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6.641
�� Electromagnetic Fields, Forces, and Motion
Spring 2005
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6.641 — Electromagnetic Fields, Forces, and Motion
Spring 2005
Problem Set 6 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 6.1
Figure 1: The potential on the surface of a dielectric cylinder in free space is imposed (Image by MIT
OpenCourseWare)
A
As always, we start by finding boundary conditions.
I. Φ(r = a) = V0 cos nφ
II. Φ for r < a is finite
III. Φ(r = ∞) = 0
Boundary conditions 1 and 2 suggest for r ≤ a
Φ(r, φ) = Ar n cos nφ
�= V �
�
�nφ
Φ(r = a, φ) = Aan�
cos�nφ
0 cos
⇒A=
V0
an
r≤a
1
Problem Set 6
6.641, Spring 2005
Boundary condition III suggests for r ≥ a
Φ(r, φ) = Br −n cos nφ
�= V �
�
�nφ
Φ(r = a, φ) = Ba−n�
cos�nφ
0 cos
⇒ B = V0 an
r≥a
So
Φ(r, φ) =
�
V0 n
an r
n
V0 arn
cos nφ r ≤ a
cos nφ r ≥ a
⎞
0
�
1
∂Φ
∂
Φ
∂Φ
−
→
⎟
⎜
+ îφ
+ îz � ⎠
E = −�Φ = − ⎝îr
∂r
r ∂φ
�∂z
⎛
B
⎧
�
�
� �
⎨ V0 n r n−1 − cos(nφ)îr + sin(nφ)îφ
r ≤ a
−
→
a
a
�
E = V n � �n+1 �
⎩
0 a
cos(nφ)îr + sin(nφ)îφ
r≥a
a
r
−
→
−
→
σf = îr · (ε0 E (r = a+ ) − ε E (r = a− ))
⎛
⎞
⎛
⎞
=
⎟
⎜
⎟
ε0 V0 n ⎜
⎜ a ⎟ cos(nφ) + εV0 n ⎜ a ⎟ cos(nφ)
⎝
⎠
⎝
a
a
a
a ⎠
����
����
=1
=1
The answer:
σf =
V0 n
(ε + ε0 ) cos(nφ)
a
From Table 6.8.1 of Haus & Melcher (an aside for cultural purposes):
−
→
−
→
σtotal = îr · (ε0 E (r = a+ ) − ε0 E (r = a− ))
2V0 nε0
cos(nφ)
a
V0 n
(ε0 − ε) cos(nφ)
σp = σtotal − σf =
a
[polarization plus free surface charge densities]
σtotal =
[polarization surface charge density]
C
�
�
�
��
∞ �
2πn
2πn
A0 �
An cos
+
φ + Bn sin
φ
Φ(r = a, φ) =
2
2π
2π
n=1
Average Φ over 2π = 0, so A0 = 0. Φ(r = a, φ) is purely even, so Bn = 0.
2
Problem Set 6
6.641, Spring 2005
D
∞
�
Φ(r = a, φ) =
An cos(nφ) =
m=1
�
V0
2
− V20
− π2 < φ < π2
π
3π
2 < φ< 2
If we multiply both sides by�cos(mφ), integrate over one period (2π) in φ, and use the orthogonality relation
� 2π
0 m �= n
cos (nφ) cos (mφ) dφ =
0
π m=n
Am =
2V0
mπ
Φ(r = a, φ) =
�
n=1,3,5,...
2V0
cos(nφ)
nπ
We omit work: nearly identical to example done in tutorial.
�
We have same boundary conditions as in (a), but boundary condition 1 is Φ(r = a, φ) = n=1,3,5...
Φ=
��
2V0
nπ
2V0
n=1,3,5,... nπ
�n=1,3,5,...
2V0
nπ
where V0 of (a) →
2V0
nπ
cos(nφ)
� r �n
cos(nφ) r ≤ a
� aa
�n
cos(nφ)
r ≤ a
r
in each term of the series.
Problem 6.2
A
�2 Φ =
1 ∂
r ∂r
�
�
∂Φ
1 ∂2Φ ∂2Φ
r
+ 2
+
∂r
r ∂φ2
∂z 2
Azimuthal symmetry ⇒
1 ∂
� Φ=
r ∂r
2
∂Φ
∂φ
= 0.
