University Electromagnetism: Laplace potent distribution and arshw's theorem

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

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

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

6. 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 : x 1 x 2 V 1 V 2 x V -c E

8. 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 Calculate V(x) and E(x) x V 0 a 0 V 0  E -V 0 a

9. 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  /  0 x/a V V 0

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

11. 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 x V x 1 x 2 V 1 V 2 -c E

12. 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: Start Earnshaw program x V x 1 x 2 V 1 V 2 1. Start with zero potential between boundaries 2. Take averages between neighbours 3. Repeat and repeat and ....

13. 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

15. Laplace / Poisson in 3 dimensions Special case 1: Cylindrical 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 

16. 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 ) the end z r x y  