) Electric Dipole Radiation:
3
Q 1m = ∫ rY m∗
1 ρd x
= ∫ rδxδy−qδz − a cos ω 0 t − qδz + a cos ω 0 t + 2qδzY m∗
1 dxdydz
0
0
m∗
= −qa cos ω 0 tY m∗
1 0, φ + Y 1 π, φ = −qaδ m0 cos ω 0 tY 1 0 + Y 1 π = 0
b) Magnetic Dipole Radiation: Since the particles move in an orbit with no area,
⃗r × ⃗
J = 0 → M 1m = 0
c) Electric Quadrupole Radiation:
3
Q 2m = ∫ r 2 Y m∗
2 ρd x
= −qδ m0 a 2 cos 2 ω 0 tY 02 θ = 0 + a 2 cos 2 ω 0 tY 02 θ = π
= −qδ m0 a 2 Y 02 0cos 2ω 0 t + 1
where I have used cos 2 ω 0 t = cos 2ω 0 t + 1/2. Thus the Fourier Series decomposition of this
moment yields terms with frequency 0, and 2ω 0 . The first term does not contribute to radiation, and
the second can be written
Q 20 t = Re−2qa 2 Y 02 0e −2iω 0 t 
so Q 20 = −2qa 2 Y 02 0 is the quantity that is used in the radiation formulas of Jackson. Using Eq.
(9.151)
dP 2, 0 = Z 0 |a2, 0|2 X
⃗ 20 2
2
dΩ
2k
and from Eq. (9.169)
a2, 0 =
where I have used Y 02 0 =
ck 4
i5 × 3
5
.
4π
3 Q 20 = ck 4
2
i5 × 3
P2, 0 =
5
4π
Thus
|a2, 0|2 =
dP 2, 0 = Z 0
dΩ
2k 2
3 −2qa 2 
2
1 c2k8q2a4
30π
1 c2k8q2a4
30π
15 sin 2 θ cos 2 θ
8π
=
1 Z k 6 c 2 q 2 a 4 sin 2 θ cos 2 θ
0
32π 2
2π Z k 6 c 2 q 2 a 4 ∫ 1 1 − x 2 x 2 dx = 2π Z k 6 c 2 q 2 a 4 × 4
0
0
15
32π 2
32π 2
−1
P2, 0 =
1 Z k6c2q2a4
60π 0
PHY 5347
Homework Set 5 Solutions – Kimel
1. The system is described by
and is azimuthally symmetric
Rθ = R 0 1 + βtP 2 cos θ; βt = β 0 cos ωt; kR << 1
Q = ∫ ρr 2 drdφd cos θ = 2π ∫
1
−1
d cos θρ ∫
Rθ
r 2 dr
0
1
= 2π ρ ∫ R 30 1 + 3βP 1 P 2 d cos θ + Oβ 2  = 4π ρR 30 → ρ = 3 3 Q
3
3
4πR 0
−1
where I’ve used the fact that 1 = P 0 . Since the system is azimuthally symmetric, Q lm = δ m0 Q l0 .
Q lm = 2πρδ m0 ∫
Using Y 0l =
1
−1
dxY 0l ∫
Rθ
0
r l+2 dr =
2πρδ m0 1
0
∫ −1 dxR l+3
0 1 + l + 3βP 2 Y l
l+3
2l+1
P l and 1 = P 0 ,
4π
Q lm =
2πρδ m0
l+3
2l + 1 R l+3 2δ + l + 3β 2 δ
0
l0
4π
2l + 1 l2
Notice that the l = 0 term is time independent and thus does not contribute to the radiation.
Next consider the l = 2 term.