�
�
∂Φ
∂2Φ
r
+
=0
∂r
∂z 2
Guess Φ = R(r)Z(z)
�
�
∂ (R(r)Z(z))
∂ 2 (R(r)Z(z))
1 ∂
r
+
=0
r ∂r
∂r
∂z 2
�
�
∂R(r)
∂ 2 Z(z)
Z(z) ∂
r
= −R(r)
r ∂r
∂r
∂z 2
Divide by Z(z)R(r):
�
�
1 ∂ 2 Z(z)
1 ∂
∂R(r)
=−
r
rR(r) ∂r
∂r
Z(z) ∂z 2
��
�
��
� �
�
only r
only z
So, to be true for all x, y, both sides must equal the same constant. For this problem, we are asked to set
that constant to zero
−
1 ∂ 2 Z(z)
∂ 2 Z(z)
=
0
⇒
=0
Z(z) ∂z 2
∂z 2
3
Problem Set 6
6.641, Spring 2005
⇒ Z(z) = Az + B
�
�
�
�
∂R(r)
∂
∂R
1 ∂
r
=0⇒
r
=0
rR(r) ∂r
∂r
∂r
∂r
�
�
∂R
C
= C ⇒ dR =
dr
⇒r
∂r
r
So, R(r) = C ln r + D, And
Φ(r, z) = (Az + B)(C ln r + D)
B
Φ(r ≤ a, l) = V0
Φ(r, z = 0) = 0
Φ(r = b, z) = 0
Φ(r = a+ , z) = Φ(r = a− , z)
Φ(r = 0, z) = finite
C
0 ≤ r ≤ a :Φ(1) (r, z) = Az + Bz ln r + C ln r + D
Φ(1) (r = 0, z) is finite ⇒ B = C = 0
Φ(1) (r, l) = V0 , Φ(1) (r, 0) = 0 ⇒ Φ(1) =
V0 z
l
a ≤ r ≤ b :Φ(2) (r, z) = Ez + F z ln r + G ln r + H
Φ(2) (b, z) = 0 ⇒ E = F (− ln b) and H = G(− ln b)
r
r
so Φ(2) (r, z) = F z ln + G ln
b
b
r
(2)
(2)
Φ (r, 0) = 0 ⇒ Φ = F z ln
b
r
V0 z
(1)
(2)
ln
Φ (r = a, z) = Φ (r = a, z) ⇒ Φ(2) =
l ln ab
b
D
⎞
0
∂Φ�
∂Φ ⎟
−
→
⎜ ∂Φ
E = −�Φ = − ⎝îr
+ îφ � + îz
⎠
∂r
∂φ
∂z
�
⎛
−
→(2)
a≤r≤b: E
V0 z
l
�
V0 z 1
V0
r
− îz
ln
l ln ab r
l ln ab
b
�
V0
z
r�
=−
î
+
î
ln
r
z
l ln ab
r
b
= −îr
−
→(2)
E
�
V0
−
→(1)
⇒E
= −îz
l
�
�
V0 z
r
� � ln
= −�
b
l ln ab
−
→(1)
0≤r<a: E
= −�
4
Problem Set 6
6.641, Spring 2005
E
�
�
→(2)
−
→(1)
−
+
−
)
−
ε
E
(r
=
a
)
= σsf
n̂
ε
E
(r
=
a
0
����
ˆir
−ε0
V0 z
+ 0 = σsf
l ln ab a
σsf = −
ε0 V0 z
al ln ab
F
→(2)
−
We need Φ(2) and E
V�
r
r0
V�
0z
0 z0
= � a ln
Φ(2) (r = r0 , z = z0 ) = � a ln
�
b
b
l� l �
l
l
�� b
�b
Equipot.: z0 ln rb0 = z ln rb
Field Lines
(2)
z
dr
Er
z
= (2) = r r =
r
ln
r
ln
dz
Ez
b
b
�
�
r
r ln dr = zdz
b
�
2
r
r � r2
z2
ln
−
=
+c
2
b
4
2
At r = r0 and z = z0
c=
r02 � r0 � r02
z2
−
ln
− 0
2
b
4
2
Are they perpendicular? For E-field:
�E-field
z0
dr ��
=
dz �(r,z)=(r0 ,z0 )
r0 ln rb0
For Equipot:
�equipot
−r0 ln rb0
dr ��
=
dz �(r,z)=(r0 ,z0 )
z0
�
�equipot � � �E-field �
dr ��
dr ��
= −1
dz �
dz �
Problem 6.3
Sphere, initial conditions: At t = 0, p(t) = 0. No charge anywhere. Dipole
�
0 t<0
p(t) =
p t≥0
5
Problem Set 6
6.641, Spring 2005
z
θ
ε0
a
p(t)
ε, σ
Figure 2: A z-directed electric dipole p(t) within a free space sphere of radius a (Image by MIT OpenCourseWare.)
