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I'm attempting to make a graph of the equipotential curves of two point charges while showing the electric field lines. I can do the electric field lines but the equipotentials are giving me trouble.

My sample for one point charge was:

equipots1 = ContourPlot[
              V1plot, {x, xmin, xmax}, {y, ymin, ymax}, 
              ContourShading -> None, Contours -> 10, AspectRatio -> ar,
              ImageSize -> Small, ContourStyle -> {{Purple, Opacity[0.6]}}
            ]
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    $\begingroup$ What's the formula for the potential for one point source? Two? (Remember about superposition.) Once you have that, with an appropriate substitution for $r$, you can just plug it into ContourPlot. $\endgroup$ – rcollyer Feb 6 '17 at 3:49
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Consider two charges, Q1 and Q2, centered at positions, pos1 and pos2, respectively.

k = 9*10^9;

Then, you can write a function which takes Q1, Q2, pos1, pos2, and an equivalent r as arguments.

plotEquipotential[Q1_, Q2_, pos1_, pos2_, r_] := 
 ContourPlot[
  k Q1/Norm[{x, y} - pos1] + k Q2/Norm[{x, y} - pos2] == k/r, 
  {x, -10, 10}, {y, -10, 10}
 ]

Example plot:

plotEquipotential[1, 2, {0,0}, {0,3}, 1]

enter image description here

To create a gif:

Export["equipotentialPlot.gif", 
 plotEquipotential[1, 2, {0, 0}, {0, 3}, #] & /@ Range[0.1, 2, 0.1], 
 "DisplayDurations" -> 0.2]

enter image description here

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  • $\begingroup$ Instead of crafting your own norm function, Norm does it all for you. $\endgroup$ – rcollyer Feb 6 '17 at 4:59
  • $\begingroup$ Didn't know that. Thank you. $\endgroup$ – Anjan Kumar Feb 6 '17 at 5:04

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