Skip to main content

physics - Electric Field Plot


I want to show an electric field of several arrangements of point charges in xy-plane. I wrote a routine that plots an electric field of a charge:


r0 = {a, b};
r1 = {-a, b};
r2 = {-a, -b};

r2 = {-a, -b};
pot[r_] := q/Norm[r - r0] + q/Norm[r - r2] - q/Norm[r - r1] - q/Norm[r - r3]
fld[r_] := (q*(r - r0)/Norm[r - r0]^3 + q*(r - r2)/Norm[r - r2]^3 - q*(r - r1)/Norm[r - r1]^3 - q*(r - r3)/Norm[r - r3]^3)
a = 2.5;
b = 2.5;
q = 1;
StreamPlot[fld[{x, y}], {x, 0, 5}, {y, 0, 5},PlotRangePadding -> None, FrameLabel -> "electric field",Epilog -> {Red, Disk[r0, 0.07], Blue , Line[{{0, 5.5}, {0, 0}, {5.5, 0}}]}]

enter image description here


Now I want to show:



a) 2 charges with inverted sign


b) 4 charges on edges of a cuboid in xy-plane (edges connect charges with inverted sign)


c) 6 randomly distributed charges with vanishing total charge by using RandomReal and initialize random generator with SeedRandom[1234567]


Could someone help me out with a,b,c ? Thank you very much!



Answer



SeedRandom[1234567];
q = RandomReal[{-1, 1}, 6];
r0 = RandomReal[{-3, 3}, {6, 2}]
q = q - Total[q]/6;
phi = Sum[

q[[i]]/Sqrt[({x, y} - r0[[i]]).({x, y} - r0[[i]])], {i, 1, 6}];
f = -Grad[phi, {x, y}]

Show[StreamPlot[Evaluate[f], {x, -4, 4}, {y, -4, 4},
StreamColorFunction -> "Rainbow",
StreamColorFunctionScaling -> False],
Graphics[Table[
If[q[[i]] < 0, {Blue, PointSize[.1*Abs[q[[i]]]],
Point[r0[[i]]]}, {Red, PointSize[.1*Abs[q[[i]]]],
Point[r0[[i]]]}], {i, 1, 6}]]]


and on a large scale


Show[StreamPlot[Evaluate[f], {x, -40, 40}, {y, -40, 40}, 
StreamColorFunction -> "Rainbow",
StreamColorFunctionScaling -> False],
Graphics[Table[
If[q[[i]] < 0, {Blue, PointSize[.1*Abs[q[[i]]]],
Point[r0[[i]]]}, {Red, PointSize[.1*Abs[q[[i]]]],
Point[r0[[i]]]}], {i, 1, 6}]]]


fig1


Comments

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...