Q 20 t = 2 π ρ 5 R 50 β = ρ = 3 3 Q 2 π ρ 5 R 50 β =
5
4πR 0 5
Q 20 t = Re
3 R 2 Qβ 0 e −iωt
0
20π
3 R 2 Qβt
0
20π
3 R 2 Qβ
0
0
20π
Q 20 =
dP2, 0
Z
⃗ 20 2
= 02 |a2, 0|2 X
dΩ
2k
a E 2, 0 =
ck 4
i5 × 3
3 Q
2 20
⃗ 20 2 = 15 sin 2 θ cos 2 θ
X
8π
dP2, 0
ck 4
= Z 02
dΩ
2k 5 × 3
3 Q
2 20
2
× 15 sin 2 θ cos 2 θ
8π
4 2
ck 
|ck 4 Q 20 |
2
1 Z0
2
= 1 Z20
sin
θ
cos
θ
=
π
160 k
160 k 2
2
=
P=
3
R 20 Qβ 0
20π
π
2
sin 2 θ cos 2 θ
9
Z 0 k 6 c 2 R 40 Q 2 β 20 sin 2 θ cos 2 θ
3200π 2
1
9
9
Z k 6 c 2 R 40 Q 2 β 20 × 2π ∫ 1 − x 2 x 2 dx =
Z 0 k 6 c 2 R 40 Q 2 β 20 × 2π × 4
2 0
15
3200π
3200π 2
−1
P=
3 Z k6c2R4Q2β2
0
0
2000π 0
PHY 5347
Homework Set 5 Solutions – Kimel
2. The system is described by
It = I 0 cos ωt = ReI 0 e −iωt 
⃗
Jt = 1a Itδr − aδcos θφ̂
⃗=I
where I determined the normalization constant 1a by the condition ∫ ⃗
J ⋅ da
⃗
Jt = Re Ia0 δr − aδcos θφ̂ e −iωt
→⃗
J = Ia0 δr − aδcos θφ̂
⃗ and E
⃗ in the radiation zone given by Eq.(9.149). Since this
We use the general expression for H
system has no net charge density and there is no intrisic magnetization, the the expansion coefficients in
these equations are given by
a E l, m =
k2
3
⃗ ⃗
∫ Y m∗
l ik r ⋅ J j l krd x
i ll + 1
a M l, m =
k2
⃗
r×⃗
J j l krd 3 x
∫ Y m∗
l ∇⋅ ⃗
i ll + 1
a) ⃗r ⋅ ⃗
J = 0 in the first equation, so there is no electric multipole radiation. In spherical coordinates
⃗r × ⃗
J = −aJθ̂
⃗ in spherical coordinates given in the back of the book,
Using the formulas for ⃗
∇⋅A
⃗
∇ ⋅ ⃗r × ⃗
J = − 1 ∂ J sin θJ = − cos θ J − ∂ J
sin θ ∂θ
sin θ
∂θ
The first term does not contribute, because cos θ = 0, while the second term can be written, using
the chain rule,
⃗
∇ ⋅ ⃗r × ⃗
J = sin θ
∂ J
∂ cos θ
The problem has azimuthal symmetry, so m = 0. Realizing derivatives of δ −functions are defined
by integration by parts,
a M l, m =
δ m0 k 2
∂ J j krd 3 x =
∫ Y 0∗l sin θ ∂ cos
l
θ
i ll + 1
ik 2
∂ sin θY 0∗  Jj l krd 3 x
∫ ∂ cos
l
θ
ll + 1
a M l, 0 =
i2πk 2 I 0 a 2 j ka ∂ sin θY 0 |
l
l cos θ=0
a
∂ cos θ
ll + 1
a M l, 0 =
i2πk 2 I 0 a 2 j l ka1 − x 2  1/2 d Y 0 x|
x=0
a
dx l
ll + 1
Since Y 0l x is either an even or odd polynomial in x, then only odd l contribute to a M l, 0. This
⃗ and E
⃗ in the radiation zone are known through
determines the expansion coefficients, and thus H
Eq.(9.149). The power distribution is given by Eq. (9.151)
b) From our previous answers, we see a E l, m = 0, and that the lowest magnetic multipole
contribution is a M 1, 0.
