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# Laplace equation

Laplace Equations

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### Laplace equation

1. 1. Special Techniques for Calculating Potential 3.1 Laplace’s Equation 3.2 The Method of Images 3.3 Separation of Variables 3.4 Multipole Expansion
2. 2. 3.1 Laplace’s Equation 3.1.1 Introduction 3.1.2 Laplace’s Equation in One Dimension 3.1.3 Laplace’s Equation in Two Dimensions 3.1.4 Laplace’s Equation in Three Dimensions 3.1.5 Boundary Conditions and Uniqueness Theorems 3.1.6 Conducts and the Second Uniqueness Theorem
3. 3. 3.1.1 Introduction The primary task of electrostatics is to study the interaction (force) of a given stationary charges. since this integrals can be difficult (unless there is symmetry) we usually calculate This integral is often too tough to handle analytically. 2 0 ˆ1 ( ) 4 F q Etest RE d R ρ τ πε = = ∫ v v v ˆˆ ˆx y zE E x E y E z= + + +××× v 0 1 1( ) 4 V d R E V ρ τ πε = ∫ = −∇ v ∴ Q
4. 4. In differential form Eq.(2.21): Poisson’s eq. Laplace’s eq or The solutions of Laplace’s eq are called harmonic function. 3.1.1 2 0 V ρ ε ∇ = − 2 0V∇ = 2 2 2 2 2 2 2 1 10 ( ) 0 i k i i i k TV A h q h q V V V x y z ∂ ∂= ∇ = ∂ ∂∏ ∂ ∂ ∂ + + = ∂ ∂ ∂  to solve a differential eq. we need boundary conditions.  In case of ρ = 0, Poisson’s eq. reduces to
5. 5. 3.1.2 Laplace’s Equation in One Dimension 2 2 0 d V V mx b dx = ⇒ = + m, b are to be determined by B.C.s e.q., 1. V(x) is the average of V(x + R) and V(x - R), for any R: 2. Laplace’s equation tolerates no local maxima or minima. ( 1) 4 1 5 ( 5) 0 5 V x m V x V x b = = = −  ⇒ ⇒ = − +  = = =  V=4 V=0 1 5 x [ ]1 2 ( ) ( ) ( )V x V x R V x R= + + −
6. 6. Method of relaxation: 3.1.2 (2)  A numerical method to solve Laplace equation.  Starting V at the boundary and guess V on a grid of interior points. Reassign each point with the average of its nearest neighbors. Repeat this process till they converge.
7. 7. 3.1.2 (3) The example of relaxation n V(x=0)V(x=1)V(x=2)V(x=3)V(x=4) 0 4 0 0 0 0 1 4 2 0 0 0 2 4 2 1 0 0 3 4 1 0 4 4 0 5 4 0 6 4 0 7 4 0 8 4 0 9 4 0 10 4 0 4 3 2 1 0
8. 8. 3.1.3 Laplace’s Equation in Two Dimensions A partial differential eq. : There is no general solution. The solution will be given in 3.3. We discuss certain general properties for now. 1. The value of V at a point (x, y) is the average of those around the point. 2. V has no local maxima or minima; all extreme occur at the boundaries. 3. The method of relaxation can be applied. 2 2 2 2 0 V V x y ∂ ∂ + = ∂ ∂ 1( , ) 2 circle V x y Vdl Rπ = ∫Ñ
