Key Concepts for this section

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Clement Atkins
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1 Key Concepts fo this section 1: Loentz foce law, Field, Maxwell s equation : Ion Tanspot, Nenst-Planck equation 3: (Quasi)electostatics, potential function, 4: Laplace s equation, Uniqueness 5: Debye laye, electoneutality Goals of Pat II: (1) Undestand when and why electomagnetic (E and B) inteaction is elevant (o not elevant) in biological systems. () Be able to analyze quasistatic electic fields in D and 3D.

2 Electostatics Steady Diffusion Φ=V 0 C=C 0 Φ=0 Gel o tissue Φ=0 (σ,ε) C=0 Gel o tissue (D) C=0 Φ=0 C=0 J Φ=0 e = σ Φ C = 0 J = D c i i i

3 Φ Φ Φ x y z Φ= + + = 0 Assume Φ ( xyz,, ) =Χ( x) ϒ( y) Ζ( z) Χ ϒ Ζ Φ=ϒΖ +ΧΖ +Χϒ = 1 Χ 1 ϒ 1 Ζ + + = 0 13 Χ x ϒ y 13 Ζ z 13 function of x x y z function of y function of z 0 Thee possibilities 1 Χ ( ), = kx x kx x kx Χ x = + e e Χ x o= k Χ ( x) = sin( kx), cos( kx) x x x o = 0 Χ ( x) = ax + b ( a, b : constants)

4 Φ= Φ =Χ ϒ 0, ( xy, ) ( x) ( y) 1 Χ 1 ϒ + = 0 Χ x ϒ y 1 Χ k X( x)~sin( kx) = Χ x sin( kl) = 0 kl = nπ ( n : intege) Eigenvalue : kn nπ = L expand Χ(x) using Fouie sine seies nπ x Χ ( x) = An sin (This satisfies B. C. at x=0, L) n L ϒ( y) nπy nπy k n y y o then, ϒ ( ) = 0 ϒ( ) ~ sinh cosh y L L Φ=V 0 Φ=0 Gel o tissue (σ,ε) Φ=0 Φ=0 nπy nπx nπ y ϒ ( y) = sinh since Φ ( x,0) = 0 Φ ( x, y) = An sin sinh L n L L Detemining A : use bounday condition n nπ x Φ ( xl, ) = V0 = An sin sinh( nπ ) n L L mπx V0 (1 cos( nπ)) opeate sin on both sides A 0 n = L nπ sinh( nπ)

5 Solving Laplace s Equation (Numeically) 1D case: D case: d Φ 0 ( x) ax b = Φ = + dx Φ x Φ y + = 0 Φ 1 ( n +, m) =Φ ( n + 1, m) Φ ( n, m) x Φ 1 ( n, m) =Φ ( n, m) Φ ( n 1, m) x Φ ( nm, + 1) Φ( n 1, m) Φ( nm, ) Φ( nm, 1) y (m) Φ ( n+ 1, m) x (n) 1 1 ( nm, ) = ( n+, m) ( n, m) =Φ ( n+ 1, m) +Φ( n 1, m) Φ( nm, ) Φ Φ Φ x x x

6 Laplace s equation In discetized fom Φ ( nm, + 1) Φ( n 1, m) Φ( nm, ) Φ( nm, 1) y (m) Φ ( n+ 1, m) Φ x Φ ( nm, ) + ( nm, ) = y Φ ( n+ 1, m) +Φ( n 1, m) +Φ ( nm, + 1) +Φ( nm, 1) 4 Φ ( nm, ) = 0 x (n) Φ ( nm, ) = Φ ( n+ 1, m) +Φ( n 1, m) +Φ ( n, m+ 1) +Φ( n, m 1) 4 Value in the middle = aveage of suounding values

7 Finite Element Method

8 Known Solutions fo Laplace equations Cylindical Coodinates ρϕz Φ 1 Φ 1 Φ Φ Φ (,, ) = = 0 ρ ρ ρ ρ ϕ z Φ ( ρϕ,, z) = R( ρ) Ψ( ϕ) Ζ( z) R( ρ) Bessel Functions ( J, N, I, K ) n n n n Ψ( ϕ) Tigonometic (sin, cos,sinh, cosh) Ζ( z) Tigonometic (sin, cos,sinh, cosh) Spheical Coodinates 1 Φ 1 Φ 1 Φ θϕ θ Φ (,, ) = 0 + sin + = 0 sinθ θ θ sin θ ϕ Φ (, θϕ, ) = R() Θ( θ) Ψ( ϕ) R() Spheical Bessel Functions Θ( θ) Legende Functions ( P (cos θ)) Ψ( ϕ) Tigonometic (sin ϕ,cos ϕ) n

9 Cell in a field E ext R σ i, ε σ 0, ε (medium) ẑ Equation to solve : J = E = Φ = Φ= Laplace s Equation e ( σ ) ( σ ) 0 0 ( ' ) 0 1 Φ 1 Φ 1 Φ θϕ θ Φ (,, ) = 0 + sin + = 0 sinθ θ θ sin θ ϕ Φ (, θϕ, ) = R() Θ() θ sepaate and solve, n 1 R () A + B n+ 1 Θ( θ) Legende Functions ( P (cos θ)) n

10 Guessing the solution E E zˆ as ext Φ= E z = E cosθ as ext ext P n (cosθ) ~ cos nθ Only n =1 tem contibutes (should be dipole field) E ext Tial Solution: 1 Φ o = A cosθ + B cos θ (fo R) θ - σ z i, ε σ 0, ε (medium) 1 Φ i = C cosθ + D cos θ (fo R) D = 0 ( Φ finite at =0) i A= E ( Φ E cos θ when ) ext o ext

11 Bounday Conditions (Fo EQS appoximation) u ( ε E) = ρe E = uu J e 0 ρ = t uu uu n E E ˆ ( ε1 1 ε ) = σs uu uu nˆ E = nˆ E ( E = E ) 1 1 tangential tangential uu uu σ s nˆ ( E E ) t σ1 1 σ = Figue (a) Diffeential contou intesecting suface suppoting suface chage density. (b) Diffeential volume enclosing suface chage on suface having nomal n. Coutesy of Heman Haus and James Melche. Used with pemission. Souce:

12 Some plots fo the solution Cell is less conductive than media Insulating Cell σ < σ 0 σ = 0 Cell is moe conductive than media σ > σ 0 Pefectly conducting Cell σ = Figue by MIT OCW.

Solutions. V in = ρ 0. r 2 + a r 2 + b, where a and b are constants. The potential at the center of the atom has to be finite, so a = 0. r 2 + b.

Solutions. V in = ρ 0. r 2 + a r 2 + b, where a and b are constants. The potential at the center of the atom has to be finite, so a = 0. r 2 + b. Solutions. Plum Pudding Model (a) Find the coesponding electostatic potential inside and outside the atom. Fo R The solution can be found by integating twice, 2 V in = ρ 0 ε 0. V in = ρ 0 6ε 0 2 + a 2