2
1/2
a M 1, 0 = i2πk I 0 aj 1 ka1 − x 2  d Y 01 x|x=0
dx
2
Using
j 1 ka → ka ;
3
d Y 0 x| =
dx 1 x=0
a M 1, 0 = i2πk 3 I 0 a 2
M l0 =
ik 3
3
1 = ik 3 2 M
l0
24π
3
1
24π
i2πk 3 I 0 a 2
2
3
4π
=
3 I πa 2
4π 0
Note that you would get the same answer, if you used Eq. (9.172) directly.
From Eq. (9.151)
dP = Z 0
dΩ
2k 2
3
2πk Fa
2
1
24π
2
3 sin 2 θ = 1 Z 0 k 4 I πa 2  2 sin 2 θ
0
8π
32π 2
If we compare this result with the one that we get for an elementary magnetic dipole, which is
given by Eq. (9.23)
⃗ /c,
with the substitution ⃗
p→m
dP = 1 Z 0 k 4 |m
⃗ |2 sin 2 θ
dΩ
32π 2
Thus we may identify
⃗ | = I 0 πa 2
|m
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 n̂ 0 = ẑ ,
dσ ̂ , n̂ , n̂  = k 4 a 6 5 − |̂ 0 ⋅ n̂ |2 − 1 |n̂ ⋅ ẑ × ̂ |2 − ẑ ⋅ n̂
0
dΩ 0 0
4
4
Orienting the system as
and using
̂ 0 = α 0 x̂ + β 0 , with |α 0 |2 + |β 0 |2 = 1
n̂ = cos θẑ + sin θx̂
then
dσ ̂ , n̂ , n̂  = k 4 a 6 5 − |α |2 sin 2 θ − 1 |β 0 |2 sin 2 θ − cos θ
0
4
4
dΩ 0 0
Using the result for the perfectly conducting sphere Eq. (10.14)
dσ n̂ , ̂, ̂ 0 , n̂ 0  = k 4 a 6 ̂ ∗ ⋅ ̂ 0 − 1 ẑ × ̂  ⋅ n̂ × ̂ ∗  2
0
2
dΩ
Using ̂ ⊥ =
dσ n̂ , ̂ ⊥ , ̂ 0 , n̂ 0  = k 4 a 6 β 0 − 1 β 0 cos θ 2 = k 4 a 6 |β 0 |2 1 − 1 cos θ
2
2
dΩ
2
Similarly ̂ || = x̂ ′
dσ n̂ , ̂ , ̂ , n̂  = k 4 a 6 α cos θ − 1 α 2 = k 4 a 6 |α |2 cos θ − 1
0
|| 0
0
0
dΩ
2 0
2
By definition
2
dσ ̂ , n̂ , n̂  = dσ n̂ , ̂ ⊥ , ̂ 0 , n̂ 0  + dσ n̂ , ̂ || , ̂ 0 , n̂ 0 
dΩ
dΩ
dΩ 0 0
= k 4 a 6 |β 0 |2 1 − 1 cos θ
2
2
+ |α 0 |2 cos θ − 1
2
2
which simplifies to
dσ ̂ , n̂ , n̂  = k 4 a 6 5 − |α |2 sin 2 θ − 1 |β |2 sin 2 θ − cos θ
0
4 0
4
dΩ 0 0
using |α 0 |2 + |β 0 |2 = 1, and cos 2 θ = 1 − sin 2 θ.