9. 9. 3.1.4 Laplace’s Equation in Three Dimensions 1. The value of V at point P is the average value of V over a spherical surface of radius R centered at P: The same for a collection of q by the superposition principle. 2. As a consequence, V can have no local maxima or minima; the extreme values of V must occur at the boundaries. 3. The method of relaxation can be applied. 2 1( ) 4 sphere V p Vda Rπ = ∫Ñ 2 2 2 0 1 , 2 cos 4 q V r R rR θ πε = = + − r r 1 2 2 22 2 0 1 [ 2 cos ] sin 44 ave q V r R rR R d d R θ θ θ ϕ πεπ − = + −∫ 2 2 0 1 2 cos 04 2 q r R rR rR πθ πε = + − ( ) ( ) at the center of the sphere 0 0 1 1 4 2 4 q q r R r R V rR rπε πε  = + − − = = 
10. 10. 3.1.5 Boundary Conditions and Uniqueness Theorems in 1D, • one end Va • the other end Vb V is uniquely determined by its value at the boundary. First uniqueness theorem : the solution to Laplace’s equation in some region is uniquely determined, if the value of V is specified on all their surfaces; the outer boundary could be at infinity, where V is ordinarily taken to be zero. V mx b= +
11. 11. 3.1.5 Proof: Suppose V1 , V2 are solutions at boundary at boundary. V3 = 0 everywhere hence V1 = V2 everywhere 2 1 0V∇ = 2 2 0V∇ = 1 2 ?V V= 3 1 2V V V= − 2 2 2 3 1 2 0V V V∇ = ∇ −∇ = 1 2 3 0V V V= ⇒ = ∴
12. 12. Proof. at boundary. Corollary : The potential in some region is uniquely determined if (a) the charge density throughout the region, and (b) the value of V on all boundaries, are specified. 3.1.5 The first uniqueness theorem applies to regions with charge. 2 1 0 V ρ ε ∇ = − 2 2 0 V ρ ε ∇ = − 3 1 2V V V= − 2 2 2 3 1 2 0 0 0V V V ρ ρ ε ε ∇ = ∇ −∇ = − + = 3 1 2 3 0 0, . ., 1 2 V V V V i e V V = − = ∴ = =
13. 13. 3.1.6 Conductors and the Second Uniqueness Theorem  Second uniqueness theorem: Proof: Suppose both and are satisfied. and 0 1 1E ε ρ∇× = v 0 1 2E ε ρ∇× = v In a region containing conductors and filled with a specified charge density ρ, the electric field is uniquely determined if the total charge on each conductor is given. (The region as a whole can be bounded by another conductor, or else unbounded.) 2E v 1E v 0 0 1 1 1 1 2 2, ith conducting ith conducting surface surface i iE da Q E da Qε ε × = × =∫ ∫ v vv v Ñ Ñ 0 0 1 1 1 2, outer bourndary outer bourndary tot totE da Q E da Qε ε × = × =∫ ∫ v vv v Ñ Ñ
14. 14. for V3 is a constant over each conducting surface define 3.1.6 3 1 2E E E= − v v v 0 0 3 1 2 1 2( ) 0E E E E E ρ ρ ε ε ∇× = ∇× − = ∇× −∇× = − = rv v v v 3 1 2 0E da E da E da× = × − × =∫ ∫ ∫ v v vv v v Ñ Ñ Ñ 2 3 3 3 3 3 3 3( ) ( ) ( )V E V E E V E∇× = ∇× + ∇ = − v v v 2 3 3 3 3 3( ) volume surface E d V E d V E daτ τ= − ∇× = − ×∫ ∫ ∫ v v v Ñ 3 3 0 surface V E da= − × =∫ v v Ñ 3 1 20E E E∴ = = v v v i.e.,
15. 15. 3.2 The Method of Images 3.2.1 The Classical Image Problem 3.2.2 The Induced Surface Charge 3.2.3 Force and Energy 3.2.4 Other Image Problems
16. 16. 3.2.1 The Classical Image Problem What is V (z>0) ? ( ) 2 2 2 2 1. 0 0 2. 0 V z V for x y z d = = → + + >> B.C. The first uniqueness theorem guarantees that there is only one solution. If we can get one any means, that is the only answer.
