Laplace Potential Distribution
and Earnshaw’s Theorem
© Frits F.M. de Mul
Laplace and Earnshaw
1
Laplace and Earnshaw
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
Laplace and Earnshaw
2
Electric Field Equations
Gauss: integral formulation:
Id. differential formulation:
1
 dV
S E  dS   0 
V

Ex

 ( x, y, z )    
 ... 
  E ( x, y, z ) 
e x  ...  Ex e x  ... 
x
0
 x 
0
Potential: integral formulation:
Id. differential formulation:
E   V
B
VB  VA   E  dl
A
 V

 

 ... V ( x, y, z )  e x
 ...  e x Ex  ...
e x
x
 x Laplaceand Earnshaw
3
Electric Field Lines and
Equipotential Surfaces
 ( x, y, z )
  E ( x, y, z ) 
0

E   V
E
V = const.
Laplace and Earnshaw
4
Laplace and Poisson: derivation
Gauss:
Potential:
 ( x, y, z)
  E ( x, y, z) 
0
E   V

  E    V   V 
0
2
   V
 2V

e x  e x  ... 
 ...  
2
x
x
x
0
Laplace and Earnshaw
 = 0: free space
(Laplace)
  0: materials
(Poisson)
5
Laplace and Poisson in 1 dimension

2
 V 
0
 2V


2
x
0
 = 0: free space (Laplace)
  0: materials (Poisson)
Calculate V(x) for  =0 by
integration of Laplace equation
V
dV
d 2V
 c   Ex
0 
2
dx
dx
E
V2
 V ( x)  cx  c'
V1
x1
x2
x
Boundary conditions:
V1 at x1 and V2 at x2 :
x  x1
V ( x)  V1 
(V2  V1 )
x2  x1
Laplace and Earnshaw
6
-c
Laplace and Poisson in 1 dimension

2
 V 
0
 2V


2
x
0
V
V2
V1
E
x
0
 2
 V ( x)  cx 
x  c'
0
-c
x2
 0: materials (Poisson)
Calculate V(x) by integration
of Poisson’s equation….
dV


c
x   Ex
dx
0

x1
 = 0: free space (Laplace)
x
1
2
Assume  =const.:
Boundary conditions at x1 and x2
 Parabolic behaviour
Laplace and Earnshaw
7
Laplace and Poisson in 1 dimension

2
 V 
0
 2V


2
x
0
 = 0: free space (Laplace)
  0: materials (Poisson)
 2
 V ( x)  cx 
x  c'
0
1
2
V
V0
Assume  =const.:
Boundary conditions at x1 and x2
Special case:
x1=0 ; V1=0 and x2= a ; V2 = V0
Calculate V(x) and E(x)
E

0
0
a
x 2 ax V0 x
V ( x)  


2 0
2 0
a

V0 
( x)   
( x  a)  8 
Laplace and E
Earnshaw
a
 0
x
-V0
a
Laplace and Poisson in 1 dimension
x 2 ax V0 x
V ( x)  


2 0
2 0
a
Assume  =const.:
Boundary conditions:
at x1 and x2
Special case:
x1=0 ; V1=0 and
x2= a ; V2 = V0
V
V0
x/a
Laplace and Earnshaw
/0
9
Laplace in 1 dimension

2
 V 
0
 2V


2
x
0
V
Earnshaw:
dV
d 2V
 c   Ex
 0
2
dx
dx
E
V2
 = 0: free space (Laplace)
  0: materials (Poisson)
 V ( x)  cx  c'
If no free charge present, then:
Boundary
conditions
Potential has no local
maxima
orat x and x :
x
x
x
x  x1
minima.
V ( x)  V 
(V  V )
V1
1
1
2
2
-c
Laplace and Earnshaw
1
x2  x1
2
1
10
Laplace in 1 dimension: Earnshaw
2
dV
dV
 c   Ex
0 
2
dx
dx
 V ( x)  cx  c '
Earnshaw:
If no free charge present, then:
Potential has no local maxima
or minima.
V
E
V2
V1
x1
x2
x
-c
Consequences:
1. V is linear function of position
2. V at each point is always in
between neighbours
Laplace and Earnshaw
11
Laplace in 1 dimension: Earnshaw
Earnshaw:
If no free charge present, then:
Potential has no local maxima
or minima.
V
V2
V1
x1
x2
x
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:
1. Start with zero potential between
boundaries
2. Take averages between
neighbours
3. Repeat and repeat and ....
Start Earnshaw program
Laplace and Earnshaw
12
Laplace in 2 dimensions: Earnshaw
Potential V=f (x,y) on S ?
V
?
Earnshaw:
If no free charge present, then:
Potential has no local maxima
or minima.
S
2
Solution of Laplace  V  0
 2
2 
 2  2  V ( x, y)  0
 x y 
will depend on boundaries.
y
x
“Partial differential equation”
Numerical solution using
“Earnshaw”-program
Laplace and Earnshaw
13
Laplace / Poisson in 3 dimensions
Spatial charge density:  =f (x,y,z)
Boundary conditions:
V1 , V2 and V3 = f (x,y,z)
Solution of Laplace/Poisson:
 2
2
2 

 2  2  2  V ( x, y )  
y
z 
0
 x
will depend on boundaries.
z
V1
V2
Potential V = f (x,y,z) ?
4-D plot needed !?

Special cases:
• cylindrical geometry
• spherical geometry
V3
x
y
Laplace and Earnshaw
14
Laplace / Poisson in 3 dimensions
Special case 1: Cylindrical geometry
z
1     1 2
2 

V 
 2 V  
r   2
2
z 
0
 r r  r  r 
2

x
If r – dependence only:
 and boundaries will be f (r).
Thus: V will be f (r) only
y
Example: V=V1 at r1 and V2 at r2 , and =0
r Calculate V = V(r )
1 d  dV 
r
0
r dr  dr 
V (r )  c. ln r  c'

dV
dV c
c 

dr
dr
r
(c and c' : const .)
r
With Laplace
boundaries
: V (r )  V1  (V2  V1 )
and Earnshaw
ln r  ln r1
15
ln r2  ln r1
Laplace / Poisson in 3 dimensions
Special case 2: Spherical geometry
2


1


1


1






2
2
 V   2 r
V 
 2
 sin 
 2 2
2
  r sin   
0
 r r  r  r sin   
z

r
y
x

If r – dependence only:
 and boundaries will be f (r).
Thus: V will be f (r) only
Example: V=V1 at r1 and V2 at r2 , and =0
Calculate V = V(r )
1 d  2 dV 
r
0
2
r dr 
dr 
1
V (r )  c.
 c'
r

r2
dV
c
dr

dV
c
 2
dr r
the end
(c and c' : const .)
WithLaplace
boundaries
: V (r )  V1  (V2  V1 )
and Earnshaw
1 / r  1 / r1
16
1 / r2  1 / r1