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Unformatted text preview: Chapter 7 Poisson’s and Laplace’s Equations Dr. Ray Chen© 7.1 Poisson’s and Laplace’s Equations
v
v
v
∇ ⋅ D = ρ v D = εE E = −∇V
v
v
∇ ⋅ D = ∇ ⋅ ( εE ) = −∇ ⋅ (ε∇V ) = ρ v
For a homogeneou region
s ρv
∇ ⋅ ∇V = −
ε ∂ε ∂ε ∂ε
=
=
=0
∂x ∂y ∂z Poisson' s Equation v ∂ Ax ∂ Ay ∂ Az
∇⋅A =
+
+
∂x
∂y
∂z ∇V = ∂V
∂V
∂V
ˆ
ˆ
ˆ
ax +
ay +
az
∂x
∂y
∂z ρ
∂ ∂V
∂ ∂V
∂ ∂V
∂ 2V ∂ 2V ∂ 2V
+
+
=− v
(
)+
(
)+ (
)=
∴ ∇ ⋅ ∇V =
∂x ∂x
∂y ∂y
∂z ∂z
ε
∂x 2
∂y 2
∂z 2
Dr. Ray Chen© 7.1 Poisson’s and Laplace’s Equations If ρ v = 0 , we have ∇ 2V = 0 , which is the Laplace' s Equation.
∇ 2 is called the Laplacian of V. If ρ v = 0 , ∇ V = 0 is always true,
2 ∇ 2 can be represented in different coordinate system,
such as cartesian, spherical, cylindrical Dr. Ray Chen© 7.1 Poisson’s and Laplace’s Equations In cartesian coordinate, Laplace' s equation is
∂ 2V ∂ 2V ∂ 2V
∇V =
+ 2 + 2
2
∂x
∂y
∂z
2 In cylindrical coordinate
1 ∂ ∂V
ρ
∇V =
ρ ∂ρ ∂ ρ 2 1 ∂ 2V ∂ 2V
+ 2 ρ ∂ φ 2 + ∂z 2 In spherical coordinate
1 ∂ 2 ∂V 1
∂ ∂V 1
∂ 2V
∇V = 2
r
+ 2 sinθ
+ 2
∂θ r sin 2 θ ∂φ 2
r ∂r ∂r r sinθ ∂θ 2 Dr. Ray Chen© 7.2 Uniqueness Theory If there are two potential functions V1 and V2 satisfying
the Laplace’ s Equation ∇ 2 V = 0 under the same boundary
condition, we have V =V2
1
Dr. Ray Chen© 7.2 Uniqueness Theory Proof: According to Laplace' s Equation
∇ 2V1 = 0 ∇ 2V2 = 0 ∇ 2 (V1 − V2 ) = 0 Each solution must also satisfy the boundary conditions,
and if we represent the given potential values on the boundaries,
then
V1 ( at the boundary) = V1b V2 ( at the boundary) = V2b
V1b = V2b = Vb or V1b − V2b = 0
Dr. Ray Chen© Next page 7.2 Uniqueness Theory Proof:
Through divergence theory, we have ∫ vol ∇ ⋅ ((V1 − V 2 )∇ (V1 − V 2 ))dv = ∫ [(V
S 1b − V 2 b )∇ (V1b v
− V 2 b )] ⋅ d S = 0 ∇ ⋅ ∇ (V1 − V 2 ) = 0 in source free region
v
v
v
∇ ⋅ (VD) = V (∇ ⋅ D) + D ⋅ (∇V ) ∴ ∇ ⋅ [(V1 − V2 )∇ (V1 − V2 )] = (V1 − V2 )[∇ ⋅ ∇ (V1 − V2 )] + ∇ (V1 − V2 ) ⋅ ∇ (V1 − V2 ) ∫ vol ∇ ⋅ [(V1 − V2 )∇ (V1 − V2 )]dv = ∫ (V1 − V2 )[∇ ⋅ ∇(V1 − V2 )] dv + ∫vol [∇(V1 − V2 )]2 dv
vol
Next page Dr. Ray Chen© 7.2 Uniqueness Theory Proof: ∴ [∇ (V1 − V 2 )]2 dv = 0 and [∇ (V1 − V 2 )]2 ≥ 0
∫vol
[∇ (V1 − V 2 )]2 has to be equal to 0 everywhere
So ∇ (V1 − V 2 ) = 0 , V1 − V2 = constant .
