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If the planes defined by pl1, pl2 are parallel then whith the following procedure we can construct parallel intermediate planes for each lambda value gr1 = Table[ParametricPlot3D[pl1[[k]] + lambda (pl2[[k]] - pl1[[k]]), {lambda, 0, 1}, PlotStyle -> {Thick, Red}], {k, 1,4}]; gr2 = Table[Graphics3D[Polygon[{pl1[[1]] + lambda (pl2[[1]] - pl1[[1]]), pl1[[2]] ...

1

So far I was able to get the moving particle but I can't get an animation for the changing E field yet. I have not followed your code, but you can get the last part to work by doing this Animate[VectorPlot3D[ Evaluate[{Ex, Ey, Ez} /. t -> t0], {x, -5, 5}, {y, -5, 5}, {z, -5, 5}], {t0, 10^-50, 1}] The above also works without using Evaluate as ...

6

Mathematica is great for Physics and simulation. One way is to use Manipulate which makes things very easy Manipulate[ ParametricPlot3D[{r Cos[w t], r Sin[w t], v t}, {t, 0, maxTime}, PlotRange -> {{-10, 10}, {-10, 10}, {0, 100}}, PerformanceGoal -> "Quality", PlotStyle -> Red], {{r, 6, "r?"}, 0.01, 10, 0.01, Appearance -> "Labeled", ...

2

f[x_, y_] := x + x/(x^2 + y^2); pnts = {{-1.8, 0.2}, {-1.8, 0.4}, {-1.8, 0.6}, {-1.8, 0.8}, {-1.8, 1.}, {-1.8, 1.2}, {-1.8, 1.4}, {-1.8, 1.6}, {-1.6, 0.2}, {-1.6, 0.4}, {-1.6, 0.6}, {-1.6, 0.8}, {-1.6, 1.}, {-1.6, 1.2}, {-1.6, 1.4}, {-1.6, 1.8}, {-1.4, 0.2}, {-1.4, 0.4}, {-1.4, 0.6}, {-1.4, 0.8}, {-1.4, 1.}, {-1.4, 1.2}, {-1.4, 1.6}, {-1....

2

You can use the new "SplineCircle" ResourceFunction to do this. Example: Graphics3D[ { Thick, ResourceFunction["SplineCircle"][{0, 0, 0}, {1,3}, {{0, 1, 0}, {0, 0, 1}}, {0, 3Pi/2}] }, Boxed->False, AxesOrigin->{0,0,0}, Axes->True, AxesEdge->{.5,.5,.5}, AxesLabel->{x, y, z} ] The syntax is ResourceFunction["...

4

A couple things. Since $\eta$ is in the denominator of your pde, your solution will have a problem when $\eta$ is 0. Luckily since zero is one of the end points we can pick a number close to zero without changing the answer much. A numerical solution is an approximation after all. Also your NDSolveValue needs the limits of $\eta$ as well as $\tau$. ...

2

Using color as the fourth dimension has severe limitations for regions that are solid in that you can only see the color of the surface of the region. While Opacity could be used to see "into" the solid region, the colors would be muddled together and you would not be able to appreciate the fourth dimension. ineq = α >= 2 β && 2 σ >= γ &&...

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