b
T ÆL
ð Þ ¼ F, b
D ÆL
ð Þ ¼ 0,
J
p
ÆL
ð Þ ¼ 0, J
n
ÆL
ð Þ ¼ 0:
ð6:55Þ
Δp and Δn must satisfy the following global charge neutrality conditions:
Z L
ÀL
Δpdz ¼ 0,
Z L
ÀL
Δndz ¼ 0:
ð6:56Þ
Only one of Eq. (6.56) is independent. The other is implied by integrating Eq. (6.53)
between –L and L and using the electric displacement boundary conditions in
Eq. (6.55), which leads to
Z L
ÀL
q Δp À Δn
ð
Þ dz ¼ 0:
ð6:57Þ
To determine the mechanical displacement and the electric potential uniquely, we
also set
u 0
ð Þ ¼ 0, φ 0
ð Þ ¼ 0:
ð6:58Þ
The polarization vector and effective polarization charge density are calculated from
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
:
ð6:59Þ
The solution to the above boundary-value problem is given by
φ ¼
b eF
b εb c þ b e
2
k cosh kL
sinh kz,
ð6:60Þ
E ¼ À
b eF
b εb c þ b e
2
cosh kL
cosh kz,
ð6:61Þ
D ¼
b eF
b c A
1
ð Þ
þ A
2
ð Þ
À
Á 1 À
cosh kz
cosh kL
,
ð6:62Þ
p ¼ p 0 À
p 0 μ
p
D
p
b eF
b εb c þ b e
2
k cosh kL
sinh kz,
ð6:63Þ
6.2 Extension of Rods
151