Presentation is loading. Please wait.

Presentation is loading. Please wait.

Laplace and Earnshaw1 Laplace Potential Distribution and Earnshaw’s Theorem © Frits F.M. de Mul.

Similar presentations


Presentation on theme: "Laplace and Earnshaw1 Laplace Potential Distribution and Earnshaw’s Theorem © Frits F.M. de Mul."— Presentation transcript:

1 Laplace and Earnshaw1 Laplace Potential Distribution and Earnshaw’s Theorem © Frits F.M. de Mul

2 Laplace and Earnshaw2 1.Electric Field equations: –Gauss’ Law and Potential Gradient Law 2.Laplace and Poisson: derivation 3.Laplace and Poisson in 1 dimension 4.Charge-free space: Earnshaw’s Theorem –Finite-Elements method for Potential Distribution 5.Laplace and Poisson in 2 and 3 dimensions

3 Laplace and Earnshaw3 Electric Field Equations Gauss: integral formulation: Potential: integral formulation: Id. differential formulation:

4 Laplace and Earnshaw4 Electric Field Lines and Equipotential Surfaces  E V = const.

5 Laplace and Earnshaw5 Laplace and Poisson: derivation Gauss: Potential:  = 0: free space (Laplace)    0: materials (Poisson)  = 0: free space (Laplace)    0: materials (Poisson)

6 Laplace and Earnshaw6 Laplace and Poisson in 1 dimension  = 0: free space (Laplace)   0: materials (Poisson) Calculate V(x) for  =0 by integration of Laplace equation Boundary conditions: V 1 at x 1 and V 2 at x 2 : x1x1 x2x2 V1V1 V2V2 x V -c E

7 Laplace and Earnshaw7 Laplace and Poisson in 1 dimension  = 0: free space (Laplace)  0: materials (Poisson) Calculate V(x) by integration of Poisson’s equation…. Assume  =const.: Boundary conditions at x 1 and x 2  Parabolic behaviour  x V x1x1 x2x2 V1V1 V2V2 E -c x0x0

8 Laplace and Earnshaw8 x V 0a 0 V0V0  E Laplace and Poisson in 1 dimension  = 0: free space (Laplace)   0: materials (Poisson) Assume  =const.: Boundary conditions at x 1 and x 2 Special case: x 1 =0 ; V 1 =0 and x 2 = a ; V 2 = V 0 -V 0 a Calculate V(x) and E(x)

9 Laplace and Earnshaw9 Laplace and Poisson in 1 dimension Assume  =const.: Boundary conditions: at x 1 and x 2 Special case: x 1 =0 ; V 1 =0 and x 2 = a ; V 2 = V 0 Assume  =const.: Boundary conditions: at x 1 and x 2 Special case: x 1 =0 ; V 1 =0 and x 2 = a ; V 2 = V 0 /0/0 x/a VV0VV0

10 Laplace and Earnshaw10 Laplace in 1 dimension  = 0: free space (Laplace)   0: materials (Poisson) Boundary conditions at x 1 and x 2 : -c V1V1 V2V2 x V x1x1 x2x2 E Earnshaw: If no free charge present, then: Potential has no local maxima or minima. Earnshaw: If no free charge present, then: Potential has no local maxima or minima.

11 Laplace and Earnshaw11 Laplace in 1 dimension: Earnshaw x V x1x1 x2x2 V1V1 V2V2 -c E Earnshaw: If no free charge present, then: Potential has no local maxima or minima. Consequences: 1. V is linear function of position 2. V at each point is always in between neighbours

12 Laplace and Earnshaw12 Laplace in 1 dimension: Earnshaw Earnshaw: If no free charge present, then: Potential has no local maxima or minima. Consequences: 1. V is linear function of position 2. V at each point is always in between neighbours Numerical method for calculating potentials between boundaries: x V x1x1 x2x2 V1V1 V2V2 1. Start with zero potential between boundaries 2. Take averages between neighbours Start Earnshaw program 3. Repeat and repeat and....

13 Laplace and Earnshaw13 Laplace in 2 dimensions: Earnshaw Earnshaw: If no free charge present, then: Potential has no local maxima or minima. Potential V=f (x,y) on S ? Solution of Laplace will depend on boundaries. “Partial differential equation” Numerical solution using “Earnshaw”-program V x y ? S

14 Laplace and Earnshaw14 Laplace / Poisson in 3 dimensions Potential V = f (x,y,z) ? Solution of Laplace/Poisson: will depend on boundaries. Boundary conditions: V 1, V 2 and V 3 = f (x,y,z) z x y V1V1 V2V2 V3V3  Spatial charge density:  =f (x,y,z) Special cases: cylindrical geometry spherical geometry 4-D plot needed !?

15 Laplace and Earnshaw15 Laplace / Poisson in 3 dimensions Special case 1: Cylindrical geometry z r x y  If r – dependence only:  and boundaries will be f (r). Thus: V will be f (r) only Example: V=V 1 at r 1 and V 2 at r 2, and  =0 Calculate V = V(r )

16 Laplace and Earnshaw16 Laplace / Poisson in 3 dimensions Special case 2: Spherical geometry If r – dependence only:  and boundaries will be f (r). Thus: V will be f (r) only Example: V=V 1 at r 1 and V 2 at r 2, and  =0 Calculate V = V(r ) z r x y   the end


Download ppt "Laplace and Earnshaw1 Laplace Potential Distribution and Earnshaw’s Theorem © Frits F.M. de Mul."

Similar presentations


Ads by Google