b) If ̂ 0 is linearly polarized making an angle φ with respect to the x axis, then
̂ 0 = α 0 x̂ + β 0
= cos φx̂ + sin φ , so α 0 = cos φ, β 0 = sin φ
Then from part a)
dσ ̂ , n̂ , n̂  = k 4 a 6 5 − |α 0 |2 sin 2 θ − 1 |β 0 |2 sin 2 θ − cos θ
4
4
dΩ 0 0
= k 4 a 6 5 − cos 2 φ sin 2 θ − 1 sin 2 φ sin 2 θ − cos θ
4
4
Using cos2φ = cos 2 φ − sin 2 φ, this expression simplifies to
dσ ̂ , n̂ , n̂  = k 4 a 6 5 1 + cos 2 θ − 3 sin 2 cos 2φ − cos θ
8
dΩ 0 0
8
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
γ
−βγ
−βγ
γ
=
=
γ2
−β 2 γ 2
γ1
−β 1 γ 1
−β 2 γ 2
γ2
−β 1 γ 1
γ1
γ2γ1 + β2γ2β1γ1
−γ 2 β 1 γ 1 − β 2 γ 2 γ 1
−γ 2 β 1 γ 1 − β 2 γ 2 γ 1
γ2γ1 + β2γ2β1γ1
So
γ = γ2γ1 + β2γ2β1γ1
βγ = γ 2 β 1 γ 1 + β 2 γ 2 γ 1
βγ
γ β γ + β2γ2γ1
β1 + β2
β= γ = 2 1 1
=
γ2γ1 + β2γ2β1γ1
1 + β2β1
Or
v = v 1 +v 1vv22
1 + 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
vu ′||
c2
1+
u⊥ =
;
u ′⊥
vu ′
γ1 + c 2|| 
and
dt = dt ′ γ1 +
vu ′||

c2
Thus taking the differential of the first equation above and using the second equation for dt,
vu ′
du ||
≡ a || =
dt
1 + c 2|| a ′|| − u ′|| + v cv2 a ′||
vu ′||
2
=
γ1 + c 2  3
1 − vc 2
vu ′||
1 + c 2  3
Similarly,
vu ′
du ⊥ ≡ a =
⊥
dt
1 + c 2||
a ′⊥ − u ′⊥ cv2 a ′||
vu ′
γ 2 1 + c 2||  3
This is equal to the expression we want to prove,
2
du ⊥ ≡ a =
⊥
dt
1 − vc 2
vu ′||
1 + c 2  3
v × a
⃗′ × ⃗
a ′⊥ + ⃗
u′ 
c2
since the BAC - CAB theorem shows that
v × a
⃗′ × ⃗
a ′⊥ + ⃗
u′  =
c2
3/2
1+
vu ′||
c2
a ′⊥ − u ′⊥ v2 a ′||
c
a ′||
PHY 5347
Homework Set 7 Solutions – Kimel
3. 11.15
⃗ =β
⃗ = 0. Then
⃗ || ẑ , so β
⃗⋅E
⃗⋅B
From Eq. (11.149), it is clear that we should take β
⃗+β
⃗
⃗×B
E ′ = γE
⃗ ′ = γB
⃗−β
⃗
⃗×E
B
The vectors in parentheses should make the same angle wrt the x axis θ ′ if they are to be parallel.
This can best be seen from the figure,
From the figure
⃗ ′ = γE 0 î − β2E 0  sin θî + β2E 0 cos θ̂ 
E
⃗ ′ = γcos θ2E 0 î + sin θ2E 0 ̂ − βE 0 ̂ 
B
Thus
tan θ ′ =
2β cos θ
2 sin θ − β
=
1 − 2β sin θ
2 cos θ
2 cos θ ⋅ 2β cos θ − 1 − 2β sin θ ⋅ 2 sin θ − β = 0
or
2β 2 sin θ − 5β + 2 sin θ = 0
This quadratic equation has the solution
β=
1
5−
4 sin θ
25 − 16 sin 2 θ
where I’ve chosen the solution which give β = 0 if θ = 0.
If θ << 1, then β → 0, and the original fields are parallel.
If θ → π/2 then β = 14 5 − 3 = 1/2. γ =
2
3
⃗ ′ = 0 + Oθ − π ̂
E
2
⃗′ = B
⃗ ′ = γ2E 0 ̂ − 1 E 0 ̂  = γE 0 3 ̂
B
2
2
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)
q
L = − 1 mu α u α − c u α A α
2
(invariant Lagrangian)
Show this Lagrangian gives the correct eqn. of motion, ie, Eq. (12.2)
du α = e F αβ u β
mc
dτ
τ2
The Action is A = ∫ Ldτ.