17. 17. Trick : Only care z > 0 z < 0 is not of concern 3.2.1 in original problem for V(z=0) = const = 0 ( )0 0E z < = v 0z ≥ ( ) ( )2 22 2 2 20 1 ( , , ) 4 q q V x y z x y z d x y z dπε   − = +   + + − + + +   ˆ 0 0 0 z x y E E z E E at z or E = − = =   ÷ =  =P v
18. 18. 3.2.2 The Induced Surface Charge Eq.(2.49) 0 ˆˆ V E z V z z σ ε ∂ = = −∇ = − ∂ v 0 0z V z σ ε = ∂ = − ∂ 0 0 z z A E A E σ σ ε ε × − = = − 0 0 Q E da at z ε × = ≅∫ v v QÑ ( ) ( )2 22 2 2 2 0 1 4 z q q z x y z d x y z d σ π =   − ∂   = −  ∂  + + − + + + 
19. 19. total induced surface charge 3.2.2 Q=-q 3 3 2 2 2 2 2 22 2 0 1 1 2( ) 2( ) 2 2 4 ( ) ( ) z z d z dq x y z d x y z d π =   − × − − × + −   = −      + + − + + +      3/2 3/22 2 2 2 2 2 2 qd qd x y d r d σ π π − − = =    + + +     01/22 2 1 qd q r d ∞− = − = −  +   2 3/20 0 2 2 2 qd Q da rdrd r d π σ θ π ∞ − = =  +   ∫ ∫ ∫
20. 20. 3.2.3 Force and Energy The charge q is attracted toward the plane. The force of attraction is With 2 point charges and no conducting plane, the energy is Eq.(2.36) ( ) ( ) ( ) 2 2 2 0 0 1 1 ˆˆ 4 4 2 q q q F z z dd dπε πε − = = −  − −  v ( ) ( ) ( ) ( ) ( ) 2 1 20 0 2 0 1 2 1 1 1 2 4 4 1 4 2 i i i i W q V p q q q q d d d d q d πε πε πε = =        = × − + − ×    +  − −    = − ∑
21. 21. 3.2.3 (2) or For point charge q and the conducting plane at z = 0 the energy is half of the energy given at above, because the field exist only at z ≥ 0 ,and is zero at z < 0 ; that is ( ) 2 0 1 4 4 q W dπε = − ( ) ( ) 2 2 0 2 2 0 0 1 4 2 1 1 4 4 4 4 d d d q W F dl dz z q q z d πε πε πε ∞ ∞ ∞ = × =   = − = − ÷ ÷   ∫ ∫ vv ˆ( )dl dlz= − v Q
22. 22. 3.2.4 Other Image Problems Stationary distribution of charge 1 q 2 q 3 q 1 q− 2 q− 3 q−
23. 23. 3.2.4 (2) Conducting sphere of radius R Image charge at R q q a ′ = − 2 R b a = ( ) 0 0 0 1 , 4 1 ˆˆˆˆ4 1 4 ˆˆˆˆ q q V r q q rr as rr bs q q a r r r s b r s r b θ πε πε πε ′  = + ÷    ′ = + ÷ − −     ÷′ = + ÷  ÷− − ÷   ′r r 1 2 2 22 cosr a ra θ = + −   r 1 2 2 22 cosr b rb θ = + −  ′r
24. 24. Force [Note : how about conducting circular cylinder?] 3.2.4 (3) ( ) 0 1 , 0 4 ˆˆˆˆ q q V R a R R r s b r s R b θ πε    ÷′ = + = ÷  ÷− − ÷   q q R b ′ = − b R q q q R a ′ = − = − a R R b = 2 R b a = ( ) ( ) 2 2 22 20 0 1 1 4 4 qq q Ra F a b a R πε πε ′ = = − − −
25. 25. 3.3 Separation of Variables 3.3.0 Fourier series and Fourier transform 3.3.1 Cartesian Coordinate 3.3.2 Spherical Coordinate
26. 26. ˆˆ ˆ, ,x y zare unique because 3.3.0 Fourier series and Fourier transform Basic set of unit vectors in a certain coordinate can express any vector uniquely in the space represented by the coordinate.e.g. Completeness: a set of function ( )xfn ( ) ( )xfCxf n n n∑ ∞ = = 1 ( )xf is complete. if for any function Orthogonal: a set of functions is orthogonal if ( ) ( ) 0=∫ dxxfxf m b a n mn ≠ =const for for mn = in 3D. Cartesian Coordinate.zVyVxViVV zyx N i i ˆˆˆˆ 1 ++== ∑=  zyx VVV ,, are orthogonal. ji ˆˆ ⋅ = i j= i j≠0 1
27. 27. 3.3.0 (2) A complete and orthogonal set of functions forms a basic set of functions. e.g. Nnk ∈, ( ) ( )sin sinkx kx− = − ( ) ( )cos coskx kx− = odd { nkif nkifdxnxkx ≠ = −∫ = 0 coscos π π π 0cossin∫− = π π dxnxkx { nkif nkifdxnxkx ≠ = −∫ = 0 sinsin π π π even sin(nx) is a basic set of functions for any odd function. cos(nx) is a basic set of functions for any even function. sin(nx) and cos(nx) are a basic set of functions for any functions.