p(t)
p0
t
Figure 3: A dipole is turned on at t = 0 to a dipole moment of p0 (Image by MIT OpenCourseWare.)
A
Since there is no charge anywhere in the system, we solve Laplace’s equation
�2 Φin = 0
�2 Φout = 0
Φin = A(t)r cos θ +
Φout =
B(t)
cos θ
r2
C(t)
cos θ
r2
We know that since there is a dipole inside, Φin (r → 0) =
the surface charge at the boundary r = a is equal to 0. So
p0 u(t) cos θ
4πε0 r 2 .
Therefore, B(t) =
I.
(εEr (r = a+ ) − ε0 Er (r = a− )) = σs = 0
�
�
�
�
p0 u(t) cos θ
p0 u(t)
ˆ
+ A(t) sin θ îθ
− A(t) cos θ ir +
Ēin = −�Φin =
2πε0 r 3
4πε0 r 3
�
�
�
�
p0 u(t)
p0 u(t)
+ A(t) sin θ îθ
=
− A(t) cos θ îr +
4πε0 r 3
2πε0 r 3
�
�
�
�
2C(t)
C(t)
Êout = −�Φout =
cos θ îr +
sin θ ˆiθ
r3
r3
6
p0 (t)
4πε0 .
At t = 0+ ,
Problem Set 6
ε0
6.641, Spring 2005
�
p0 u(t)
− A(t)
2πε0 a3
�
=ε
2C(t)
a3
(1)
II. Also, Φ is continuous at r = a, so
p0 u(t)
C(t)
+ A(t) = 3
a
4πε0 a3
Add 1 and 2 to get
3p0 u(t)
4π�
a3
=
(2)
(2ε+ε0 )C(t)
.
a3
�
Substitute C(t) =
3p0 u(t)
4π(2ε+ε0 ) .
p0 u(t)
3p0 u(t)
+ A(t) =
4π(2ε + ε0 )a3
4πε0 a3
Therefore A(t) =
So at t = 0+
�
Φin =
Φout =
p0 (ε0 −ε)u(t)
2πε0 (ε0 +2ε)a3
p0
p0 (ε0 − ε)r
+
4πε0 r 2
2πε0 (ε0 + 2ε)a3
3p0
u(t) cos θ
4π(ε0 + 2ε)r 2
�
u(t) cos θ
r≤a
r≥a
B
For t → ∞ at steady state,
∂σs
∂t
=0
�
∂σs
0
(σout Eout,r − �
=0
σ in Ein,r )�r=a = −
∂t
(σ = 0 inside sphere)
Therefore in steady state C(t) = 0. If C(t) = 0, then
steady state
�
�
p0 r
p0
Φin =
−
u(t) cos θ r ≤ a
4πε0 a3
4πε0 r 2
p0 u(t)
4πε0 a3
p0 u(t)
+ A(t) = 0, A(t) = − 4πε
3 . So for t → ∞ in
0a
Φout = 0 r ≥ a
C
Recall that Φ(t) =
Φss
����
steady state potential
⎛
⎜
+ ⎝ Φ(t = 0+ ) −
� �� �
initial potential
so τ =
2ε+ε0
2σ .
7
⎟ − τt
⎠ e . We already have these
steady state potential
terms. We have to find the time constant τ .
