4
$\begingroup$

I have to plot the Electrical Field of half infinity charged plates.

The mathematical Function to this is:

Efeld[x_, z_] := {\[Sigma] (Log[x^2 + (z - d)^2] - Log[x^2 + (z + d)^2]), \[Sigma] (\[Pi] Sign[z - d] + 2 ArcTan[x/(z - d) ] - \[Pi] Sign[z + d] - 2 ArcTan[x/(z + d) ])}

This are my StreamPlots:

strp = Table[{y, 0}, {y, -5, 7, .5}]; additionalstrp = Join[strp, {{6.`, 4.8}, {6.`, 5.6}, {6.`, 6.4}, {6.`, 7.15}, {6.`, 7.9}, {6.8`, 8}, {6.`, -4.8}, {6.`, -5.6}, {6.`, -6.4}, {6.`, -7.15}, {6.`, -7.9}, {6.8`, -8}, {-4.5`, 7.6}, {-4.9`, 8.}, {-4.5`, -7.6}, {-4.9`, -8.}, {0.125, \[Pi]}}];

This are some Parameters:

plotsizeminx := 5 plotsizemaxx := 7 plotsizeminy := 8.5 plotsizemaxy := 8.5

This is the Plot:

 strplot =StreamPlot[ Efeld[-x, -y] /. {\[Sigma] -> 1, d -> \[Pi]}, {x, -plotsizeminx, +plotsizemaxx}, {y, -plotsizeminy, plotsizemaxy}, Epilog -> {Black, Thickness[0.01], Line[{{-plotsizeminx, Pi}, {0, Pi}}], Line[{{-plotsizeminx, -Pi}, {0, -Pi}}]}, StreamScale -> None, StreamStyle -> {Thickness[0.002], Black}, StreamPoints -> additionalstrp, PlotRangePadding -> -0.3, AspectRatio -> 1/1.5]

This is picture of the plot:

enter image description here

As you can see i have highligted some areas where the streaming arrows stop "too early".

Mainly i defined the Streampoints depending on the horizontal symmetry axis, so that it looks simular to another picture.

Even tho i have definde the StreamPoints by myself i do not have enough control to get rid of this "gaps" at the and of a "StreamLine".

Do you have any Idea how to get rid of this gaps.

Thank you.

Improve this question
$\endgroup$
2
  • $\begingroup$ have already found a solution $\endgroup$ Commented Apr 10, 2017 at 15:49
  • $\begingroup$ Please write up your solution? $\endgroup$ Commented Apr 11, 2017 at 19:37

1 Answer 1

Reset to default
2
$\begingroup$

Ok, so some time ago a other user helped me with a similar problem. at least i took this solution and manipulated it to fit to the acutal problem.

strp2 = Table[{y, 0}, {y, -4.5, 15, .5}] params = {\[Sigma] -> 1, d -> \[Pi]}; ode = {x'[t], y'[t]} == (Efeld[x, y] /. {v : y | x :> v[t]} /. params) {Derivative[1][x][t], Derivative[1][y][t]} == {Log[x[t]^2 + (-\[Pi] - y[t])^2] - Log[x[t]^2 + (\[Pi] - y[t])^2], -2 ArcTan[x[t]/(-\[Pi] - y[t])] + 2 ArcTan[ x[t]/(\[Pi] - y[t])] + \[Pi] Sign[-\[Pi] - y[t]] - \[Pi] Sign[\[Pi] - y[t]]} psol = ParametricNDSolveValue[ { ode, {x[0], y[0]} == {a, b}, WhenEvent[ Abs[x[t]] < plotsizemaxx || Abs[x[t]] > plotsizeminx || Abs[y[t]] < plotsizemaxy || Abs[y[t]] > plotsizeminy, "StopIntegration"], WhenEvent[Abs[y[t] - \[Pi]] < 0.01 && x[t] < 0, "StopIntegration"] }, {x, y}, {t, -100, 100}, {a, b}, Method -> {"ExplicitRungeKutta", "StiffnessTest" -> True}, MaxSteps -> 10000, MaxStepSize -> 0.01]; sols = psol @@@ strp2; // AbsoluteTiming (*{25.573434066961003`,Null}*)

its a numerical solution. so i totally bypass the function StreamPlot. i apply this method to every streampoint so i get every line upto the detail needed. with Graphics[] you can plot them alltogether.

Graphics[ {Black, Thickness[0.002], Line[Transpose@Through[#["ValuesOnGrid"]] & /@ sols], {Black, Thickness[0.01], Line[{{-plotsizeminx, \[Pi]}, {0, \[Pi]}}], Black, Thickness[0.01], Line[{{-plotsizeminx, -\[Pi]}, {0, -\[Pi]}}]}} , PlotRange -> {{-plotsizeminx, +plotsizemaxx}, {-plotsizeminy, plotsizemaxy}}]

at this point i´m no expert to this (acutually im having a mathematica lecture right now, until now i had to learn this by myself), so i hope this is helpful.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I suppose you mean this answer? (When the help came from Q&A on this site, the common practice is to include a link. And often people link to references off-site, too.) $\endgroup$ Commented Apr 12, 2017 at 22:44
  • $\begingroup$ yes, i meant this one ^^ thx again :) $\endgroup$ Commented Apr 15, 2017 at 10:28

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.