28. 28.      for any ( )xf ( ) ( ) ( ) 2 xfxf xg −− = ( ) ( ) ( ) 2 xfxf xh −+ =odd ( ) ( ) ( )xhxgxf += even 3.3.0 (3) , ↑ odd ↑ even [sinkx] [coskx] Fourier transform and Fourier series ( ) ( ) 0 (*) sin cosn n n f x A nx B nx ∞ = = +∑ ( ) ( ) 1 sinnA f x nx dx π ππ − = ∫ ( ) ( ) 1 cosnB f x nx dx π ππ − = ∫ ( ) 0 2 1 00 == ∫− AdxxfB π ππ } ∞⋅⋅⋅⋅= ,2,1n ;
29. 29. 3.3.0 (4) Proof ( ) ( ) ( ) * sin sin kx dx f x kxdx π π π π − − × ⇒ × ∫ ∫ ( ) ( ) 0 sin cos sinn n n A nx B nx kx dx π π ∞ − = = +∑ ∫ ( ) ( ) 0 sin sinn n A nx kx dx π π ∞ − = = ∑ ∫ ( ) ( )0 1 sinkA f x kx dx π π ∴ = ∫ if 0kA kπ= ≠
30. 30. 0 cos cosn n B nx kx dx π π ∞ − = = ∑ ∫ ( ) ( ) ( ) ( ) * cos cos kx dx f x kx dx π π π π − − × ⇒ ∫ ∫ ( ) ( ) 0 sin cos cosn n n A nx B nx kx dx π π ∞ − = = +∑ ∫ ( )0 1 2 B f x dx π ππ − = ∫ {= ( ) ( ) 1 cosnB f x nx dx π ππ − ∴ = ∫ 3.3.0 (5) B0 2π for k = 0 Bk π for k = 1, 2, …
31. 31. 3.3.1 Cartesian Coordinate Use the method of separation of variables to solve the Laplace’s eq. Example 3 0)( )()0( 0)( 0)0( 0 →∞→ == == == xV yVxV yV yV π Find the potential inside this “slot”? Laplace’s eq. 2 2 2 2 0 V V x y ∂ ∂ + = ∂ ∂ set )()(),( yYxXyxV = 2 2 2 2 0 X Y Y X x y ∂ ∂ + = ∂ ∂
32. 32. 0 11 )( 2 2 )( 2 2 = ∂ ∂ + ∂ ∂  onlyydepedentygonlyxdepedentxf y Y Yx X X f and g are constant 1 2 0C C+ = 12 2 1 )( C dx Xd X xf == 2 22 1 ( ) d Y g y C Y dy = = 2 21 kCC =−=set so Xk dx Xd 2 2 2 = 2 2 2 d Y k Y dy = − kyDkyCyYBeAexX kxkx cossin)(,)( +=+= − =),( yxV )( kxkx BeAe − + )cossin( kyDkyC + 3.3.1 (2) 1 2, are constant ( ) ( ) 0 C C f x g y+ =
33. 33. 3.3.1 (3) B.C. (iv) ( ) 0 0, 0V x A k→ ∞ → ⇒ = > B.C. (i) ( 0) 0 0V y D= = ⇒ = =),( yxV kx e− )cossin( kyDkyC + ( , ) sinkx V x y Ce ky− = B.C. (ii) ( ) 0 sin 0 1,2,3V y k kπ π= = ⇒ = = L B.C. (iii) )()0( 0 yVxV == =)(0 yV 1 sinkx k k C e ky ∞ − = ∑ A fourier series for odd function 00 2 ( )sinkC V y ky dyπ π = ∫ The principle of superposition =),( yxV 1 sinkx k Ce ky ∞ − = ∑
34. 34. 3.3.1 (4) For antconstVyV == 00 )( 0 0 0 4 0 2 sin 0 2 (1 cos ) k V k V C ky dy if k even V k k if k odd π π π π π = =  = − =  = ∫ sin10 0 sinh 1,3,5, 4 21 ( , ) sin tan ( ) yk x k V V V x y e ky kπ π − − = = =∑ L