�
∂ �
σEr (r = a+ ) +
εEr (r = a+ ) − ε0 Er (r = a− ) = 0
∂t
�
�
�
�
∂
2C(t) p0 u(t)
2σC(t)
+
ε 3 −
+ ε0 A(t) = 0
a3
∂t
a
2πa3
ε0 C(t) p0 u(t)
ε0 A(t) =
−
a3
4πa3
�
�
�
�
� �
2ε + ε0 ∂
∂ 3p0 u(t)
2σ
C(t)
=
C(t)
+
a3
∂t
a3
∂t 4πε0 a3
Φss
����
⎞
Problem Set 6
6.641, Spring 2005
D
�
�
�0
�
σs (t) = σs (t → ∞) + �
σs�
(t �
=�
0+ ) − σs (t → ∞) e−t/τ
�
�
= σs (t → ∞) 1 − e−t/τ
0
σs (t → ∞) = εE
�r (r = a+ , t → ∞) − ε0 Er (r = a− , t → ∞)
�
∂Φin ��
= ε0
∂r �r=a− ,t→∞
�
3p0 cos θ �
2ε + ε0
−t/τ
=−
1
−
e
,τ=
3
4πε0 a
2σ
Problem 6.4
As no free current inside and outside the sphere, Maxwell’s equations for MQS gives:
� · B̄ = 0, � × H̄ = 0
Here
H̄ = −�χ
So
�2 χ = 0
The general solution for Laplace’s equation in spherical coordinates is
χ = Ar cos θ
r<R
χ = (Dr + C/r 2 ) cos θ
r>R
The magnetic field is
H̄ = −�χ = −
�
∂χ ¯
1 ∂χ ¯
1 ∂χ ¯
ir +
iθ +
iφ
∂r
r ∂θ
r sin θ ∂φ
= −A(cos θ īr − sin θ īθ ) = −Aīz
r<R
�
3
= −(D − 2C/r ) cos θ īr + (D + C/r 3 ) sin θ īθ
r≥R
The boundary conditions are
H̄(r = ∞) = 0 ⇒ D = 0
Hθ (r = R+ ) = Hθ (r = R− ) ⇒ A = D + C/R3
Br (r = R+ ) = Br (r = R− ) ⇒ Hr (r = R+ ) = Hr (r = R− ) + M0 cos θ
with solutions
A = M0 /3
C = M0 R3 /3
8
Problem Set 6
6.641, Spring 2005
The scalar potential and the magnetic field are:
χ=
M0 r
cos θ
3
r<R
M0 R3
cos θ r > R
3r 2
M0
M0 ¯
H̄ = −
(cos θ īr − sin θ īθ ) = −
iz
3
3
χ=
=
M0 R3
2M0 R3
cos
θ
ī
+
sin θ īθ
r
3r 3
3r 3
r<R
r>R
Problem 6.5
A
From Gauss’ law � · D̄ = ρf ⇒ n̄ · (D̄1 − D̄2 ) = σsf , the polarization surface charge is
−� · P̄ = ρp ⇒ −n̄ · (P̄1 − P̄2 ) = σsp
so σsp = Pr (r = a− ) = P0 sin φ.
B
As no free charge inside and outside the cylinder, Maxwell’s equations for EQS gives:
� · D̄ = 0, � × Ē = 0
Here Ē = −�Φ, So �2 Φ = 0. The general solution for Laplace’s equation in cylindrical coordinates is
Φ = Ar sin φ
r≤a
= (Br + C/r) sin φ
r≥a
The electric field is
Ē = −�Φ = −
�
1 ∂Φ ¯
∂Φ ¯
∂Φ ¯
ir +
iφ +
iz
∂r
r ∂φ
∂z
= −A(sin φīr + cos φīφ )
�
r≤a
2
= −(B − C/r ) sin φīr − (B + C/r 2 ) cos φīφ
r≥a
The boundary conditions are
Ē(r = ∞) = 0 ⇒ B = 0
Φ(r = a+ ) = Φ(r = a− ) ⇒ Aa = C/a
ε0 Er (r = a+ ) = ε0 Er (r = a− ) + P0 sin φ
with solutions
A=
P0
,
2ε0
C=
P 0 a2
2ε0
9
Problem Set 6
6.641, Spring 2005
The scalar potential and the electric field are
Φ=
=
P0 r
sin φ
2ε0
P0 a2
sin φ
2ε0 r
r≤a
r ≥ a
¯ = − P0 (sin φīr + cos φīφ ) r < a
E
2ε0
=
P 0 a2
P0 a2
sin
φ
ī
−
cos φīφ
r
2ε0 r 2
2ε0 r 2
r > a
10