How can I show Electric Potential due to a Dipole as a 3D surface?

Hi,

I'm very new to matlab and have been picking my brain over how to 3D plot the electric potetial due to a dipole.

Here is the code I have currently come up with:

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clear;clf
[X,Y] = meshgrid(-2: .4: 2, -2: .4: 2)
r = sqrt(X.^2+Y.^2)
Z = (1/2)*1./(r.^2)
surf(X,Y,Z)

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Where Z represents the Electric Potential scalar field. NOTE: I have purposely excluded cos(theta) in the numerator for simplicity (and because I cannot seem to successfully include it). I kind of have a few questions:

• How do I avoid from dividing by 0? (This leaves a gap in the tip of the plot)
• How can I successfully include the cosine argument ( since it is necessary to provide an accurate plot)

I would really appreciate some help.

[KSSV]

Where you want to include the cosine argumnet?

[Luca]

It needs to be in the numerator of Z

Answer 1

[David Goodmanson]

Hi Luca,

Since you can't show the values of the potential at every point in the 3d space, this is most commonly done by showing a surface of constant potential. The following code first plots a sphere of radius 1 as a demo, with x,y,z defined by the usual two angles in spherical coordinates

Ignoring some constants, the potential is phi = cos(theta)/r^2. Suppose phi = +1 is the constant value. then r = sqrt(cos(theta)). In the upper hemisphere cos(theta) is positive so the sqare root works. In the lower hemisphere cos(theta) is negative, so in order to plot that, one can use r = sqrt(abs(cos(theta))) which works, at the cost of the potential coming out at a constant -1 in the lower hemisphere, which is actually physically correct. (If you only want the upper lobe then th1 below can be given an upper limit of pi/2).

th1 = linspace(0,pi,31);
phi1 = linspace(0,2*pi,41);
[th phi] = meshgrid(th1,phi1);
R = 1;                         % sphere
% R = sqrt(abs(cos(th)));        % uncomment this line for the dipole
x = R.*sin(th).*cos(phi);
y = R.*sin(th).*sin(phi);
z = R.*cos(th);
surf(x,y,z)
axis equal