35. 35. 3.4 Multipole Expansion 3.4.1 Approximate Potentials at Large distances 3.4.2 The Monopole and Dipole Terms 3.4.3 Origin of Coordinates in Multipole Expansions 3.4.4 The Electric Field of a Dipole
36. 36. 1 1 2 1 cos 24 0 ( ) cos ( ) d r qd r V p θ πε θ− ≅ ≅ r r+ - 3.4.1 Approximate Potentials at Large distances Example 3.10 A dipole (Fig 3.27). Find the approximate potential at points far from the dipole. ⇒ 1 4 0 ( ) ( ) q q V p πε + − = − r r 2 2 2 2 ( ) cos d r rd θ± = + mr 2 2 24 (1 cos ) d d r r r θ= ± + 2 (1 cos ) r d d r r θ >> ≅ m 1 1 1 12 2 (1 cos ) (1 cos ) d d r r r r θ θ − ± ≅ ≅ ±m r
37. 37. 3.4.1 Example 3.10 For an arbitrary localized charge distribution. Find a systematic expansion of the potential? 0 1 1 ( ) 4 V p dρ τ πε = ∫ r 2 2 2 2 2 2 cos [1 ( ) 2( )cos ] r r r r r r rr r θ θ ′ ′ ′ ′= + − = + − r 1 ( )( 2cos ) r r r r r θ ′ ′ = +∈ ∈= −r
38. 38. 3.4.1(2) 1 1 3 52 2 3 3 2 8 16 [1 ( )( 2cos ) ( ) ( 2cos ) ( ) ( 2cos ) ] r r r r r r r r r r r r r θ θ θ ′ ′ ′ ′ ′ ′ = − − + − − − +L 1 3 1 5 32 2 3 3 2 2 2 2 [1 ( )cos ( ) ( cos ) ( ) ( cos cos ) ] r r r r r r r θ θ θ θ ′ ′ ′ = + + − + − +L 1 0 ( ) (cos ) r n nr r n P dθ ρ τ ∞ ′ = ′= ∑ Monopole term Dipole term Quadrupole term Multipole expansion 1 1 1 1 1 3 52 32 2 8 16 (1 ) (1 ) r r − ∈ ∈ ∈= +∈ = − + − +L r 1 ( 1)4 0 0 1 1 1 1 3 12 2 34 2 20 1 ( ) ( ) (cos ) [ cos ( ) ( cos ) ] n nn n r r r V r r P d r d r d r d πε πε θ ρ τ ρ τ θ ρ τ θ ρ τ ∞ + = ′= ′ ′= + + − + ∑ ∫ ∫ ∫ ∫ L
39. 39. 3.4.2 The Monopole and Dipole Terms monopole dipole dominates if r >> 1 dipole moment (vector) A physical dipole is consist of a pair of equal and opposite charge, 1 4 0 ( ) Q mon r V p πε = 1 1 24 0 1 1 24 0 ( ) cos ˆ dip r r P V p r d r r d πε πε θ ρ τ ρ τ ′= ′= × ∫ ∫ v v 14243 ˆcosr r rθ′ ′= × v ˆ1 24 0 1 ( ) p r dip r n i i i V p p r d q r πε ρ τ × = = ′ ′= = ∑∫ v v v v q± ( )p qr qr q r r qd+ − + −′ ′ ′ ′= − = − = vv v v v v
40. 40. 3.4.3 Origin of Coordinates in Multipole Expansions if ( ) p r d r d d r d d d p dQ ρ τ ρ τ ρ τ ρ τ ′= ′= − ′= − = − ∫ ∫ ∫ ∫ vv vv vv 0Q p p= = v
41. 41. 3.4.4 The Electric Field of a Dipole A pure dipole ˆ cos 2 24 40 0 2 cos 34 0 1 sin 34 0 1 sin 34 0 ( , ) 0 ˆˆ( , ) (2cos sin ) P r P dip r r V P r r r V P r r V r P dip r V r E E E E r r θ πε πε θ πε θ θ θ πε ϕ θ ϕ πε θ θ θ θ θ × ∂ − ∂ ∂ − ∂ ∂ − ∂ = = = = = = = = = + v v