How Do You Calculate the Electric Potential of an Ellipsoid?

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To calculate the electric potential of an ellipsoid, the initial approach involves parametrizing the ellipse and transforming coordinates to simplify the problem to a spherical case. The calculation becomes complex due to the need for fundamental vector products and the conversion of angles in the new coordinates. The discussion highlights the importance of considering the distance from the ellipsoid, as being "far away" allows for the assumption that the observation point's distance from the origin is greater than the surface points' distances. This is crucial because proximity to the surface may complicate the use of spherical harmonics due to the non-uniform radius of the ellipsoid. The problem was posed by a professor, requiring a comparison of solutions for both spherical and ellipsoidal shapes.
Amit Kumar Basistha
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Homework Statement
Recently I came across the following problem:

Suppose ##\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1## is an ellipse with surface charge density ##\sigma=\sigma_0\sin(\theta)\cos(\phi)## where ##\theta## is the angle with the ##z-## axis and ##\phi## is with the ##x-## axis. Find the potential and multipole moments at a point far away from the ellipse.
Relevant Equations
Maxwell's Equations
Spherical Harmonics
My initial idea was to first parametrize the ellipse as ##(a\sin(\theta')\cos(\phi'),b\sin(\theta')\sin(\phi'),c\cos(\theta'))## and then calculate ##\theta,\phi## in terms of these coordinates. I then did the coordinate transform ##x\to\frac{x}{a},y\to\frac{y}{b},z\to\frac{z}{c}## to convert it to the sphere case where you can find the potential and multipole moments using spherical harmonics. But the whole calculation is messy because you have to find the fundamental vector product and all those stuff for the coordinate change and the expression for the angles in terms of these coordinates.
 
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I think you need to consider how far is "far away". This looks like it needs an approximation. What does an elliptical (American) football look like next to a spherical (rest-of-the-world) football from 100 m away?
 
My understanding why the question mentions far away is because if the point is near the surface of the ellipsoid then there might be issues with the spherical harmonics expansion as the radius of the ellipsoid is not uniform. So, far away means you can safely assume that it's distance from the origin (Center of the Ellipsoid) is more that the distance of any point on the surface of the Ellipsoid from the origin
 
In this kind of problem, you have a point of observation at ##\vec r## from the origin and a charges at ##\vec r^{~'}##. At least to me (and I've been wrong before), "far away" usually means ##|\vec r|>>|\vec r^{~'}|## which in this case would translate to ##|\vec r|>>a,b,c.## However, if that is the case, why mention the ellipsoid? Do you have a reference where you found this problem? You just may have to put up with the messy calculation.
 
No I don't have a reference. Our Prof asked us to solve the problem in the case of the sphere and then asked us as a separate question to do it for the Ellipsoid
 
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