at the boundary V1 − V 2 = V1b − V2 b = 0
∴ V1 = V 2 Dr. Ray Chen© 7.2 Uniqueness Theory The uniqueness theorem also applies to Poisson' s equation,
for if
∇ 2V1 = − ρ v / ε
∇ 2V2 = − ρ v / ε ∇ 2 (V1 − V2 ) = 0 , V1b = V2b ⇒ V1 = V 2 Once we can find any method of solving Laplace’s
or Poisson’s equation subject to boundary
conditions, we have solved our problem once for all. Dr. Ray Chen© 7.3 Examples of the solution of Laplace’s equation Example 1:
Solve a 1  D Laplace equation of ∇ 2V = 0 with x as the variable .
v
what are the value of E , D N , Q and C if x = 0 , V = 0 and x = d , V = V 0 as
shown in the figure. x Solution: V = V0 ∂ 2V
∂V
=0 →
= A → V = Ax + B
∂x
∂x 2 V = 0 , at x = 0 , V = V0 , at x = d , ⇒ B = 0 , A = d V0
d V =0 ∴ V = (V0 / d ) x + 0 = (V0 / d ) x
v
v
V
V0
ˆ
ˆ x , D = −ε 0 a x
∴ E = −∇ V = − a
d
d
V
V
D N = −ε 0 = ρ s , → Q = ∫ ρ s ds = ρ s S = −ε 0 S
d
d
Dr. Ray Chen© C= ε Q
V0 = εS
d 7.3 Examples of the solution of Laplace’s equation Example 2:
A coaxial cable with V = V 0 at ρ = a and V = 0 at ρ = b ,
v v
determine the V , E , D , Q & C . Solution:
∇ 2V = ∂ ∂V dV
ρ
=0 →ρ
=A → ∂ρ ρ ∂ρ dρ 1 ⋅ V0 = A ln a + B , A = V0 / ln A ρ dρ , V = Alnρ + B
0 = A ln b + B , → B = − A ln b , ∫ dV = ∫ a
b V ln ρ
V ln b V0 ln(ρ / b )
V = 0
− 0
=
ln(a / b ) ln(a / b )
ln(a / b )
Dr. Ray Chen© a b
L 7.3 Examples of the solution of Laplace’s equation Solution(continued):
From − ∂ ln
∂ρ ρ
b = ∂ ln b ρ ∂ρ − ∂ ln
=
∂ b ρ b ρ ∂
⋅ ρ b
1
ρ
=− ⋅ 2 =−
∂ρ
ρ
b ρ ρ ∂ ln
v
V0 ∂V
b
ˆ
−
E = −∇ V = −
aρ =
∂ρ
∂ρ
ln(a / b ) v
v
DN = ε E ρ =a =− εV0
a ⋅ b 1 = − V0 ⋅
ˆ
aρ ρ ln(a / b ) 1
ˆ
aρ
ln(a / b ) − ε V 0 2π aL
1 εV 0 Q = 2π a ⋅ L ⋅ −
=
⋅
a ln (a / b )
a ln (a / b ) C = − ε V 0 2π aL
2πε L
Q
=
=
,
V
a ln (a / b ) / V 0
ln (b / a ) Dr. Ray Chen© C = 2πε
per unit length
ln (b / a ) 7.3 Examples of the solution of Laplace’s equation Example 3:
Two concentric spheres with radii of a and b ,
V = V0 at r = a and V = 0 at r = b
v v
Determine the values of V, E, D, Q and C V=V0 r=a r=b
V=0 Dr. Ray Chen© 7.3 Examples of the solution of Laplace’s equation V=V0 r=a Solution to example 3:
According to Laplace' s equation : ∇ 2V = 0 ,
Set r 2 dV
A
= A , dV = 2 dr,
dr
r V=− so ∇ 2V = A
+B
r From V = 0 at r = b, V = V0 at r = a, we get
∴A= V0
,
1 1
−
a b B=− 1 d 2 dV r
=0
r 2 dr dr r=b
V=0 A
+ B = 0 → A = −Bb,
b A
+ B = V0
a A − V0 / b
=
1 1
b
−
a b 1 1
−
A
