# Error with VectorPlot [closed]

Clear["Global'*"]
dipole = 1/Sqrt[(x - .5)^2 + y^2] - 1/Sqrt[(x - .5)^2 + y^2];
plot1 = VectorPlot[{x/CubeRoot[(x - .5)^2 + y^2], y/
CubeRoot[(x - .5)^2 + y^2]},
{x, -3, 3}, {y, -3, 3}, VectorPoints \[RightArrow] 8,
VectorScale \[RightArrow] Small];
plot2 = CoutourPlot[dipole, {x, -3, 3}, {y, -3, 3},
DisplayFunction \[RightArrow] Identity,
PlotPoints \[RightArrow] 20, Contours \[RightArrow] 5];
Show[{plot1, plot2}],
Epilog \[RightArrow] {{Rue[0.3], Disk[{.5, 0}, 0.1]},
{Rue[0.95], Disk[{.5, 0}, 0.1]}},
DisplayFunction \[RightArrow] \$DisplayFunction


This returns the error: "VectorPlot: Options expected beyond position 3." I'm not 100% sure what I did wrong (this is my first time using Mathematica), and I'd appreciate some help!

• Replace \[RightArrow]s with ->? – kglr Nov 7 '18 at 21:01

V = 1/Sqrt[(x - .5)^2 + y^2] -
1/Sqrt[(x + .5)^2 + y^2]; EV = -Grad[V, {x, y}];

plot1 = VectorPlot[{(-0.5 + x)/((-0.5 + x)^2 + y^2)^(3/2) - (
0.5 + x)/((0.5 + x)^2 + y^2)^(3/2),
y/((-0.5 + x)^2 + y^2)^(3/2) - y/((0.5 + x)^2 + y^2)^(
3/2)}, {x, -3, 3}, {y, -3, 3}, StreamPoints -> Fine,
VectorPoints -> Fine, StreamStyle -> LightGray];
plot2 = ContourPlot[
1/Sqrt[(x - .5)^2 + y^2] - 1/Sqrt[(x + .5)^2 + y^2], {x, -2,
2}, {y, -2, 2}, Contours -> 120,
ColorFunction -> "BlueGreenYellow", PlotRange -> {-10, 10}];
Show[{plot2, plot1,
Graphics[{{Hue[0.3], Disk[{.5, 0}, 0.1]}, {Hue[0.95],
Disk[{-.5, 0}, 0.1]}}]}] 