τ1
δA = 0 yields the Lagrange equations of motion
d ∂L − ∂L = 0
dτ ∂u α
∂x α
d ∂L = −m d u α − q ∂A α dx μ
c ∂x μ dτ
dτ
dτ ∂u α
∂L = − q u ∂ α A β
c β
∂x α
∂L
∂L
− ∂x
= 0 yields
So dτd ∂u
α
α
q
q
q
m d u α = c u β ∂ α A β − c ∂ μ A α u μ = c u β ∂ α A β − ∂ β A α 
dτ
or
d u α = q F αβ u
β
mc
dτ
PHY 5347
Homework Set 8 Solutions – Kimel
3. 12.2 (a)
L ′ = L + d λt, ⃗
x
dt
x  − λt 1 , ⃗
x = 0 → L and L ′ yield the same Euler-Lagrange Eqs. of Mot.
δ ∫ L ′ − Ldt = δλt 2 , ⃗
t2
t1
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.
⃗ − eφ with A μ = φ, A
⃗
⃗⋅A
L = −m 1 − u 2 + eu
If A α → A α + ∂ α Λ, then
φ → φ + ∂0Λ
⃗ →⃗
A
a−⃗
∇Λ
where the minus sign in the second equation should be noticed. Thus
⃗ − eφ − eu
⃗⋅A
⃗⋅⃗
L → −m 1 − u 2 + eu
∇Λ − e∂ 0 Λ
Now
μ
d Λt, ⃗
x = ∂ μ Λt, ⃗
x ∂x = ∂ Λ + ⃗
∇Λ ⋅ ⃗
u
dt
∂x
∂t
∂t
Or
⃗ − eφ − e d Λt, ⃗
⃗⋅A
L → −m 1 − u 2 + eu
x
dt
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
⃗ 0 along z, ⃗
v 0 along y
a) Take E
Generally
⃗
dp
v ×B
⃗+ ⃗
⃗ ;
=e E
c
dt
dE = ev⃗ ⋅ E
⃗
dt
⃗ = 0, and E
⃗ = E 0 ẑ
Since B
dp z
= eE 0
dt
dp y
=0
dt
The initial condition ⃗
p0 = mv 0 and the above equations show the subsequent motion is in the
y − z plane. Consistent with this initial condition, we have
p y t = mv 0 ; p z t = eE 0 t
Et =
⃗
p 2 tc 2 + m 2 c 4 =
m 2 v 20 c 2 + m 2 c 4 + ceE 0 t 2 =
ω 20 + ceE 0 t 2
Using
⃗
p = mγv⃗ = E2 ⃗
v
c
v y t =
yt =
ω0
where ρ ≡ ceE
. So
0
p y t
=
Et/c 2
mv 0 c 2 t
∫
ceE 0 0
mv 0 c 2
ω 20 + ceE 0 t 2
mv 0 c
dt
=
sinh −1 t/ρ
eE
2
2
0
ρ +t
mv 0 c
tceE
sinh −1  ω 0 0 
eE 0
yt =
Eq. (1)
Similarly
p z t
=
Et/c 2
v z t =
eE 0 tc 2
ω 20 + ceE 0 t 2
Thus
2
zt = eE 0 c
ceE 0
∫0
t
tdt
=c
ρ2 + t2
ρ 2 + t 2  − ρ
Eq.(2)
b) From Eq. (1)
t=
ω0
eE y
eE
sinh mv00 c  = ρ sinhky, with k = mv 0c
0
ceE 0
Then from Eq.(2)
z = cρ
Let us plot
sinh 2 ky + 1 − 1
Eq.(3)
sinh 2 x + 1 − 1
20
18
16
14
12
10
8
6
4
2
0
1
2
x
3
4
5
For small t : t/ρ << 1, and ky << 1. Thus we can taylor expand Eq.(3) and get
z = cρk 2 y 2 /2
which is quadratic in y giving a parabolic shape.
