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6.642 Continuum Electromechanics
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Fall 2008
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Fall 2008
6.642 — Continuum Electromechanics
Problem Set 2 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 1
Prob. 2.16.5
Gauss’ law and E = −�Φ requires that if there is no free charge
��2 Φ + �� · �Φ = 0
(1)
For the given exponential dependence of the permittivity, the x dependence of the coefficients in this expres­
sion factors out and it again reduces to a constant coefficient expression
∂2Φ ∂2Φ ∂2Φ
∂Φ
+
+
+ 2η
=0
∂x2
∂y 2
∂z 2
∂x
(2)
In terms of the complex amplitude forms from Table 2.16.1, Eq. 2 requires that
d2 Φ̃
dΦ̃
+ 2η
− k2 Φ̃ = 0
(3)
dx2
dx
�
Thus, solutions have the form exp(px) where p = −η ± λ, λ = k + η .
2
2
˜ be Φ̃α and Φ̃β on the upper and lower
The linear combination of these that satisfies the conditions that Φ
surfaces respectively is as given in the problem. The displacement vector is then evaluated as
�
�
−η sinh λx + λ cosh λx
−η sinh λ(x − Δ) + λ cosh λ(x − Δ)
D̃ = −�β Φ̃α eη(x+Δ)
− Φ̃β eηx
(4)
sinh λΔ
sinh λΔ
Evaluation of this expression at the respective surfaces then gives the transfer relations summarized in the
problem.
Courtesy of James R. Melcher. Used with permission. Solution to problem 2.16.5 in
Melcher, James. Solutions Manual for Continuum Electromechanics. 1982, p. 2.28.
Problem 2
Prob. 4.3.3
Figure 1: Force per unit area on middle charged surface (c/d) is found using the Maxwell stress tensor (Image
by MIT OpenCourseWare.)
Courtesy of James R. Melcher. Used with permission. Solution to problem 4.3.3 in
Melcher, James. Solutions Manual for Continuum Electromechanics. 1982, pp. 4.3-4.4.
1
Problem Set 2
6.642, Fall 2008
With positions as designated in the sketch, the total force per unit area is
< fz >z =< Dxc Ezc − Dxd Ezd >z =
1
�(D̃xc Ẽzc∗ − D̃xd Ẽzd∗ )
2
(5)
With the understanding that the surface charge on the sheet is a given quantity, boundary conditions
reflecting the continuity of tangential electric field at the three surfaces and that Gauss’ law be satisfied
through the sheet are
Φ̃a given;
Φ̃b given;
Φ̃c = Φ̃d ;
D̃xc − D̃xd = σ̃f given
Bulk relations are given by Table 2.16.1. In the upper region
� a �
�� a �
�
D̃x
− coth kd sinh1 kd
Φ̃
= �0 k
− sinh1 kd coth kd
D̃xc
Φ̃c
and in the lower
� d �
�
D̃x
− coth kd
= �0 k
b
˜
− sinh1 kd
Dx
1
sinh kd
coth kd
��
Φ̃d
˜b
Φ
�
(6)
(7)
(8)
In view of Eq. 6, Eq. 5 becomes
< fz >z =
1
�[−jkσ̃f Φ̃c∗ ]
2
(9)
so what is now required is the amplitude Φ̃c . The surface charge, given by Eq. 6, as the difference D̃xc − D̃xd ,
follows in terms of the potentials from taking the difference of Eqs. 7 (second row) and 8 (first row). The
resulting expression is solved for
Φ̃c =
Φ̃a + Φ̃b
σ̃f
+
2�0 k coth kd 2 cosh kd
(10)
Substituted into Eq. 9 (where the self terms in σ̃f σ̃f∗ are imaginary and can therefore be dropped) the force
is expressed in terms of the given excitations
�
�
1
Φ̃a∗ + Φ̃b∗
< fz >z = k� −jσ̃f
(11)
2
2 cosh kd
b)
Translation of the given excitations into complex amplitudes gives
σ̃f = −σ0 ejωt ejkδ ,
Φ̃a = V0 ejωt ,
Φ̃b = ±V0 ejωt
(12)
Thus, with the even excitation, where Φa = Φb
< fz >z = −
kV0 σ0
sin kδ
2 cosh kd
(13)
and with the odd excitation, < fz >z = 0.
c)
This is a specific case from part (b) with ω = 0 and δ = λ/4. Thus,
< fz >z = −
kV0 σ0
2 cosh kd
(14)
The sign is consistent with the sketch of charge distribution on the sheet and electric field due to the potentials
on the walls sketched.
