Notes 19 Gradient and Laplacian

Transcription

1 ECE 3318 Applied Electricity and Magnetism Spring 218 Prof. David R. Jackson Dept. of ECE Notes 19 Gradient and Laplacian 1

2 Gradient Φ ( x, y, z) =scalar function Φ Φ Φ grad Φ xˆ + yˆ + zˆ x y z We can write this as grad Φ= xˆ + yˆ + zˆ Φ x y z Hence grad Φ = Φ 2

3 Directional Derivative Property r We look at how a function changes from one point to a nearby point. d ( xyz,, ) dr (,, ) r + dr x + dx y + dy z + dz dl = dr Recall: ( ) ( ) dφ=φ r + dr Φ r Φ Φ Φ = dx + dy + dz x y z (from calculus) Φ Φ Φ Φ xˆ + yˆ + zˆ x y z ( ) ( ) ( ) dr = xˆ dx + yˆ dy + zˆ dz Hence dφ = Φ dr 3

4 Directional Derivative Property (cont.) r d ( xyz,, ) dr (,, ) r + dr x + dx y + dy z + dz dl = dr Then Use dφ = Φ dr dr = d ( ) dφ = Φ d This gives us the directional derivative: dφ = Φ d Directional derivative 4

5 Physical Interpretation Gradient dφ Φ = Φ d θ Hence dφ = Φ d cosθ dφ Φ=constant = d Direction we march in along path We maximize the directional derivative when we march along in the direction of the gradient (θ = ). The magnitude of the gradient vector gives us the directional derivative when we go in the direction of the gradient. The gradient is perpendicular to a level curve of the function (θ = π / 2). 5

6 Mountain Example Topographic map: Φ(x, y) = height of the landscape at any point. y Φ Φ = + x ( xy, ) xˆ yˆ Φ y Φ x Φ = 1 [m] Φ = [m] Φ = -1 [m] 6

7 Summary of Gradient Formulas Rectangular Φ Φ Φ Φ = xˆ + yˆ + zˆ x y z Cylindrical ˆ 1 Φ = ˆ ρ Φ + φ Φ + ẑ Φ ρ ρ φ z Spherical ˆ1 ˆ 1 Φ = rˆ Φ + θ Φ + φ Φ r r θ rsinθ φ 7

8 Relation Between E and Φ Recall: AB ( ) ( ) V =Φ A Φ B E dr B A Also, from calculus ( ) ( ) Φ A Φ B = dφ = Φ dr A B A B Hence, from the above two results we have B A B E dr = Φ dr = Φ dr A B A 8

9 Relation Between E and Φ (cont.) B A B E dr = Φ dr This must be true for any path. A Assume a small path in the x direction: A x B x B B ( ˆ ) x x x A A x x x B E dr = E x dx = E dx E dx = E x A x B A Similarly, for the second integral: B A ( ) Φ dr Φ x x 9

10 Relation Between E and Φ (cont.) Hence: E x ( ) = Φ x Similarly, using paths in the y and z directions, we have E E y z ( ) ( ) = Φ = Φ y z Hence, we have E = Φ This gives us a new way to find the electric field, by first calculating the potential (illustrated next with examples). Note: The choice of R (the reference point) does not affect E (the gradient of a constant is zero). 1

11 Example Find E from the point charge z x Φ= q 4πε r q [C] [ V] y E Φ ˆ1 Φ ˆ 1 Φ = Φ = rˆ + θ + φ r r θ rsinθ φ = Φ rˆ r q = rˆ 2 4πεr E q = rˆ 2 4πεr [ V/m] 11

12 Line Charge Example y Find E from the line charge ˆ 1 Φ = ˆ ρ Φ + φ Φ + ẑ Φ ρ ρ φ z 2 1 Φ ρ b l Φ= ln 2πε ρ Arbitrary reference point b From previous calculation: Φ = -1 [V] E = E x ρl b = Φ = ˆ ρ ln ρ 2 πε ρ ρl = ˆ ρ ( ln b ln ρ) 2πε ρ = ρ ˆ 2 ρl ˆ ρ 2 πε ρ l ρ πε ρ [V/m] 12

13 Example z Find: E (,, z) (,, z) a R y On the z axis: (,, ) = ˆ (,, ) E z ze z z E = Φ x (,, z) Φ = ρ l [C/m] From previous calculation: 2ε ρ z a + a 2 2 E z Φ = z 13

14 Example (cont.) E z (,, z) so (,, ) ρ dφ z d a = = dz dz 2 2 2ε z + a ρ a 1 Ez z z a z 2ε 2 ( ) ( 2 2) 3/2,, = + ( 2 ) We thus have E z ( z) ρ a ( 2 2 z + a ) [ ],, = V/m 3/2 2ε z 14

15 Vector Identity ( ψ ) = Proof: ( ψ ) xˆ yˆ zˆ x y z ψ ψ ψ x y z = ψ ψ ψ ψ ψ ψ = xˆ yˆ + zˆ yz zy xz zx xy yx = 15

16 Curl Property in Electrostatics (revisited) E = Φ (in statics) E = Φ so = ( ) = Φ ( ) E = 16

17 Equivalent Statements of Path Independence In Statics E = Φ Path Independence E = C E dr = 17

18 Poisson Equation This is a differential equation that the potential satisfies. (This is useful for solving boundary value problems that involve conductors or dielectrics.) Start with the electric Gauss law: D = ρ v ε E = ρ ( ) ( ) v ε Φ = ρ Φ = ( ) v ρ ε v 18

19 Poisson Equation (cont.) Define the Laplacian : Φ Φ Φ ( Φ ) = xˆ + yˆ + zˆ x y z Φ Φ Φ = + + x y z Φ Φ Φ Lap Φ ( Φ ) = + + x y z Poisson s Eq.: Lap Φ= ρv ε 19

20 Poisson Equation (cont.) Del-operator notation for Laplacian: = xˆ + yˆ + zˆ x y z x y z x y z x y z = = xˆ + yˆ + zˆ xˆ + yˆ + zˆ = so x y z Φ= + + Φ = Laplacian operator Hence 2 Lap Φ ( Φ ) = Φ 2

21 Poisson Equation (cont.) Hence, we have 2 Φ= ρ v ε Poisson Equation If ρ v = then 2 Φ= Laplace Equation Note: Φ=Φ ( xyz,, ), ρ = ρ ( xyz,, ) v v 21

22 Poisson Equation (cont.) Siméon Denis Poisson Pierre-Simon Laplace (from Wikipedia) 22

23 Laplacian Rectangular Φ Φ Φ Φ= x y z Cylindrical ρ ρ ρ ρ φ z Φ 1 Φ Φ Φ= ρ Spherical r Φ sinθ Φ Φ Φ= + + r r r r sinθ θ θ r sin θ φ 23

24 Summary of Formulas: Electrostatic Triangle One nice way to summarize all of the equation of electrostatics into one nice visual display is the electrostatic triangle (courtesy of Prof. Donald R. Wilton). 24

25 Electrostatic Triangle ρ v Φ= V ρ v 4πε R dv ( ε E) = ρ v E = V ρ Rˆ v 4πε R S 2 dv D nˆ ds = Q encl 2 ρ Φ= v ε Φ E = Φ E ( ) ( ) Φ r =Φ R E dr r R 25