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I have seen Contour diagrams for Equipotentials . That are drawn like so:

I also saw One image for these contours that was in 3D (Negative Point Charge) :

enter image description here

I was Wondering If there's any equation that represents these Contours In 3D?

Because I wanted to recreate these 3D Contour On Mathematica. I Know A little about 3d functions but not much.

But I do Know that this 3D Function Looks A bit like the function

$$f(x)=\frac{1}{x^2}$$ and $$f(x)=\frac{-1}{x^2}$$ But I couldn't get it to taper at the tip.

Could Anyone tell me how it could be made into 3d or how i could make it taper into a pointed tip at the top/bottom.

BTW:{It Also resembles the physics.stackexchange.com Logo but isn't differentiable at the top}

I tried to make it 3D using the same equation but i got something like :

For this $$f(x)=\frac{-1}{x^2+y^2}$$ I got:
but it isnt as steep as the countour i wanted.

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2  
Migration of software part of the question to mathematica.SE? –  Qmechanic May 6 '12 at 9:13
    
i just want the equation that would give me the required plot. –  The-Ever-Kid May 6 '12 at 9:26
    
Well you can easily change range of your plots in Mathematica, have you tried with that? –  dingo_d May 6 '12 at 9:47
    
@Qmechanic NO..... maybe its $\frac{kQ}{x^2+y^2}$ dingo_d the range of my function is infinity i want the function to change a bit so that it tapers more steeply and makes a point fairly earlier than infinity –  The-Ever-Kid May 6 '12 at 10:29
    
This seems to be a question about how to reproduce the desired plot in Mathematica, rather than a plot about making electric equipotential lines (which you've shown you're able to do), so I'll see if the Mathematica mods want it. –  David Z May 7 '12 at 5:17
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closed as off topic by Qmechanic, David Z May 7 '12 at 5:17

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1 Answer

up vote 3 down vote accepted

The plot you're trying to imitate seems to not go to infinity. I'd suggest you play around with potentials of the form $$V=\frac{V_0}{\sqrt{x^2+y^2+\delta^2}}.$$

E.g. Something like

Plot3D[-(1/Sqrt[x^2 + y^2 + \[Delta]^2]) /. {\[Delta] -> 1/10}, {x, -1, 1}, {y, -1,
1}, PlotRange -> Full, AspectRatio -> 1, ColorFunction -> Hue]

givesenter image description here

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This looks right, you can change the z range so it will look more 'steeper': PlotRange -> {{-1, 1}, {-1, 1}, {0, -20}} You can also change opacity and play with Mesh to get circles. –  dingo_d May 6 '12 at 14:49
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