D r ¼
1
2
ffiffi
r
p eC 3 À εC 5 À eC 4 À εC 6
ð
Þ
1
r
1 À e
Àkr
À
Á þ 2 eC 4 À εC 6
ð
Þ ke
Àkr
h
i
sin
θ
2
,
D θ ¼
1
2
ffiffi
r
p eC 3 À εC 5 þ eC 4 À εC 6
ð
Þ
1
r
1 À e
Àkr
À
Á
h
i
cos
θ
2
:
ð2:65Þ
It can be verified that the above solution satisfies the boundary conditions at the
crack faces. When C 2 , C 4 , and C 6 are all equal to zero, the solution reduces to that of
a piezoelectric dielectric.
2.5 Antiplane Waves in a Plate
Consider shear-horizontal (SH) waves in a plate of crystals of class (6mm) [6]. The
c-axis is along x 3 . The plate is unelectroded. In this problem, the electric field in the
free space is considered (Fig. 2.11).
This is an antiplane problem. The governing equations for the fields in the plate
are from Eqs. (1.32) and (1.33):
c∇
2 u þ e∇
2
φ ¼ ρ€ u,
e∇
2 u À ε∇
2 φ ¼ q Δp À Δn
ð
Þ ,
ð2:66Þ
∂
∂t
Δp
ð Þ ¼ p 0 μ
p
∇
2 φ þ D
p
∇
2 Δp
ð Þ,
∂
∂t
Δn
ð Þ ¼ Àn 0 μ
n
∇
2 φ þ D
n
∇
2 Δn
ð Þ:
ð2:67Þ
In the free space above and below the plate, the electric potential is governed by
∇
2 φ ¼ 0, x 2
j j > h,
φ ! 0, x 2 ! Æ1:
ð2:68Þ
The free space electric displacement is given by
x2
x1
2h
Free space
Crystal
Free space
Fig. 2.11 A piezoelectric
semiconductor plate and
coordinate system
2.5 Antiplane Waves in a Plate
27