Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 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.

share|cite|improve this question

closed as off topic by Qmechanic, David Z May 7 '12 at 5:17

Questions on Physics Stack Exchange are expected to relate to physics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

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
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

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.