r b , E = ∇ V = Q , V = Q 1 − 1 ,
V = − + B = V0 ab
2
1 1
4πε a b r
4πε r
−
a b
Q
4πε
=
, when b → ∞ , C = 4πε a which is point charge capacitanc e
C=
1 1
Vab
−
a b
Dr. Ray Chen© 7.4 Example of solution of Poisson’s equation
A p  n junction between two halves of a semiconductor bar
extending in the x  direction, assume that x
a x
a ρ v = 2 ρ v 0 sec h tanh and ρ v , max = ρ v 0 at x = 0.881a, N d = N a
v
dQ
Use Poisson' s Equation to determine E, V, Q and
dV p n 0
Dr. Ray Chen© x 7.4 Example of solution of Poisson’s equation p ρ n 0 x Ex Fig. 7.3 Dr. Ray Chen© V 7.4 Example of solution of Poisson’s equation Solution:
Poisson equation. : ∇ 2V = − ρv
ε 2ρ
d 2V x x
In one dimension case, we have 2 = − v 0 sec h tanh ε
dx
a
a ex − e −x
ex + e −x
sinh( x )
From Math, sinh(x) =
, cosh(x) =
, tanh( x ) =
, coth( x ) = (tanh( x ))−1
2
2
cosh( x )
sec h( x ) = (cosh( x ))−1 , c sec h( x ) = (sinh( x ))−1 , ∴ ∫ sec h( Ex = d sec h( x )
d tanh( x )
= − sec h( x ) ⋅ tanh( x ),
= sec h( x )
dx
dx x
x
x
x
x
x
) tanh( )dx = a ∫ sec h( ) tanh( )d ( ) = − a sec h( ) + C
a
a
a
a
a
a − 2ρ v0a ε sec h( x
) + C1
a Boundary condition x → ±∞ , E = 0 ∴ C1 = 0 , Ex = − 2 ρυ 0 a ε sec h( x
)
a Next page
Dr. Ray Chen© 7.4 Example of solution of Poisson’s equation Solution(continued):
V = − ∫ E x dx =
∞ where ∫ sec h(
0 2 ρυ 0 a ε ⋅ a∫ 4 ρυ 0 a 2 ∞
ρυ 0 a 2
x
x
x
x
−1 x / a
sec h( )d ( ) =
∫ sec h( a )d ( a ) = 4 ε tan e + C 2
a
ε
a
0 x
x
)d ( ) = tan −1 e x / a + C
a
a ρυ 0 a 2
0=4
tan −1 e x / a + C 2
ε
π at x = 0, V = 0, ρυ 0 a 2
V =4
(tan −1 e x / a − )
ε
4
−1 x → ∞ , tan e x/a = π
2 ; ∴ V 0 = V( x → ∞ ) − V ( x → − ∞ ) = Dr. Ray Chen© x → −∞ , tan −1 e x / a = 0
2πρ υ 0 a 2 ε Next page 7.4 Example of solution of Poisson’s equation Solution(continued):
v
Q = ∫ ∇ ⋅ D dυ ,
∞ ∴Q = ∞ ∞ ∫ ∫ ∫ 2 ρυ
0 0 v
v
d
( εE x )
∇ ⋅ D = ∇ ⋅ε E =
dx 0 0 x
x
sec h( ) tanh( )dxdydz ,
a
a ∞ = 2 ρ υ 0 Sa ∫ sec h(
0 V ε a= 0 2πρ v0 C= dQ
=
dV0 Dr. Ray Chen© ∴Q = S ρ v 0ε
εS
⋅S =
,
2πV0
2πa dV0
dQ
I=
=C
dt
dt ∞ ∫ ∫
0 0 x
x
x
) tanh( )d ( ) = 2 ρ υ 0 Sa
a
a
a 1/ 2 , ∞ 2 ρ v 0 εV 0 π dydz = S , the cross  section of the p  n junction ...
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This note was uploaded on 08/15/2009 for the course EE 325 taught by Professor Brown during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Brown

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