For large t : t/ρ >> 1, and we see the sinh term dominates in Eq.(3) and we get
z
which is an exponential shape.
cρe ky
2
PHY 5347
Homework Set 9 Solutions – Kimel
2. 12.5
a) The system is described by
⃗ and E
⃗ . We want E
⃗ ′⊥ = 0 = γE + u⃗ × B
⃗ .
Background: particle having m, e. Choose ⃗
u ⊥ to B
c
⃗ = −E
⃗ ; now B
⃗× ⃗
⃗ = cE
⃗×B
⃗ . Using BAC - CAB on the lhs of the equation gives
Thus u⃗c × B
u×B
⃗ ⃗
⃗
u = c E×B
2 . Thus from the figure,
B
⃗ ⃗
⃗
u = c E ×2 B = c E ẑ
B
B
Then, using Eq. 11. 149
⃗ || = 0;
E
⃗ ′⊥ = 0;
E
⃗ ′⊥ = 1 B
⃗ ′ = 0; B
⃗
B
||
γ =
1−
u
c
2
⃗
B
So
⃗ ′⊥ =
B
1−
E
B
2
⃗ =
B
B 2 − E 2 B̂
2
B2
Now from the class notes,
′
⃗′ = ⃗
du
⃗ B ′ , where ω
⃗ B ′ = eB′ , where in this case E ′ is the energy of the particle.
u′ × ω
′
dt
E
I’ll choose the same boundary conditions as in class, decribed in the figure.
So
⃗
u ′⊥ = ω B ′ acosω B ′ t ′ ̂ 3 − sinω B ′ t ′ ̂ 1 
where u ′⊥ t = 0 = ω B ′ a (ie, the BC determine a.
⃗
x ′ t ′  = u ′|| t ′ ̂ 2 + a̂ 3 sin ω B ′ t ′ + ̂ 1 cos ω B ′ t ′ 
Consider the inverse Lorentz transformation between the frames,
γ
βγ 0 0
ct ′
βγ
γ
0 0
z′
x
0
0
1 0
x′
y
0
0
0 1
y′
ct
z
=
or
t = γt ′ + βγz ′ /c = γt ′ + βγa sin ω B ′ t ′ /c ≡ ft ′  → t ′ = f −1 t
So
zt = βγcf −1 t + γa sin ω B ′ f −1 t
xt = a cos ω B ′ f −1 t
′ −1
f t
yt = u ||0
⃗ field alone. Then the
b) If |E| > |B|, one can transform to a frame where the field is a static E
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 = − 1 ∂ α A β ∂ α A β − 1c J α A α
8π
which can be rewritten
L = − 1 ∂ β A α ∂ β A α − 1c J α A α
8π
Using the Euler-Lagrange equations of motion,
∂β
∂L
− ∂L = 0
∂A α
∂∂ β A α 
Noting
∂L
= − 1 ∂βAα
4π
∂∂ β A α 
∂L = − 1 J α
c
∂A α
The Euler-Lagrange equations of motion are
α
∂ β ∂ β A α  = ∂ β ∂ β A α  = 4π
c J
or
α
∂ β ∂ β A α − ∂ α A β + ∂ α A β  = ∂ β F βα + ∂ β ∂ α A β = 4π
c J
If we assume the Lorentz gauge, ∂ β A β = 0, then the above reduces to
α
∂ β F βα = 4π
c J
Maxwell’s equations, given by Eq. (11.141).
b) Eq. 12. 85 gives
−
1 F F αβ  − 1 J α A α
c
16π αβ
The term in parentheses can be written
F αβ F αβ = 2∂ α A β ∂ α A β − 2∂ α A β ∂ β A α  + 2A β ∂ β ∂ α A α
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 ∂ α A β ∂ β A α .
