PHY4324/Fall 09: EM II
HOMEWORK ASSIGNMENT #5: SOLUTIONS
due by 11:45 p.m. Wednesday 10/07
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly. Every (algebraic) final result
must be supplemented by a check of units. Without such a check, no more
than 75% of the credit will be given even for an otherwise correct solution.
1. Assuming that the cables in Example 7.13 are held at potential difference V and carry current I, find the power
transported down the cables. [25 points]
The magnetic field in between the cylinders at distance s from the origin is B = µ0 I φ̂/2πs. Assuming that
the inner cylinder carries charge λ per unit length, the electric field in the same region is E = λŝ/2πsǫ0 . The
Poynting vector S = E × B/µ0 = λI ẑ/4π 2 s2 ǫ0 . The amount of energy transported per unit time through the
area in between the cylinders
Z b
Z b
b
λI
ds
λI
ln
=
dssS =
P = 2π
2πǫ
s
2πǫ
a
0 a
0
a
The charge per unit length can be expressed via voltage using
Z b
Z b
V =
dsE = (λ/2πǫ0 )
ds/s = (λ/2πǫ0 ) ln(b/a).
a
a
Therefore, P = V I.
2. G8.4 [25 points] a) The z-component of the force acting on the unit area of the equidistant surface is
(T · da)z = Tzx dax + Tzy day + Tzz daz .
Now, dax = day = 0 whereas daz = −sdsdφ. The electric field at the equidistant surface has obviously only a
radial component:
E=
s
1
s
q
2q
√
ŝ =
ŝ
2
2
2
2
2
4πǫ0 s + a
2πǫ0 (s + a2 )3/2
s +a
1
1
Tzz = ǫ0 (Ez2 − E 2 ) = −ǫ0 E 2 .
2
2
The total force is obtained by integrating over the area
2 Z ∞
q
s2
Fz = 2π
dss 2
2πǫ0
(s + a2 )3
0
Solve the integral via a substitution t = s2 :
Z
Z
Z
Z ∞
1 ∞
1 ∞ t + a2 − a2
1 ∞
a2
t
1
s2
=
=
=
−
dt
dt
dt
dss 2
(s + a2 )3
2 0
(t + a2 )3
2 0
(t + a2 )3
2 0
(t + a2 )2
(t + a2 )3
0
1 1
1
1
=
− 2 = 2
2
2 a
2a
4a
Substituting this result into Fz , we obtain
Fz =
q2
4πǫ0 (2a)2
(1)
2
b) In this case, the radial components of the fields due to point charges cancel while the z-component add up.
Assuming that the positive charge is above the plane,
E = −2
a
a
q
q
√
ẑ = −
ẑ
2
2
2
2
2
4πǫ0 (s + a ) s + a
2πǫ0 (s + a2 )3
Now,
ǫ
1
Tzz = ǫ0 (Ez2 − E 2 ) = Ez2 .
2
2
Integrating over s with the help of the same substitution, we obtain the same result as before but with an
opposite sign.
3. G8.5 [25 points]
a) If the upper plate carriers charge σ > 0 per unit area, the electric field in the between the plates is
Ex = 0, Ey = 0, Ez = −σ/ǫ0 . All off-diagonal components of Tij are zeroes, while the diagonal ones are
equal to
Txx = Tyy = −
ǫ0
σ2
σ2
ǫ0 2
, Tzz = Ez2 =
Ez = −
2
2ǫ0
2
2ǫ0
b) The total force on the upper plate
Z
Z
Z
σ2
Fz = (T · da)z = Tzz daz = − Tzz dxdy = −
A,
2ǫ0
2
σ
.
where A is the plate area. The force per unit area is fz = − 2ǫ
0
c) −Tzz = σ 2 /2ǫ0 is the momentum crossing per unit are per unit time of the upper plate.
d) The recoil force is the momentum delivered per unit time, so the force per unit area on the top plate is
-σ 2 ẑ/2ǫ0 .
4. G9.9 [25 points]
a) For a wave moving in the negative x direction, k = −(ω/c)x̂ and k · r = −(ω/c)x. The polarization vector
n̂ = ẑ. The electric field of the wave
ω
E(x, t) = E0 cos( x + ωt)ẑ.
c
The magnetic field is along the y axis
B = (x, t) =
ω
E0
cos( x + ωt)ŷ
c
c
b)
ω
k = √ (x̂ + ŷ + ẑ)
3c
Since the polarization vector is parallel to the xz plane, it must be of the form ax̂ +√
bẑ, where a and b are
some coefficients. On the other hand, since n̂ must be orthogonal to k, n̂ · k = (ω/ 3c)(a + b) = 0, that
is, a = −b. Finally, since n̂ is a unit vector,
x̂ − ẑ
n̂ = √
2
The direction of the magnetic field is found from
1
k × n̂ = √ (−x̂ + 2ŷ − ẑ).
6
3
The electric and magnetic fields are given by
x̂ − ẑ
ω
√ (x + y + z) + ωt
√
3c
2
E0
−x̂ + 2ŷ − ẑ
ω
√
B(x, y, z, t) =
cos √ (x + y + z) + ωt
c
3c
6
E(x, y, z, t = E0 cos
(2)
(3)
5. Bonus: G8.12 [25 points]–we did a DA estimate in class but do a real computation this time.
The field of the electric charge
qe
bf r
4πr3
E=
The field of the magnetic charge
B=
where r′ = r − dẑ and r′ = r2 + d2 − 2rd cos θ
The linear momentum density
1/2
µ0 qm r′
µ0 qm r′
=
4π r3
4π (r′ )3
.
′
~ = 1 S = ǫ0 E × B = µ0 qe qm r × r
P
2
2
3
′3
c
16π r r
The angular momentum density
′
~ = r×P
~ = µ0 qe qm r × (r × r )
L
16π 2
r3 r′3
The triple cross product reduces to
r × (r × r′ ) = r(r · r′ ) − r2 r′ = dr2 (ẑ − r̂ cos θ)
~ vanish, whereas the z component becomes
Upon integration, the x and y components of L
µ0 qe qm dr2 (1 − cos2 θ)
µ0 qe qm d(1 − cos2 θ)
µ0 qe qm (r × (r × r′ ))z
=
=
2
3
′3
2
3
′3
16π
r r
16π
r r
16π 2
rr′3
Lz =
The total z component of the angular momentum
Lz =
Z
Lz dτ =
Z
∞
Z
0
π
0
Z
0
2π
µ0 qe qm
Lz r sin θdrθdφ =
8π
Z
2
π
0
Z
0
∞
r
(r2
+
d2
− 2dr cos θ)
3/2
dr (1 − cos2 θ) sin θdθ
Denote t ≡ cos θ and do the integral over r first noticing that
r
(r2
+
d2
− 2drt)
3/2
=
d
rt − d
1
√
2
2
d(1 − t ) dr r + d2 − 2drt
Therefore,
Z
∞
0
Now do the integral over θ
∞
d
rt − d
1
1
√
dr
=
=
2 ) dr
3/2
2 + d2 − 2drt 0
2
2
d(1
−
t
d(1
− t)
r
(r + d − 2drt)
r
Z
1
−1
1 − t2
dt =
1−t
Z
1
(1 + t)dt = 2.
−1
Finally,
L=
µ0 qe qm
ẑ.
4π
!