q
∂
∂t
Δp
ð Þ ¼ À
∂J
p
∂z
,
q
∂
∂t
Δn
ð Þ ¼
∂J
n
∂z
:
ð7:90Þ
We consider the static extensional deformation of an electrically isolated free rod.
The boundary conditions are
b
T ÆL
ð Þ ¼ 0, b
D ÆL
ð Þ ¼ 0,
J
n
ÆL
ð Þ ¼ 0, J
p
ÆL
ð Þ ¼ 0:
ð7:91Þ
Δp and Δn satisfy the following global charge neutrality conditions:
Z L
ÀL
Δpdz ¼ 0,
Z L
ÀL
Δndz ¼ 0:
ð7:92Þ
Only one of Eq. (7.92) is independent. To determine the mechanical displacement
and the electric potential uniquely, we set
u 0
ð Þ ¼ 0, φ 0
ð Þ ¼ 0:
ð7:93Þ
Solving Eqs. (7.88), (7.89), and (7.90) under Eqs. (7.91), (7.92), and (7.93) does
not present any mathematical challenge. Once the equations are solved, the polarization vector and effective polarization charge density can be calculated from their
definitions:
P ¼ D À ε 0 E ¼
b
D
A
1
ð Þ
þ A
2
ð Þ
À ε 0 E,
ρ
P
¼ ÀP k,k ¼ À
dP
dz
:
ð7:94Þ
The expressions of the fields are found to be
φ ¼
e b λ þ b pb c
θ
b εb c þ e 2
ð
Þk
sinh kz
cosh kL
,
ð7:95Þ
E ¼ À
e b λ þ b pb c
θ
b εb c þ e 2
ð
Þ
cosh kz
cosh kL
,
ð7:96Þ
u ¼ À
e e b λ þ b pb c
θ
b εb c þ e 2
ð
Þ b ck
sinh kz
cosh kL
þ
b λθ
b c
z,
ð7:97Þ
200
7 Thermal Effects