2
Problem Set 2
6.642, Fall 2008
Figure 2: Potentials Φa = Φb = −V0 cos kz shown with surface charge σf = σ0 sin kz (Image by MIT
OpenCourseWare.)
Problem 3 (Zahn, Problem 23, Chapter 1)
a)
Cartesian Cylindrical Spherical
hx = 1
hr = 1
hr = 1
hy = 1
hφ = r
hθ = r
hz = 1
hz = 1
hφ = r sin θ
b)
df =
∂f
∂f
∂f
du +
dv +
dw
∂u
∂v
∂w
= �f · d�
= �f · [hu duīu + hv dv īv + hw dwīw ]
1 ∂f
1 ∂f
1 ∂f
; (�f )v =
; (�f )w =
hu ∂u
hv ∂v
hw ∂w
1 ∂f ¯
1 ∂f ¯
1 ∂f
¯
�f =
iu +
iv +
iw
hu ∂u
hv ∂v
hw ∂w
(�f )u =
c)
dSu = hv hw dvdw;
dSv = hu hw dudw;
dSw = hu hv dudv
dV = hu hv hw dudvdw
d)
Divergence
�
�
¯
Φ=
A · dS =
S
+
�
2,v+Δv
Au hv hw dvdw −
1,u
Av hu hw dudw −
�
Au hv hw dvdw
1� ,u−Δu
�
Av hu hw dudw
2� ,v
3
Problem Set 2
+
�
6.642, Fall 2008
Aw hu hv dudv −
3,w+Δw
�
Aw hu hv dudv
3� ,w
�
Au hv hw |u − Au hv hw |u−Δu
Av hu hw |v+Δv − Av hu hw |v
Aw hu hv |w+Δw − Aw hu hv |w
=
+
+
Δu
Δv
Δw
�
�
A¯ · dS
A¯ · dS
S
S
=
� · Ā =
lim
Δu→0,Δv→0,Δw→0
ΔV
hu hv hw ΔuΔvΔw
�
�
1
∂(hv hw Au ) ∂(hu hw Av ) ∂(hu hv Aw )
=
+
+
hu hv hw
∂u
∂v
∂w
Figure 3: Contour for determining (� × Ā)u (Image by MIT OpenCourseWare.)
Curl
(� × Ā)u =
�
lim
Δv→0,Δw→0
�
Ā · d�
L
hv hw ΔvΔw
Ā · d� = [Av hv Δv|w − Av hv Δv|w+Δw ] + [Aw hw Δw|v+Δv − Aw hw Δw|v ]
L
(� × Ā)u =
=
lim
Δv→0,Δw→0
1
hv hw
�
Av hv |w − Av hv |w+Δw
Aw hw |v+Δv − Aw hw |v
+
Δw
Δv
�
�
�
1
∂(hw Aw ) ∂(hv Av )
−
hv hw
∂v
∂w
Similarly
�
�
∂(hu Au ) ∂(hw Aw )
1
(� × Ā)v =
−
hu hw
∂w
∂u
�
�
∂(hv Av ) ∂(hu Au )
1
−
(� × Ā)w =
hu hv
∂u
∂v
e)
�
�
�
�
�
�
��
1
∂ hu hw ∂f
∂
∂ hv hw ∂f
hu hv ∂f
� f = � · (�f ) =
+
+
hu hv hw ∂u
hu ∂u
∂v
hv ∂v
∂w
hw ∂w
2
4
�
ΔuΔvΔw