Test3d13, 13th 'benchmark test' data file for CPO3D

Field at a circular hole in an infinite sheet

 

The problem is simulated by putting a hole of diameter 1 in an electrode of diameter 5, and enclosing the regions on both sides of the electrode with cylinders of length 10. With 192 segments the errors are less than 0.004 for the potential and field on the axis, but go up to 0.014 for the more difficult region near the aperture edge (where the fields become infinitely large at the edge itself). With N = 768 these maximum errors become 0.0014 and 0.009 respectively.


Detailed description:

 

The following data were obtained when the memory and speed of PC's was much more limited than at present, so the available number of segments was small and the requested inaccuracies were fairly high to give a quick demonstration.

 

The problem is simulated by putting a hole of diameter 1 in an electrode of diameter 5, and enclosing the regions on both sides of the electrode with cylinders of length 10. The potentials and field components are sampled at 3 points on the axis and 3 points near to the edge of the aperture (at a distance of 0.1 from the edge).

 

Taking E1 and E2 as the asymptotic fields for z<0 and z>0 respectively, then for a hole of diameter R the exact expression for the potential along the axis is

 V(0,z) = (E2-E1)*(z/pi)*(arctan(z/R)+(R/z)) + (E2+E1)*(z/2)

In the present example, E1 = 0, E2 = -1, R = 1.

 

The equation for the off-axis potential is given by P W Hawkes and E Kasper, Principles of Electron Optics. Further details and an algorithm are given in equation details.

 

The table below shows the comparison between the results obtained with the present simulation and the exact values of the potentials and field components. The 'exact' values are those for the present geometry, and not for a hole in an infinite sheet, and have been obtained by using CPO2D (see the test file test2d15.dat) and extrapolating the number of segments N to infinity.


z=

-1.0

 0.0

 1.0

-0.1

 0.0

 0.1

r=

 0.0

 0.0

 0.0

 1.0

 0.9

 1.0

potl: present, N=192

 0.0711

 0.3207

 1.0682

 0.0606

 0.1421

 0.1606

                    N=768 

 0.0692

 0.3194

 1.0687

 0.0593

 0.1404

 0.1593

                    exact 

 0.0683

 0.3183

 1.0683

 0.0582

 0.1388

 0.1582

ez: present, N=192

-0.0901

-0.4995

-0.9087

 0.1068

-0.4995

-1.1059

                  N=768

-0.0907

-0.4996

-0.9085

 0.1109

-0.4996

-1.1053

                  exact 

-0.0909

-0.5000

-0.9092

 0.1152

-0.4999

-1.1152

er: present, N=192

 0.0

 0.0

 0.0

 0.4554

 0.6462

 0.4549

                  N=768

 0.0

 0.0

 0.0

 0.4493

 0.6455 

 0.4523

                 exact

 0.0

 0.0

 0.0

 0.4436

 0.6572

 0.4436


It can be seen from the table that the present simulation with N = 192 gives errors that are less than 0.004 for the potential and field on the axis, but go up to 0.013 for the more difficult region near the aperture edge (where the fields become infinitely large at the edge itself). With N = 768 these maximum errors become 0.0011 and 0.013 respectively. Extrapolation to N = infinity does not seem to be possible in this example because the potentials and fields do not seem to change smoothly enough as N is changed (although extrapolation is successful with the 2-dimensional version of this test -see test2d15.dat). For example the behaviour of the potential at z = 0, which should be 1/pi = 0.31831, is:

 

N=

96

144

192

288

384

432

480

576

672

768

(pot-1/pi)*1000=

5.08

4.27

2.42

2.18

0.95

0.79

0.63

0.53

0.78

1.10

(pot-1/pi)*N**1.5=

4.8

7.4

6.4

10.7

7.1

7.1

6.6

7.3

13.6

23.4

 

For N = 144 to 576 the error in the potential is approximately proportional to N**1.5, and the extrapolation to N=infinity gives the potential 0.31819 +/- 0.00023. . Comparing this with the value .31813 obtained from test2d15 (thereby taking into account the finite size of the shielding cylinder), the error is 0.00006 +/- 0.00023. The values of the potential at the two highest values of N have larger errors (for example .0013 at N=768), for reasons that are not understood at present.