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
⃗
⃗ x⃗, t = eβ |ret
A
R
φx⃗, t = e |ret ,
R
Let us assume that we observe the radiation close enough to source so R/c << 1 and t ′ ≅ t. Then
in the radiation zone
⃗ = −n̂ ∂ × A
⃗ = eβ̇ × n̂
⃗ = n̂ ∂ × A
⃗ =⃗
B
∇×A
′
cR
∂R
c∂t
⃗ =B
⃗ × n̂
E
dPt
⃗ 2 = e 2 β̇ × n̂ 2 = e 2 |v̇ |2 sin 2 θ
= c RB
4π
4πc
dΩ
4πc 3
where θ is the angle between n̂ and β̇ (assuming here the particle is moving linearly)
2
Pt = 2e 3 |v̇ |2
3c
Let the time-average be defined by
τ
⟨ft⟩ ≡ 1τ ∫ ftdt
0
Then
dPt
dΩ
=
e 2 sin 2 θ |v̇ |2
4πc 3
2
⟨Pt⟩ = 2e 3 |v̇ |2
3c
a) Suppose ⃗
xt = ẑ a cos ω 0 t. Then v̇ = ddt 2z = −aω 20 cos ω 0 t
2
|v̇ |2
τ
= aω 20  2 1τ ∫ cos 2 ω 0 tdt = aω 20  /2
0
So
dPt
dΩ
=
e 2 aω 2  2 sin 2 θ
0
8πc 3
2
2
⟨Pt⟩ = e 3 aω 20 
3c
b) Suppose ⃗
xt = Rî cos ω 0 t +
sin ω 0 t. Then
2
v̇ t = −Rω 20 î cos ω 0 t +
sin ω 0 t
k̂
î
n̂ × v̇ =
sin θ cos φ
sin θ sin φ
cos θ
−Rω 20 cos ω 0 t
−Rω sin ω 0 t
0
2
dPt
= e 3 Rω 20  2 cos 2 θ sin 2 ω 0 t + cos 2 ω 0 t + sin 2 θ sin 2 ω 0 t + φ 
dΩ
4πc
dPt
dΩ
⟨Pt⟩ =
=
e 2 Rω 2  2 1 + cos θ
0
2
4πc 3
2
e 2 Rω 2  2 2π ∫ 1 1 + x  dx = 2e 2 Rω 2  2
0
0
2
−1
3c 3
4πc 3
2
PHY 5347
Homework Set 11 Solutions – Kimel
2. 14.12
a) From Jackson, Eq (14.38)
dP = e 2
dΩ
4πc
⃗ × β̇
n̂ × n̂ − β
⃗ 5
1 − n̂ ⋅ β
2
Using azimuthal symmetry, we can choose n̂ in the x-z plane.
