Chapter 7 - 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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

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.

Ask a homework question - tutors are online