Potential due to an Ellipsoid

  • Thread starter Amit Kumar Basistha
  • Start date
  • Tags
    Ellipse
  • #1
Amit Kumar Basistha
3
0
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.
 
Last edited:
Physics news on Phys.org
  • #2
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?
 
  • #3
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
 
  • #4
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.
 
  • #5
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
 

Similar threads

  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
427
  • Advanced Physics Homework Help
Replies
1
Views
723
  • Advanced Physics Homework Help
Replies
4
Views
448
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
925
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
564
  • Advanced Physics Homework Help
Replies
7
Views
1K
Back
Top