From the figure
n̂ = cos θẑ + sin θx̂
⃗t ′  = − a ω 0 sin ω 0 t ′ ẑ = −β sin ω 0 t ′ ẑ
β
c
2
aω
β̇ t ′  = − c 0 cos ω 0 t ′ ẑ = −ω 0 β cos ω 0 t ′ ẑ
⃗ × β̊ = 0
Using β
⃗ × β̇
n̂ × n̂ − β
2
= n̂ × β̇
2
and
n̂ × β̇ =
ω 0 β sin θ cos ω 0 t ′
So
sin θ cos 2 ω 0 t ′
dP t ′  = e cβ
dΩ
4πa 2 1 + β cos θ sin ω 0 t ′  5
2
Defining φ = ω 0 t ′ ,
4
2
sin 2 θ cos 2 φ
e 2 cβ 4 1 2π
⟨ dP t ′ ⟩ =
dφ
∫
dΩ
4πa 2 2π 0 1 + β cos θ sin φ 5
Or, doing the integral,
4 + β 2 cos 2 θ
e 2 cβ 4
⟨ dP ⟩ =
sin 2 θ
2
7/2
2
2
dΩ
32πa 1 − β cos θ
c)
4+.05 2 cos 2 θ
sin 2 θ
7/2
1−.05 2 cos 2 θ
5
4
3
2
1
0
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
4+.95 2 cos 2 θ
sin 2 θ
7/2
1−.95 2 cos 2 θ
300
250
200
150
100
50
0
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:
⟨T⟩ = 1 ⟨ dV ⟩r
2 dr
If V = ar n , then
⟨T⟩ = n ⟨V⟩
2
In our case V = 12 kr 2 , with k = mω 20 , so n = 2 and
⟨T⟩ = ⟨V⟩
Or,
⟨ dV ⟩ = E
r
dr
We are given
dE = − τ ⟨ dV
m
dr
dt
2
⟩
This can be rewritten
dE = − τ kE
m
dt
So
τ
E = E 0 e − m kt = E 0 e −τω 0 t = E 0 e −Γt
2
Similarly,
⃗
τ ⟨ 1 dV ⟩L
dL
⃗
= −m
r dr
dt
But 1r dV
= mω 20 , so
dr
L = L 0 e −τω 0 t = L 0 e −Γt
2
PHY 5347
Homework Set 12 Solutions – Kimel
2. 16.2 V = −, q = −e
dE = − τ ⟨ dV
m
dt
dr
2
⟩
If V = ar n , then the Virial theorem tells us
⟨T⟩ = n ⟨V⟩
2
In the present case, n = −1, so
2
E = ⟨T⟩ + ⟨V⟩ = 1 ⟨V⟩ = − Ze
2r
2
dV = Ze 2
dr
r2
Now
dE = − τ ⟨ dV
m
dr
dt
2
⟩
gives
d 1 = 2Ze 2 τ
dr rt
mr 4 t
or
r 2 dr = −2Ze 2 τdt/m
2
But τ = 23 ce3 m , so
r 2 dr = −3Zcτ 3 τt
Integrating both sides gives
r 3 t = r 30 − 9Zcτ 3 τt
b) At this point, for simplicity of notation, I’m going to take c =  = 1. Then from problem
14.21,
1 = 2 e 2 Ze 2  4 m
T
3
n5
We are given
2
r = n a0
Z
2
Where a 0 = me1 2 = Bohr radius, and τ = 23 em
Z
− dn = − Z dr = Z 32 Zτ 2 = Z 3
dt
2a 0 n dt
2a 0 n r
2a 0 n
a0n2
2
2
Z 2 em
3
2
4
6
= 2 Zm e 5
3
n
in agreement with the result of problem 14.21.
c) From part b
But rt =
t=
r 30 − r 3 t
9Zτ 2
−
n 2f a 0
Z
n 2f a 0
n2a
, r 0 = iZ 0
Z
t=
n 2i a 0
Z
3
9Zτ
3
n6 − n6
= 1 a 30 i 4 2 f
9
Z τ
2
In our present case Z = 1, so
t= 1
9
1
me 2
3
n 6i − n 6f
2 e2
3 m
2
= 1 e −10 n 6i − n 6f 
4m
In these units, (from the particle data book) MeV −1 = 6. 6 × 10 −22 s. e 2 = α = 1/137, and
m = 207 ×. 511MeV.
t=
1 × 6. 6 × 10 −22 s
n 6i − n 6f  = 7. 53 × 10 −14 n 6i − n 6f s
4 × 207 ×. 5111/137 5
For the cases desired,
t 1 = 7. 53 × 10 −14 10 6 − 4 6 s = 7. 5 × 10 −8 s
t 2 = 7. 53 × 10 −14 10 6 − 1 6 s = 7. 530 0 × 10 −8 s
Just as a check on working with these units, notice
2
1
τ = 2 em = 2
6. 6 × 10 −22 s = 6. 29 × 10 −24 s
3 . 511 × 137
3
in agreement with what we found before.